The Mathematical Intelligencer volume 32 issue 4

Acknowledgement

IIn the Summer issue of The Mathematical Intelligencer,vol. 32, no. 2, we presented, with permission, a geo-metric-combinatoric pattern due to Anthony Hill. This

was an array of 66 six-segment graphs, whose significance,and extraordinary origin, were explained on p. 3 of theissue; the array appeared on the cover. Its creator, Anthony

Hill, has pointed out to us that the colors in the coverdesign have no role in its status as solution of a certaincombinatorial problem, and he asks that we apologize fordeparting from his concept by using color. We take fullresponsibility for this design decision, and we regret ourfailure to carry out his intentions in this respect. We hopeMr. Hill is comforted by the fact that every careful reader ofthe explanation we published will appreciate the meaningof his discovery and the irrelevance of the colors thereto.

Chandler Davis and Marjorie Senechal

� 2010 Springer Science+Business Media, LLC, Volume 32, Number 4, 2010 1

Letter to the Editors

Maria TeresaCalapso’s HyperbolicPythagoreanTheorem

The Mathematical Intelligencer encourages comments

about the material in this issue. Letters to the editor

should be sent to either of the editors-in-chief, Chandler

Davis or Marjorie Senechal.

IIread with interest Paolo Maraner’s recent MathematicalIntelligencer note ‘‘A Spherical Pythagorean Theorem’’(Vol. 32, No. 3, Fall 2010, 46–50, DOI:10.1007/s00283-010-9152-9). In it, the author shows that a proper

generalization of the Pythagorean theorem that wouldrender it true in an absolute setting, cannot stay with thehypothesis that one angle of the triangle be right, but ratherhas to relax it to state that one angle, say A, should be thesum of the other two, say B and C. Also the Pythagoreantheorem should state that the areas of the circles with sidesb and c as radii should be equal to the area of the circle withside a as radius (let us denote the latter area by sa). I wouldlike to point out that precisely this form of the Pythagoreantheorem was stated and proved in the hyperbolic plane byMaria Teresa Calapso in [2], where it is shown that theconverse holds as well, that is, that we have sa = sb +

sc only in triangles in which A = B + C holds. In [8] it wasshown that the generalized Pythagorean formula, valid inany hyperbolic triangle, is �a ¼ �bþ sinðA�BÞ

sin C �c, and[1, 3, 7] contain like-minded forms of the generalizedPythagorean theorem. As its title indicates, the main noveltyin Paolo Maraner’s paper is the fact that this version of thePythagorean theorem holds in the spherical setting as well.Even in the hyperbolic case, the paper has the merit of

providing a short proof that appeals only to basic formulasof hyperbolic geometry, simpler than the proofs in [1, 2].

The author also asks for the proper axiomatic setting inwhich the above-stated Pythagorean theorem would hold.Given that areas of circles demand the full axiomatic importof the real numbers, the version referring to areas of circlesfalls outside elementary (first-order logic) axiomatic con-siderations. If a version regarding triangles satisfying A =

B + C turns out to be true in Hilbert’s absolute geometry(axiomatized by the axioms I 1–3, II, and III of [5] or by theaxioms A1–A9 of [6]), then it must involve areas of polygonson the sides a, b, c (perhaps right isosceles triangles withequal sides having the length of the triangle side on whichthey are erected), with the Hilbert definition of area equalityas equivalence by completion (Erganzungsgleichheit).

Relevant for an absolute version of the Pythagoreantheorem is also the absolute version of the IntersectingChords Theorem (III.36 in Euclid’s Elements) in [4].

REFERENCES

[1] Familiari-Calapso, M. T., Le theoreme de Pythagore en geometrie

absolue. C. R. Math. Acad. Sci. Paris. Ser. A-B. 263 (1966), A668–

A670.

[2] Familiari-Calapso, M. T., Sur une classe di triangles et sur le

theoreme de Pythagore en geometrie hyperbolique. C. R. Acad.

Sci. Paris Ser. A–B 268 (1969), A603–A604.

[3] Calapso, M. T., Ancora sul teorema di Pitagora in geometria

assoluta. Atti Accad. Peloritana Pericolanti Cl. Sci. Fis. Mat. Natur.

50 (1970), 99–107.

[4] Hartshorne, R., Non-Euclidean III.36. Amer. Math. Monthly 110

(2003), 495–502.

[5] Hilbert, D., Grundlagen der Geometrie, 12. Auflage. Teubner,

Stuttgart, 1977.

[6] W. Schwabhauser, W. Szmielew, and A. Tarski, Metamathemat-

ische Methoden in der Geometrie. Springer-Verlag, Berlin, 1983.

[7] Vranceanu, G., Sopra la geometria noneuclidea. Atti Accad.

Peloritana Pericolanti Cl. Sci. Fis. Mat. Natur. 50 (1970), 119–123.

[8] Vranceanu G. G., Sur la trigonometrie noneuclidienne. Rend. Circ.

Mat. Palermo (2) 20 (1971), 254–262

Victor Pambuccian

Division of Mathematical and Natural Sciences

Arizona State University—West Campus

Phoenix, AZ 85069-7100

USA

e-mail: [email protected]

2 THE MATHEMATICAL INTELLIGENCER � 2010 SPRINGER SCIENCE+BUSINESS MEDIA, LLC

http://dx.doi.org/10.1007/s00283-010-9152-9

http://dx.doi.org/10.1007/s00283-010-9152-9

Note

Even Hilbert Nods…BOB LLOYD

WWe often describe a text as ‘‘authoritative’’ when wedo not expect there to be much question aboutthe content, so that students can safely be referred

to it. Nevertheless, there is a downside to this concept. Weexpect to make our own mistakes, but in dealing with thework of an authority, there can be a reluctance to ques-tion, so mistakes can persist. In mathematics the supremeexample of this is probably Aristotle’s claim that space canbe completely filled by packing cubes or tetrahedra. Twomillennia passed before it was pointed out that this is nottrue for tetrahedra.

i

I would like to draw attention to another mistake whichhas persisted, though only for three-quarters of a century.Anschauliche Geometrie, by David Hilbert and StephanCohn-Vossen,ii appeared in 1932, was published in Englishas Geometry and the Imagination in 1952,iii and reissued in1999. A second German edition came out in 1995.iv Despiteits age, the book has clearly been in demand, and thecomment that ‘‘many of us for years have been pushingthe classic Geometry and the Imagination (to graduate oradvanced undergraduate students)’’v suggests that it is a textwith authority and influence. The following note is notintended to be critical; rather, the sense is that, ‘‘If it canhappen to him, then there’s hope for the rest of us!’’

The problem comes in the discussion of the symmetriesof the Platonic solids, and concerns the diagram of a cubewithin a dodecahedron; this may be older than Aristotle,vi

though the first description is in Euclid.vii Figure 1 repro-duces two diagrams from Hilbert and Cohn-Vossen.viii

These diagrams were used to discuss the relationshipsbetween the point groups of the three different solid fig-ures shown, using the pure rotation groups rather than thefull point symmetries; the same approach will be used here.These rotation groups are frequently represented by thesymbols T or 332 for tetrahedral symmetry, and O or 432 foroctahedral, the symmetry of the cube. The dodecahedronhas icosahedral symmetry, I or 532.

In the left-hand diagram in Figure 1, the tetrahedra andthe cube have different symmetries. I consider just one of thetetrahedra, though the argument is unaffected by workingwith two independent tetrahedra. Combining two objectsof different symmetry often gives a lower symmetry;combinations with decreasing or increasing symmetry are

discussed in Cromwell.ix The rotation axes of the tetrahe-dron/cube combination are the same as those of an isolatedtetrahedron, and the 4-fold axes of the cube are absent in thecombination. The symmetry of this combination is thus T(332), and it can be constructed as shown because T is asubgroup of O (432). This is essentially the argument pre-sented by Hilbert and Cohn-Vossen,x though they do notmention the symmetry of the combination.

Hilbert and Cohn-Vossen extend this argument to theright-hand diagram in Figure 1. In the English version,xi

they claim that:

‘‘Similarly it turns out that the octahedral group is asubgroup of the icosahedral group. This is the reasonwhy a cube can be inscribed in a dodecahedron in thesame way as the tetrahedron can be inscribed in a cube.’’

The English is a precise translation of the original Ger-man (see note I), and the German text has remainedunchanged in the new edition.

The first sentence of this extract is clearly erroneous.The operations of the group O include 4-fold rotations. Theoperations of I (532) include 5-fold and 3-fold rotations,but no 4-fold rotations, so O cannot be a subgroup of I.The second sentence contains a different error. Unlike thetetrahedron-cube combination, the inscription of a cubewithin a dodecahedron does not depend on the symmetryof one body being that of a subgroup of the other. A specialcase has been used to make a more general argument.

Table 1 shows a correlation of the elements of the threegroups concerned. Here, Cn denotes an axis of 360�/nrotation symmetry, and the table gives the numbers of theserotation axes for each of the groups T, O, and I. It is evidentthat although O is not a subgroup of I, T is a subgroup ofboth I and O.

The rotations transforming the cube/dodecahedron com-bination into itself are the operations of the four C3 axesthrough opposite corners of the cube, and of the three C2

axes through opposite face centers of the cube. There areno other rotation operations, so the combination of the twosolids has symmetry T. The diagram can be constructed, notbecause the symmetry of one body is a subgroup of that ofthe other, but because the bodies separately have O and Isymmetry, and these have a subgroup, T, in common.xii

The diagram of a tetrahedron in a cube is a special case ofthis, where the combination happens to have the samesymmetry as one of the two bodies being combined, butthis is not always the case. The combinations of an octa-hedron with a cubexiii are even more special, since here thesymmetry groups of the two solids are identical with that ofthe combination.

A remarkably similar mistake occurs in a much morerecent work.xiv This also discusses the cube-tetrahedronand dodecahedron-cube diagrams, and claims that:


‘‘…every symmetry of the cube is also a symmetry of thedodecahedron.’’This is argued from the observation that the vertices of

the cube are a subset of those of the dodecahedron.However, as above, the 4-fold rotation axes (‘‘symmetries’’)of the cube have disappeared in the combination. Thebook does not reference Hilbert and Cohn-Vossen here,and the argument is expressed in the language of the fullgroups rather than that of the rotation groups (see note II),so it seems that the same error has occurred independently.Readers who are involved with the teaching of geometrymight consider warning students about this problem.

ACKNOWLEDGMENT

I thank Springer Science+Business Media for permission toreproduce the two diagrams.

NotesI. The original reads, ‘‘Ebenso erweist sich nun die Ok-

taedergruppe als Untergruppe der Ikosaedergruppe.Aus diesem Grunde kann man einen Wurfel in eineDodekaeder in gleicher Weise hineinstellen wie einTetraeder in einen Wurfel.’’

II. Hilbert and Cohn-Vossen’s book is claimed as ‘‘aninspiration’’ in the bibliography, and is referenced atother points. In the full groups, the symmetries are Oh,Ih, and Th for the combination.

REFERENCESiHeath, T. L., Mathematics in Aristotle, Oxford, Clarendon Press, 1949,

pp. 177–178.iiHilbert, D. and Cohn-Vossen, S., Anschauliche Geometrie, Die

Grundlehren der Mathematischen Wissenschaften Band XXXVII,

Berlin, Julius Springer, 1932.iiiHilbert, D. and Cohn-Vossen, S., Geometry and the Imagination,

translated by P. Nemenyi. New York, Chelsea Publishing Co.

1952.ivHilbert, D. and Cohn-Vossen, S., Anschauliche Geometrie, mit einem

Geleitwort von Marcel Berger (2. Aufl.), Berlin, Springer, 1995.vBanchoff, T., Bulletin of the American Mathematical Society, 34, 1,

January 1997, p. 34.viAltmann, B., Euclid–The Creation of Mathematics, New York,

Springer-Verlag, 1999, p. 285.viiRef. vi, Euclid, Book XIII; see ref. vi, p. 294.viiiRef. ii, p. 83.ixCromwell, P. R., Polyhedra, Cambridge, Cambridge University Press,

1996, pp. 359–385.xRef. ii, p. 83; Ref. iii, p. 92.xiRef. iii, p. 92.xiiRef. ix, pp. 361–362.xiiiRef. ii, p. 82.xivSmith, J. T. Methods of Geometry, New York, Chichester, John

Wiley & Sons, Inc., 2000, p. 404.

School of Chemistry

Trinity College

Dublin 2

Ireland

e-mail: [email protected]; [email protected]

Figure 1. Two different ways of inscribing a tetrahedron in a cube, and a cube inscribed in a dodecahedron. (Reproduced

from ii.)

Table 1. A correlation of elements in the groups I, T, and O

I (532) 6C5 10C3 – – 15C2

T (332) – 4C3 – – 3C2

O (432) – 4C3 3C2 3C4 6C2

4 THE MATHEMATICAL INTELLIGENCER

The BilinskiDodecahedronand AssortedParallelohedra,Zonohedra,Monohedra,Isozonohedra,and OtherhedraBRANKO GRUNBAUM

FFifty years ago Stanko Bilinski showed that Fedorov’senumeration of convex polyhedra having congruentrhombi as faces is incomplete, although it had been

acceptedasvalid for theprevious 75 years. Thedodecahedronhe discovered will be used here to document errors by severalmathematical luminaries. It also prompted an examination ofthe largely unexplored topic of analogous nonconvex poly-hedra, which led to unexpected connections and problems.

BackgroundIn 1885 Evgraf Stepanovich Fedorov published the results of sev-eral years of research under the title ‘‘Introduction to the Study ofFigures’’ [9], in which he defined and studied a variety of conceptsthatare relevant toour story.Thisbook-longwork is consideredbymany tobeoneof themilestones ofmathematical crystallography.For a long time this was, essentially, inaccessible and unknown toWestern researchers except for a summary [10] in German.1

1The only somewhat detailed description of Fedorov’s work available in English (and in French) is in [31]. Fedorov’s book [9] was never translated to any Western

language, and its results have been rather inadequately described in the Western literature. The lack of a translation is probably at least in part to blame for ignorance of

its results, and an additional reason may be the fact that it is very difficult to read [31, p. 6].


Several mathematically interesting concepts were intro-duced in [9]. We shall formulate them in terms that arecustomarily used today, even though Fedorov’s originaldefinitions were not exactly the same. First, a parallelohe-dron is a polyhedron in 3-space that admits a tiling of thespace by translated copies of itself. Obvious examples ofparallelohedra are the cube and the Archimedean six-sidedprism. The analogous 2-dimensional objects are calledparallelogons; it is not hard to show that the only polygonsthat are parallelogons are the centrally symmetric quad-rangles and hexagons. It is clear that any prism with aparallelogonal basis is a parallelohedron, but we shallencounter many parallelohedra that are more complicated.It is clear that any nonsingular affine image of a parallelo-hedron is itself a parallelohedron.

Another new concept in [9] is that of zonohedra. Azonohedron is a polyhedron such that all its faces arecentrally symmetric; there are several equivalent defini-tions. All Archimedean prisms over even-sided bases are

zonohedra, but again there are more interesting examples.A basic result about zonohedra is:

Each convex zonohedron has a center.This result is often attributed to Aleksandrov [1] (see [5]),

but in fact is contained in a more general theorem2 ofMinkowski [27, p. 118, Lehrsatz IV]. Even earlier, this wasTheorem 23 of Fedorov ([9, p. 271], [10, p. 689]), althoughFedorov’s proof is rather convoluted and difficult to follow.

We say that a polyhedron is monohedral (or is amonohedron) provided its faces are all mutually congruent.The term ‘‘isohedral’’—used by Fedorov [9] and Bilinski[3]—nowadays indicates the more restricted class of poly-hedra with the property that their symmetries acttransitively on their faces.3 The polyhedra of Fedorov andBilinski are not (in general) ‘‘isohedra’’ by definitions thatare customary today. We call a polyhedron rhombic if all itsfaces are rhombi. It is an immediate consequence of Euler’stheorem on polyhedra that the only monohedral zonohe-dra are the rhombic ones.

One of the results of Fedorov ([9, p. 267], [10, p. 689]) iscontained in the claim:

There are precisely four distinct types of monohedralconvex zonohedra: the rhombic triacontahedron T, therhombic icosahedron F, the rhombic dodecahedron K,and the infinite family of rhombohedra (rhombic hexa-hedra) H.‘‘Type’’ here is to be understood as indicating classes of

polyhedra equivalent under similarities. The family ofrhombohedra contains all polyhedra obtained from thecube by dilatation in any positive ratio in the direction of abody-diagonal.

These polyhedra are illustrated in Figure 1; they aresometimes called isozonohedra. The polyhedra T and K goback at least to Kepler [23], whereas F was first described byFedorov [9]. I do not know when the family H was firstfound — it probably was known in antiquity.

An additional important result from Fedorov [9] is thefollowing; notice the change to ‘‘combinatorial type’’ fromthe ‘‘affine type’’ that is inherent in the definition.

Every convex parallelohedron is a zonohedron of one ofthe five combinatorial types shown in Figure 2. Con-versely, every convex zonohedron of one of the fivecombinatorial types in Figure 2 is a parallelohedron.4

.........................................................................

AU

TH

OR BRANKO GRUNBAUM received his PhD

from the Hebrew University in Jerusalem

in 1957. He is Professor Emeritus at the

University of Washington, where he has

been since 1966. His book ‘‘Convex Poly-topes’’ (1967, 2003) has been very popular,

as was the book ‘‘Tilings and Patterns’’

(coauthored by G. C. Shephard) published in

1986. He hopes that ‘‘Configurations of

Points and Lines’’ (2009) will revive the

interest in this exciting topic, which was

neglected during most of the twentieth

century. Grunbaum’s research interests aremostly in various branches of combinatorial

geometry.

Department of Mathematics

University of Washington 354350

Seattle, WA 98195-4350

USA


2Minkowski’s theorem establishes that a convex polyhedron with pairwise parallel faces of the same area has a center; the congruence of the faces in each pair follows,

regardless of the existence of centers of faces (which is assumed for zonohedra).3The term ‘‘gleichflachig’’ (= with equal surfaces) was quite established at the time of Fedorov’s writing, but what it meant seems to have been more than the word

implies. As explained in Edmund Hess’s second note [21] excoriating Fedorov [10] and [11], the interpretation as ‘‘congruent faces’’ (that is, monohedral) is mistaken.

Indeed, by ‘‘gleichflachig’’ Hess means something much more restrictive. Hess formulates it in [21] very clumsily, but it amounts to symmetries acting transitively on the

faces, that is, to isohedral. It is remarkable that even the definition given by Bruckner (in his well-known book [4, p. 121], repeating the definition by Hess in [19] and

several other places) states that ‘‘gleichflachig’’ is the same as ‘‘monohedral’’ but Bruckner (like Hess) takes it to mean ‘‘isohedral.’’ Fedorov was aware of the various

papers that use ‘‘gleichflachig,’’ and it is not clear why he used ‘‘isohedral’’ for ‘‘monohedral’’ polyhedra. In any case, this led Fedorov to claim that his results disprove

the assertion of Hess [19] that every ‘‘gleichflachig’’ polyhedron admits an insphere. Fedorov’s claim is unjustified, but with the rather natural misunderstanding of

‘‘gleichflachig’’ he was justified to think that his rhombic icosahedron is a counterexample. This, and disputed priority claims, led to protests by Hess (in [20] and [21]),

repeated by Bruckner [4, p. 162], and a rejoinder by Fedorov [11]. Neither side pointed out that the misunderstanding arises from inadequately explained terminology;

from a perspective of well over a century later, it seems that both Fedorov and Hess were very thin-skinned, inflexible, and stubborn.4In different publications Fedorov uses different notions of ‘‘type.’’ In several (e.g., [10, 12]) he has only four ‘‘types’’ of parallelohedra, since the rhombic dodecahedron

and the elongated dodecahedron ((c) and (b) in Figure 2) are of the same type in these classifications. Since we are interested in combinatorial types, we accept

Fedorov’s original enumeration illustrated in Figure 2.


Fedorov’s proof is not easy to follow; a more accessibleproof of Fedorov’s result can be found in [2, Ch. 8].

Bilinski’s Rhombic DodecahedronFedorov’s enumeration of monohedral rhombic isohedra(called isozonohedra by Fedorov and Bilinski, and by Cox-eter [7]) mentioned previously claimed that there areprecisely four distinct types (counting all rhombohedra asone type). Considering the elementary character of such anenumeration, it is rather surprising that it took three-quartersof a century to find this to be mistaken.5 Bilinski [3] foundthat there is an additional isozonohedron and proved:

Up to similarity, there are precisely five distinct convexisozonohedra.The rhombic monohedral dodecahedron found by

Bilinski shall be denoted B; it is not affinely equivalent toKepler’s dodecahedron (denoted K) although it is of thesame combinatorial type. Bilinski also proved that there areno other isozonohedra. To ease the comparison of B and K,both are shown in Figure 3.

Bilinski’s proof of the existence of the dodecahedron Bis essentially trivial, and this makes it even more mysterious

how Fedorov could have missed it.6 The proof is based ontwo observations:

(i) All faces of every convex zonohedron are arranged inzones, that is, families of faces in which all membersshare parallel edges of the same length; and

(ii) All edges of such a zone may be lengthened orshortened by the same factor while keeping thepolyhedron zonohedral.

(a) (b)

(c) (d) (e)

Figure 2. Representatives of the five combinatorial types of

convex parallelohedra, as determined by Fedorov [9]. (a) is

the truncated octahedron (an Archimedean polyhedron); (b)

is an elongated dodecahedron (with regular faces, but not

Archimedean); (c) is Kepler’s rhombic dodecahedron K (a

Catalan polyhedron); (d) is the Archimedean 6-sided prism;

and (e) is the cube.

BK

Figure 3. The two convex rhombic monohedra (isozonohe-

dra): Kepler’s K and Bilinski’s B.

FT

HK

Figure 1. The four isozonohedra (convex rhombic monohe-

dra) enumerated by Fedorov. Kepler found the triacon-

tahedron T and the dodecahedron K, whereas Fedorov

discovered the icosahedron F. The infinite class H of rhombic

hexahedra seems to have been known much earlier.

5This is a nice illustration of the claim that errors in mathematics do get discovered and corrected in due course. I can only hope that if there are any errors in the present

work they will be discovered in my lifetime.6A possible explanation is in a tendency that can be observed in other enumerations as well: After some necessary criteria for enumeration of objects of a certain kind

have been established, the enumeration is deemed complete by providing an example for each of the sets of criteria––without investigating whether there are more than

one object per set of criteria. This failure of observing the possibility of a second rhombic dodecahedron (besides Kepler’s) is akin to the failure of so many people that

were enumerating the Archimedean solids (polyhedra with regular faces and congruent vertices, i.e., congruent vertex stars) but missed the pseudorhombicu-

boctahedron (sometimes called ‘‘Miller’s mistake’’); see the detailed account of this ‘‘enduring error’’ in [13].


In particular, all such edges on one zone can be deleted(shrunk to 0). Performing such a zone deletion—a processmentioned by Fedorov—starting with Kepler’s rhombictriacontahedron T yields (successively) Fedorov’s icosahe-dronF,Bilinski’s dodecahedronB, and two rhombohedra, theobtuse Ho and the acute Ha. This family of isozonohedra thatare descendants of the triacontahedron is shown in Figure 4.The proof that there are no other isozonohedra is slightlymore complicated and is not of particular interest here.

The family of ‘‘direct’’ descendants of Kepler’s rhombicdodecahedron K is smaller; it contains only one rhombo-hedron H*o (Fig. 5). However, one may wish to include inthe family a ‘‘cousin’’ H*a—consisting of the same rhombias H*o, but in an acute conformation.

One of the errors in the literature dealing with Bilinski’sdodecahedron is the assertion by Coxeter [7, p. 148] that thetwo rhombic dodecahedra—Kepler’s and Bilinski’s—areaffinely equivalent. To see the affine nonequivalence of thetwo dodecahedra (easily deduced even from the drawingsin Fig. 3), consider the long (vertical) body-diagonal ofBilinski’s dodecahedron (Fig. 3b). It is parallel to four ofthe faces and in each face to one of the diagonals. In twofaces this is the short diagonal, in the other two the longone. But in the Kepler dodecahedron the correspondingdiagonals are all of the same length. Since ratios of lengths

of parallel segments are preserved under affinities, thisestablishes the nonequivalence.

If one has a model of Bilinski’s dodecahedron in hand,one can look at one of the other diagonals connectingopposite 4-valent vertices, and see that no face diagonal isparallel to it. This is in contrast to the situation with Kepler’sdodecahedron.

By the theorems of Fedorov mentioned previously, sinceBilinski’s dodecahedron B is a zonohedron combinatoriallyequivalent to Kepler’s, it is a parallelohedron. This can beeasily established directly, most simply by manipulatingthree or four models of B. It is strange that Bilinski does notmention the fact that B is a parallelohedron.

In this context we must mention a serious error com-mitted by A. Schoenflies [30, pp. 467 and 470] and veryclearly formulated by E. Steinitz. It is more subtle than thatof Coxeter, who may have been misguided by the follow-ing statement of Steinitz [34, p. 130]:

The aim [formulated previously in a different form] is todetermine the various partitions of the space into con-gruent polyhedra in parallel positions. Since an affineimage of such a partition is a partition of the same kind,affinely related partitions are not to be considered asdifferent. Then there are only five convex partitions ofthis kind. [My translation and comments in brackets].How did excellent mathematicians come to commit such

errors? The confusion illustrates the delicate interactionsamong the concepts involved, considered by Fedorov,Dirichlet, Voronoi, and others. A correct version of Stei-nitz’s statement would be (see Delone [8]):

Every convex parallelohedron P is affinely equivalent to aparallelohedron P0 such that a tiling by translates of P0

K

H*o H*a

Figure 5. Kepler’s rhombic dodecahedron K and its descen-

dant, rhombohedron H*o. The rhombohedron H*a is ‘‘related’’

to them since its faces are congruent to those of the other two

isozonohedra shown; however, it is not obtainable from K by

zone elimination.

T

Ho Ha

B

F

Figure 4. The triacontahedron and its descendants: Kepler’s

triacontahedron T, Fedorov’s icosahedron F, Bilinski’s

dodecahedron B, and the two hexahedra, the obtuse Ho and

the acute Ha. The first three are shown by .wrl illustrations in

[25] and other web pages.


coincides with the tiling by the Dirichlet-Voronoi regionsof the points of a lattice L0. The lattice L0 is affinely relatedto the lattice L associated with one of the five Fedorovparallelohedra P00. But P0 need not be the image of P00

under that affinity. Affine transformations do not com-mute with the formation of Dirichlet-Voronoi regions.In particular, isozonohedra other than rhombohedra are

not mapped onto isozonohedra under affine transforma-tions that are not similarities.

As an illustration of this situation, it is easy to see thatBilinski’s dodecahedron B is affinely equivalent to a poly-hedron B0 that has an insphere (a sphere that touches all itsfaces). The centers of a tiling by translates of B0 form alattice L0 such that this tiling is formed by Dirichlet-Voronoiregions of the points of L0. The lattice L0 has an affine imageL such that the tiling by Dirichlet-Voronoi regions of thepoints of L is a tiling by copies of the Kepler dodecahedronK. However, since the Dirichlet domain of a lattice is notaffinely associated with the lattice, there is no implicationthat either B or B0 is affinely equivalent to K.

A simple illustration of the analogous situation in theplane is possible with hexagonal parallelogons (as men-tioned earlier, a parallelogon is a polygon that admits atiling of the plane by translated copies). As shown inFigure 6, the tiling is by the Dirichlet regions of a lattice ofpoints. This lattice is affinely equivalent to the latticeassociated with regular hexagons, but the tiling is obviouslynot affinely equivalent to the tiling by regular hexagons.

It is appropriate to mention here that for simple paral-lelohedra (those in which all vertices have valence 3) thattile face-to-face Voronoi proved [38] that each is the affineimage of a Dirichlet-Voronoi region. For various strength-enings of this result see [26].

Nonconvex ParallelohedraBilinski’s completion of the enumeration of isozonohedraneeds no correction. However, it may be of interest to

examine the situation if nonconvex rhombic monohedraare admitted; we shall modify the original definition andcall them isozonohedra as well. Moreover, there are variousreasons why one should investigate—more generally—nonconvex parallelohedra.

It is of some interest to note that the characterization ofplane parallelogons (convex or not) is completely trivial. Aversion is formulated as Exercise 1.2.3(i) of [16, p. 24]: Aclosed topological disk M is a parallelogon if and only if it ispossible to partition the boundary of M into four or six arcs,with opposite arcs translates of each other. Two examplesof such partitions are shown in Figure 7.

Another reason for considering nonconvex parallelohe-dra is that there is no intrinsic justification for their exclusion,whereas—as we shall see—many interesting forms becomepossible, and some tantalizing problems arise. The crosses,semicrosses, and other clusters studied by Stein [32] andothers provide examples of such questions and results.7 Italso seems reasonable that the use of parallelohedra inapplications need not be limited to convex ones.

It is worth noting that by Fedorov’s Definition 24 (p. 285of [9], p. 691 of [10]) and earlier ones, a parallelohedronneed not be convex, nor do its faces need to be centrallysymmetric.

Two nonconvex rhombic monohedra (in fact, isohedra)have been described in the nineteenth century; see Coxeter[7, pp. 102–103, 115–116]. Both are triacontahedra, and areself-intersecting. This illustrates the need for a precisedescription of the kinds of polyhedra we wish to considerhere.

Convex polyhedra discussed so far need little explana-tion, even though certain variants in the definition arepossible. However, now we are concerned with widerclasses of polyhedra regarding which there is no generallyaccepted definition.8 Unless the contrary is explicitly noted,in the present note we consider only polyhedra with sur-face homeomorphic to a sphere and adjacent faces notcoplanar. We say they are of spherical type. There areinfinitely many combinatorially different rhombic mono-hedra of this type—to obtain new ones it is enough to‘‘appropriately paste together’’ along common faces two ormore smaller polyhedra. This will interest us a little bit later.

The two triacontahedra mentioned above are notaccepted in our discussion. However, a remarkable

Figure 6. An affine transform of the lattice of centers at left

leads to the lattice of the tiling by regular hexagons. The

Dirichlet domains of the points of the lattice are transformed

into the hexagons at right, which clearly are not affinely

equivalent to regular hexagons.

Figure 7. Planigons without center have boundary parti-

tioned into 4 or 6 arcs, such that the opposite arcs are

translates of each other.

7Recent results on crosses and semicrosses can be found in [14].8Many different classes of nonconvex polyhedra have been defined in the literature. It would seem that the appropriate definition depends on the topic considered, and

that a universally accepted definition is not to be expected.


nonconvex rhombic hexecontahedron of the spherical typewas found by Unkelbach [37]; it is shown in Figure 8. Itsrhombi are the same as those in Kepler’s triacontahedron T.It is one of almost a score of rhombic hexecontahedradescribed in the draft of [15]; however, all except U are notof the spherical type.

For a more detailed investigation of nonconvex isozono-hedra, we first restrict attention to rhombic dodecahedra. Westart with the two convex ones—Kepler’s K and Bilinski’sB—and apply a modification we call indentation. Anindentation is carried out at a 3-valent vertex of an isozono-hedron. It consists of the removal of the three incident facesand their replacement by the three ‘‘inverted’’ faces—that is,the triplet of faces that has the same outer boundary as theoriginal triplet, but fits on the other side of that boundary.This is illustrated in Figure 9, where we start from Kepler’sdodecahedron K shown in (a), and indent the nearest3-valent vertex (b). It is clear that this results in a nonconvexpolyhedron. Since all 3-valent vertices of Kepler’s dodeca-hedron are equivalent, there is only one kind of indentationpossible. On the other hand, Bilinski’s dodecahedron B inFigure 10(a) has two distinct kinds of 3-valent vertices, so theindentation construction leads to two distinct polyhedra; seeparts (b) and (c) of Figure 10.

Returning to Figure 9, we may try to indent one of the3-valent vertices in (b). However, none of the indentationsproduces a polyhedron of spherical type. The minimaldeparture from this type occurs on indenting the vertexopposite to the one indented first; in this case the twoindented triplets of faces meet at the center of the originaldodecahedron (see Fig. 9c). We may eliminate this coin-cidence by stretching the polyhedron along the zonedetermined by the family of parallel edges that do notintrude into the two indented triplets. This yields a paral-lelogram-faced dodecahedron that is of spherical type (butnot a rhombic monohedron); see Figure 9(d). A relatedpolyhedron is shown in a different perspective asFigure 121 in Fedorov’s book [9].

It is of significant interest that all the isozonohedra inFigures 9 and 10—even the ones we do not quite accept,shown in Figures 9(c) and 10(e)—are parallelohedra. Thiscan most easily be established by manipulating a fewmodels; however, graphical or other computational verifi-cation is also readily possible.

To summarize the situation concerning dodecahedralrhombic monohedra, we have the following polyhedra ofspherical type:

Two convex dodecahedra (Kepler’s and Bilinski’s);Three simply indented dodecahedra (one from Kepler’spolyhedron, two from Bilinski’s);One doubly indented dodecahedron (from Bilinski’spolyhedron).We turn now to the two larger isozonohedra, Fedorov’s

icosahedron F and Kepler’s triacontahedron T. Since eachhas 3-valent vertices, it is possible to indent them, and sincethe 3-valent vertices of each are all equivalent under sym-metries, a unique indented polyhedron results in each case(Fig. 11).

The icosahedron F admits several nonequivalent doubleindentations (see Fig. 12); two are of special interest, and

Figure 8. Unkelbach’s hexecontahedron. It has pairs of dis-

joint, coplanar but not adjacent faces, which are parts of the

faces of the great stellated triacontahedron. All its vertices are

distinct, and all edges are in planes of mirror symmetry.

(b)(a)

(d)(c)

Figure 9. Indentations of the Kepler rhombic dodecahedron

K, shown in (a). In (b) is presented the indentation at the vertex

nearest to the observer; this is the only indentation arising from

(a). A double indentation of the dodecahedron in (a), which is

a single indentation of (b), is shown in (c); it fails to be a

polyhedron of the spherical type, since two distinct vertices

coincide at the center; hence it is not admitted. By stretching

one of the zones, as in (d), an admissible polyhedron is

obtained—but it is not a rhombic monohedron.


we shall denote them by D1 and D2. There are many othermultiple—up to sixfold—indentations; their precise num-ber has not been determined. An eightfold indentation ofthe triacontahedron T is shown in [39, p. 196]; it admitsseveral additional indentations.

The double indentations D1 and D2 of F shown inFigure 12 are quite surprising and deserve special mention:They areparallelohedra! Again, the simplest way to verify thisis by using a few models and investigating how they fit. Thiscontrastswith the singly indented icosahedron,which is not aparallelohedron. None of the other isozonohedra obtainableby indentation of F or T seems to be a parallelohedron.

A different construction of isozonohedra is through theunion of two or more given ones along whole faces, butwithout coplanar adjacent faces; clearly this means that allthose participating in the union must belong to the samefamily of rhombic monohedra—either the family of thetriacontahedron, or of Kepler’s dodecahedron, or of rhom-bohedra (with equal rhombi) not in either of these families.

Besides a brief notice of this possibility by Fedorov, the onlyother reference is to the union of two rhombohedra men-tioned by Kappraff [22, p. 381].9

For an example of this last construction, by attaching tworhombohedra in allowablewaysone canobtain threedistinctdecahedra, one of which is shown in Figure 13. Another ischiral, that is, comes in two mirror-image forms. This con-struction can be extended to arbitrarily long chains ofrhombohedra; from n rhombohedra there results a parallel-ohedron with 4n + 2 faces; see Figure 13 for n = 3. Foranother example, from three acute and one obtuse rhom-bohedra of the triacontahedron family, that share an edge,one can form a decahexahedron E. It is chiral, but it has anaxis of 2-fold rotational symmetry. By suitable unions of oneof these decahexahedron with a chain of n rhombohedra(n C 2), one can obtain isozonohedra with 4n + 16 faces. Allisozonohedra mentioned in this paragraph happen to beparallelohedra aswell.Hence there are rhombic monohedralparallelohedra for all even k C 6 except for k = 8.

The isozonohedra just described show that there existrhombic monohedral parallelohedra with arbitrarily longzones. However, there is a related open problem:

Given an even integer k C 4, is there a rhombicmonohedral parallelohedron such that every zone hasexactly k faces?The cube has k = 4, the rhombic dodecahedra K and B

have k = 6, and the doubly indented icosahedra D1 and D2

(b)(a)

(c) (d)

Figure 11. (a) Icosahedron F and (b) its indentation; (c)

Triacontahedron T and (d) its indentation.

(a)

(c)(b)

(e)(d)

Figure 10. Indentations of the Bilinski dodecahedron shown

in (a). The two different indentations are illustrated in (b) and

(c), the former at an ‘‘obtuse’’ 3-valent vertex, the latter at an

‘‘acute’’ vertex. The double indentation of (a), resulting from a

single indentation of (b), is presented in (d); (e) shows an

additional indentation of (c) which, however, is not a

polyhedron in the sense adopted here, since two faces overlap

in the gray rhombus.

9In carrying out this construction we need to remember that adjacent faces may not be coplanar. This excludes the ‘‘semicrosses’’ of Stein [32] and other authors,

although it admits the (1,3) cross. For more information see [33].


are examples with k = 8. No information is available for anyk C 10.

Although the number of examples of nonconvex isoz-onohedra and parallelohedra could be increased indefi-nitely, in the next section we shall propose a possibleexplanation of which isozonohedra are parallelohedra.10

Remarks(i) The parallelohedra discussed previously lack a center

of symmetry, which was traditionally taken as present inparallelohedra and more generally—in zonohedra. Convexzonohedra have been studied extensively; they have many

interesting properties, among them central symmetry.11

However, the assumption of central symmetry (of the faces,and hence of the polyhedra) amounts to putting the cartbefore the horse if one wishes to study parallelohedra—that is, polyhedra that tile space by translated copies

In fact, the one and only immediate consequence ofthe assumed property of polyhedra that allow tilings bytranslated copies is that their faces come in pairs that aretranslationally equivalent. For example, the octagonalprism in Figure 14 is not centrally symmetric, and its baseshave no center of symmetry either. But even so, it clearly is aparallelohedron. The dodecahedra in Figures 9(b) and10(b),(c) have no center of symmetry although their faces arerhombi and have a center of symmetry each. On the otherhand, the doubly indented polyhedron is Figure 10(d)has a center. As mentioned before, each of these is aparallelohedron.

We wish to claim that central symmetry is a red herringas far as parallelohedra are concerned. The reason that therequirement of central symmetry may appear to be naturalis that studies of parallelohedra have practically withoutexception been restricted to convex ones. Now, for convexpolyhedra the pairing of parallel faces by translationimplies that they have equal area, whence by a theorem ofMinkowski (see Footnote 2) the polyhedron has a center,which implies that the paired faces coincide with theirimage by reflection in a point—that is, are necessarilycentrally symmetric, and therefore are zonohedra. But thisargument is not valid for nonconvex parallelohedra, hencesuch polyhedra need not have a center of symmetry.

In his first short description of nonconvex parallelohe-dra, Fedorov writes (§83 in [9, p. 306]):

The preceding deduction of simple [that is, centrallysymmetric polyhedra with pairwise parallel and equalfaces] convex parallelohedra is equally applicable tosimple concave [that is, non-convex] ones, and hence webring here only illustrations. We do not show the concave

(a) F

(c) D2(b) D1

(a) (b)

Figure 12. (a) The Fedorov rhombic icosahedron F; (b) A

double indentation of the F yields a nonconvex rhombic

icosahedron D1 of the spherical type that is a parallelohe-

dron; (c) A different double indentation D2 is also a

parallelohedron.

Figure 13. Isozonohedra with 10 and 14 faces.

Figure 14. A nonconvex parallelohedron without a center of

symmetry.

10Crystallographers are interested in parallelohedra far more general than the ones considered here: The objects they study in most cases are not polyhedra in the

sense understood here, but object combinatorially like polyhedra but with ‘‘faces’’ that need not be planar. The interested reader should consult [29] and [24] for more

precise explanations and details.11It is worth mentioning that Fedorov did not require any central symmetry in the definition of zonohedra ([9, p. 256], [10, p. 688]). However, he switched without

explanation to considering only zonohedra with centrally symmetric faces. As pointed out by Taylor [36], this has become the accepted definition.


tetraparallelohedron [the hexagonal prism] since this issimply a prism with a concave par-hexagon as basis.Fig. 121 presents the ordinary, and Fig. 122 the elongatedconcave hexaparallelohedron [the rhombic dodecahe-dron and the elongated dodecahedron]; Fig. 123 showsthe concave heptaparallelohedron [the truncated octa-hedron]. Obviously, there exists no concave triparallel-ohedron [cube]. (My translation and bracketed remarks)

Fedorov’s parallelohedron in Figure 121 of [9] is isomor-phic to the polyhedron shown in our Figure 9(d). A mono-hedral rhombic dodecahedron combinatorially equivalent toit is shown in our Figure 10(d) and is derived from theBilinski dodecahedron.

However, Fedorov does not provide any proof for hisassertion, and in fact it is not valid in general. For example,his Figure 123 does not show a polyhedron of spherical type,since one of the edges is common to four faces. This can beremedied by lengthening the short horizontal edges, butshows the need for care in carrying out the construction.

(ii) The study of nonconvex parallelohedra necessitatesthe revision of various well-established facts concerningconvex parallelohedra. For example, one of the crucialinsights in the enumeration of parallelohedra (and parallel-otopes in higher dimensions) is the property that every zonehas either four or six faces. This is not true for nonconvexparallelohedra. For example, the double indentation D1 ofFedorov’s F shown in Figure 12(b) is a parallelohedron—even though all zones of D1 have 8 faces.

For another example, in some cases changing of thelengths of edges of a zone has limitations if the sphericaltype is to be preserved.

At present, there seems to be no clear understanding ofthe requirements on a polyhedron of spherical type to be aparallelohedron. As mentioned earlier, the three indentedpolyhedra in Figures 9(b) and 10(b),(c) are parallelohedra;They can be stacked like six-sided prisms. In fact, with agrain of salt added, starting with suitably chosen six-sidedprisms, they may be considered as examples of Fedorov’ssecond construction of nonconvex polyhedra [9, p. 306]:

If we replace one or several faces of a parallelohedron,or parts of these, by some arbitrary surfaces supportedon these same broken lines, in such a way that a closedsurface is obtained, and observing that precisely thesame [translated] replacement is made in parallel posi-tion on the faces that correspond to the first ones or theirparts, then, obviously the new figure will be a parallel-ohedron, though without a center….It seems clear that Fedorov did not consider this con-

struction important or interesting, since he did not provideeven a single illustration. But it does lead to parallelohedrawith some or all faces triangular, in contrast to the convexcase; an example is shown in Figure 15. A more elaborateexample of a nonconvex parallelohedron with some trian-gular faces, that does not admit a lattice tiling, is described bySzabo [35].

Another difference between convex and nonconvexparallelohedra is that the convex ones can be decomposedinto rhombohedra; this is of interest in various contexts—see, for example, Hart [18] and Ogawa [28]. In general, such

decomposition is not possible for nonconvex parallelohe-dra. For example, the doubly indented dodecahedron inFigure 10(d) is not a union of rhombohedra.

(iii) Examination of the various isozonohedra that are—or are not—parallelohedra, together with the observationthat questions of central symmetry appear irrelevant in thiscontext, lead to the following conjecture:

ConjectureLet P be a sphere-like polyhedron, with no pairs ofcoplanar faces. If the boundary of P can be partitionedinto pairs of non-overlapping ‘‘patches’’ {S1, T1}; {S2, T2};…; {Sr, Tr}, each patch a union of contiguous faces, suchthat the members in each pair {Si, Ti} are translates ofeach other, and the complex of ‘‘patches’’ is topologicallyequivalent as a cell complex to one of the parallelohedrain Figure 2, then P is a parallelohedron. Conversely, ifno such partition is possible then P is not aparallelohedron.As illustrations of the conjecture, we can list the fol-

lowing examples:

(a) The three singly indented dodecahedra in Figures 9 and10 satisfy the conditions, with the patches S1, T1 formedby the triplet of indented faces and their opposites, andthe other pairs formed by pairs of opposite faces. Thenthis cell complex is topologically equivalent to the cellcomplex of the faces of the six-sided prism (Fig. 2d). Aswe know, these dodecahedra are parallelohedra. Notethat the fact that they are combinatorially equivalent tothe convex dodecahedra K and B is irrelevant, since thecomplex of pairs of faces of the indented polyhedra isnot isomorphic to that of the un-indented ones: Somepairs {Si, Ti} of parallel faces are separated by only asingle other face, whereas in K and B they are separatedby two other faces.

(b) The doubly indented dodecahedron in Figure 10(d)complies with the requirements of the conjecture in adifferent way: Each pair {Si, Ti} consists of just a pair ofparallel faces; the complex so generated is isomorphicto the one arising from Kepler’s K.

(c) The doubly indented icosahedron D1 of Fedorov’s F,shown in Figure 12(b), provides additional support forthe conjecture. Two of the pairs—say {S1, T1} and {S2,T2}—are formed by the indented triplets and their

Figure 15. A monohedral parallelohedron with triangles as

faces.


opposites. The other pairs {Si, Ti} are the remaining fourpairs of parallel faces. The complex they form isisomorphic to the face complex of the elongateddodecahedron shown in Figure 2(b). The same situa-tion prevails with the doubly indented icosahedron D2

of Figure 12(c). Other double indentations of theicosahedron F, as well as the single indentation of F,fail to satisfy the assumptions of the conjecture and arenot parallelohedra.

(d) No indentation of the rhombic triacontahedron satisfiesthe assumptions of the conjecture, and in fact none is aparallelohedron.

(e) The decahexahedron E mentioned previously has adecomposition into pairs {Si, Ti} that is isomorphic tothe complex of the faces of the cube. The samesituation prevails with regard to the chains of rhombo-hedra mentioned previously.

(iv) The present article leaves open all questions regard-ing parallelohedra that are not rhombic monohedra. Inparticular, it would be of considerable interest to generalizethe above conjecture to these parallelohedra. Such anextension would also have to cover the results on ‘‘clusters’’of cubes such as the crosses and semicrosses investigated byS. K. Stein and others [32, 33, 14]. One can also raise thequestion of what are analogues for suitably defined ‘‘clus-ters’’ of rhombohedra, or other parallelohedra.

(v) There just possibly may be a prehistory to the Bilinskidodecahedron. As was noted by George Hart [17, 18], a netfor a rhombic dodecahedron was published by John LodgeCowley [6] in the mid-eighteenth century; see Figure 16. Therhombi in this net appear more similar to those of the Bilinskidodecahedron than to the rhombi of Kepler’s. However,

these rhombi do not have the correct shape and cannot befolded to form any polyhedron with planar faces. (Since theangles of the rhombi are, as close as can be measured, 60�and 120�, the obtuse angles of the shaded rhombus would beincident with two other 120� angles—which is impossible.)An Internet discussion about the net mentioned the possi-bility that the engraver misunderstood the author’sinstructions; however, it is not clear what the author actuallyhad in mind, since no text describes the polyhedron.The later edition of [6] mentioned by Hart [17] was notavailable to me.

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http://www.georgehart.com/virtual-polyhedra/dodecahedra.html

http://www.georgehart.com/virtual-polyhedra/dodecahedra.html

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http://www.orchidpalms.com/polyhedra/rhombic/RTC/RTC.htm

Mathematically Bent Colin Adams, Editor

The proof is in the pudding.

Opening a copy of The Mathematical Intelligencer

you may ask yourself uneasily, ‘‘What is this

anyway—a mathematical journal, or what?’’ Or

you may ask, ‘‘Where am I?’’ Or even ‘‘Who am I?’’

This sense of disorientation is at its most acute when

you open to Colin Adams’s column.

Relax. Breathe regularly. It’s mathematical, it’s a

humor column, and it may even be harmless.

� Column editor’s address: Colin Adams,

Department of Mathematics,

Bronfman Science Center, Williams College,

Williamstown, MA 01267, USA


Group TherapyCOLIN ADAMS

DDr. Stew: Hello, come on in. You’re Hank, right? I’dlike you to meet the group. This is Karen, and Bill,and Amanda and Sylvia. I’m Dr. Johnson, but you

can call me Dr. Stew. Is this your first experience with grouptherapy?Hank: Uh, yes, it is.Dr. Stew: Well I think you will find it very helpful. Whydon’t you take a seat over there between Karen and Bill.They’re married, but they don’t mind sitting apart.Hank: Married? You allow two people who are married inthe same group therapy session?Dr. Stew: Why not? Since they have a child, we know theycan multiply.

(Laughter from the group. Hank laughs along nervously.)Dr. Stew: Now, the basic idea of group therapy is to asso-ciate our issues and problems with concepts in grouptheory.Hank: What?Dr. Stew: You know, the properties of abstract groups.Hank: I’m confused.Dr. Stew: You are here for group therapy, aren’t you?Hank: But group therapy refers to the fact that there is agroup of us here in the room.

(Everyone laughs.)Dr. Stew: Hardly. It refers to a group. You know, a set ofelements with a multiplicative operation, inverses, etc. Youhave had Abstract Algebra, haven’t you?Hank: Yes, but …Dr. Stew: Good.Well, let’s get started. Sylvia, youwere talkinglast week about how your mother favors you over your sister.Do you think she has put a partial ordering on your wholefamily?Sylvia: No question. Often the poset structure becomesmore important than the algebraic structure.Dr. Stew: Does that bother you?Sylvia: Yes, it seems misguided. Why should a set-theoreticconstruct supersede an algebraic one? Over many years, wehave built up a sophisticated set of relations that haveallowed us to interpret our family relationships as a group.To give up all that structure seems counterproductive.Hank: Wait a minute. How sophisticated an algebraic objectcan it be? After all, your family is a finite set.Sylvia: Are you implying that finite groups can’t be inter-esting? What about the general linear group of dimension nover a finite field? What about the Weyl groups? I find yourattitude quite condescending.Hank: I’m sorry. I didn’t mean to …


Sylvia: My family happens to be isomorphic to the quasid-ihedral group. And I am proud of that.Dr. Stew: Please everyone. Let’s try to be there for eachother. Hank, we have a rule here. ‘‘Our support shouldalways have measure 1.’’Hank: Sorry, I just …Dr. Stew: Okay, let’s turn to Karen. How are things with youthis week?Karen: Well, we had family over for the holidays, and what adisaster that was. My younger sister Emily, who was recentlydivorced, showed up with her new boyfriend Frank. Oh,was he obnoxious. He made lewd jokes, insulted mygrandmother to her face, and accused my mother of inten-tionally giving him the smallest pork chop. Then, afterdinner, my older sister Claire showed up with her new beauand it was none other than Emily’s ex-husband Craig.Everyone was stunned. Especially since we all couldn’tstand Craig when he was married to Emily.Sylvia: What happened?Karen: Amazingly enough, Frank and Craig hit it off. Theytook turns insulting the decor, the food and members of thefamily. They were awful.Dr. Stew: Well, the subgroup generated by Craig and Frankseems to be a problem.Hank: All families have subgroups like this. Sounds normalenough to me.Karen: What? Are you kidding? There is no way this sub-group is normal. Just conjugate it by Emily, and you don’tget the same subgroup.

Bill: Hank, where exactly did you take group theory?Sylvia: Are you confusing conjugate with conjugal?Hank: That wasn’t what I meant.Dr. Stew: If the subgroup generated by Frank and Craigwere normal, then the family could quotient out by thesubgroup and they would have a perfectly functioningfamily group again. But unfortunately, the subgroup is notby any means normal.Hank: I didn’t mean normal in the group-theoretic sense.You’re making this all so complicated. It seems to me itshould be simple.Bill: Well, it’s one thing to acknowledge that the subgroupgenerated by Frank and Craig is not normal, but to claimnone of the proper subgroups is normal, well that’s anothermatter.Karen: (Angry.) What makes you think that there isn’t asingle proper subgroup in my entire family that is normal?You don’t even know my family. It’s incredibly presump-tuous on your part.Hank: No, I didn’t mean …Dr. Stew: Okay, I think we had better move on. Hank,please try to be considerate of other group member’s situ-ations. Let’s do some free association. I say a word, you saywhat you think of. We’ll go around the room, starting withAmanda. Blue.Amanda: Carrot.Dr. Stew: Good. Sylvia, weasel.Sylvia: Chocolate.Dr. Stew: Good. Bill, clarinet.Bill: Horse.Dr. Stew: Good. And for Hank, wingnut.

Hank: Excuse me, but I’m confused. These associationsdon’t seem to have anything to do with the words you aresaying.Dr. Stew: That’s right.Hank: But then I don’t understand why people are sayingthem.Dr. Stew: Well, Hank, you have to remember. Everything’sassociative in a group.Hank: Oh, come on …Dr. Stew: Anyway, enough of that. Amanda, what’s beengoing on with you?

Amanda: Well, my father has been commuting back andforth between Boston and New York. He’s just home on theweekends.Dr. Stew: By himself? Does he drive?

Amanda: He usually rides with my uncle and theneighbor down the street. They have the same situation.Karen: That must be tough for you.Bill: You must miss him a lot.Hank: Isn’t someone going to say something about thecommutator subgroup, or the group being abelian becauseeveryone commutes?Dr. Stew: No. What does that have to do with anything?Hank: I am so confused.Dr. Stew: Let’s talk about that, Hank. It sounds like you arehaving an identity crisis.Hank: It does?Dr. Stew: Yes, it’s unclear who is the identity in your familygroup.Hank: I’m not following you.Dr. Stew: Who is it when multiplied by any other member ofthe family yields that same member of the family?Hank: You know, I am having some trouble interpretingthis analogy with a group. What exactly is multiplication oftwo people?Dr. Stew: Well, what would you like it to be?Hank: How about something I can understand, not justsome nebulous ill-defined concept created so this piss pooranalogy can be sustained ad nauseam.Dr. Stew: I sense some hostility from you.Hank: Well, yes, I am a little frustrated.Dr. Stew: Maybe we can figure out where this hostility iscoming from. It probably goes back a ways. Do you havesiblings?Hank: Yes. I have two older brothers, Jeff and Tom, and anolder sister Caroline. And then a much younger sister Liz.She was really brought up more by my three older siblingsthan by my parents.Dr. Stew: I see. Now, tell us. If you were going to give aword in the generators that are your mother, father, andsiblings that best describes you, what would it be?Hank: Excuse me?Dr. Stew: You know. Let M denote your mother, F yourfather, J for Jeff, T for Tom, C for Caroline and L for Liz. Thenmake a word from these generators and their inverses thatbest describes you, that encompasses what parts of youcome from these generators. We are all a product of ourfamilies.Hank: I would say you’re kidding, but I am guessing youare not. Okay, I’ll play along, How about … um …


M F C-1JTC L-1C F -1M -1?Bill: Ha!Hank: What? What now?Bill: That’s the trivial word.Hank: No it’s not.Bill: Yes it is. You said before that Liz was a product of Jeff,Tom and Caroline so L equals JTC, meaning JTCL-1 is trivial.That was in the middle of your word. Once we trivializethat, the remaining generators and inverses cancel and theword collapses. So yes that is the trivial word.

Dr. Stew: Hank, I find it indicative of your feelings of selfloathing that you would pick a trivial word to describeyourself. Clearly an identity crisis, as you see yourself as theidentity element.Hank: (Stands up.) That’s it. I am out of here. You peopleare crazy.Dr. Stew: We don’t use the word ‘‘crazy,’’ Hank. We say‘‘topologically mixing.’’ But, you know, perhaps it is best ifyou leave. You aren’t ready for group therapy. I think youneed one-on-one help, probably on a continuous basis. Youshould see an analyst.


Years Ago David E. Rowe, Editor

Two Great Theoremsof Lord Brounckerand His Formulabðs�1Þbðsþ1Þ ¼ s2;

bðsÞ ¼ s þ 12

2s þ 32

2sþ 522sþ

. ..

ð1Þ

SERGEY KHRUSHCHEV

Spring of 1655

AAGoogle search for ‘Spring of 1655’ returned a list ofevents, including the Insurrection of March 1655against Cromwell [30], the discovery of a satellite of

Saturn by Christiaan Huygens [16, pp.14–16], and, by theway, a reference to my own paper on Brouncker’scontinued fraction [19]. These events, with the exceptionof my paper, took place in March of 1655, when Brounckermade his greatest discovery in mathematics.

My paper had the modest purpose of restoring historicaljustice to William Brouncker, one of the brilliant minds inEngland in those times.

Cromwell’s contribution reduced to suppression of Roy-alists, which on the one hand kicked Brouncker out of BigPolitics for about 15 years, to which he finally returned onlyin 1660, and on the other hand promoted John Wallis to theposition of Professor in Oxford, since he belonged to theopposite political camp and, according to historical records,even helped the Parliamentarians decode Royalist messages.SinceBrouncker achievednothingmore inmathematics after1660, I have the strong conviction that without Cromwell

there would be no Brouncker’s formula, especially becausethere would, possibly, be no Wallis’s formula either.

Christiaan Huygens didn’t believe Brouncker’s formulaand asked one question. We discuss his role in more detaila bit later.

To clarify the role of Saturn whose symbol is oppositethat of Jupiter I cite a website on astrology: ‘‘In astrology,Saturn is associated with restriction and limitation. WhereJupiter expands, Saturn constricts. Although the themesof Saturn seem depressing, Saturn brings structure andmeaning to our world. Saturn knows the limits of time andmatter. Saturn reminds us of our boundaries, our responsi-bilities, and our commitments. It brings definition to ourlives. Saturn makes us aware of the need for self-control andof boundaries and our limits.’’1

Arithmetica InfinitorumBy the end of 1654, John Wallis, a good friend of Brounc-ker, had almost finished his book Arithmetica Infinitorum.A few words to explain the title: The area under the graphof y = x over the segment [0, 1] of the real line can beobtained by approximating the graph from below withinscribed rectangles with bases [k/n, (k + 1)/n], k = 1, ...,n - 1. Their total area

1

nþ 2

nþ . . .þ n� 1

n

� �� 1

n¼ nðn� 1Þ

2n2¼ 1

2� 1

2n

approaches 1/2 as n? +?, which is the area of the trian-gle. This method obviously extends to ‘parabolas’ y = x p, pbeing a positive integer, by the arithmetic formulas

Xn�1

k¼1

k p ¼ n pþ1

pþ 1þ O n pð Þ:

So, the title Arithmetica Infinitorum meant a new Arith-metics of Infinities. At the beginning of 1655, Wallis’s mainproblem with this book was that he couldn’t complete itwith an arithmetical formula for the area under the arc of thecircle y ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffi1� x2p

over [0, 1]. He knew, of course, Viete’sformula

2

p¼

ffiffiffi2p

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2þ

ffiffiffi2pp

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2þ

ffiffiffi2ppq

2� . . . ; ð2Þ

but it didn’t fit his understanding of Arithmetics of Infinities,since the number of radicals in (2) increases with everynew step to the right.

Moreover, Wallis had another restriction. He planned toobtain this new formula using his method of interpolation,presented for the first time in his book. In modern notations

� Send submissions to David E. Rowe,

Fachbereich 08, Institut fur Mathematik,

Johannes Gutenberg University,

D-55099 Mainz, Germany.

e-mail: [email protected] 1http://www.cafeastrology.com/saturn.html


http://www.cafeastrology.com/saturn.html

Wallis wanted to find an arithmetic expression for theintegral

p4¼Z 1

0

ffiffiffiffiffiffiffiffiffiffiffiffiffi1� x2p

dx:

MotivatedbyViete’s formula, he introduced a family I(p, q) ofreciprocals to the integrals, which he was able to compute:

Iðp; qÞ ¼ 1R 1

0 ð1� x1=pÞqdx¼ ðpþ 1Þðpþ 2Þ. . .ðpþ qÞ

1 � 2 � . . . � q

¼ pþ q

p

� �: ð3Þ

Here p and q are positive integers. For p = 1/2, q = n wehave

I1

2;n

� �¼ 1 � 3 � . . .ð2nþ 1Þ

2 � 4 � . . . � ð2nÞ ¼1

vn: ð4Þ

Clearly, vn decreases with increasing n. Since this sequence isobtained by a very simple law, one may hope that it may benaturally interpolated to positive numbers and, in particular,to n = 1/2. The value of v0 is 1 and of v1 is 2/3 =

0.66. . . The value of interest v1/2 = p/4 = 0.78. . . is regularlyplaced between v0 and v1. In other words, the results of thisnumerical experiment can only be explained by some simpleformula for v1/2. But what is the formula? This questionbothered Wallis a lot and, being convinced that he was onlyone step away from the solution (which was, in fact, thecase), he continued his tremendous efforts to find the for-mula. Finally, good luck entered on his side (possibly underthe influence of Saturn) and Wallis not only found the infiniteproduct

1

2Ið1=2; 1=2Þ

¼ 2

p¼ 1 � 3

2 � 2 �3 � 54 � 4 �

5 � 76 � 6 � . . . � ð2n� 1Þ � ð2nþ 1Þ

2n � 2n� . . .; ð5Þ

now bearing his name, he also proved its convergence to2/p.

On the one hand, according to Stedall [33], pp. xviii-xix,formula (5) appeared after February 28, 1655. On the otherhand, already at the beginning of April, 1655, Wallisresponded to Hobbes’s threats to reveal a quadrature of theunit circle by publishing some excerpts from [35]. It followsthat (5) was most likely proved at the beginning of March,1655.

Now put yourself in Wallis’s place. What would you do ifone way or another you obtained such a brilliant result? Yes,Wallis did the same thing. He wrote a letter to his friend inmathematics and music, William Brouncker, see [22]. Soon,Brouncker responded with the formula (1) and mentionedthat b(1) = 4/p. In modern literature, the latter formula isnamed for him. This is not completely correct, since theidentity b(1) = 4/p can be obtained quite elementarily [10].It is much more difficult to prove that b(s - 1)b(s + 1) = s2,which actually was Brouncker’s great contribution. How-ever, there are exceptions: See, for instance, [5], where onecan find a version of the story of Brouncker’s formula quiteclose to that promoted by this paper.

Wallis’s PuzzleThings are not so simple with Brouncker’s proof. In the lastsection 191 of [35], Wallis writes that he tried several timesto convince Brouncker to publish the proof, but all hisattempts were in vain. One can only guess why Brounckerdidn’t want to publish it. It is quite possible that—bearing inmind his position as a true Royalist, especially in view of theInsurrection of March 1655—Brouncker just didn’t wantto be involved in the severe controversy between Wallisand Hobbes [3], which also started in March of 1655. Thetruth in this controversy was, of course, on Wallis’s side,since Hobbes was a man completely unable to understandmathematics. However, very often such public disputes,especially with educated people, result in big trouble.

Being a great patriot ofBritishmathematics, Wallis couldn’tleavewhathegot fromBrounckerunpublished.Therefore, heundertook an attempt to present this in his book, providingexplanations of his own. I think that he understood that hiscomments were not complete and required some furtherstudy. But as is clear from the lastwords of his book, hehopedthat later this would be explained in full detail. As furtherdevelopments showed, it was a very good decision which, infact, doubled the value of [35].

The puzzle Wallis left didn’t escape the careful attentionof Euler, who even took his copy of Arithmetica Infinito-rum to St. Petersburg. In [12, 17] Euler writes:

This theorem, which explicitly presents values of thecontinued fraction as integral formulas, deserves men-tion the more as it be less obvious. . . . Therefore, forquite a long time I have undertaken great efforts toprove this Theorem so that its proof a priori can berelated to this function; this research, in my opinion, is

.........................................................................

AU

TH

OR SERGEY KHRUSHCHEV was born in

Leningrad, USSR, graduated from the.

Leningrad State University, Department of

Mathematics and Mechanics, in 1972, and

received his D.Sc. degree from the Steklov

Institute. (Leningrad branch) in 1982. From

1987 to 1995 he directed the creation

of the Euler International MathematicalInstitute in Leningrad/St.Petersburg. After

seven years as full professor at Atilim

University in Ankara, Turkey, he is now full

professor at the Eastern Mediterranean

University in North Cyprus. His research

interests include classical analysis, operator

theory and probability theory. In his spare

time he climbs mountains.


Eastern Mediterranean University

North Cyprus via Mercin 10

Gazimagusa

Turkey




more difficult, but I believe it may result in great bene-fits. While any such research was condemned to failure,I regret most of all the fact that Brouncker’s method wasnowhere present and most likely fell into oblivion.Euler couldn’t find a solution to Wallis’s puzzle, but his

proofs of (1) led him to the discovery of the theory of theGamma and Beta functions [20].

Huygens’s QuestionWhen [35] was finally published in 1656, Wallis distributeda number of copies of the book among working mathe-maticians. One copy of [35] reached Huygens, who foundBrouncker’s formula in its last section. There is no doubtthat Huygens understood nothing from Wallis’s comments,since he demanded a numerical confirmation. This was, bythe way, not such an easy task since, as Euler later showedin [10], the convergents to b(1) are nothing but

1

1þ

12

2 þ

32

2 þ��þ

ð2n� 1Þ2

2¼ Qn

Pn¼Xn

k¼0

ð�1Þk

2k þ 1! p

4; ð6Þ

where to save paper we use for continued fractions Roger’snotations

Pn

Qn¼ 1þ 12

2 þ

32

2 þ��þ

ð2n� 1Þ2

2:

Similar to the case of sumsP

k=1n , continued fractions can

be also written as

Pn

Qn¼ 1þ K

n

k¼1

ð2k � 1Þ2

2

!:

Here K stands for German ‘‘Kettenbruche’’. The alternatingseries in (6) converges to p

4 ¼ arctanð1Þ but not veryquickly. In addition, if one does not know even formula(6), which was the case with Huygens, the prospect ofevaluating the continued fraction with, say 20 simpleterms ð2k�1Þ2

2 called partial fractions, does not look veryencouraging.

Needless to say, Wallis redirected Huygens’s question toBrouncker. And Brouncker shortly found an ingenioussolution. We can rewrite Brouncker’s formula as follows:

bðsÞbðs þ 2Þ ¼ ðs þ 1Þ2: ð7Þ

On the one hand, a look at Brouncker’s continued fraction(1) shows that its evaluation for large s requires fewerpartial fractions. On the other hand, formula (7) suggests away to relate b(1) with, say, b(4n + 1) and b(4n + 3):

bð4nþ 1Þ ¼ 22

1 � 3 �42

3 � 5 � . . . � ð2nÞ2

ð2n� 1Þð2nþ 1Þ �4

p� ð2nþ 1Þ;

bð4nþ 3Þ ¼ 1 � 322� 3 � 5

42� . . . � ð2n� 1Þð2nþ 1Þ

ð2nÞ2� ð2nþ 1Þp:

ð8Þ

If n = 6, then the continued fraction b(4 � 6 + 1) =

b(25) has partial denominators 2 � 25 = 50, which consid-erably improves its convergence. Thus, we obtain thefollowing bounds for p:

b0 �Q2kþ1

P2kþ1\p\b0 �

Q2k

P2k;

where

b0 ¼ 4 � 22 � 42 � 62 � 82 � 102 � 122

32 � 52 � 72 � 92 � 112

¼ 78:602424992035381646. . .;

and P /Q are convergents to b(25). Putting k = 0, 1, 2 in theabove formula, we find that

k ¼ 0 3:14158373269526\p\3:14409699968142

k ¼ 1 3:14159265194782\p\3:14159274082066

k ¼ 2 3:14159265358759\p\3:14159265363971 :

Notice that already the first convergent to b(25) gives fourtrue places of p. The fifth convergent without tedious cal-culations gives 11 true places. This was the first algebraiccalculation of p. Viete in 1593 couldn’t use his formula (2)and instead applied the traditional method of Archimedes toobtain 9 decimal places. The same, by the way, is true forWallis’s infinite product (5). In 1596, Ludolph van Ceulenobtained 20 decimal places by using a polygon with 60 9 229

sides. The cumbersome calculations made by Ludolphare incomparable with Brouncker’s short and beautifulcalculations. A detailed historical account of Brouncker’scalculations can be found in [31]. It seems that this achieve-ment of Brouncker’s remained unnoticed, and even hisformulas (8) were later rediscovered by Euler.

In 1654 Huygens published a book [18] in which he pre-sented his geometric method, which considerably improvedthat of Archimedes. However, although Huygens got fromBrouncker the numerical confirmation he demanded, hedefinitely didn’t realize its importance. As is clear fromabove, one can find with Brouncker’s method as many truedecimal places of p as necessary. In [27, pp.75–77] there is atable of achievements in finding true decimal places of p. Ihave no doubt that Brouncker with his formula could get allthe places obtained in the period of 1596–1793 in just acouple of evenings. Still, no book includes him in the lists ofwinners . . .

How Was It Done?Now the time has come to reveal the secrets of Brouncker’sproof. To begin with, it is important to realize, as I confirmin the next section, that Brouncker developed the theory ofcontinued fractions with positive terms. Next, we knowfrom § that Brouncker got formula (5) from Wallis.What would a person like Brouncker who developed abeautiful theory do in this case? First, he would rewrite thepartial products of Wallis’s infinite product as a continuedfraction:

1 � 32 � 2

� 3 � 54 � 4 �

5 � 76 � 6 � . . . � ð2n� 1Þ � ð2nþ 1Þ

2n � 2n

¼ 1 � 30 þ

2 � 20 þ

3 � 50 þ...þ

ð2n� 1Þ � ð2nþ 1Þ0 þ

ð2nÞ � ð2nÞ1

:

ð9Þ


The problem with the continued fraction in (9) is that itszero partial denominators cannot be extended ad infini-tum, since if one does this, the convergents will alter-nate between 0 and ?. (Wallis, in his comments, refers tosuch a continued fraction as oscillating.) Another look at(9) shows that the odd numerators are products of the forms(s + 2) = s2 + 2s = (s + 1)2 - 1, where s is odd. Now, atthe very beginning of §191 in [35] we find: ‘‘The NobleGentleman noticed that two consecutive odd numbers, ifmultiplied together, form a product which is the square ofthe intermediate even number minus one. . . . He asked,therefore, by what ratio the factors must be increased toform a product, not equal to those squares minus one, butequal to the squares themselves’’.

This suggests increasing s to b(s) and s + 2 to b(s + 2)for odd s in the right-hand side of (9) so that they satisfy (7).Then, to keep (9) valid, odd zero partial denominators inthe right-hand side of (9) must become positive. That isexactly what we need to complete the proof. The factthat s + 1 is even is also helpful, since it may providenecessary cancellations. Now, using (7) repeatedly, we maywrite

bð1Þ¼ 22

bð3Þ¼22

42bð5Þ¼22

42

62

bð7Þ¼22

42

62

82bð9Þ¼ . . .

¼22

42

62

82

102

122. . .ð4n�2Þ2

ð4nÞ2bð4nþ1Þ

¼12

22

32

42. . .ð2n�1Þ2

ð2nÞ2bð4nþ1Þ

¼1 �322�3 �5

42�5 �7

62� . . . � ð2n�1Þð2nþ1Þ

ð2nÞ2�bð4nþ1Þð2nþ1Þ :

ð10Þ

Combined with Wallis’s formula, this implies

bð1Þ¼ 2

pþoð1Þ

� ��bð4nþ1Þð2nþ1Þ ; ð11Þ

where o(1) denotes a function approaching 0 at ?.Since s + 2 \ b(s + 2) and b(s)b(s + 2) = (s + 1)2, we

have

s\bðsÞ\s2 þ 2s þ 1

s þ 2¼ s þ 1

2þ s; ð12Þ

which together with (11) give

bð1Þ ¼ limn

2

pþ oð1Þ

� �� lim

n

bð4nþ 1Þð2nþ 1Þ ¼

4

p: ð13Þ

It remains to find a formula for b(s).Let us analyze how one obtains the regular continued

fraction for a real number, say p. First, the number isexpanded into an infinite decimal fraction

p ¼ 3:1415926535. . .

¼ 3þ 1

10þ 4

102þ 1

103þ 5

104þ 9

105þ 2

106þ 6

107. . . :

Then this infinite decimal fraction is cut at, say, the 10thplace. After that, the decimal fraction obtained is converted

into a simple fraction, and, by the Euclidean algorithm, canfinally be developed into a regular continued fraction:

3:1415926535 ¼ 31415926535

10000000000

¼ 3þ 1

7þ 88514255

1415926535

¼ 3þ 1

7þ 1

15þ88212710

88514255

¼ 3þ 1

7 þ

1

15þ

1

1þ

301545

88212710

¼ 3þ 1

7 þ

1

15 þ

1

1 þ

1

292 þ

161570

301545:

This gives us the first seven true terms [3; 7, 15, 1, 292, 1, 1]of the regular continued fraction for p. The later one cutsoff the decimal expansion of p, the more true terms of thecontinued fraction are obtained. Motivated by (12), we cansimilarly represent b(s) as a sum:

bðsÞ ¼ s þ c0 þc1

2sþ c2

s2þ c3

s3þ . . . : ð14Þ

The coefficients c0, c1, c2, . . ., can be found inductivelyusing (7). By (12), c0 = 0. To find c1 we assume that

bðsÞ ¼ s þ c1

sþ . . .

and then determine c1 from the equation

s2 þ 2s þ 1 ¼ bðsÞbðs þ 2Þ ¼ s2 þ 2s þ 2c1 þ oð1Þ; s! þ1;

implying that c1 = 1/2. It follows that

bðsÞ ¼ s þ 1

2sþ c2

s2. . . :

Similarly, c2 = 0, c3 = -9/8, c4 = 0, c5 = 153/16, c6 = 0and, therefore,

bðsÞ ¼ s þ 8s4 � 18s2 þ 153

16s5þ c7

s7. . . :

Cutting the above formula at c6s�6 � 0 and applying theEuclidean algorithm to the quotient of polynomials, wehave

8s4 � 18s2 þ 153

16s5¼ 1

2s þ 9ð4s3 � 34sÞ8s4 � 18s2 þ 153

¼ 1

2s þ 9

8s4 � 18s2 þ 153

4s3 � 34s

¼ 1

2s þ 9

2s þ 25ð2s2 þ 153=25Þ4s3 � 34s

¼ 1

2s þ 9

2s þ 252s þ . . .

:

A remarkable property of these calculations is that 12 = 1,32 = 9, 52 = 25, etc., appear automatically as common


divisors of the coefficients of the polynomials in Euclid’salgorithm. The fraction 153/25 appears only because wedidn’t find the exact value of c7. Increasing the number ofterms in (14), we naturally conclude that

bðsÞ ¼ s þ 12

2s þ

32

2s þ

52

2s þ

72

2s þ ... þ

ð2n� 1Þ2

2s þ...: ð15Þ

If you think that it was already enough to finish theproof for a mathematician working in 1655, when evenNewton’s calculus was not available, then you make a bigmistake, because following Wallis’s writings you discoverthe following formulas:

P0ðsÞQ0ðsÞ

P0ðs þ 2ÞQ0ðs þ 2Þ � ðs þ 1Þ2 ¼ sðs þ 2Þ � ðs þ 1Þ2 ¼ ð�1Þ ;

P1ðsÞQ1ðsÞ

P1ðs þ 2ÞQ1ðs þ 2Þ � ðs þ 1Þ2 ¼ 4s4 þ 16s3 þ 20s2 þ 8s þ 9

4s2 þ 8s

� 4s4 þ 16s3 þ 20s2 þ 8s

4s2 þ 8s¼ 9

4s2 þ 8s;

P2ðsÞQ2ðsÞ

P2ðs þ 2ÞQ2ðs þ 2Þ � ðs þ 1Þ2

¼ 16s6 þ 96s5 þ 280s4 þ 480s3 þ 649s2 þ 594s

16s4 þ 64s3 þ 136s2 þ 144s þ 225� ðs þ 1Þ2

¼ �225

16s4 þ 64s3 þ 136s2 þ 144s þ 225: ð16Þ

One can find these very formulas in [33, pp. 169–170],where Wallis writes after the last formula: ‘. . . which is lessthan the square F2 + 2F + 1. And thus it may be done asfar as one likes; it will form a product which will be (inturn) now greater than, now less than, the given square(the difference, however, continually decreasing, as isclear), which was to be proved.’ [In Wallis’s notations s = F.]

To clarify these comments of Wallis, we sketch brieflyBrouncker’s theory of continued fractions with positiveterms.

The Euler-Wallis FormulasTo evaluate a finite continued fraction, for instance thefourth convergent to the regular continued fraction for p:

3þ 1

7þ

1

15þ

1

1þ

1

292;

one must rewrite it as an ascendant continued fraction

11

11

292þ 1þ 15

þ 7þ 3 ¼ 355

113;

and make all arithmetic operations in the direction from thebottom to the top. The result is Metius’s approximation forp. It is obvious that such a method leads to tremendousnumerical difficulties. Brouncker found another way to dothis. For every n we write

Pn

Qn� b0 þ

a1

b1 þ

a2

b2 þ �� þ

an

bn;

and define Pn and Qn formally by ascendant continuedfractions. Then fPngn>0 and fQngn>0 satisfy

Pn ¼ bnPn�1 þ anPn�2;

Qn ¼ bnQn�1 þ anQn�2;ð17Þ

PnQn�1 � Pn�1Qn ¼ ð�1Þn�1a1. . .an ð18Þ

where

P�1 ¼ 1; Q�1 ¼ 0; P0 ¼ b0; Q0 ¼ 1:

As soon as formulas (17) and (18) are stated they can beeasily proved by induction. If you ask why P-1 = 1 andQ-1 = 0, then the answer is given by the following theo-rem (which again was definitely known to Brouncker).

THEOREM 1 (BROUNCKER, 1655) If ak and bk are all

positive, then

Pn

Qn� Pn�1

Qn�1¼ ð�1Þn�1a1. . .an

QnQn�1; n ¼ 1; 2; . . .; ð19Þ

b0 ¼P0

Q0\ � � �\P2k

Q2k\ � � �\P2kþ1

Q2kþ1\ � � �\P1

Q1\

P�1

Q�1¼ þ1:

ð20Þ

Since Wallis included (17) in [35], and Euler wrote achapter on continued fractions in [11], formulas (17) arenow called the Euler-Wallis formulas. Formulas (19–20)lead to a simple criterion for the convergence of a contin-ued fraction with positive terms.

COROLLARY 2 A continued fraction with positive terms

converges to a finite value if and only if

a1a2 . . . an

QnQn�1! 0: ð21Þ

THEOREM 3 Brouncker’s continued fraction (1) con-

verges for every s [ 0.

PROOF. To remove factorials, let Qn = (2n + 1)!!Dn. By

Corollary 2, the continued fraction (1) converges if and

only if

a1a2 . . . an

QnQn�1¼ ð2n� 1Þ!!2ð2nþ 1Þ!!ð2n� 1Þ!!DnDn�1

¼ 1

ð2nþ 1ÞDnDn�1! 0:

By (17)

Dn ¼2s

2nþ 1Dn�1 þ

2n� 1

2nþ 1Dn�2; ð22Þ


which implies that

ð2nþ 1ÞDnDn�1 ¼ 2sD2n�1 þ ð2n� 1ÞDn�1Dn�2: ð23Þ

Since D0 = 1, D-1 = 0, the iteration of (23) leads to a niceformula,

ð2nþ 1ÞDnDn�1 ¼ 2s D20 þ D2

1 þ . . .þ D2n�1

� �: ð24Þ

For even n we have

D2n [

1

nþ 1: ð25Þ

It is true for n = 2, since Dn(0) = 3/5 by (22) and 9/25 [1/3. Now, if D2

n�2 [ ðn� 1Þ�1, then by (22)

D2n [

2n� 1

2nþ 1

� �2

D2n�2 [

2n� 1

2nþ 1

� �2 1

n� 1[

1

nþ 1;

as elementary algebra shows. Now, by (24) and (25),

ð2nþ 1ÞDnDn�1 [ 2s 1þ 1

3þ 1

5þ � � � þ 1

2mþ 1

� �;

where m is the greatest number satisfying 2m6n� 1.Since the series

X1m¼0

1

2mþ 1¼ 1

diverges, Brouncker’s continued fraction converges for s [ 0.

Let us observe that for the case of s> 1; which is all thatrequired for the evaluation of p, the proof of Theorem 3 canbe completed in a very elementary way. Indeed, if s = 1,then, by (22), Dn lies between Dn-1 and Dn-2, implying thatthe whole sequence lies between D0 = 1 and D1 = 2/3. Ifs [ 1, then Dn(s) [ Dn(1), since by (22) all coefficients of thepolynomials Dn are positive. It follows that for s>1

PnðsÞQnðsÞ

� Pn�1ðsÞQn�1ðsÞ

\ 3

4nþ 2:

The Functional EquationRecall that we derived Brouncker’s continued fraction fromthe single assumption that it satisfies the functional equa-tion (7). This, of course, hints that b(s) indeed satisfies (7),but how can we prove it? Wallis’s notes at the end of §explain this. Since by Theorem 3 Brouncker’s continuedfraction converges, one should only compare the values ofthe convergents at s and s + 2. Passing to the polynomials,one may notice that

PnðsÞPnðs þ 2Þ � ðs þ 1Þ2QnðsÞQnðs þ 2Þ ¼ bn ð26Þ

is a constant at least for the first values of n = 0, 1, 2; seeformulas (16). If we know that bn does not depend on S, thenit is not difficult to find it. Putting s = -1 in (26), we obtainbn = Pn(-1)Pn(1). By (17), polynomial Pn(s) is odd for evenn and is even for odd n. Moreover, Pn(1) = (2n + 1)!!It follows that bn ¼ ð�1Þn�1Pnð1Þ2 ¼ ð�1Þn�1ð2nþ 1Þ!!2.Assuming that (26) holds for every n with bn = -(-1)n

[(2n + 1)!!]2, we obtain, for s [ 0,

P2kðsÞQ2kðsÞ

� P2kðs þ 2ÞQ2kðs þ 2Þ\ðs þ 1Þ2\P2kþ1ðsÞ

Q2kþ1ðsÞ� P2kþ1ðs þ 2ÞQ2kþ1ðs þ 2Þ:

By Theorems 1 and 3, then, b(s) satisfies (7). It remains toprove (26). By (19)

PnðsÞQnþ1ðsÞ � Pnþ1ðsÞQnðsÞ ¼ �ð�1Þn½ð2nþ 1Þ!!�2 ¼ bn;

ð27Þ

implying that the polynomials in the left-hand sides of (26)and (27) coincide. It follows that

PnðsÞfPnðs þ 2Þ �Qnþ1ðsÞg¼ QnðsÞfðs þ 1Þ2Qnðs þ 2Þ � Pnþ1ðsÞg ð28Þ

is equivalent to (26). By (27), the polynomials Pn and Qn

cannot have common factors. It follows that Pn(s + 2)-Qn+1(s) = ln(s)Qn(s), where ln(s) is a linear function. Tofind ln let us observe that the polynomial part of Qn+1/Qn is2s by (17). It follows also from (17) that Pn(s) = 2nsn+1 +

asn-1 + ... and Qn(s) = 2nsn + bsn-2 + .... Hence, thepolynomial part of Pn(s + 2)/Qn(s) is s + 2n + 2, implyingthat ln(s) = 2n + 2 - s. The proof now is completed by atechnical lemma.

LEMMA 4 Let Pn(s)/Qn(s) be the nth convergent to

Brouncker’s continued fraction (15). Then

ðs þ 1Þ2Qnðs þ 2Þ ¼ Pnþ1ðsÞ þ ð2nþ 2� sÞPnðsÞ; ð29Þ

Pnðs þ 2Þ ¼ Qnþ1ðsÞ þ ð2nþ 2� sÞQnðsÞ: ð30Þ

PROOF. The Euler-Wallis formulas (17) for convergents

P/Q look as follows

PnðsÞ ¼ 2sPn�1ðsÞ þ ð2n� 1Þ2Pn�2ðsÞ;P0ðsÞ ¼ s;P�1ðsÞ ¼ 1;

QnðsÞ ¼ 2sQn�1ðsÞ þ ð2n� 1Þ2Qn�2ðsÞ;Q0ðsÞ ¼ 1;Q�1ðsÞ ¼ 0;

and can be used to establish (29) and (30) for n = 0, 1, 2.Let us check, for instance, (30). A natural idea is to showthat the right-hand side of (30) satisfies the Euler-Wallisequation for Pn (s + 2). Then assuming that (30) is true forevery n \ k, we can write

Pkðs þ 2Þ ¼ 2ðs þ 2ÞPk�1ðs þ 2Þ þ ð2k � 1Þ2Pk�2ðs þ 2Þ¼ 2ðs þ 2ÞQkðsÞ þ 2ðs þ 2Þð2k � sÞQk�1ðsÞþ ð2k � 1Þ2Qk�1ðsÞ þ ð2k � 1Þ2ð2k � 2� sÞQk�2ðsÞ¼ 2sQkðsÞ þ ð2k þ 1Þ2Qk�1ðsÞ � ðð2k þ 1Þ2

� ð2k � 1Þ2ÞQk�1ðsÞ þ 4QkðsÞþ 2ðs þ 2Þð2k � sÞQk�1ðsÞþ ð2k � 1Þ2ð2k � 2� sÞQk�2ðsÞ¼ Qkþ1ðsÞ þ ð2k þ 2� sÞQkðsÞ � 8kQk�1ðsÞ� ð2k � 2� sÞQkðsÞþ 2ðs þ 2Þð2k � sÞQk�1ðsÞþ ð2k � 1Þ2ð2k � 2� sÞQk�2ðsÞ¼ Qkþ1ðsÞ þ ð2k þ 2� sÞQkðsÞþ ð2k þ 2� sÞf2sQk�1ðsÞ þ ð2k � 1Þ2Qk�2ðsÞ � QkðsÞg¼ Qkþ1ðsÞ þ ð2k þ 2� sÞQkðsÞ:

A similar calculation proves (29).


Formula (30) can be used to estimate the accuracy ofBrouncker’s algebraic method for the evaluation of p; see §

. It follows from (30) that

PnðsÞ[Qnþ1ðs � 2ÞPnþ1ðs � 2ÞPnþ1ðs � 2Þ; s [ 2:

Observing that QnðsÞ/Pn(s) ? 1/b(s) as n??, we obtainthat

Pnð25Þ[ cPnþ12ð1Þ ¼ cð2nþ 25Þ!!

for some c [ 0. Now, by (18)

b0

Qn

Pnð25Þ � p

\b0

Qn

Pnð25Þ � Qn�1

Pn�1ð25Þ

¼ b0

½ð2n� 1Þ!!�2

Pnð25ÞPn�1ð25Þ ¼ O1

n25

� �:

Wallis’s ProductThe functional equation (7), which finally resulted in (15),can easily be used to develop b(s) into an infinite product:

bðsÞ ¼ ðs þ 1Þ2

ðs þ 3Þ2bðs þ 4Þ ¼ ðs þ 1Þ2

ðs þ 3Þ2� ðs þ 5Þ2

ðs þ 7Þ2bðs þ 8Þ

¼ ðs þ 1Þ2

ðs þ 3Þ2� ðs þ 5Þ2

ðs þ 7Þ2� . . . � ðs þ 4n� 3Þ2

ðs þ 4n� 1Þ2bðs þ 4nÞ

¼ ðs þ 1Þðs þ 1Þðs þ 5Þðs þ 3Þ2

� . . .�

ðs þ 4n� 3Þðs þ 4nþ 1Þðs þ 4n� 1Þ2

bðs þ 4nÞðs þ 4nþ 1Þ:

Multipliers are grouped in accordance to the rule of Wallis’sformula:


¼ 1� 4

ðs þ 4n� 1Þ2;

which provides the convergence of the product, at least fors [ -3. Since by Theorem 3 the continued fraction (1)converges, we can combine Brouncker’s ideas to obtain thefollowing theorem which is fair to attribute to him.

THEOREM 5 Let y(s) be a function on (0, +?) satisfying

(7) and the inequality s \ y(s) for s [ C, where C is a

constant. Then

yðsÞ ¼ ðs þ 1ÞY1n¼1


¼ s þ K1

n¼1

ð2n� 1Þ2

2s

!ð31Þ

for every positive s.

If we put s = 1 in (31), we obtain Wallis’s product (5).Nowadays, the proof of Wallis’s formula can be shortenedto a few lines. Integration by parts shows that

Z p=2

0

sin2n hdh¼p2�1 �3 �5 � . . . � ð2n�1Þ

2 �4 �6 � . . . �2n¼p

2�un;

Z p=2

0

sin2nþ1 hdh¼ 2 �4 �6 � . . . �2n

3 �5 �7 � . . . � ð2nþ1Þ¼vn:

ð32Þ

Combining (32) with the trivial inequalities

Z p=2

0

sin2n hdh[Z p=2

0

sin2nþ1hdh[Z p=2

0

sin2nþ2hdh ð33Þ

(observe that sin2n h[ sin2nþ1h[ sin2nþ2h on (0, p/2)), weimmediately obtain

un

vn[

2

p[

un

vn1� 1

2nþ2

� �;

which shows that

0\1 �32 �2 �

3 �54 �4 �

5 �76 �6 � . . . �

ð2n�1Þ � ð2nþ1Þ2n �2n

�2

p\

3

pð2nþ2Þ;

ð34Þ

implying (5).This now standard proof is, in fact, a small improvement

(use of the inequalities (33) was Euler’s idea [14, Ch. IX, §356]) over Wallis’s original arguments. Notice that in 1654–1655, when Wallis worked on his book, neither integrationby parts nor the change of variable formula were available.Instead, Wallis made his discoveries using a simple relationof integrals with areas as well as his method of interpolation.

One can also observe that there is no direct relationbetween Wallis’s and Brouncker’s proofs. Therefore, it isunlikely that Brouncker consulted Wallis when he tried tofind his own proof. Moreover, this shows that Wallis’s proofwas the first.

Ramanujan’s Formula2

Daniel Bernoulli and Goldbach posed the problem offinding a formula extending the factorial n ? n! = 1 � 2. . . �n to real values of n. In his letter of October 13, 1729

to Goldbach, Euler solved this problem. There are nodoubts that Euler’s solution was motivated by Wallis’sinterpolation method [35].

Arguing by analogy with Brouncker, one can seek anextension CðxÞ for Cðnþ 1Þ ¼ n! as a solution to

Cðx þ 1Þ ¼ xCðxÞ; x [ 0: ð35Þ

If 0 \ x \ 1, then iterating (35) we obtain that

CðxÞ ¼ Cðx þ nþ 1Þxðx þ 1Þ. . .ðx þ nÞ; n> 0: ð36Þ

Now with convexity arguments one can easily obtainEuler’s formula which leads to Euler’s definition of theGamma function:

CðxÞ ¼ limn!1

nxn!

xðx þ 1Þ. . .ðx þ nÞ: ð37Þ

2See [29], noticed by Hardy.


Euler’s formula is not of Wallis’s type, but it can be easilyrearranged into the Newman-Schlomilch formula

1

CðxÞ ¼ xecxY1j¼1

1þ x

j

� �e�x=j

� �; ð38Þ

c = 0.577215... being the Euler-Mascheroni constant. See[17] for details. Counting the zeros and poles of b and C, wearrive at the Ramanujan formula:

THEOREM 6 (Ramanujan) For every s [ 0

bðsÞ ¼ s þ K1

n¼1

ð2n� 1Þ2

2s

!¼ 4

Cð3þs4 Þ

Cð1þs4 Þ

�2

¼ RðsÞ:

As soon as Ramanujan’s formula is found, it is easy toprove it. By Theorem 5, it is sufficient to check that R(s)[ sfor big s. This is equivalent to

C2ðs þ 12Þ

C2ðsÞ[ s � 1

4:

Stirling’s formula implies

Cðs þ 1=2ÞCðsÞ ¼

ffiffisp

1� 1

8sþ 1

128s2þ . . .

� �; ð39Þ

which proves the required inequality if s? +?.

FermatIn 1657, Arithmetica Infinitorum reached Pierre de Fermatin Toulouse, France. Fermat, interested mostly in numbertheory, didn’t read the book carefully (see [33, pp. xxvi-xxvii]). However, Fermat challenged Wallis to solve theDiophantine equation

x2 ¼ 1þ y2D ð40Þ

in positive integers x and y if D is not a perfect square. IfD = P2 with integer P, then (40) obviously does not havepositive integer solutions since x + yP, which is an integerfactor of x2 - y2P2, cannot divide 1. I omit the details of theinitial misunderstanding of this problem on the part ofBrouncker and Wallis. (They may be found in [32] and in[6].) Instead, I mention a mystery here. Although Fermatnever looked in [35, 191], one of his two challenges was theproblem (40), which can be solved by the method pre-sented in [35, 191].

In the fifth century BC, the Pythagorean Hippasus ofMetapontum solved an important geometry problem.Namely, he showed that if AB?AD , then x1 = |AB| =

|AD| and x0 = |BD| are not commensurable. Hippasus’sgeometrical construction is remarkably similar to the con-struction of regular continued fractions (see Fig. 1). Indeed,x0 [ x1 [ x2 = |ED|, where E is defined so that|AB| = |BE|. Computations with angles in DABE;DAEFand DFED show that |AF| = |FE| = |ED|. Hence

x0 ¼ 1 � x1 þ x2 ;

x1 ¼ 2 � x2 þ x3; jA1Dj ¼ x3\x2 :

Observing that DABD�DEFD, we have x2 = 2� x3 + x4. Theconstruction can now proceed by induction, and it will neverstop (notice that An never equals D). The result is that x0/x1

can be represented by an infinite continued fraction

ffiffiffi2p¼ x0

x1¼ 1þ 1

2 þ

1

2 þ

1

2 þ

1

2 þ...: ð41Þ

Since rational numbers are the values of finite regularcontinued fractions and the development into a regularcontinued fraction is unique, this shows that

ffiffiffi2p¼ jBDj=

jADj is an irrational number.In the 17th century, Descartes’s method of coordinates

was very popular, and the above arguments make a goodillustration of its algebraic nature. As we see later, Brounckertranslated Descartes’s paper on musical scales into Englishand even wrote an addendum to it [1]. Taking for grantedBrouncker’s skills demonstrated in the proof of (1) and hisinterest in Descartes, it is natural to assume that Brounckerknew the periodic continued fraction (41).

Let D = 2. The first solutions to (40) can be found byinspection:

x ¼ 3 17 99y ¼ 2 12 70

ð42Þ

To begin with, let us observe that equation (40), at leastformally, looks very much like equation (26), solved byBrouncker to find his formula at Wallis’s request. Thisobservation hints that continued fractions can possibly beused here too. The table of convergents to

ffiffiffi2p

is this:

1 2 2 2 2 2 2 2 . . .

1

0;

1

1;

3

2;

7

5;

17

12;

41

29;

99

70;

239

169; . . .:

ð43Þ

It is easy to see that the quotients x/y from (42) are the firstodd convergents in (43). Using (17), we easily find the nextpair in (43):

x ¼ 2� 239þ 99 ¼ 577

y ¼ 2� 169þ 70 ¼ 408

and by a direct calculation obtain that

1þ 2� 4082 ¼ 332929 ¼ 5772:

Figure 1. x1 = 2x2 + x3.


The only conclusion which one may derive from this is thatthe solutions to (40), at least for D = 2, are given by thenumerators and denominators of odd convergents to

ffiffiffiffiDp

.

There is no direct evidence that Brouncker argued thisway. However, the form of the solution he sent to Wallis (see[32, pp. 321–322]) indicates that most likely he found it fol-lowing arguments similar to those he used to prove (1):

2� Q : 2� 51

1¼ 12; 12� 5

5

6¼ 70 ; 70� 5

29

35¼ 408. . .

ð44Þ

2�Q : 2� 51

1� 5

5

6� 5

29

35� 5

169

204� . . . : ð45Þ

To break the code of (45), let Qn be the denominator of thenth convergent to

ffiffiffi2p

. Then Q1 = 2, Q3 = 12, Q5 = 70, ....Clearly, (44) relates Qn to Qnþ2 ¼ 2. Then (45) represents thesolutions y as partial products of the infinite product ofWallis’s type:

Q1 �Q3

Q1� Q5

Q3� Q7

Q5�Q9

Q7� . . . :

The repeating constant 5 in (45) is explained by an ele-mentary lemma.

LEMMA 7 The recurrence Qn+2 = 6Qn - Qn-2 holds for

n>1 .

PROOF.

þQnþ2 ¼ 2Qnþ1 þQn

2Qnþ1 ¼ 4Qn þ 2Qn�1

�Qn ¼ �2Qn�1 � Qn�2

8><>: ð46Þ

The proof follows by adding the three formulas in (46).

Adding the first two equations in (46) results inQn+2 = 5Qn + 2Qn-1, which together with Lemma 7 implythat 5 \ Qn+2/Qn \ 6, as is clearly indicated in (45). Now,Lemma 7 hints that

Qnþ2

Qn¼ 6� 1

6 �

1

6 �

1

6 �

1

6 � ��! a;

where a [ (5, 6) is the solution to the quadratic equation

X ¼ 6� 1

X; that is, X ¼ 3þ 2

ffiffiffi2p¼ 5:82842712474619. . .:

Notice that 3 = x and 2 = y is the minimal solution to(40) with D = 2, whereas the decimal values of the frac-tions in (45) are

55

6¼ 5:83. . . ; 5

29

35¼ 5:82857. . . ; 5

169

204¼ 5:828431. . .:

Using Lemma 7, we now can prove that odd convergents toffiffiffi2p

give solutions to equation (40) if D = 2. Let us assumethat this is true for all indexes 2k - 1 with k6n. Then, byLemma 7

P22nþ1 � 2Q2

2nþ1 ¼ ð6P2n�1 � P2n�3Þ2 � 2ð6Q2n�1 �Q2n�3Þ2

¼ 1þ 36� 12ðP2n�1P2n�3 � 2Q2n�1Q2n�3Þ:

Forfirst valuesofn, the combination in the thirdparentheses is

P2n�1P2n�3 � 2Q2n�1Q2n�3 ¼ 3: ð47Þ

Compare, by the way, (47) with (26). So, we may incor-porate (47) into the induction hypotheses and obtain that

P2nþ1P2n�1 � 2Q2nþ1Q2n�1

¼ ð6P2n�1 � P2n�3ÞP2n�1 � 2ð6Q2n�1 � Q2n�3ÞQ2n�1

¼ 6� ðP2n�1P2n�3 � 2Q2n�1Q2n�3Þ ¼ 3;

which completes the construction.For D = 3, Brouncker gives the following solution:

3� Q : 1� 31

1� 3

3

4� 3

11

15� 3

41

56� . . . :

We easily find that

ffiffiffi3p¼ 1þ 1

1 þ

1

2 þ

1

1 þ

1

2 þ

1

1 þ

1

2 þ...;

and the convergents with odd indexes

x ¼ 2 7 26 97y ¼ 1 4 15 56

ð48Þ

again satisfy the equation x2 - y2D = 1. In this case,

Qn

Qn�2¼ 4� 1

4�

1

4�...�! 2þ 1 �

ffiffiffi3p¼ 3:732050807568877. . .

and

33

4¼ 3:75 ; 3

11

15¼ 3:7333. . . ; 3

41

56¼ 3:73214285714. . .;

since Qn+2 = 4Qn - Qn-2.For D = 7, this law must be modified, since x/y =

3/1 = P1/Q1 is not a solution to x2 - y2D = 1. However,P3/Q3 = x/y = 8/3 is a solution.

In general, if x1, y1 is a solution to (40), then xn and yn in

xn þ yn

ffiffiffiffiDp¼ ðx1 þ y1

ffiffiffiffiDpÞn ð49Þ

are also solutions. Indeed, sinceffiffiffiffiDp

is irrational, (49) is stillvalid with + replaced by -. Then

x2n� y2

nD¼ ðx1þ y1

ffiffiffiffiDpÞnðx1� y1

ffiffiffiffiDpÞn ¼ ðx2

1 � y21DÞn ¼ 1 :

These formulas however are not so convenient for practicalcomputations of the solutions starting with the minimalone. Here is a simple theorem on continued fractionswhich solves this problem. We put x0 = 1, y0 = 0, which isalso a solution to (40).

THEOREM 8 The solutions fðxn; ynÞgn>1 to equation (40)

satisfy

xnþ1 ¼ ð2x1Þxn � xn�1 ;

ynþ1 ¼ ð2x1Þyn � yn�1 ;ð50Þ


and the fractions fyn=xngn>0 are the convergents to thecontinued fraction

1ffiffiffiffiDp ¼ y1

x1� 1

2x1 �

1

2x1 �

1

2x1 �...: ð51Þ

PROOF. By (49) for n = 0, 1, ...

xnþ1 ¼ x1xn þ y1Dyn ; ynþ1 ¼ y1xn þ y1yn:

Iterating these formulas, we obtain

xnþ1 ¼ x1xn þ y21Dxn�1 þ x1y1yn�1D

¼ x1xn þ x21xn�1 þ x1y1yn�1D � xn�1

¼ x1xn þ x1ðx1xn�1 þ y1Dyn�1Þ � xn�1

¼ ð2x1Þxn � xn�1;

which proves the first identity in (50). Similar calculationsprove the second. Now, (50) implies that yn/xn are theconvergents to the continued fraction (51), which convergeto 1=

ffiffiffiffiDp

as n!1, by

1ffiffiffiffiDp � yn

xn¼ 1

xnðxn þ yn

ffiffiffiffiDpÞ¼ 1

xnðx1 þ y1

ffiffiffiffiDpÞn�!0:

Analyzing the correspondence of Wallis and Brounckeron Fermat’s question, Whitford presents in [36, p.52] exactlythe same formulas as in (50). This makes it natural toconjecture that Brouncker, in fact, used Theorem 8 for hissolution to Fermat’s question.

Theorem 8 shows that Brouncker’s method works notonly for particular values of D such as D = 2, 3, 7, but alsofor any D, provided a minimal solution (x1, y1) exists.Indeed, we may write

D � Q : y1 �y2

y1� y3

y2� y4

y3� y5

y4� . . . :

By (50), yn+1 = x1yn + y1xn, which implies that

ynþ1

yn¼ x1 þ y1

xn

yn�! x1 þ y1

ffiffiffiffiDp

:

Application of (50), as in the case of D = 2, leads to thesame conclusion. Brouncker’s formulas (50) convenientlylists infinitely many solutions, provided one is known. By(51), y1 divides every yn.

The problem of the minimal solution to (40), for a givenvalue D, can be solved similarly to the proof of formula (26).Let us observe that if P2

n � Q2nD ¼ 1, then, by (20), the integer

n is odd. Hence, P2n � Q2

nD ¼ 1 if and only if (see (18))

P2n � Q2

nD ¼ PnQn�1 � Pn�1Qn

or equivalently,

PnðPn � Qn�1Þ ¼ QnðQnD � Pn�1Þ:

By (18), the greatest common divisor of Pn and Qn is 1. Itfollows that

Pn ¼ knQn þ Qn�1;

QnD ¼ knPn þ Pn�1:ð52Þ

By Theorem 1 (recall that n is odd),

0\Pn

Qn�

ffiffiffiffiDp

\P1

Q1�

ffiffiffiffiDp

\1; ð53Þ

sinceffiffiffiffiDp

is irrational. Then, by the first equation of (52),

Pn

Qn¼ kn þ

Qn�1

Qn: ð54Þ

Let [x] be the greatest integer not exceeding x. Then, by (53)and (54), we obtain that

kn ¼Pn

Qn

�¼ ½

ffiffiffiffiDp�: ð55Þ

Therefore, the problem of finding the minimal odd n suchthat P2

n �Q2nD ¼ 1 reduces to the search of the minimal odd

n, satisfying

Pn ¼ ½ffiffiffiffiDp�Qn þQn�1: ð56Þ

Later, Euler [13] proved that the minimal solution to (56)(equivalently to (40)) in the form of (Pk, Qk) exists if theregular continued fraction of

ffiffiffiffiDp

is periodic with period d. Ifd is even, then k = d - 1. For instance, if D = 7, then d = 4,implying that (P3, Q3) is the minimal solution. If d is odd,then k = 2d - 1. For instance, if D = 2, then d = 1, and,therefore, (P1, Q1) is the minimal solution. This paper,written in 1765, appeared only in 1767. After that, Lagrangeproved in [23] and [24] that if x2 - y2D = 1, then x/y is anodd convergent to

ffiffiffiffiDp

and also that the regular continuedfraction of any quadratic irrationality is periodic. This com-pleted the proof that Brouncker’s method lists all thesolutions, as well as that each Pell’s equation has infinitelymany solutions. Later, Euler presented his results in his book[15], which was translated into French by Lagrange. Lagrangeincluded his theory as an addendum to this translation.

In contrast to the case of Wallis’s product, this timeWallis, taking Brouncker’s hints, found his own solution toFermat’s problem. It is now called the English method; see[6] and [32] for details. In spite of his comments on con-tinued fractions in [35, §191], Wallis didn’t follow the linesindicated above.

It was Euler who named equation (40) after Pell, in hisfirst papers on this subject (see, for instance, [8, §15] or[7]). Most likely, the correspondence between Brounckerand Wallis was unavailable in St. Petersburg. Therefore,Whitford’s opinion [36] that Euler mentioned Pell, becausePell included the Diophantine equation x = 12y2 - z2 inthe English translation of Rahn’s Algebra [28, p.134], looksvery convincing. There is some evidence that the firstappearance of this problem goes back to Archimedes’sCattle Problem, which reduces to Pell’s equation

x2 � 4729494y2 ¼ 1:

Due to Hippasus’s problem and Brouncker’s solution ofFermat’s Challenge, this may look reasonable. But theminimal solution to the Pell’s equation has thousands ofplaces. Therefore, it is not clear how Archimedes couldhave written the minimal positive solution himself. See [6],[21, p.3], [34] and [36] for the history of Pell’s equation and arecent paper [25] for history and applications.

HarmonyThe Weber-Fechner law says that a human being’s responseto physical phenomena obeys a logarithmic law (see [26,


pp. 111–112). By converting exponential growth to a linearscale, this ability makes people less sensitive to changes inthe outside world and reduces their reactions to the mostsignificant ones. We cannot control too many parameters atthe same time, and the Weber-Fechner law reflects this fact.In particular, our ear compares not the heights of pitches butthe logarithms of their quotients.

The main problem of a musical scale is to arrange asystem of quotients of pitches creating an impression ofharmony under the logarithmic law of response. In practice,this means finding a step of linear scale such that the loga-rithms of the quotients chosen can be well approximated byinteger multiples of this step. The human ear can normallyhear pitches in the range 20 Hz to 20 kHz. Notice that 20 �24 and 20000 � 214. Applying the logarithms, we see that29 = 512 Hz corresponds to the center of the logarithmicscale. If a string of length l creates a pitch of the frequencyx = 512 Hz, then the string of length l/2 doubles the fre-quency to 2x. The logarithmic base a is chosen so as tonormalize the following number as a unit:

loga

2xx

� �¼ loga 2 ¼ 1;

which implies that a = 2. The ratio 2x : x = 2 determinesthe interval (x, 2x) called the octave. The ratio 3x/2 : xcorresponding to the half of the interval (x, 2x) (the fre-quency 3x/2 is generated by the string of length 2l/3), iscalled the perfect fifth. The ear hears this ratio as

log2

3

2x : x

� �¼ log2 3� 1:

Our ear hears the perfect fifth the best, and, therefore, itmust be approximated the best possible way. The conver-gents to the continued fraction

log2 3� 1 ¼ 0; 584962500721. . .

¼ 1

1 þ

1

1 þ

1

2 þ

1

2 þ

1

3 þ

1

1 þ

1

5 þ

1

2 þ

1

23 þ...

make the series

1;1

2;

3

5;

7

12;

24

41; . . . : ð57Þ

Approximations 1 and 1/2 are too crude. Approximation 3/5 is used in Eastern music. Approximation 7/12 is the best.It divides the musical scale into 12 semitones, and 7 suchsemitones correspond to the fifth.

If the interval between two notes is a ratio of small inte-gers, these two notes are called consonant. Otherwise, theyare called dissonant.3 This happens again due to therestricted abilities of human beings. Computers would haveanother opinion on this matter. There are seven intervalswhich are commonly considered to be consonant (they hadalready appeared inDescartes’s table; see [1, p.13]). Themostimportant among them are 3/2 (the perfect fifth) and 5/4 (themajor third), since the binary logarithms of other consonantintervals are linear combinations of 1, log2 3/2 and log2 5/4with the coefficients in {0, 1, -1}. Hence, the error of the

approximation by a uniform scale is completely determinedby those for log2 3/2, and log2 5/4 and cannot exceed themaximum of the two. Now,

log2

5

4

� �¼ 0:32192809488736234787. . .

¼ 1

3 þ

1

9 þ

1

2 þ

1

2 þ

1

4 þ

1

6 þ...

shows that 1/3 = 4/12 is a convergent to log2 5/4. Thisguarantees that the equal temperament system of 12 uniformsemitones gives two good rational approximations7/12 and 4/12 to two basic intervals 3/2 and 5/4, and, hence,to all seven consonant intervals. See [4] and [20] for a moredetailed discussion.

In the addendum to [1], which was published two yearsbefore his first great discovery in continued fractions,Brouncker analyzed the scale of 17 equal semitones fromthe point of view of the Descartes theory. He didn’t applycontinued fractions then, but, as is clear from the above,continued fractions are important for the analysis of har-mony. Simple calculations show that

log2 3=2 ¼ 7

12þ 0:00162. . . ¼ 10

17� 0:00327;

which implies that the scale of 17 equal semitones doublesthe error of approximation for the perfect fifth comparedwith the scale of 12 equal semitones. As to the approxi-mation of log2 5/4, the 17-based scale also almost doublesthe error compared with the 12-based scale:

log2 5=4 ¼ 4

12� 0:011. . . ¼ 5

17þ 0:027. . . : :

In my opinion, it was the study of problems of musicalscales which finally led Brouncker to positive continuedfractions. Therefore, it looks like Wallis’s question on theexistence of other arithmetic formulas for p, similar to hisinfinite product, fell on ground carefully prepared byBrouncker.

EpilogueI give another citation from the astrological site: ‘‘Jupiter,the King of the Gods, is the ruler of Sagittarius. In Astrology,Jupiter is a planet of plenty. It is tolerant and expansive. Thefirst of the social planets, Jupiter seeks insight throughknowledge. Some of this planet’s keywords include morality,gratitude, hope, honor and the law. Jupiter is a planet ofbroader purpose, reach and possibility’’.

Things changed for Brouncker when in 1658 Cromwelldied, and Brouncker started to move gradually from theprotection of Saturn to that of Jupiter . Already, in 1660, hewas elected as a member of Parliament. In 1662 Brounckerwas promoted by King Charles II to an important position. In1663, the Royal Society of London was created, and LordBrouncker was nominated as its first president. See [2] forother details of Brouncker’s remarkable career. However,things changed for him in mathematics, too. He left forever avery fruitful and important area that he had discovered. Later,

3Euler developed an original theory of sound classification by ‘‘degree of pleasure’’ in his monograph [9].


Euler developed this area into the theory of special functions.Stieltjes introduced his theory of moments. One may observethat the proof of Brouncker’s formula is the starting point ofStieltjes’s theory. Chebychev discovered orthogonal poly-nomials. Brouncker’s polynomials fPngn>�1 turned out to beorthogonal with respect to the weight

dl ¼ 1

8p3C

1þ it

4

� �4

dt;

interestingly related to the Gamma function. These polyno-mials are placed at the very center of the family of Wilsonpolynomials, which include all the classical orthogonalpolynomials. More can be found in my book [20].

I have a question for the reader. Which protection do youprefer, Saturn’s or Jupiter’s ? Please think before giving ananswer. Brouncker enjoyed them both, but which was bet-ter? Where is the yacht constructed by Brouncker for KingCharles II? Nobody knows, but his formula and two GreatTheorems are in front of you.

REFERENCES

[1] W. Brouncker. ‘‘Animadversions upon the Musick-Compendium

of Descartes’’ (London, 1653).

[2] J. J. O’Connor and E. F. Robertson. ‘‘William Brouncker’’, (The

MacTutor History of Mathematics Online Archive, 2002). http://

www.history.mcs.st-and.ac.uk

[3] M. Jesseph Douglas. ‘‘Squaring the Circle: The War between

Hobbes and Wallis’’ (University of Chicago, Chicago, 2000).

[4] E. G. Dunne and M. McConnell. Pianos and continued fractions,

Mathematics Magazine 72:2 (1999), pp. 104–115.

[5] J. Dutka. ‘‘Wallis’s product, Brouncker’s continued fraction, and

Leibniz’s series’’, Archive for History of Exact Sciences 26:2

(1982), pp. 115–126.

[6] H. M. Edwards. ‘‘Fermat’s Last Theorem: A Generic Introduction

to Algebraic Number Theory’’ (Springer, New York, 1977).

[7] L. Euler. ‘‘Euler’s letter to Goldbach on August 10, 1730’’, OO723

(in the Euler Archive, http://www.math.dartmouth.edu/*euler/).

[8] L. Euler. ‘‘De solutione problematum diophanteorum per numeros

integros’’, Commentarii academiae scientiarum Imperials Petro-

politanae VI (1738), 175–188 (presented on May 29, 1733);

reprinted in Opera Omnia, Ser. 1, Vol 2, pp. 6–17; E029

[9] L. Euler. ‘‘Tentamen novae theoriae musicae ex certissismis harmo-

niae principiis dilucide expositae’’ (Petropoli, Academiae Scientiarum,

1739); reprinted in Opera Omnia, Ser. 3, Vol. 1, pp. 197–427; E033.

[10] L. Euler. ‘‘De fractionibus continuus, dissertatio’’, Commentarii

Academiae Scientiarum Imperials Petropolitane IX(1744) for 1737,

98–137 (presented on February 7, 1737); reprinted in Opera Omnia,

Ser. 1, Vol. 14, pp. 187–216; E071; translated into English: Math-

ematical Systems Theory (1985) 4:18.

[11] L. Euler. ‘‘Introductio in analysin infinitorum’’ (Apud Marcum –

Michaelem Bousquet & Socios, Lausanne, 1748); E101.

[12] L. Euler. ‘‘De fractionibus continuus, observationes’’, Commen-

tarii Academiae Scientiarum Imperials Petropolitane XI (1750b) for

1739, pp. 32–81 (presented on January 22, 1739); reprinted in

Opera Omnia, Ser. I, Vol. 14, pp. 291–349; E123.

[13] L. Euler. ‘‘De usu novi algorithmi in Problemate Pelliano solven-

do’’, Novi Commentarii academiae scientiarum Petropolitane.,

11, (1767), pp. 29–66; reprinted in Opera Omnia, Ser. 1, Vol. 3,

pp. 73-111; E323.

[14] L. Euler. ‘‘Integral Calculus’’, Vol. I (Impeofis Academiae Imperialis

Scientiarum, St. Petersburg, 1768). Russian translation: Moscow,

GITTL, 1956; E342.

[15] L. Euler. ‘‘Vollstandige Anleitung zur Algebra’’ (Leipzig, 1770);

E387–E388.

[16] A. Van Helden. ‘‘Huygens’s Ring, Cassini’s Division and Saturn’s

Children’’ (Smithsonian Institution Libraries, Washington D. C.,

2006), http://www.sil.si.edu/silpublications/dibner-library-lectures/

2004-VanHelden/2004_VanHelden.pdf.

[17] J. Havil. ‘‘Gamma. Exploring Euler’s Constant’’ (Princeton

University Press, Princeton, 2003).

[18] C. Huygens. ‘‘De circuli magnitudine inventa. Accedunt eiusdem

problematum quorundam illustrium constructiones’’ (J. and D.

Elzevier, Leiden, 1654).

[19] S. Khrushchev. ‘‘A recovery of Brouncker’s proof for the quad-

rature continued fraction’’, Publicacions Matematiques 50 (2006),

pp. 3–42.

[20] S. Khrushchev. ‘‘Orthogonal Polynomials and Continued Frac-

tions: From Euler’s point of view’’ (Cambridge University Press,

Cambridge, 2008).

[21] H. Koch. ‘‘Number Theory. Algebraic Numbers and Functions’’

(AMS, Providence, 2000).

[22] F. D. Kramar. Integration Methods of John Wallis, in: Historico-

mathematical Research 14, pp. 11–100, in Russian (FizMatGiz,

Moscow, 1961).

[23] J. L. Lagrange ‘‘Solution d‘un probleme d’arithmetique’’, Mis-

cellanea Taurinensia, 4 (1766–1769); = Oeuvres I pp. 671–731

Paris, Gauthier-Villars, MDCCCLXVII.

[24] J. L. Lagrange. ‘‘Sur la solution des problemes indetermines du

second degre’’, Memoires de l’Academie royale des sciences et

belles-lettres (de Berlin), annee 1767; Oeuvres II, pp 377–538,

Paris, Gauthier-Villars, MDCCCLXVIII.

[25] H. W. Lenstra Jr. ‘‘Solving the Pell equation’’, Notices of the

American Mathematical Society 49:2 (2002), pp. 182–192.

[26] E. Maor. ‘‘e: The Story of a Number.’’ (Princeton University Press,

Princeton, 1994).

[27] A. S. Posamentier and I. Lehmann. ‘‘A Biography of the World’s

Most Mysterious Number’’ (Prometheus Books, New York,

2004).

[28] J. H. Rahn. ‘‘An introduction to algebra, translated out of the High

Dutch into English by Thomas Brancker, M.A. Much altered and

augmented by D.P.’’ (Moses Pitt, London, 1668).

[29] G. H. Hardy, Seshu Aiyar, B. M. Wilson (eds.). ‘‘Collected Papers

of Srinivasa Ramanujan’’ (Chelsea Publishing Co./American

Mathematical Society, 2000).

[30] Sir Reginald R. F. D., Palgrave, K.C.B.. ‘‘Oliver Cromwell H. H.,

the Lord Protector and the Royalist Insurrection Against His

Government of March, 1655’’ (Sampson, Low, Marston and Co.

Ltd., London, 1903).

[31] J. A. Stedall ‘‘Catching proteus: the collaborations of Wallis and

Brouncker I. Squaring the circle’’, Notes and Records of the Royal

Socitey of London 54:3 (2000), pp. 293–316.


http://www.math.dartmouth.edu/~euler/

http://www.sil.si.edu/silpublications/dibner-library-lectures/2004-VanHelden/2004_VanHelden.pdf

http://www.sil.si.edu/silpublications/dibner-library-lectures/2004-VanHelden/2004_VanHelden.pdf

[32] J. A. Stedall ‘‘Catching proteus: the collaborations of Wallis and

Brouncker II. Number Problems’’, Notes Rec. R. Soc. London

54:3 (2000), pp. 317–331.

[33] J. A. Stedall. (English translation) The Arithmetics of Infinitesimals:

John Wallis, 1656’’ (Springer-Verlag, New York, 2004).

[34] I. Vardi. ‘‘Archimedes’ Cattle Problem’’, The American Mathe-

matical Monthly 105:4 (1998), pp. 305–319.

[35] J. Wallis. ‘‘Arithmetica Infinitorum’’ (Typis Leon: Lichfield Acade-

mia Typographi, Icnpenfis Tho. Robinson, London, 1656).

[36] E. E. Whitford. ‘‘The Pell Equation’’ (College of the City of New York,

New York, 1912), available online in the University of Michigan

Historical Math Collection (http://www.hti.umich.edu/u/umhistmath/).


http://www.hti.umich.edu/u/umhistmath/

Mathematical Entertainments Michael Kleber and Ravi Vakil, Editors

Tilings*FEDERICO ARDILA

�, AND RICHARD P. STANLEY�

This column is a place for those bits of contagious

mathematics that travel from person to person in the

community, because they are so elegant, surprising, or

appealing that one has an urge to pass them on.

Contributions are most welcome.

CConsider the following puzzle. The goal is to coverthe region

using the following seven tiles.

The region must be covered entirely without any over-lap. It is allowed to shift and rotate the seven pieces in anyway, but each piece must be used exactly once

One could start by observing that some of the piecesfit nicely in certain parts of the region. However,the solution can really only be found through trial anderror.

17

56 2

34

For that reason, even though this is an amusing puzzle, itis not very intriguing mathematically.

This is, in any case, an example of a tiling problem. Atiling problem asks us to cover a given region using a givenset of tiles completely and without any overlap. Such acovering is called a tiling. Of course, we will focus ourattention on specific regions and tiles that give rise tointeresting mathematical problems.

Given a region and a set of tiles, there are many differentquestions we can ask. Some of the questions that we willaddress are the following:

• Is there a tiling?• How many tilings are there?• About how many tilings are there?• Is a tiling easy to find?• Is it easy to prove that a tiling does not exist?• Is it easy to convince someone that a tiling does not exist?• What does a ‘‘typical’’ tiling look like?• Are there relations among the different tilings?• Is it possible to find a tiling with special properties, such

as symmetry?

Is There a Tiling?From looking at the set of tiles and the region we wish tocover, it is not always clear whether such a task is even

� Please send all submissions to the

Mathematical Entertainments Editor,

Ravi Vakil, Stanford University,

Department of Mathematics, Bldg. 380,

Stanford, CA 94305-2125, USA


12 3 4

5

6 7

* This paper is based on the second author’s Clay Public Lecture at the IAS/Park City Mathematics Institute in July, 2004

� Supported by the Clay Mathematics Institute

� Partially supported by NSF grant #DMS-9988459, and by the Clay Mathematics Institute as a Senior Scholar at the IAS/Park City Mathematics Institute


possible. The puzzle at the beginning of this article is sucha situation. Let us consider a similar puzzle where the tiles,Solomon Golomb’s polyominoes, are more interestingmathematically.

A pentomino is a collection of five unit squares arrangedwith coincident sides. Pentominoes can be flipped orrotated freely. The figure shows the 12 different pentomi-noes. Since their total area is 60, we can ask, for example: Isit possible to tile a 3 9 20 rectangle using each one of themexactly once?

This puzzle can be solved in at least two ways. Onesolution is shown above. A different solution is obtained ifwe rotate the shaded block by 180�. In fact, after spendingsome time trying to find a tiling, one discovers that these(and their rotations and reflections) are the only two pos-sible solutions.

One could also ask whether it is possible to tile two6 9 5 rectangles using each pentomino exactly once. Oneway of doing it is shown below. There is only one other suchtiling, obtained by rearranging two of the pentominoes; it is anice puzzle for the reader to find those two tiles.

Knowing that, one can guess that there are several tilingsof a 6 9 10 rectangle using the 12 pentominoes. However,one might not predict just how many there are. Anexhaustive computer search has found that there are 2,339such tilings.

.........................................................................................................................................................

AU

TH

OR

S FEDERICO ARDILA was born and grew

up in Bogota, Colombia. He received his

Ph.D. from MIT under the supervision of

Richard Stanley. He is an assistant professor

at San Francisco State University and anadjunct professor at the Universidad de Los

Andes in Bogota. He studies objects in

algebra, geometry, topology and phylo-

genetics by understanding their underlying

combinatorial structure. He leads the

SFSU–Colombia Combinatorics Initiative,

a research and learning collaboration

between students in the United Statesand Colombia. When he is not at work,

you might find him on the futbol field,

treasure hunting in little record stores,

learning a new percussion instrument, or

exploring the incredible San Francisco Bay

Area.

Department of MathematicsSan Francisco State University

San Francisco, CA 94132

USA


RICHARD STANLEY is currently the Norman

Levinson Professor of Applied Mathematics

at MIT. His main research interest is

combinatorics and its connection with such

other areas of mathematics as algebraic

topology, commutative algebra and

representation theory. He is the author of

three books, including Enumerative Combi-natorics, Volumes 1 and 2, and over 150

research papers. He has supervised over 50

Ph.D. students and maintains on his web page

a dynamic list of exercises related to Catalan

numbers, including over 175 combinatorial

interpretations.


Massachusetts Institute of Technology

Cambridge, MA 02139

USA



These questions make nice puzzles, but are not the kindof interesting mathematical problem that we are lookingfor. To illustrate what we mean by this, let us consider aproblem that is superficially somewhat similar, but that ismuch more amenable to mathematical reasoning.

Suppose we remove two opposite corners of an 8 9 8chessboard, and we ask: Is it possible to tile the resultingfigure with 31 dominoes?

Our chessboard would not be a chessboard if its cellswere not colored dark and white alternatingly. As it turnsout, this coloring is crucial in answering the question athand.

Notice that, regardless of where it is placed, a dominowill cover one dark and one white square of the board.Therefore, 31 dominoes will cover 31 dark squares and 31white squares. However, the board has 32 dark squares and30 white squares in all, so a tiling does not exist. This is anexample of a coloring argument; such arguments are verycommon in showing that certain tilings are impossible.

A natural variation of this problem is to now remove onedark square and one white square from the chessboard, asshaded above. Now the resulting board has the samenumber of dark squares and white squares; is it possible totile it with dominoes?

Let us show that the answer is yes, regardless of whichdark square and which white square we remove. Considerany closed path that covers all the cells of the chessboard,like the following one.

Now start traversing the path, beginning with the pointimmediately after the dark hole of the chessboard. Cover thefirst and second cell of the path with a domino; they arewhite and dark, respectively. Then cover the third and fourth

cells with a domino; they are also white and dark, respec-tively. Continue in this way, until the path reaches thesecond hole of the chessboard. Fortunately, this second holeis white, so there is no gap between the last domino placedand this hole. We can, therefore, skip this hole and continuecovering the path with successive dominoes. When the pathreturns to the first hole, there is again no gap between the lastdomino placed and the hole. Therefore, the board is entirelytiled with dominoes. We now illustrate this procedure.

What happens if we remove two dark squares and twowhite squares? If we remove the four squares closest to acorner of the board, a tiling with dominoes obviouslyexists. On the other hand, in the example below, a dominotiling does not exist, since there is no way for a domino tocover the upper left square

.This question is clearly more subtle than the previous

one. The problem of describing which subsets of thechessboard can be tiled by dominoes leads to some verynice mathematics. We will say more about this topic in theSection ‘‘Demonstrating That a Tiling Does Not Exist’’below.

Let us now consider a more difficult example of a col-oring argument, to show that a 10 9 10 board cannot betiled with 1 9 4 rectangles.

Giving the board a chessboard coloring gives us noinformation about the existence of a tiling. Instead, let ususe four colors, as shown above. Any 1 9 4 tile that weplace on this board will cover an even number (possiblyzero) of squares of each color


.Therefore, if we had a tiling of the board, the total

number of squares of each color would be even. But thereare 25 squares of each color, so a tiling is impossible.

With these examples in mind, we can invent many similarsituations where a certain coloring of the board makes atiling impossible. Let us now discuss a tiling problem thatcannot be solved using such a coloring argument.

Consider the region T(n) consisting of a triangular arrayof n(n + 1)/2 unit regular hexagons.

T(1)

T(3)T(4)

T(2)

Call T(2) a tribone. We wish to know the values of n forwhich T(n) can be tiled by tribones. For example, T(9) canbe tiled as follows.

Sinceeach tribone covers 3 hexagons, n(n + 1)/2must bea multiple of 3 for T(n) to be tileable. However, this does notexplain why regions such as T(3) and T(5) cannot be tiled.

Conway and Lagarias [3, 21] showed that the triangulararray T(n) can be tiled by tribones if and only if n = 12k,12k + 2, 12k + 9 or 12k + 11 for some k� 0: Thesmallest values of n for which T(n) can be tiled are 0, 2,9, 11, 12, 14, 21, 23, 24, 26, 33 and 35. Their proof uses acertain nonabelian group that detects information aboutthe tiling that no coloring can detect, while coloringarguments can always be rephrased in terms of abeliangroups. In fact, it is possible to prove that no coloringargument can establish the result of Conway and Lagarias[16].

Counting Tilings, ExactlyOnce we know that a certain tiling problem can besolved, we can go further and ask: How many solutionsare there?

As we saw earlier, there are 2,339 ways (up to symme-try) to tile a 6 9 10 rectangle using each one of the 12pentominoes exactly once. It is perhaps interesting that thisnumber is so large, but the exact answer is not so inter-esting, especially since it was found by a computer search.

The first significant result on tiling enumeration wasobtained independently in 1961 by Fisher and Temperley [7]and by Kasteleyn [12]. They found that the number of tilingsof a 2m 9 2n rectangle with 2mn dominoes is equal to

4mnYmj¼1

Ynk¼1

cos2 jp2mþ 1

þ cos2 kp2nþ 1

� �:

Here P denotes product, and p denotes 180�, so the numberabove is given by 4mn times a product of sums of twosquares of cosines, such as

cos2p5¼ cos 72� ¼ 0:3090169938. . . :

This is a remarkable formula! The numbers we are multi-plying are not integers; in most cases, they are not evenrational numbers. When we multiply these numbers wemiraculously obtain an integer, and this integer is exactlythe number of domino tilings of the 2m 9 2n rectangle.

For example, for m = 2 and n = 3, we get:

46ðcos236� þ cos225:71. . .�Þ � ðcos236� þ cos251:43. . .�Þ� ðcos236� þ cos277:14. . .�Þ � ðcos272� þ cos225:71. . .�Þ� ðcos272� þ cos251:43. . .�Þ � ðcos272� þ cos277:14. . .�Þ¼ 46ð1:4662. . .Þð1:0432. . .Þð0:7040. . .Þ� ð0:9072. . .Þð0:4842. . .Þð0:1450. . .Þ ¼ 281:

Skeptical readers with a lot of time to spare are invited tofind all domino tilings of a 4 9 6 rectangle and check thatthere are, indeed, exactly 281 of them.

Let us say a couple of words about the proofs of thisresult. Kasteleyn expressed the answer in terms of a certainPfaffian, and reduced its computation to the evaluation of arelated determinant. Fisher and Temperley gave a differentproof using the transfer matrix method, a technique oftenused in statistical mechanics and enumerative combi-natorics.

There is a different family of regions for which thenumber of domino tilings is surprisingly simple. The Aztecdiamond AZ(n) is obtained by stacking successive centeredrows of length 2, 4, ..., 2n, 2n, . . ., 4, 2, as shown.

AZ(3) AZ(7)AZ(2)AZ(1)

The Aztec diamond AZ(2) of order 2 has the following 8tilings:


Elkies, Kuperberg, Larsen and Propp [6] showed that thenumber of domino tilings of AZ(n) is 2n(n+1)/2. The fol-lowing table shows the number of tilings of AZ(n) for thefirst few values of n

1 2 3 4 5 6

2 8 64 1024 32768 2097152

Since 2(n+1)(n+2)/2/2n(n+1)/2 = 2n+1, one could try toassociate 2n+1 domino tilings of the Aztec diamond oforder n + 1 to each domino tiling of the Aztec diamondof order n, so that each tiling of order n + 1 occursexactly once. This is one of the four original proofs foundin [6]; there are now around 12 proofs of this result. Noneof these proofs is quite as simple as the answer 2n(n+1)/2

might suggest.

Counting Tilings, ApproximatelySometimes we are interested in estimating the number oftilings of a certain region. In some cases, we will want todo this, because we are not able to find an exact formula.In other cases, somewhat paradoxically, we might preferan approximate formula over an exact formula. A goodexample is the number of domino tilings of a rectangle.We have an exact formula for this number, but this for-mula does not give us any indication of how large thisnumber is.

For instance, since Aztec diamonds are ‘‘skewed’’squares, we might wonder: How do the number of dominotilings of an Aztec diamond and a square of about the samesize compare? After experimenting a bit with these shapes,one notices that placing a domino on the boundary of anAztec diamond almost always forces the position of severalother dominoes. This almost never happens in the square.This might lead us to guess that the square should havemore tilings than the Aztec diamond.

To try to make this idea precise, let us make a definition.If a region with N squares has T tilings, we will say that it

hasffiffiffiffiffiffiffiffiffiffi½N �T

pdegrees of freedom per square. The motivation,

loosely speaking, is the following: If each square coulddecide independently how it would like to be covered, and

it hadffiffiffiffiffiffiffiffiffiffi½N �T

ppossibilities to choose from, then the total

number of choices would be T.The Aztec diamond AZ(n) consists of N = 2n(n + 1)

squares, and it has T = 2n(n+1)/2 tilings. Therefore,the number of degrees of freedom per square in AZ(n)is:

ffiffiffiffiffiffiffiffiffiffi½N �T

p¼

ffiffiffiffiffi42p

¼ 1:189207115. . .

For the 2n 9 2n square, the exact formula for the numberof tilings is somewhat unsatisfactory, because it does notgive us any indication of how large this number is. For-tunately, as Kasteleyn, Fisher and Temperley observed,one can use their formula to show that the number ofdomino tilings of a 2n 9 2n square is approximatelyC4n^2, where

C ¼ eG=p

¼ 1:338515152. . .:

Here G denotes the Catalan constant, which is defined asfollows:

G ¼ 1� 1

32þ 1

52� 1

72þ � � �

¼ 0:9159655941. . .:

Thus, our intuition was correct. The square board is‘‘easier’’ to tile than the Aztec diamond, in the sense that ithas approximately 1.3385. . . degrees of freedom persquare, while the Aztec diamond has 1.1892. . ..

Demonstrating That a Tiling Does Not ExistAs we saw in the Section entitled ‘‘Is There a Tiling?’’,there are many tiling problems where a tiling exists, butfinding it is a difficult task. However, once we havefound it, it is very easy to demonstrate its existence tosomeone: We can simply show them the tiling!

Can we say something similar in the case where a tilingdoes not exist? As we also saw in the Section entitled ‘‘IsThere a Tiling?’’, it can be difficult to show that a tiling doesnot exist. Is it true, however, that if a tiling does not exist,then there is an easy way of demonstrating that tosomeone?

In a precise sense, the answer to this question is almostcertainly no in general, even for tilings of regions using1 9 3 rectangles [1]. Surprisingly, though, the answer is yesfor domino tilings!

Before stating the result in its full generality, let usillustrate it with an example. Consider the following regionconsisting of 16 dark squares and 16 white squares. (Theshaded cell is a hole in the region.)

One can use a case-by-case analysis to become con-vinced that this region cannot be tiled with dominoes.Knowing this, can we find an easier, faster way to convincesomeone that this is the case?

One way of doing it is the following. Consider the 6dark squares marked with a •. They are adjacent to a totalof 5 white squares, which are marked with an *. Wewould need 6 different tiles to cover the 6 marked darksquares, and each one of these tiles would have to coverone of the 5 marked white squares. This makes a tilingimpossible.


** *

* *

Philip Hall [10] showed that in any region that cannotbe tiled with dominoes, one can find such a demonstra-tion of impossibility. More precisely, one can find k cellsof one color which have fewer than k neighbors. There-fore, to demonstrate to someone that tiling the region isimpossible, we can simply show them those k cells andtheir neighbors!

Hall’s statement is more general than this and is com-monly known as the marriage theorem. The name comesfrom thinking of the dark cells as men and the white cellsas women. These men and women are not very adven-turous: They are only willing to marry one of theirneighbors. We are the matchmakers; we are trying to findan arrangement in which everyone can be happily mar-ried. The marriage theorem tells us exactly when such anarrangement exists.

Tiling Rectangles with RectanglesOne of the most natural tiling situations is that of tiling arectangle with smaller rectangles. We now present threebeautiful results of this form.

The first question we wish to explore is: When can anm 9 n rectangle be tiled with a 9 b rectangles (in anyorientation)? Let us start this discussion with some moti-vating examples.

Can a 7 9 10 rectangle be tiled with 2 9 3 rectangles?This is clearly impossible, because each 2 9 3 rectanglecontains 6 squares, while the number of squares in a 7 9 10rectangle is 70, which is not a multiple of 6. For a tiling tobe possible, the number of cells of the large rectangle mustbe divisible by the number of cells of the small rectangle. Isthis condition enough?

Let us try to tile a 17 9 28 rectangle with 4 9 7 rectan-gles. The argument of the previous paragraph does notapply here; it only tells us that the number of tiles needed is17. Let us try to cover the left-most column first.

Our first attempt failed. After covering the first 4 cells ofthe column with the first tile, the following 7 cells with thesecond tile, and the following 4 cells with the third tile,there is no room for a fourth tile to cover the remaining twocells. In fact, if we manage to cover the 17 cells of the firstcolumn with 4 9 7 tiles, we will have written 17 as a sum of4 s and 7 s. But it is easy to check that this cannot be done,so a tiling does not exist. We have found a second reasonfor a tiling not to exist: It may be impossible to cover thefirst row or column, because either m or n cannot bewritten as a sum of a s and b s.

Is it then possible to tile a 10 9 15 rectangle using 1 9 6rectangles? In fact, 150 is a multiple of 6, and both 10 and 15can be written as a sum of 1 s and 6 s. However, this tilingproblem is still impossible!

The full answer to our question was given by de Bruijnand by Klarner [4, 13]. They proved that an m 9 n rectanglecan be tiled with a 9 b rectangles if and only if:

• mn is divisible by ab,• the first row and column can be covered; i.e., both m and

n can be written as sums of a s and b s, and• either m or n is divisible by a, and either m or n is

divisible by b.

Since neither 10 nor 15 is divisible by 6, the 10 9 15rectangle cannot be tiled with 1 9 6 rectangles. There arenow many proofs of de Bruijn and Klarner’s theorem. Aparticularly elegant one uses properties of the complexroots of unity [4, 13]. For an interesting variant with four-teen (!) proofs, see [20].

The second problem we wish to discuss is the following.Let x [ 0, such as x ¼

ffiffiffi2p

. Can a square be tiled withfinitely many rectangles similar to a 1 9 x rectangle (in anyorientation)? In other words, can a square be tiled withfinitely many rectangles, all of the form a 9 ax (where amay vary)?

For example, for x = 2/3, some of the tiles we can useare the following.

1.5

1 3

2 4

6 3π

2π

They have the same shape, but different sizes. In thiscase, however, we only need one size, because we can tilea 2 9 2 square with six 1 9 2/3 rectangles.

1 1

x = 2/3

2/3

2/3

For reasons which will become clear later, we point outthat x = 2/3 satisfies the equation 3x - 2 = 0. Notice alsothat a similar construction will work for any positiverational number x = p/q.?


Let us try to construct a tiling of a square with similarrectangles of at least two different sizes. There is a tilingapproximately given by the picture below. The rectanglesare similar because 0.7236. . ./1 = 0.2/0.2764. . ...

1/5

.2764...

.7236...

1

How did we find this configuration? Suppose that wewant to form a square by putting five copies of a rectanglein a row, and then stacking on top of them a larger rect-angle of the same shape on its side, as shown. Assume thatwe know the square has side length 1, but we do not knowthe dimensions of the rectangles. Let the dimensions of thelarge rectangle be 1 9 x. Then the height of each smallrectangle is equal to 1 - x. Since the small rectangles aresimilar to the large one, their width is x(1 - x). Sittingtogether in the tiling, their total width is 5x(1 - x), whichshould be equal to 1.

Therefore, the picture above is a solution to our prob-lem if x satisfies the equation 5x(1 - x) = 1, which werewrite as 5x2 - 5x + 1 = 0. One value of x that satisfiesthis equation is

x ¼ 5þffiffiffi5p

10¼ 0:7236067977. . .;

giving rise to the tiling illustrated above.But recall that any quadratic polynomial has two zeros;

the other one is

x ¼ 5�ffiffiffi5p

10¼ 0:2763932023. . .;

and it gives rise to a different tiling that also satisfies theconditions of the problem.

It may be unexpected that our tiling problem has asolution for these two somewhat complicated values of x.In fact, the situation can get much more intricate. Let us finda tiling using 3 similar rectangles of different sizes.

1

x = .5698...

.4302...

.2451... .7549...

Say that the largest rectangle has dimensions 1 9 x.Imitating the previous argument, we find that x satisfies theequation

x3 � x2 þ 2x � 1 ¼ 0:

One value of x that satisfies this equation is

x ¼ 0:5698402910. . . :

For this value of x, the tiling problem can be solved asabove. The polynomial above has degree three, so ithas two other zeros. They are approximately 0:215þ1:307

ffiffiffiffiffiffiffi�1p

and 0:215� 1:307ffiffiffiffiffiffiffi�1p

. These two complexnumbers do not give us real solutions to the tiling problem.

In the general situation, Freiling and Rinne [8] andLaczkovich and Szekeres [14] independently gave the fol-lowing amazing answer to this problem. A square can betiled with finitely many rectangles similar to a 1 9 x rect-angle if and only if:

• x is a zero of a polynomial with integer coefficients, and• for the polynomial of least degree satisfied by x, any zero

aþ bffiffiffiffiffiffiffi�1p

satisfies a [ 0.

It is very surprising that these complex zeros, that seemcompletely unrelated to the tiling problem, actually play afundamental role in it. In the example above, a solution fora 1 9 0.5698. . . rectangle is only possible because 0.215. . .is a positive number. Let us further illustrate this result withsome examples.

The value x ¼ffiffiffi2p

does satisfy a polynomial equationwith integer coefficients, namely, x2 - 2 = 0. However,the other root of the equation is �

ffiffiffi2p

\0 . Thus, a squarecannot be tiled with finitely many rectangles similar to a1�

ffiffiffi2p

rectangle.On the other hand, x ¼

ffiffiffi2pþ 17

12 satisfies the quadraticequation 144x2 - 408x + 1 = 0, whose other root is�

ffiffiffi2pþ 17

12 ¼ 0:002453 � � � [ 0. Therefore, a square can betiled with finitely many rectangles similar to a 1� ð

ffiffiffi2pþ 17

12Þrectangle. How would we actually do it?

Similarly, x ¼ffiffiffi23p

satisfies the equation x3 - 2 = 0. Theother two roots of this equation are �

ffiffi23p

2 �ffiffi23p ffiffi

3p

2


: Since�ffiffi23p

2 \0, a square cannot be tiled with finitely many rect-angles similar to a 1�

ffiffiffi23p

rectangle.Finally, let r/s be a rational number, and let x ¼ r

s þffiffiffi23p

.One can check that this is still a zero of a cubic polynomial,whose other two zeros are:

r

s�

ffiffiffi23p

2

� ��

ffiffiffi23p ffiffiffi

3p

2


:

It follows that a square can be tiled with finitely many rect-angles similar to a 1� ðrs þ

ffiffiffi23pÞ rectangle if and only if

r

s[

ffiffiffi23p

2:

As a nice puzzle, the reader can pick his or her favoritefraction larger than

ffiffiffi23p

=2 , and tile a square with rectanglessimilar to a 1� ðrs þ

ffiffiffi23pÞ rectangle.

For other tiling problems, including interesting algebraicarguments, see [18].

The third problem we wish to discuss is motivated bythe following remarkable tiling of a rectangle into ninesquares, all of which have different sizes. (We will soon seewhat the sizes of the squares and the rectangle are.) Suchtilings are now known as perfect tilings.


a

b c

de

f

h

ig

To find perfect tilings of rectangles, we can use theapproach of the previous problem. We start by proposing atentative layout of the squares, such as the pattern shown,without knowing what sizes they have. We denote the sidelength of each square by a variable. For each horizontal lineinside the rectangle, we write the following equation: Thetotal length of the squares sitting on the line is equal to thetotal length of the squares hanging from the line. Forexample, we have the ‘‘horizontal equations’’ a + d =

g + h and b = d + e. Similarly, we get a ‘‘vertical equation’’for each vertical line inside the rectangle, such as a = b + dor d + h = e + f. Finally, we write the equations that saythat the top and bottom sides of the rectangle are equal, andthe left and right sides of the rectangle are equal. In this case,they are a + b + c = g + i and a + g = c + f + i. It thenremains to hope that the resulting system of linear equationshas a solution and, furthermore, is one where the values ofthe variables are positive and distinct. For the layout pro-posed above, the system has a unique solution up to scaling:(a, b, c, d, e, f, g, h, i) = (15, 8, 9, 7, 1, 10, 18, 4, 14). Thelarge rectangle has dimensions 32 9 33.

Amazingly, the resulting system of linear equationsalways has a unique solution up to scaling, for any proposedlayout of squares. (Unfortunately, the resulting ‘‘side lengths’’are usually not positive and distinct.) In 1936, Brooks, Smith,Stone and Tutte [2] gave a beautiful explanation of this result.They constructed a directed graph whose vertices are thehorizontal lines found in the rectangle. There is one edge foreach small square, which goes from its top horizontal line toits bottom horizontal line. The diagram below shows theresulting graph for our perfect tiling of the 32 9 33 rectangle.

158

1814

104

71

9

We can think of this graph as an electrical network of unitresistors, where the current flowing through each wire isequal to the length of the corresponding square in the tiling.The ‘‘horizontal equations’’ for the side lengths of the squaresare equivalent to the equations for conservation of current inthis network, and the ‘‘vertical equations’’ are equivalent toOhm’s law. Knowing this, our statement is essentiallyequivalent to Kirchhoff’s theorem: The flow in each wire isdetermined uniquely, once we know thepotential differencebetween some two vertices (i.e., up to scaling).

Brooks, Smith, Stone and Tutte were especially interestedin studying perfect tilings of squares. This also has a niceinterpretation in terms of the network. To find tilings ofsquares, we would need an additional linear equation statingthat the vertical and horizontal side lengths of the rectangleare equal. In the language of the electrical network, this isequivalent to saying that the network has total resistance 1.

While this correspondence between tilings and networksis very nice conceptually, it does not necessarily make it easyto construct perfect tilings of squares, or even rectangles. Infact, after developing this theory, Stone spent some timetrying to prove that a perfect tiling of a square was impossi-ble. Roland Sprague finally constructed one in 1939, tiling asquare of side length 4205 with 55 squares. Since then, mucheffort and computer hours have been spent trying to findbetter constructions.Duijvestijn andhis computer [5] showedthat the smallest possible number of squares in aperfect tilingof a square is 21; the only such tiling is shown below.

16

15

50

27

423733

29 25 189

11

7

35

19

4

6

24

2

178

What Does a Typical Tiling Look Like?Suppose that we draw each possible solution to a tiling prob-lem ona sheet of paper, put these sheets of paper in a bag, andpick one of them at random. Can we predict what we will see?


The random domino tiling of a 12 9 12 squareshown, with horizonal dominoes shaded darkly andvertical dominoes shaded lightly, exhibits no obviousstructure. Compare this with a random tiling of the Aztecdiamond of order 50. Here, there are two shades ofhorizontal dominoes and two shades of vertical domi-noes, assigned according to a certain rule not relevanthere. These pictures were created by Jim Propp’s TilingsResearch Group.

.

This very nice picture suggests that something inter-esting can be said about random tilings. The tiling isclearly very regular at the corners, and gets more chaoticas we move away from the edges. There is a well definedregion of regularity, and we can predict its shape. Jock-usch, Propp and Shor [11] showed that for very large n,and for ‘‘most’’ domino tilings of the Aztec diamondAZ(n), the region of regularity ‘‘approaches’’ the outsideof a circle tangent to the four limiting sides. Sophisticatedprobability theory is needed to make the terms ‘‘most’’and ‘‘approaches’’ precise, but the intuitive meaningshould be clear.

This result is known as the Arctic Circle theorem. Thetangent circle is the Arctic Circle; the tiling is ‘‘frozen’’ outsideof it. Many similar phenomena have since been observedand (in some cases) proved for other tiling problems.

Relations Among TilingsWhen we study the set of all tilings of a region, it is oftenuseful to be able to ‘‘navigate’’ this set in a nice way.Suppose we have one solution to a tiling problem, and wewant to find another one. Instead of starting over, it isprobably easier to find a second solution by making smallchanges to the first one. We could then try to obtain a thirdsolution from the second one, then a fourth solution, andso on.

In the case of domino tilings, there is a very easyway to do this. A flip in a domino tiling consists ofreversing the orientation of two dominoes forming a2 9 2 square.

This may seem like a trivial transformation to get fromone tiling to another. However, it is surprisingly powerful.Consider the two following tilings of a region.

Although they look very different from each other, onecan, in fact, reach one from the other by successively flip-ping 2 9 2 blocks.

Thurston [21] showed that this is a general phenome-non. For any region R with no holes, any domino tiling ofR can be reached from any other by a sequence of flips.

This domino flipping theorem has numerous applica-tions in the study of domino tilings. We point out that thetheorem can be false for regions with holes, as shown bythe two tilings of a 3 9 3 square with a hole in the middle.Here, there are no 2 9 2 blocks and, hence, no flips at all.There is a version, due to Propp [17], of the domino flip-ping theorem for regions with holes, but we will notdiscuss it here.

Confronting InfinityWe now discuss some tiling questions that involve arbi-trarily large regions or arbitrarily small tiles.

The first question is motivated by the following identity:

1

1 � 2þ1

2 � 3þ1

3 � 4þ � � � ¼ 1:


Consider infinitely many rectangular tiles of dimensions1� 1

2 ;12� 1

3 ;13� 1

4 ; . . .: These tiles get smaller and smaller,and the above equation shows that their total area is exactlyequal to 1. Can we tile a unit square using each one of thesetiles exactly once?

1

1/2 1/31/2 1/3

1/41/4

1/51/5

1/6 ...

1

1

This seems to be quite a difficult problem. An initialattempt shows how to fit the first five pieces nicely. How-ever, it is difficult to imagine how we can fit all of the piecesinto the square without leaving any gaps.

1

1/2

1/3

1/2

1/3

1/4

1/5

1/4

1/51/6

To this day, no one has been able to find a tiling orprove that it does not exist. Paulhus [16] has come veryclose; he found a way to fit all these rectangles into asquare of side length 1.000000001. Of course, Paulhus’spacking is not a tiling as we have defined the term, sincethere is leftover area.

Let us now discuss a seemingly simple problemthat makes it necessary to consider indeterminately largeregions. Recall that a polyomino is a collection of unitsquares arranged with coincident sides.

Let us call a collection of polyominoes ‘‘good’’ if it ispossible to tile the whole plane using the collection as tiles,and ‘‘bad’’ otherwise. A good and a bad collection ofpolyominoes are shown below.

**

*

badgood

It is easy to see why it is impossible to tile the wholeplane with the bad collection shown above. Once we laydown a tile, the square(s) marked with an asterisk cannotbe covered by any other tile.

However, we can still ask: How large of a square regioncan we cover with a tiling? After a few tries, we will findthat it is possible to cover a 4 9 4 square.

It is impossible, however, to cover a 5 9 5 square. Anyattempt to cover the central cell of the square with a tile willforce one of the asterisks of that tile to land inside thesquare as well.

In general, the question of whether a given collection ofpolyominoes can cover a given square is a tremendouslydifficult one. A deep result from mathematical logic statesthat there does not exist an algorithm to decide the answerto this question.1

An unexpected consequence of this deep fact is thefollowing. Consider all the bad collections of polyominoesthat have a total of n unit cells. Let L(n) be the side lengthof the largest square that can be covered with one of them.The bad collection of our example, which has a total of22 unit squares, shows that Lð22Þ� 4:

One might expect L(22) to be reasonably small. Givena bad collection of tiles with a total of 22 squares,imagine that we start laying down tiles to fit togethernicely and cover as large a square as possible. Since thecollection is bad, at some point we will inevitably form ahole that we cannot cover. It seems plausible to assumethat this will happen fairly soon, since our tiles are quitesmall.

Surprisingly, however, the numbers L(n) are incredi-bly large! If f(n) is any function that can be computedon a computer, even with infinite memory, thenL(n) [ f(n) for all large enough n. Notice that

1A related question is the following: Given a polyomino P, does there exist a rectangle that can be tiled using copies of P? Despite many statements to the contrary in

the literature, it is not known whether there exists an algorithm to decide this.


computers can compute functions that grow veryquickly, such as

f ðnÞ ¼ nn; f ðnÞ ¼ nnn

;or

f ðnÞ ¼ nn : : :n

ða tower of lengthnÞ; . . .:

In fact, all of these functions are tiny in comparison withcertain other computable functions. In turn, every com-putable function is tiny in comparison with L(n).

We can give a more concrete consequence of this result.There exists a collection of polyominoes with a modestnumber of unit squares2, probably no more than 100, withthe following property: It is impossible to tile the wholeplane with this collection; however, it is possible to com-pletely cover Australia3 with a tiling.

A very important type of problem is concerned withtilings of infinite (unbounded) regions, in particular, tilingsof the entire plane. This is too vast a subject to touch onhere. For further information, we refer the reader to the700-page book [9] by Grunbaum and Shephard devotedprimarily to this subject.

REFERENCES

[1] D. Beauquier, M. Nivat, Remila E., E. Robson. Tiling figures of the

plane with two bars. Comput. Geom. 5 (1995), 1–25.

The authors consider the problem of tiling a region with horizontal

n 9 1 and vertical 1 9 m rectangles. Their main result is that, for n

C 2 and m [ 2, deciding whether such a tiling exists is an NP-

complete question. They also study several specializations of this

problem.

[2] R. Brooks, C. Smith, A. Stone and W. Tutte. The dissection of

rectangles into squares. Duke Math. J. 7 (1940), 312–340.

To each perfect tiling of a rectangle, the authors associate a

certain graph and a flow of electric current through it. They show

how the properties of the tiling are reflected in the electrical net-

work. They use this point of view to prove several results about

perfect tilings, and to provide new methods for constructing them.

[3] J. Conway and J. Lagarias. Tiling with polyominoes and combi-

natorial group theory. J. Combin. Theory Ser. A 53 (1990), 183–

208.

Conway and Lagarias study the existence of a tiling of a region in

a regular lattice in R2 using a finite set of tiles. By studying the way

in which the boundaries of the tiles fit together to give the

boundary of the region, they give a necessary condition for a tiling

to exist, using the language of combinatorial group theory.

[4] N. de Bruijn. Filling boxes with bricks. Amer. Math. Monthly 76

(1969), 37–40.

The author studies the problem of tiling an n-dimensional box of

integer dimensions A1 9 . . . 9 An with bricks of integer dimen-

sions a1 9 . . . 9 an. For a tiling to exist, de Bruijn proves that every

ai must have a multiple among A1, . . . , An. The box is called a

multiple of the brick if it can be tiled in the trivial way. It is shown

that, if a1|a2, a2|a3, . . ., an-1|an, then the brick can only tile boxes

that are multiples of it. The converse is also shown to be true.

[5] A. Duijvestijn. Simple perfect squared square of lowest order.

J. Combin. Theory Ser. B 25 (1978), 240–243.

The unique perfect tiling of a square using the minimum possible

number of squares, 21, is exhibited.

[6] N. Elkies, G. Kuperberg, M. Larsen and J. Propp. Alternating sign

matrices and domino tilings I, II. J. Algebraic Combin 1 (1992),

111–132, 219–234.

It is shown that the Aztec diamond of order n has 2n(n+1)/2 domino

tilings. Four proofs are given, exploiting the connections of this

object with alternating-sign matrices, monotone triangles, and the

representation theory of GL(n). The relation with Lieb’s square-ice

model is also explained.

[7] M. Fisher and H. Temperley. Dimer problem in statistical

mechanics—an exact result. Philosophical Magazine 6 (1961),

1061–1063.

A formula for the number of domino tilings of a rectangle is given

in the language of statistical mechanics.

[8] C. Freiling and D. Rinne. Tiling a square with similar rectangles.

Math. Res. Lett 1 (1994), 547-558.

The authors show that a square can be tiled with rectangles

similar to the 1 9 u rectangle if and only if u is a zero of a poly-

nomial with integer coefficients, all of whose zeros have positive

real part.

[9] B. Grunbaum and G. Shephard. Tilings and patterns. W.H.

Freeman and Company, New York (1987).

This book provides an extensive account of various aspects of

tilings, with an emphasis on tilings of the plane with a finite set of

tiles. For example, the authors carry out the task of classifying

several types of tiling patterns in the plane. Other topics dis-

cussed include perfect tilings of rectangles and aperiodic tilings of

the plane.

[10] P. Hall. On representatives of subsets. J. London Math. Soc 10

(1935), 26-30.

Given m subsets T1, . . . , Tm of a set S, Hall defines a complete

system of distinct representatives to be a set of m distinct ele-

ments a1, . . . , am of S such that ai [ Ti for each i. He proves that

such a system exists if and only if, for each k = 1, . . . , m, the

union of any k of the sets contains at least k elements.

[11] W. Jockusch, J. Propp and P. Shor P. Random domino tilings

and the Arctic Circle theorem, preprint, 1995, arXiv:math.CO/

9801068.

In a domino tiling of an Aztec diamond, the diamond is partitioned

into five regions: Four outer regions near the corners where the tiles

are neatly lined up, and one central region where they do not follow

a predictable pattern. The authors prove the Arctic circle theorem:

In a random tiling of a large Aztec diamond, the central region is

extremely close to a perfect circle inscribed in the diamond.

[12] P. Kasteleyn. The statistics of dimers on a lattice I. The number of

dimer arrangements on a quadratic lattice. Physica 27 (1961),

1209-1225.

Kasteleyn proves exact and asymptotic formulas for the number

of domino tilings of a rectangle, with edges or with periodic

boundary conditions. He then discusses the relationship between

this problem and the Ising model of statistical mechanics.

2Say ‘‘unit squares’’ have a side length of 1 cm.3which is very large and very flat


[13] D. Klarner. Packing a rectangle with congruent n-ominoes. J.

Combin. Theory 7 (1969), 107-115.

Klarner investigates the problem of tiling a rectangle using an odd

number of copies of a single polyomino. He also characterizes the

rectangles that can be tiled with copies of an a 9 b rectangle, and

the rectangles that can be tiled with copies of a certain octomino.

[14] M. Laczkovich and G. Szekeres. Tilings of the square with similar

rectangles. Discrete Comput. Geom 13 (1995), 569-572.

The authors show that a square can be tiled with rectangles

similar to the 1 9 u rectangle if and only if u is a zero of a poly-

nomial with integer coefficients, all of whose zeros have positive

real part.

[15] I. Pak. Tile invariants: New horizons. Theoret. Comput. Sci. 303

(2003), 303-331.

Given a finite set of tiles T, the group of invariants G(T) consists of

the linear relations that must hold between the number of tiles of

each type in tilings of the same region. This paper surveys what is

known about G(T). These invariants are shown to be much

stronger than classical coloring arguments.

[16] M. Paulhus. An algorithm for packing squares. J. Combin. Theory

Ser. A 82 (1998), 147–157.

Paulhus presents an algorithm for packing an infinite set of

increasingly small rectangles with total area A into a rectangle of

area very slightly larger than A. He applies his algorithm to three

known problems of this sort, obtaining extremely tight packings.

[17] J. Propp. Lattice structure for orientations of graphs, preprint,

1994, arXiv: math/0209005.

It is shown that the set of orientations of a graph that have the

same flow-differences around all circuits can be given the struc-

ture of a distributive lattice. This generalizes similar constructions

for alternating-sign matrices and matchings.

[18] S. Stein and S. Szabo. Algebra and tiling. Homomorphisms in the

service of geometry. Mathematical Association of America:

Washington, DC, 1994.

This book discusses the solution of several tiling problems using

tools from modern algebra. Two sample problems are the fol-

lowing: A square cannot be tiled with 30�–60�–90� triangles, and

a square of odd integer area cannot be tiled with triangles of unit

area.

[19] W. Thurston. Conway’s tiling groups. Amer. Math. Monthly 97

(1990), 757–773.

The author presents a technique of Conway’s for studying tiling

problems. Sometimes it is possible to label the edges of the tiles

with elements of a group, so that a region can be tiled if and only if

the product (in order) of the labels on its boundary is the identity

element. The idea of a height function that lifts tilings to a three-

dimensional picture is also presented. These techniques are

applied to tilings with dominoes, lozenges, and tribones.

[20] S. Wagon. Fourteen proofs of a result about tiling a rectangle.

Amer. Math. Monthly 94 (1987), 601–617.

Wagon gives 14 proofs of the following theorem: If a rectangle

can be tiled by rectangles, each of which has at least one integral

side, then the tiled rectangle has at least one integral side.


A World Recordin Atlantic Cityand the Lengthof the Shooter’sHand at CrapsS. N. ETHIER AND FRED M. HOPPE*

IIt was widely reported in the media that, on 23 May2009, at the Borgata Hotel Casino & Spa in AtlanticCity, Patricia DeMauro (spelled Demauro in some

accounts), playing craps for only the second time, rolledthe dice for four hours and 18 minutes, finally sevening outat the 154th roll. Initial estimates of the probability of a runat least this long (assuming fair dice and independent rolls)ranged from one chance in 3.5 billion [5] to one chance in1.56 trillion [10]. Subsequent computations agreed on onechance in 5.6 (or 5.59) billion [2, 6, 9].

This established a new world record. The old record washeld by the late Stanley Fujitake (118 rolls, 28 May 1989,California Hotel and Casino, Las Vegas) [1]. One might askhow reliable these numbers (118 and 154) are. In Mr.Fujitake’s case, casino personnel replayed the surveillancevideotape to confirm the number of rolls and the durationof time (three hours and six minutes). We imagine that thesame happened in Ms. DeMauro’s case.

There is also a report that Mr. Fujitake’s record wasbroken earlier by a gentleman known only as The Captain(148 rolls, July 2005, Atlantic City) [8, Part 4]. However, thisincident is not well documented (specifically, the exact dateand casino name were not revealed) and was unknown to

Borgata officials. In fact, a statistical argument has beenoffered [4, p. 480] suggesting that the story is apocryphal.

Our aim in this article is not simply to derive a moreaccurate probability, but to show that this apparently prosaicproblem involves some interesting mathematics, includingMarkov chains, matrix theory, generating functions, andGalois theory.

BackgroundCraps is played by rolling a pair of dice repeatedly. Formost bets, only the sum of the numbers appearing on thetwo dice matters, and this sum has distribution

pj :¼ 6� j j � 7j36

; j ¼ 2; 3; . . .; 12: ð1Þ

The basic bet at craps is the pass-line bet, which is definedas follows. The first roll is the come-out roll. If 7 or 11appear (a natural), the bettor wins. If 2, 3, or 12 appears (acraps number), the bettor loses. If a number belonging to

P :¼ f4; 5; 6; 8; 9; 10g

appears, that number becomes the point. The dice continueto be rolled until the point is repeated (or made), in which

*Supported by NSERC.

44 THE MATHEMATICAL INTELLIGENCER � 2010 Springer Science+Business Media, LLC

case the bettor wins, or 7 appears, in which case the bettorloses. The latter event is called a seven out. The first rollfollowing a decision is a new come-out roll, beginning theprocess again.

A shooter is permitted to roll the dice until he or she sevensout. The sequence of rolls by the shooter is called the shooter’shand. Notice that the shooter’s hand can contain winning 7s andlosingdecisionsprior to the sevenout.The lengthof the shooter’shand (i.e., the number of rolls) is a random variable we willdenote by L. Our concern here is with

tðnÞ :¼ PðL�nÞ; n� 1; ð2Þ

the tail of the distribution of L. For example, t(154) is theprobability of achieving a hand at least as long as that ofMs. DeMauro. As can be easily verified from (3), (6), or (9)below, t(154)&0.178 882 426 9 10-9; to state it in the waypreferred by the media, this amounts to one chance in 5.59billion, approximately. The 1 in 3.5 billion figure came froma simulation that was not long enough. The 1 in 1.56 trillionfigure came from (1 - p7)

154, which is the right answer tothe wrong question.

Two MethodsWe know of two methods for evaluating the tail probabil-ities (2). The first is by recursion. As pointed out in [3],t (1) = t (2) = 1 and

tðnÞ ¼ 1�Xj2P

pj

!tðn� 1Þ þ

Xj2P

pjð1� pj � p7Þn�2

þXj2P

pj

Xn�1

l¼2

ð1� pj � p7Þl�2pj tðn� lÞ ð3Þ

for each n C 3. Indeed, for the event that the shooter sevensout in no fewer than n rolls to occur, consider the result ofthe initial come-out roll. If a natural or a craps numberoccurs, then, beginning with the next roll, the shooter mustseven out in no fewer than n - 1 rolls. If a point numberoccurs, then there are two possibilities. Either the point isstill unresolved after n - 2 additional rolls, or it is made atroll l for some l [ {2, 3, . . ., n - 1} and the shooter subse-quently sevens out in no fewer than n - l rolls.

The second method, first suggested, to the best of ourknowledge, by Peter A. Griffin in 1987 (unpublished) and

rediscovered several times since, is based on a Markovchain. The state space is

S :¼ fco; p4-10; p5-9; p6-8; 7og � f1; 2; 3; 4; 5g; ð4Þ

whose five states represent the events that the shooter iscoming out, has established the point 4 or 10, has estab-lished the point 5 or 9, has established the point 6 or 8, andhas sevened out. The one-step transition matrix, which canbe inferred from (1), is

P :¼ 1

36

12 6 8 10 03 27 0 0 64 0 26 0 65 0 0 25 60 0 0 0 36

0BBBB@

1CCCCA: ð5Þ

The probability of sevening out in n - 1 rolls or fewer isthen just the probability that absorption in state 7o occursby the (n - 1)th step of the Markov chain, starting in stateco. A marginal simplification results by considering the 4by 4 principal submatrix Q of (5) corresponding to thetransient states. Thus, we have

tðnÞ ¼ 1� ðPn�1Þ1;5 ¼X4

j¼1

ðQn�1Þ1;j : ð6Þ

Clearly, (3) is not a closed-form expression, and we do notregard (6) as being in closed form either. Is there a closed-form expression for t(n)?

Positivity of the EigenvaluesWe begin by showing that the eigenvalues of Q are posi-tive. The determinant of

Q� zI ¼ 1

36

12� 36z 6 8 10

3 27� 36z 0 0

4 0 26� 36z 0

5 0 0 25� 36z

0BBB@

1CCCA

is unaltered by row operations. From the first row, subtract6/(27 - 36z) times the second row, 8/(26 - 36z) times thethird row, and 10/(25 - 36z) times the fourth row, can-celling the entries 6/36, 8/36, and 10/36 and making the(1,1) entry equal to 1/36 times

.........................................................................................................................................................

AU

TH

OR

S S. N. ETHIER is professor of mathematicsat the University of Utah, and he specializes

in applied probability. His book, The Doc-trine of Chances: Probabilistic Aspects ofGambling, will be published this year by

Springer.


University of UtahSalt Lake City, UT 84112

USA


FRED M. HOPPE is professor of mathe-matics and statistics, and associate faculty of

the Booth School of Engineering Practice, at

McMaster University. At times, he has pub-

lished on branching processes, population

genetics, probability bounds, nuclear risk, and

lotteries.

Department of Mathematics and StatisticsMcMaster University

Hamilton, ON L8S 4K1

Canada



12� 36z � 36

27� 36z� 4

8

26� 36z� 5

10

25� 36z: ð7Þ

The determinant of Q� zI, and therefore the characteristicpolynomial q(z) of Q, is then just the product of thediagonal entries in the transformed matrix, which is(7) multiplied by (27 - 36z)(26 - 36z)(25 - 36z)/(36)4.Thus,

qðzÞ ¼ ½ð12� 36zÞð27� 36zÞð26� 36zÞð25� 36zÞ� 18ð26� 36zÞð25� 36zÞ� 32ð27� 36zÞð25� 36zÞ� 50ð27� 36zÞð26� 36zÞ�=ð36Þ4:

We find that q(1), q(27/36), q(26/36), q(25/36), and q(0)alternate signs, and therefore the eigenvalues are positiveand interlaced between the diagonal entries (ignoring theentry 12/36). More precisely, denoting the eigenvalues by1 [ e1 [ e2 [ e3 [ e4 [ 0, we have

1[e1 [27

36[e2 [

26

36[e3 [

25

36[e4 [0:

The matrix Q, which has the structure of an arrowheadmatrix, is not symmetric, but is positive definite. Anonsymmetric matrix is positive definite if and only if itssymmetric part is positive definite. This is easily seen to bethe case for Q by applying the same type of row operationsto the symmetric part A¼ 1

2 ðQþQTÞ to show that theeigenvalues of A interlace its diagonal elements (except 12/36), and hence are positive.

A Closed-Form ExpressionThe eigenvalues of Q are the four roots of the quarticequation q(z) = 0 or

23328z4 � 58320z3 þ 51534z2 � 18321z þ 1975 ¼ 0; ð8Þ

while P has an additional eigenvalue, 1, the spectral radius.We can use the quartic formula (or Mathematica) to findthese roots. We notice that the complex number

a :¼ f1=3 þ 9829

f1=3;

where

f :¼ �710369þ 18iffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1373296647

p;

appears three times in each root. Fortunately, a is positive,as we see by writing f in polar form, that is, f = reih. Weobtain

a ¼ 2ffiffiffiffiffiffiffiffiffiffi9829p

cos1

3cos�1 � 710369

9829ffiffiffiffiffiffiffiffiffiffi9829p

� �� :

The four eigenvalues of Q can be expressed as

e1 :¼ eð1; 1Þ;e2 :¼ eð1;�1Þ;e3 :¼ eð�1; 1Þ;e4 :¼ eð�1;�1Þ;

where

eðu; vÞ :¼ 5

8þ u

72

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi349þ a

3

r

þ v

72

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi698� a

3� 2136u

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3

349þ a

rs:

Next we need to find right eigenvectors correspondingto the five eigenvalues of P. Fortunately, these eigenvectorscan be expressed in terms of the eigenvalues. Indeed, withrðxÞ defined to be the vector-valued function

�5þ ð1=5Þx�175þ ð581=15Þx � ð21=10Þx2 þ ð1=30Þx3

275=2� ð1199=40Þx þ ð8=5Þx2 � ð1=40Þx3

1

0

0BBBBBB@

1CCCCCCA

we find that right eigenvectors corresponding to eigen-values 1, e1, e2, e3, e4 are

ð1; 1; 1; 1; 1ÞT; rð36e1Þ; rð36e2Þ; rð36e3Þ; rð36e4Þ;

respectively. Letting R denote the matrix whose columnsare these right eigenvectors and putting L :¼ R�1, the rowsof which are left eigenvectors, we know by (6) and thespectral representation that

tðnÞ ¼ 1� fR diagð1; en�11 ; en�1

2 ; en�13 ; en�1

4 ÞLg1;5:

After much algebra (and with some help from Math-ematica), we obtain

tðnÞ ¼ c1en�11 þ c2en�1

2 þ c3en�13 þ c4en�1

4 ; ð9Þ

where the coefficients are defined in terms of the eigen-values and the function

f ðw; x; y; zÞ :¼ ð�25þ 36wÞ�4835� 5580ðx þ y þ zÞ

þ 6480ðxy þ xz þ yzÞ � 7776xyz�=

38880ðw � xÞðw � yÞðw � zÞ½ �

as follows:

c1 :¼ f ðe1; e2; e3; e4Þ;

c2 :¼ f ðe2; e3; e4; e1Þ;

c3 :¼ f ðe3; e4; e1; e2Þ;

c4 :¼ f ðe4; e1; e2; e3Þ:

Of course, (9) is our closed-form expression.Incidentally, the fact that t (1) = t (2) = 1 implies that

c1 þ c2 þ c3 þ c4 ¼ 1 ð10Þ

and

c1e1 þ c2e2 þ c3e3 þ c4e4 ¼ 1:

In a sequence of independent Bernoulli trials, each withsuccess probability p, the number of trials X needed to


achieve the first success has the geometric distribution withparameter p, and

PðX �nÞ ¼ ð1� pÞn�1; n� 1:

It follows that the distribution of L is a linear combinationof four geometric distributions. It is not a convex combi-nation: (10) holds but, as we will see,

c1 [ 0; c2\0; c3\0; c4\0:

In particular, we have the inequality

tðnÞ\c1en�11 ; n� 1; ð11Þ

as well as the asymptotic formula

tðnÞ� c1en�11 as n!1: ð12Þ

Another way to derive (9) is to begin with the recursiveformula (3). The generating function of the tail probabilities(2) is

T ðzÞ :¼X1n¼3

tðnÞzn�1;

and by (3) we have

T ðzÞ ¼ 1�Xj2P

pj

!zðz þ T ðzÞÞ

þXj2P

pjð1� pj � p7Þz2

1� ð1� pj � p7Þz

þXj2P

p2j z2

1� ð1� pj � p7Þzð1þ z þ T ðzÞÞ:

Solving for T (z) using (1), we find that

T ðzÞ ¼ z2ð20736� 33828z þ 16346z2 � 1975z3Þ23328� 58320z þ 51534z2 � 18321z3 þ 1975z4

;

the denominator of which can be written (cf. (8)) as

23328ð1� e1zÞð1� e2zÞð1� e3zÞð1� e4zÞ:

A partial-fraction expansion leads to (9), except that f isreplaced by

f ðw; x; y; zÞ :¼ � 1975� 16346w þ 33828w2 � 20736w3

23328w2ðw � xÞðw � yÞðw � zÞ :

Using Vieta’s formulas, this alternative version of (9) can beshown to be equivalent to the original one; in fact, yetanother version uses

f ðw;x;y;zÞ :¼ 1975� 16346wþ 33828w2� 20736w3

3w2ð6107� 34356wþ 58320w2 � 31104w3Þ ;

which has the advantage of depending only on w.

Numerical ApproximationsRounding to 18 decimal places, the non-unit eigenvalues ofP are

e1 � 0:862 473 751 659 322 030;

e2 � 0:741 708 271 459 795 977;

e3 � 0:709 206 775 794 379 015;

e4 � 0:186 611 201 086 502 979;

and the coefficients in (9) are

c1 � 1:211 844 812 464 518 572;

c2 � �0:006 375 542 263 784 777;

c3 � �0:004 042 671 248 651 503;

c4 � �0:201 426 598 952 082 292:

These numbers will give very accurate results over a widerange of values of n.

The result (12) shows that the leading term in (9) may beadequate for large n; it can be shown that

1\c1en�11 =tðnÞ\1þ 10�m

for m = 3 if n C 19; for m = 6 if n C 59; for m = 9 ifn C 104; and for m = 12 if n C 150.

Crapless CrapsIn crapless craps [7, p. 354], as the name suggests, there areno craps numbers and 7 is the only natural. Therefore, theset of possible point numbers is

P0 :¼ f2; 3; 4; 5; 6; 8; 9; 10; 11; 12g

but otherwise the rules of craps apply. More precisely, thepass-line bet is won either by rolling 7 on the come-out rollor by rolling a number other than 7 on the come-out rolland repeating that number before 7 appears.

With L0 denoting the length of the shooter’s hand, theanalogues of (4)–(6) are

S0 :¼ fco; p2-12; p3-11; p4-10; p5-9; p6-8; 7og� f1; 2; 3; 4; 5; 6; 7g;

P0 :¼ 1

36

6 2 4 6 8 10 0

1 29 0 0 0 0 6

2 0 28 0 0 0 6

3 0 0 27 0 0 6

4 0 0 0 26 0 6

5 0 0 0 0 25 6

0 0 0 0 0 0 36

0BBBBBBBBBBB@

1CCCCCCCCCCCA

;

and

t0ðnÞ :¼ PðL0�nÞ ¼ 1� ðPn�10 Þ1;7:

There is an interesting distinction between this gameand regular craps. The non-unit eigenvalues of P0 are theroots of the sextic equation

0 ¼ 15116544z6 � 59206464z5 þ 93137040z4

� 73915740z3 þ 30008394z2 � 5305446z þ 172975;


and the corresponding Galois group is, according to Maple,the symmetric group S6. This means that our sextic is notsolvable by radicals. Thus, it appears that there is noclosed-form expression for t0(n).

Nevertheless, the analogue of (9) holds (with six terms).All non-unit eigenvalues belong to (0, 1) and all coefficientsexcept the leading one are negative. Thus, the analogues of(11) and (12) hold as well. Also, the distribution of L0 is alinear combination of six geometric distributions. Theseresults are left as exercises for the interested reader.

Finally, t0(154) & 0.296 360 068 9 10-10, which is to saythat a hand of length 154 or more is only about one-sixth aslikely as at regular craps (one chance in 33.7 billion,approximately).

ACKNOWLEDGMENTS

We thank Roger Horn for pointing out the interlacing

property of the eigenvalues of Q. We also thank a referee

for suggesting the alternative approach via the generating

function T(z).

REFERENCES

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Hawaii, 1 May 2008 and 1 June 2008. http://www.aroundhawaii.

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html and http://www.aroundhawaii.com/lifestyle/travel/2008-06-

the-man-with-the-golden-arm-part-ii.html

[2] Bialik, C. Crunching the numbers on a craps record. The

Numbers Guy, Wall Street Journal blog. 28 May 2009. http://

blogs.wsj.com/numbersguy/crunching-the-numbers-on-a-craps-

record-703/

[3] Ethier, S. N. A Bayesian analysis of the shooter’s hand at craps.

In: S. N. Ethier and W. R. Eadington (eds.) Optimal Play:

Mathematical Studies of Games and Gambling, pp. 311–322.

Institute for the Study of Gambling and Commercial Gaming,

University of Nevada, Reno, 2007.

[4] Grosjean, J. Exhibit CAA. Beyond Counting: Exploiting Casino

Games from Blackjack to Video Poker. South Side Advantage

Press, Las Vegas, 2009.

[5] Paik, E. Denville woman recalls setting the craps record in AC.

Newark Star-Ledger, 27 May 2009. http://www.nj.com/news/

local/index.ssf/2009/05/pat_demauro_remembers_only_one.html

[6] Peterson, B. A new record in craps. Chance News 49 (2009).

http://www.causeweb.org/wiki/chance//index.php/Chance_

News_49

[7] Scarne, J. and Rawson, C. Scarne on Dice. The Military

Service Publishing Co., Harrisburg, PA, 1945.

[8] Scoblete, F. The Virgin Kiss and Other Adventures. Research

Services Unlimited, Daphne, AL, 2007.

[9] Shackleford, M. Ask the Wizard! No. 81. 1 June 2009.

http://wizardofodds.com/askthewizard/askcolumns/askthewizard

81.html

[10] Suddath, C. Holy craps! How a gambling grandma broke the

record. Time, 29 May 2009. http://www.time.com/time/nation/

article/0,8599,1901663,00.html


http://www.aroundhawaii.com/lifestyle/travel/2008-05-the-man-with-the-golden-arm-part-i.html



http://www.aroundhawaii.com/lifestyle/travel/2008-06-the-man-with-the-golden-arm-part-ii.html

http://www.aroundhawaii.com/lifestyle/travel/2008-06-the-man-with-the-golden-arm-part-ii.html

http://blogs.wsj.com/numbersguy/crunching-the-numbers-on-a-craps-record-703/



http://www.nj.com/news/local/index.ssf/2009/05/pat_demauro_remembers_only_one.html

http://www.nj.com/news/local/index.ssf/2009/05/pat_demauro_remembers_only_one.html

http://www.causeweb.org/wiki/chance//index.php/Chance_News_49

http://www.causeweb.org/wiki/chance//index.php/Chance_News_49

http://wizardofodds.com/askthewizard/askcolumns/askthewizard81.html

http://wizardofodds.com/askthewizard/askcolumns/askthewizard81.html

http://www.time.com/time/nation/article/0,8599,1901663,00.html

http://www.time.com/time/nation/article/0,8599,1901663,00.html

How to Win WithoutOvertly Cheating:The Inverse SimpsonParadoxORA E. PERCUS AND JEROME K. PERCUS

AAnyone contemplating a statistical analysis is warned,at an early stage of the game, ‘‘But don’t combine thestatistics of monkey wrenches and watermelons,’’ or

the equivalent. Failure to heed this instruction – at a moresophisticated level to be sure – gives rise frequently toSimpson’s Paradox: if choice A is ‘‘better on average’’ thanchoice B in each of two differing circumstances, it maynevertheless happen that merging the two sets of dataproduces the opposite conclusion. We are going to look atthis familiar pitfall, and then analyze the less familiar dangerthat it may occur ‘‘accidentally on purpose.’’

SimpsonConsider the following specially constructed example forthe sake of illustration:

Two workers A and B are evaluated on performance inone easy task (#1) and one hard task (#2).

Task#1Worker A: 20 tries, mean success rate 0.8

Worker B: 80 tries, mean success rate 0.6

�

Task#2Worker A: 80 tries, mean success rate 0.4

Worker B: 20 tries, mean success rate 0.2

�

Here 0.8[ 0.6, 0.4 [ 0.2, showing A’s superiority onboth tasks. But if we made the mistake of considering thetotal number of successes out of 100 tries for each worker,

we would see 16 + 32 \ 48 + 4, and B might seem pref-erable, just because B has de-emphasized the hard task.

Or, more generally, if we tally successes (S) for A and Bin tasks #1 and #2, and find Worker A has SA1 successes outof NA1 tries on the first, SA2 successes out of NA2 tries on thesecond task, with similar notation for Worker B, and if weset SA = SA1 + SA2, NA = NA1 + NA2, etc., it may very well

happen thatSA

NA\

SB

NBeven though

SA1

NA1[

SB1

NB1;SA2

NA2[

SB2

NB2:

This phenomenon is well known and well documented[3, 5–9] – but hope springs eternal. Only recently [1] a drugmanufacturer, whose current potential blockbuster drug(Xinlay) failed to better a placebo in two clinical trials withuncorrelated protocols, proposed to a regulatory agency topool the two sequences. If that criterion had been used,their drug would have appeared to outperform the pla-cebo, allowing them to move forward. The regulatoryagency panel was not unaware of the possibility of para-dox, and denied the reinterpretation of the data.

Inverse SimpsonThe Simpson Paradox is data driven. It may, or may not, holdin a given situation; that is, data sets which indicate a similarstatistical conclusion, when combined, may or may not pointto the opposite conclusion. In fact, if the component data setsare sufficiently similar (say, as an extreme, identical), thenpooling them surely will not reverse the conclusion.

We will speak of the Inverse Simpson Paradox in case westart with a comparison between two large data sets – say,


successes and failures with drug A, and similarly with drugB – and ask if deceptive conclusions can result fromdecomposing it into two comparisons. Instead of pooling or‘‘aggregating’’ two experiments, we askwhether it is possibleto concoct a decomposition or ‘‘de-aggregation’’ into twoexperiments and give apparently opposite conclusions. Theend result will then be the same old Simpson Paradox, only itarose by imposing a decomposition not a merging.

There is a variety of purposes one may have in mind:

a) Most directly and legitimately, it may be realized that datafrom two sources were combined for simplicity, and so aunique natural decomposition is called for, which may beinstructive even if it does reverse the conclusion. Thisappears to be the case in the oft-quoted Berkeley sexdiscrimination controversy [5] discussed below.

b) Least directly and least legitimately – but perhaps aneffective strategy in litigation – one can ask for thatdecomposition that maximally reverses the conclusion,and then artfully invent excuses for lumping the data togive those subsets.

c) Putting a different spin on b), one can ask for thatdecomposition that maximally comes jointly to eitherconclusion, and use this as an investigative tool torecognize hidden properties of significant subsets of data.

We see at once that artificially inducing Inverse Simpsonis indeed generally possible.

If we break down our given pool of data into two arti-ficial tasks and arrange that Task #1 contains no failures forA and Task #2 contains no successes for B, then to be sureWorker A will be dramatically superior to Worker B. Any-one scanning the results would sound the alarm.

But this isonlyasuspiciouslyextremeversionof thestrategy.A clever manipulator can make results look more reasonable.

To put it in context, let us consider the well-knownBerkeley sex discrimination case [5]. We exhibit the phe-nomenon with simplified numbers for clarity. See Table 1.The aggregate figures show 41 males admitted out of 100applicants, but only 29 females out of 100 applicants. Theappearance of discrimination against females is strong, but

dubious. The success rate for either sex in Dept. 1 was .2; inDept. 2, .5. Women had been applying to the tougherdepartment. Then combining the two departments created astatistical artifact.

We are entitled to wonder whether a sneaky administratorcould cover up a case of out-and-out discrimination, by takingwhat was in reality one big pool and making assignments ofapplicants to one or another department so as to make theimbalance seem attributable to this kind of artifact. Let us seewhat this would involve in a general situation.

We are given NA and PA = SA/NA, NB and PB = SB/NB, forwhich, without loss of generality, PA [PB. The cover-up is toconsist in compartmentalizing the A-pool as NA1 = aNA,NA2 = (1 - a)NA, and the B-pool as NB1 = bNB, NB2 = (1 - b)NB; the successes in the various subsets will be denotedSA1 = PA1NA1, SA2 = PA2NA2, SB1 = PB1NB1, SB2 = PB2NB2.Denote by a and b, respectively, the proportions of the A andB total data to be allocated to compartment # 1. The questionthen is whether they can be chosen so that

PA1 ¼ PB1 � k

PA2 ¼ PB2 � lð2:1Þ

indicating no advantage to A or B in either case. No problem!Since SA1 = akNA, SA2 = (1 - a)lNA, SB1 = bkNB, SB2 =

(1 - b)lNB, the condition is

PA ¼ akþ ð1� aÞlPB ¼ bkþ ð1� bÞl

ð2:2Þ

Thus PA and PB are both averages of k and l, whichtherefore must lie outside the interval (PB, PA) as in the

Table 1. Simplified Berkeley Admission Data

Dept. 1 Dept. 2

Male Applicants 30 70

Males Admitted 6 35

Female Applicants 70 30

Females Admitted 14 15

Total Male Admissions/Applicants 41/100 = .41

Total Female Admissions/Applicants 29/100 = .29

.........................................................................................................................................................

AU

TH

OR

S ORA E. PERCUS received an M.Sc. in

Mathematics at Hebrew University, Jerusa-

lem, and a Ph.D. in Mathematical Statistics

from Columbia University in 1965. She has

been active in several areas of mathematics,

including probability, statistics, and combi-

natorics.

Courant Institute of Mathematical Sciences

New York University

New York, NY 10012

USA


JEROME K. PERCUS received a B.S. in Elec-

trical Engineering, an M.A. in Mathematics, and

a Ph.D. in Physics, in 1954, from Columbia

University. He has worked in numerous areas

of applied mathematics, primarily in chemicalphysics, mathematical biology, and medical

statistics.

Courant Institute of Mathematical Sciences

New York University

New York, NY 10012

USA


figure. In fact, if they are chosen to lie outside this intervalthen the desired decomposition specified by a and b isuniquely determined. Explicitly,

a ¼ l� PA

l� k; b ¼ l� PB

l� k

1� a ¼ PA � kl� k

; 1� b ¼ PB � kl� k

ð2:3Þ

The deceptive administrator would be prudent to invent adecomposition in which k is roughly in the middle of the(0, PB) interval, and l roughly in themiddleof (PA,1), in orderto allay suspicion. In the (simplified) Berkeley example,where PA = 0.41, PB = 0.29, a = 0.3, b = 0.7, we see thatthe observed k = .2, l = .5 do satisfy this criterion.

With (2.3), we find that a suitable decompositionremoves the apparent bias against females: no assertion ofdiscrimination can then be made. But we also saw that avery simply constructed decomposition can lead to anapparent success rate much higher for A than for B. What iswrong with that construction, aside from looking awfullysuspicious? Nothing, but to see whether the extremebehavior is rightly suspect we should attend to the statis-tical significance of the new assertions, a point that wasemphasized by the FDA panel cited above. This is thesubject of the following discussion, see also Zidek [11].

Statistical Significance of the Inverse SimpsonParadoxStatistical significance is customarily quantified [3] by attachinga confidence level to the assertion made. In particular, quitegenerally if our initial data is characterized by success num-bers SA, SB, and total numbers NA, NB with NA + NB = N, andif we define the success rates PA = SA/NA, PB = SB/NB, thenthe confidence level with which we can assert that the pro-cess underlying the observations had probabilities pA and pB

satisfying pA C pB is given, in the large sample limit, by

/ðN 1=2CABÞ; where /ðxÞ ¼ 1ffiffiffiffiffiffi2pp

Z x

�1e�y2

dy

CAB ¼ ðPA � PBÞ=rAB [ 0

r2AB ¼

PAð1� PAÞNA=N

þ PBð1� PBÞNB=N

� r2A þ r2

B:

ð3:1Þ

Our objective is to supply a decomposition into two hypo-thetical trials (SA1, NA1, SB1, NB1) and (SA2, NA2, SB2, NB2)which reverse the original conclusion at a common level ofconfidence. Hence, if

C 0i ¼ ðPBi � PAiÞ=ri [ 0; i ¼ 1; 2

where PAi ¼ SAi=NAi; PBi ¼ SBi=NBi

r2i ¼

PAið1� PAiÞNAi=N

þ PBið1� PBiÞNBi=N

;

ð3:2Þ

we want the two pairs of trials to reverse the initial assertionat a common level of confidence

ðPB1 � PA1Þ=r1 ¼ C 0 ¼ ðPB2 � PA2Þ=r2 ð3:3Þwith C 0[ 0. To start, we need to find the restrictions on C 0

under which the required (PA1, PA2, PB1, PB2) satisfying(3.2) can be found.

The solution is direct but algebraically cumbersome, andis presented in detail in [2]. Using the notation

if 0� x� 1; then �x � 1� x; ð3:4Þ

the conclusion is that if a C b, then

C 0 � min�b �PA � �a �PB

�arB;�b �PA � �a �PB

�brA

;

�aPB � bPA

arB;aPB � bPA

brA

�:

ð3:5Þ

Since we require C 0 C 0, this implies that

a=b�PA=PB � 1; �b=�a� �PB= �PA� 1: ð3:6Þ

The expression is a bit involved and, even worse, containsthe unknown parameters pAi, pBi implicitly. But it can besimplified by reducing its right-hand side and therebystrengthening the requirement on C 0 a bit. This is alsocarried out in [2], resulting in:

THEOREM If a C b then

PA þ PB�1 : C 0 �2ðc�cÞ1=2ðð �PB= �PAÞ2 � 1ÞðPA � PBÞðPB=PAÞ=

PA

PB

�PB

�PA

� �2

�1

" #

PA þ PB�1 : C 0 �2ðc�cÞ1=2ððPA=PBÞ2 � 1ÞðPA � PBÞð �PA= �PBÞ=

PA

PB

�PB

�PA

� �2

�1

" #

where c¼ NA=N ð3:7Þ

are sufficient to carry out the apparent reversal of rankingof A and B.

Let us take a simple example that has been previouslyquoted [4, 8]. We will paraphrase it and use rounded-off data.Hospitals A and B specialize in treating a certain deadlydisease. NA = 1000patients are treated at A and NB = 1000 atB. Of these, SA = 900 recover, whereas SB = 800 recover, sothat PA = .9, PB = .8 and Hospital A is apparently the placeto go. In fact, one computes CAB = .05, so that this conclu-sion is supported at the .05 9 (2000)1/2 = 2.24 standarddeviation level. Detailed investigation shows that matters arenot so simple. Some patients enter in otherwise good shape,others inpoor shape.Of the former, NA1 = 900 enter hospitalA, and 870 recover; of the latter, NA2 = 100 enter and 30recover, so PA1 = .967, PA2 = .3.

On the other hand, NB1 = 600 enter Hospital B in goodshape and SB1 = 590 recover, whereas NB2 = 400, SB2 =

210. Thus, PB1 = .983, PB2 = .55. We see that by not mix-ing the two classes of patients, Hospital B is superior foreach class – at levels C 01 = .038 (1.7 standard deviations)

BP AP

0 1

Figure 1. Placement of Averaging Parameters k and l.


and C 02 = .176 (7.9 standard deviations). Simpson is cer-tainly exemplified.

Equally, however, if only the combined data have beenrecorded, the person controlling the presentation of theevidence may be tempted to engage in Inverse Simpson.The criteria as to which patients entered in good shape,which in poor shape, are inevitably a bit fuzzy, after all.Given the aggregate data, the decomposition into the twoclasses could be planned with the intention of most con-vincingly asserting the opposite of the conclusion from theaggregate data. If this had been done according to theprescription of the Theorem, then with the same input data,we would have found a = .935, b = .738 (not far from thea = .9, b = .6 corresponding to the additional data pre-sented in Table 2) and would have concluded with thesuperiority of Hospital B at a confidence level corre-sponding to C 0 B .107 or 4.79 standard deviations for eachclass of patients.

Concluding RemarksThe Simpson paradox, one of the simplest examples of thecommon misuse of statistics, has received increasingattention, because its consequences can be quite drastic(and sometimes profitable). In the classic Simpson Para-dox, the only question is whether or not to combine data

from different sources (and trying to justify the decision tocombine). What we have seen here is that the InverseSimpson paradox, even in its most ‘‘sophisticated’’ versionin which mean differences are weighted by appropriatestandard deviations, is nearly universally applicable. Thiscan be an effective tool of analysis, but it is also a dan-gerous technique for distorting statistical data.

REFERENCES

[1] Abboud L. (2005) ‘‘Abbott Seeks to Clear Stalled Drug’’. Wall

Street Journal, Sept. 12.

[2] ArXiv: 0801.4522.

[3] Berger J. O. (1985) Statistical Decision Theory and Bayesian

Analysis. Springer-Verlag, New York.

[4] Bickel, P. J., Hammel, E. A., and O’Connell, J. W. (1975). ‘‘Sex

Bias in Graduate Admissions: Data from Berkeley.’’ Science 187,

398–404.

[5] Blyth C.R. (1972) ‘‘On Simpson’s Paradox and Sure-Thing

Principle,’’ JASA 67, No. 338, 364-366.

[6] Capocci, A. and Calaion, F. (2006). ‘‘Mixing Properties of

Growing Networks and Simpson’s Paradox.’’ Phys. Rev. E74,

026122.

[7] Moore, D. S. and McCabe, G. P. (1998). Introduction to the

Practice of Statistics, 100 – 201. W. Freeman and Co., New York.

[8] Moore, T. ‘‘Simpson and Simpson-like Paradox Examples.’’ see

http://www.math.grinnell.edu/*mooret/reports/SimpsonExamples.

pdf

[9] Saari, D. (2001). Decisions and Elections, Cambridge University

Press, Cambridge.

[10] Simpson, E. H. (1951). ‘‘The Interpretation of Interaction in

Contingency Tables.’’ J. Roy. Stat. Soc. B13, 238–241.

[11] Zidek, J. (1984). ‘‘Maximal Simpson-Disaggregation of 2 9 2

Tables.’’ Biometrica 71, No. 1, 187–190.

Table 2. Simplified Hospital Recovery Data

Good Shape Poor Shape

Admissions to Hospital A 900 100

Recovered in Hospital A 870 30

Admissions to Hospital B 600 400

Recovered in Hospital B 590 210

Total Recovered/Admissions in A: 900/1000 = .9

Total Recovered/Admissions in B: 800/1000 = .8


http://www.math.grinnell.edu/~mooret/reports/SimpsonExamples.pdf

http://www.math.grinnell.edu/~mooret/reports/SimpsonExamples.pdf

ConfoundedLawrence M. Lesser

3 of 8 poemsI submittedto the classic journalwere accepted,while 1 of 3 my rival did were, so I won.2 of 3 poems I sent to the modern journal were

accepted,while my rival had 3 of 5, so I won.

But overall,my rival had half of hers accepted and I did not,so she wonafter all. I wasconfounded! I foundthat numbers don’t lie,but don’t explainwhy. Whytry comparingif comparison can bereversedwith a Peterson rollby underdog wrestlingdata, rival, or self?When my parts aresummed,am I less than someof my parts?

The University of Texas at El Paso

El Paso, USA.



UnderstandingCoin-TossingJAROSLAW STRZALKO, JULIUSZ GRABSKI, ANDRZEJ STEFANSKI,PRZEMYSLAW PERLIKOWSKI AND TOMASZ KAPITANIAK

IIt is commonly known that a toss of a fair coin is arandom event and this statement is fundamental in theclassical probability theory [1, 2, 3]. On the other hand,

the dynamics of the tossed coin is described by determin-istic equations, with no external source of random influence[4, 5, 6], so one can expect predictability in the results. It ispossible to construct a mapping of the initial conditions(position, configuration, momentum, and angular momen-tum at the beginning of the coin motion) to a final observedconfiguration, that is, the coin terminates with its head (tail)side up or on its edge. The initial conditions which aremapped onto heads create a heads basin of attraction whilethose mapped onto tails create a tails basin of attraction [4].The boundary which separates heads and tails basinsconsists of initial conditions mapped onto the coin standingon the edge. The structure of these boundaries has a sig-nificant impact on the problem of the coin-tossing predict-ability, that is, smooth basin boundaries allow predictabilitywhile fractal boundaries can lead to unpredictability [10, 11].However, the precise structure of the heads-tails basinboundaries for a realistic model of a coin-tossing isunknown.

Here, we show that heads-tails basin boundaries aresmooth, so the outcome of the coin-tossing is predictable.We have found that an increase in the number of impactsin the period when the coin bounces on the floor makesthe basin boundaries more complex, and in the limitingcase of an infinite number of impacts the behavior of thecoin is chaotic and the basins of heads and tails becomeintermingled [12, 13, 14, 15, 16]. Our results demonstratethat although the coin-tossing is predictable, it can alsoapproximate the random process and can serve as thefoundation for understanding the behavior of physical(mechanical) randomizers [17, 18, 19]. We expect ourresults to be a new point in the discussion of the nature ofrandom processes [17, 20].

The Coin ModelA coin can be modeled as a rigid body, namely a cylinderwith a radius r and height h as shown in Figure 1. Weconsider a nonsymmetrical coin (the so-called cheat coin)for which the center of mass C is located at the distancenC = 0, gC = 0, fC = 0 from the geometrical center B.Any arbitrary position of a rigid body with respect to thefixed reference frame Oxyz can be described by a combi-nation of the position of the origin of the local referenceframe x 0 y 0 z 0 and the orientation (angular position) of thisframe n, g, f [21, 22]. The local reference frame x 0 y 0 z 0 isrigidly attached to the body and its axes are parallel to thexyz frame; n, g, f is the frame embedded and fixed in thebody. For the origin of the local frames it is convenient tochoose the geometric center of the body model B.

In our studies, we consider the following motion of thecoin. We assume that the coin is thrown at the height z0

with the initial conditions U0 ¼ fx0; y0; z0; _x0; _y0; _z0;w0; m0;/0;xw0;xm0;x/0g, that is, the initial position of the center ofmass is {x0, y0, z0}, its initial velocity f _x0; _y0; _z0g, the coin’sinitial orientation {w0, m0, /0}, and the initial angular velocity{xw0, xm0, x/0}. After a free fall when the z coordinateis zero, the coin collides with the parallel base (floor). It isassumed that at the collision, a portion of the coin energy isdissipated, that is, the collision is described by the restitu-tion coefficient v\ 1. The friction at the contact betweenthe coin and the floor is described by the friction coeffi-cient lfr [23]. After the collision, the coin’s center of massmoves to height z1 at which the total mechanical energy ofthe coin E is equal to its total energy in the moment afterthe collision E 0 minus the energy dissipated because of airresistance. Next, the coin moves on until it collides with thefloor again. The calculations terminate when, after n-thcollision, the total mechanical energy of the coin is smallerthan the potential energy at the level of the coin’s centerof mass (approximately mgr, where g is the gravitational


acceleration), as this condition prevents the coin fromflipping over. We also consider rotations of the coin on thefloor. Full details of our model are given in [9].

The equations of motion describing the tossed coin areNewton’s equations, with no external source of randominfluence, that is, fluctuations of air, thermodynamic orquantum fluctuations of the coin. These equations arediscontinuous, so in analysing them one cannot apply con-tinuity theorems or direct calculations of Lyapunov expo-nents. If the outcome of a long sequence of the coin-tossingsis to give a random result, it can only be because the initialconditions vary sufficiently from toss to toss. The flow givenby the equations of motion maps all possible initial condi-tions into one of the final configurations. The set of initialconditions which is mapped onto the heads configurationcreates a heads basin of attraction b(H) while the set of initialconditions mapped onto the tails configuration creates a tailsbasin of attraction b(T). The boundary separating the headsand tails basins consists of initial conditions mappedonto thecoin-standing-on-edge configuration [24]. For an infinitelythin coin, this set is a set of zero measure and thus withprobability 1 the coin ends up either heads or tails. For anonzero thinness of the coin this measure is not zero, but theprobability that the edge configuration is stable is low.

Assume that one can set the initial conditions U0 withuncertainty �, where � is small. If a ball B in the phase spacecentered at U0 contains only the points which go to one ofthe final states, the outcome is predictable and repeatable.If in the ball B there are points leading to different finalstates (denote the set of points leading to heads as b0(H)

and the set points leading to tails as b0(T)), then the result oftossing is not predictable. One can calculate the probabilityof heads (tails) as probðheadsÞ ¼ lðb0ðHÞÞ=lðBÞ ðprobðtailsÞ ¼ lðb0ðT ÞÞ=lðBÞÞ, where l is a measure of the setsb0(H), b0(T) and B.

The possibility that heads-tails basin boundaries arefractal [10], riddled [12, 13, 14, 15], or intermingled [12, 16], isworth investigating. Near a given basin boundary, if theinitial conditions are given with uncertainty �, a fraction f(�)of the initial conditions give an unpredictable outcome. Inthe limit �? 0, f(�) � �a where a\ 1 for fractal and a = 1 forsmooth boundaries. Fractal basins’ boundaries are discon-tinuous (for example an uncountable sequence of disjointstripes) or continuous (a snowflake structure) [11]. From thepoint of view of the predictability of the coin-toss the pos-sibility of intermingled basins is the most interesting.

Let us briefly explain the meaning of the term inter-mingled basins of attraction. The basin b(H) is said to beriddled by the basin b(T) when it satisfies the followingconditions: (i) it has a positive Lebesgue measure, (ii) forany point in b(H), a ball in the phase space of arbitrarilysmall radius has a nonzero fraction of its volume in thebasin b(T). The basin b(T) may or may not be riddled bythe basin b(H). If the basin b(T) is also riddled by the basinb(H), the basins are said to be intermingled. In this case, inany neighborhood of the initial condition leading to headsthere are initial conditions which are mapped to tails, thatis, there does not exist an open set of initial conditionswhich is mapped to one of the final states: an infinitelysmall inaccuracy in the initial conditions makes the state ofthe coin tossing unpredictable.

In our numerical calculations we consider the followingcoin data: m = 20 grams, r = 1.25 cm, h = 0.2 cm (formerPolish 1 PLN coin made of a light aluminum-based alloy)and nC = 0.1 cm, gC = 0.1 cm, fC = -0.02 cm. We con-sidered the air resistance acting on the coin in bothtangential and normal directions and described by thefollowing coefficients kn = 0.8, ks = 0.2 [9]. The frictionbetween the coin and the floor during the impact isdescribed by friction coefficient lfr = 0.2.

Results and DiscussionFigure 2 (a-d) shows the basins of attraction of heads andtails calculated for various coin models. The dark regionscorrespond to heads and the white ones to tails. The case ofthe coin terminating on the soft floor (restitution coefficient

.........................................................................

AU

TH

OR

S JAROSLAW STRZALKO teaches classi-

cal and analytical mechanics and is a co-

author of a number of text books. In his

free time he likes to go fishing.

Division of Dynamics

Technical University of Lodz

Stefanowskiego 1/15, 90-924 LodzPoland


Figure 1. 3-dimensional model of the coin and its orientation

in space.

.........................................................................

JULIUSZ GRABSKI teaches classical and

analytical mechanics at all levels and pur-

sues research in these areas. He loves

jogging and skiing.



Stefanowskiego 1/15, 90-924 Lodz

Poland


.....


v = 0) is shown in Figure 2(a). The case which allows thebouncing of the coin on the floor surface (v = 0.6) isshown in Figure 2(b). The structure of the basin bound-aries for the case without bouncing on the floor is similar tothe structure in the Keller model [7]. One can notice that thestructure of the basin boundaries is more complex (looksfractal or intermingled) when the coin is allowed to bounceon the floor as can be seen in Figure 2(b). To check thepossibility that these basins are fractal (or intermingled), theappropriate enlargements are presented in Figure 2(c,d). Itcan be seen that apart from the graininess because of thefinite number of points, the boundaries are smooth (seeFig. 2d). Under further magnification no new structure canbe resolved, that is, no evidence of intermingled or evenfractal basin boundaries is visible. The same conclusion hasbeen reached in the studies of simple one- or two-dimen-sional models [8, 4, 5]. Figure 2 (a-d) is based on the results

obtained from numerically integrated equations of motion.We fixed all initial conditions except two, namely: theposition of the coin mass center z0 and the angular velocityxn0. We check that similar structures of the basin bound-aries are observed when different initial conditions areallowed to vary. The two-dimensional sections of the phasespace presented in Figure 2 (a-d) are a good indication ofwhat happens in the entire phase space. We point out thatthe same structure of basins of attraction has been observedfor the symmetrical coin [9].

This allows us to state our main result: for any initialcondition U0 there exists �[ 0 such that a ball with radius �centered at U0 contains points which belong either to theset b(H) or the set b(T). In other words, if one can settle theinitial condition with appropriate accuracy, the outcome ofthe coin-tossing procedure is predictable and repeatable.

Now we try to explain why for particularly small (but notinfinitely small) � the coin-tossing procedure can approxi-mate a random process. A sequence of coin-tosses will berandom if the uncertainty � is large in comparison to thewidth W of the stripes characterizing the basins of attraction,so the condition �[[ W is essential for the outcome to berandom [4]. It is interesting to note that the uncertainty �depends on the mechanism of coin tossing while thequantity W is determined by the parameters of the coin.

In the case of the coin bouncing on the floor thestructure of the heads and tails basin boundaries becomescomplex (Figure 2b). In Figure 3(a-c) we show the calcu-lations of these basins for a different number of impacts n.One can observe the face of the coin which is up after then-th collision. Figure 3(a-c) shows the results for respec-tively 0, 3 and 10 collisions. With the increase of thecollision numbers it is possible to observe that the com-plexity of the basin boundaries increases with the num-ber of impacts. With the finite graininess (resolution) ofFigure 3(a-c) these basin boundaries look fractal and onecan speak about a fractalization-like process which can beobserved with an increasing number of impacts.

To explain this process, consider the limiting case of aninfinite number of impacts. Such a case neglects air resis-tance and assumes elastic impacts, that is, v = 1, andcannot be realized in a real experiment. Consider the map

Figure 2. Basins of attraction of heads (black) and tails (white);

(a) coin lands on the soft surface, (b-d) coin bounces on the

floor, (c,d) enlargements of (b). The following parameters have

been used: x0 ¼ y0 ¼ 0; _x0 ¼ _y0 ¼ _z0 ¼ 0;u0 ¼ w0 ¼ 0;J0 ¼7p=180 rad, xf0 = 0, xg0 = 40.15 rad/s.

.........................................................................

ANDRZEJ STEFANSKI is also a professor

of mechanics. He works on nonlinear

dynamics, particularly the synchronization

of chaotic systems. He loves good drinksand plays soccer.




Poland


.........................................................................

PRZEMYSLAW PERLIKOWSKI is a post-

doc at Humboldt University of Berlin (he

received his Ph.D. from the Technical

University of Lodz, however). He works

on dynamical systems, particularly systemswith time delay. During the summer holi-

days he goes trekking and camping in

Croatia.

Institute of Mathematics

Humboldt University of Berlin

Unter der Linden 6, 10099 Berlin

Germanye-mail: [email protected]

......


U : [0, 2p] ? [0, 2p] which maps the point /n on the edge ofthe coin, which hits the floor at the nth impact, to the point/n+1 which hits the floor at the (n + 1)st impact. Analysisof the time series of points (/1, /2, …) shows that thedynamics of U is chaotic when the largest Lyapunovexponent is positive (it has been numerically estimatedfrom the time series to be 0.08). In this limiting case(n ??) the basins of heads and tails are intermingled andthe outcome of the coin-tossing is unpredictable. Numeri-cally, this can be observed when in the successiveenlargements of the heads-tails basin boundaries the newstructure is visible, as in Figure 4 (a-c) where the basins ofheads and tails are calculated for n = 1000 impacts. Theprobability (we consider 106 different initial conditions)that a coin side which is up initially will still be up after 15impacts is equal to 0.50987 and after 1000 impacts to

0.50006. This indicates that in the case of n ? ? thisprobability tends to 0.5.

In a real experiment, such a very large number of impactscannot be realized because of the dissipation (inelasticimpacts and air resistance) so the fractalization-like processhas to stop. In our experiments [9] we observed that a typicalcoin falling from the height of 186 cm bounces on a woodenfloor about 8-14 times. The existence of the chaotic processdescribed by the map U introduces a time-sensitive depen-dence on initial conditions characterized by the positivemaximum temporal Lyapunov exponent [25, 26, 27]. Thissensitivity is responsible for the ‘‘fractalization’’ shown inFigure 3(a-c) and explains why the coins behave in practiceas perfect randomizers.

ACKNOWLEDGMENTS

This study has been partially supported by the Polish

Department for Scientific Research (DBN) under project

No. N N501 0710 33.

REFERENCES

[1] W. Feller: An Introduction to Probability: Theory and Examples,

Wiley, New York, 1957.

[2] E.T. Jaynes: Probability Theory: The Logic of Science, Cam-

bridge University Press, Cambridge, 1996.

[3] J.E. Kerrich: An Experimental Introduction to the Theory of

Probability, J. Jorgensen, Copenhagen, 1946.

[4] V.Z. Vulovic and R.E. Prange: Randomness of true coin toss.

Physical Review, A33/1: 576 (1986).

[5] P. Diaconis, S. Holmes, and R. Montgomery: Dynamical Bias in

the Coin Toss, SIAM Rev., 49, 211 (2007).

[6] T. Mizuguchi and M. Suwashita: Dynamics of coin tossing,

Progress in Theoretical Physics Supplement, 161, 274 (2006).

[7] J.B. Keller: The probability of heads, Americam Mathematical

Monthly, 93, 191 (1986).

[8] Y. Zeng-Yuan and Z. Bin: On the sensitive dynamical system

and the transition from the apparently deterministic process to

the completely random process, Appl. Math. Mech., 6, 193

(1985).

[9] J. Strzalko, J Grabski, A. Stefanski, P. Perlikowski, and T.

Kapitaniak: Dynamics of coin tossing is predictable, Phys. Rep.,

469, 59 (2008).

[10] C. Grebogi, S.W. McDonald, E. Ott, and J.A. Yorke: Metamor-

phosis of basins boundaries, Phys. Lett., 99A, 415 (1983).

[11] B.B. Mandelbrot: Fractal Geometry of Nature, Freeman, San

Francisco, 1982.

[12] J.C. Alexander, J.A. Yorke, Z. You, and I. Kan: Riddled basins,

Int. J. Bifur. Chaos 2, 795 (1992).

[13] J.C. Sommerer and E. Ott: Riddled basins, Nature 365, 136 (1993).

[14] E. Ott, J.C. Alexander, I. Kan, J.C. Sommerer, and J.A. Yorke:

The transition to chaotic attractors with riddled basins, Physica

D, 76, 384 (1994).

[15] T. Kapitaniak, Yu. Maistrenko, A. Stefanski, and J. Brindley:

Bifurcations from locally to globally riddled basins, Phys. Rev.

E57, R6253 (1998).

[16] T. Kapitaniak: Uncertainty in coupled systems: Locally intermin-

gled basins of attraction. Phys. Rev. E53, 53 (1996)

.........................................................................

TOMASZ KAPITANIAK is a professor of

mechanics and a head of the Division of

Dynamics; his research is concentrated on

nonlinear dynamics and chaos theory. He

paints in his free time.




Poland


Figure 4. Basins of attraction of heads (black) and tails (white)

in the case when the dissipation of energy is neglected;

n = 1000 impacts, (b,c) are enlargements of (a). The same

parameters as in Figure 2 have been used.

Figure 3. Basins of attraction indicating the face of the coin

which is up after the n-th collision: (a) n = 0, (b) n = 3, (c)

n = 10, heads and tails are indicated in black and white,

respectively. The same parameters as in Figure 2 have been used.


[17] H. Poincare: Calcul de Probabilites, George Carre, Paris, 1896.

[18] E. Hopf: On causality, statistics and probability, Journal of

Mathematical Physics, 13, 51 (1934).

[19] E. Hopf: Uber die Bedeutung der Willkurlincken Funktionen fur

die Wahrscheinlichkeitstheorie, Jahresbericht der Deutschen

Mathematiker-Vereinigung, 46, 179 (1936).

[20] J. Ford: How random is a coin toss, Physics Today, 40, 3 (1983).

[21] H. Goldstein: Classical Mechanics, Addison-Wesley, Reading,

1950.

[22] J.E. Marsden and T.S. Ratiu: Introduction to Mechanics and

Symmetry, Springer, New York, 1994.

[23] J.I. Nejmark and N.A. Fufajev: Dynamics of Nonholonomic

Systems, Translations of Mathematical Monographs, (American

Mathematical Society, vol. 33, 1972).

[24] D.B. Murray and S.W. Teare, Probability of a tossed coin landing

on edge, Phys. Rev. E 48, 2547 (1993).

[25] T. Kapitaniak: Distribution of transient Lyapunov exponents of

quasi-periodically forced systems, Prog. Theor. Phys., 93, 831

(1995).

[26] T. Kapitaniak: Generating strange nonchaotic attractors, Phys.

Rev. E, 47, 1408 (1993).

[27] T. Tel: Transient chaos, J Phys A: Math Gen, 22, 691 (1991).


The Mathematical Tourist Dirk Huylebrouck, Editor

A Walk ThroughMathematical TurinSANDRO CAPARRINI

Does your hometown have any mathematical tourists

attractions such as statues, plaques, graves, the cafe

where the famous conjecture was made, the desk where

the famous initials are scratched, birthplaces, houses, or

memorials? Have you encountered a mathematical sight

on your travels? If so, we invite you to submit an essay to

this column. Be sure to include a picture, a description

of its mathematical significance, and either a map or

directions so that others may follow in your tracks.

� Please send all submissions to Mathematical Tourist Editor,

Dirk Huylebrouck, Aartshertogstraat 42,

8400 Oostende, Belgium


UUntil recent years, Turin (Torino) had the dubiousdistinction of being one of the few historic Italiancities rarely visited by tourists. Its traditional image

was that of an industrial area, mainly known for theautomobile industry and related activities. Yet there aremany things that make Turin stand apart from other Italiancities. While most of the places of interest in Italy are histor-ically connected with Roman history or with the Renaissance,Turin flourished during the nineteenth century, when itbecame an example of economy propelled by science andtechnology. The rich cultural heritageof the city is reflected inthe varieties of its museums. Turin has one of the fewautomobile museums in the world, and an Egyptian museumdisplaying what is perhaps the oldest collection of its kind.There is also a spectacular cinema museum showing that,before World War I, this was one of the most importantcenters of the cinema industry in Europe.

Turin’s architecture will be of particular appeal to visitorsof mathematical inclination. The plan of the city is remark-ably regular, revealing the work of Ancient Rome’s militaryarchitects and the later influence of the eighteenth-centuryFrench Enlightenment. The streets are usually wide andstraight, intersecting at right angles and punctuated withpublic squares geometrically regular in shape. Most of thehistorical buildings date back to the Baroque period. How-ever, there is a streak of fine madness running through thisapparently tranquil and orderly city. Scattered throughoutTurin there are some wildly imaginative, early twentieth-century Art Nouveau buildings that rival those of Barcelona.One of the weirdest examples of eccentricity in architectureis the Casa Scaccabarozzi, popularly called Fetta di polenta(‘‘Slice of polenta’’) because of its yellow color, designed bythe architect Alessandro Antonelli in 1881 (Fig. 1). This five-storey building stands on a tiny right triangle having one side(along Corso San Maurizio) of 4 m and the hypotenuse(along Via Giulia di Barolo) of 21 m. If you happen to be inTurin, this little gem is worth a visit.

Figure 1. The Fetta di polenta.


While to mathematically minded people Turin is usuallyassociated with Joseph-Louis Lagrange (1736–1813), manyother important figures in the history of mathematics alsospent long periods here. Indeed, the Turin scientific schoolranks high in Italy. As early as the second half of the six-teenth century, Giovanni Battista Benedetti (1530–1590), animportant forerunner of Galileo in mechanics, was courtmathematician to the Duke of Savoy. In the eighteenthcentury the physicist Giambattista Beccaria (1716–1781),one of the founders of the scientific study of electricity anda correspondent of Benjamin Franklin, was a professor atthe university of Turin; Lagrange, the greatest scientistTurin ever produced, was one of his students. Physics andchemistry in Turin are also well represented by the AbbeNollet (1700–1770), who came to Turin in 1739, and byAmedeo Avogadro (1776–1856), known for the Avogadronumber. Among nineteenth-century mathematicians whoworked in Turin there were Giovanni Plana (1781–1864),Augustin-Louis Cauchy (1789–1857) and Luigi FedericoMenabrea (1809–1896). Around the turn of the century, themost productive period for mathematics in Turin, thenames of Giuseppe Peano (1858–1932), Mario Pieri (1860–1913), Corrado Segre (1863–1924), Vito Volterra (1860–1940) and Cesare Burali-Forti (1861–1931) must be cited. Inthe twentieth century there were Guido Fubini (1879–1943)and Francesco Tricomi (1897–1978). A lively description ofthe University of Turin in 1900 was given by J. L. Coolidge[1]. There are many interesting remarks on the intellectuallife in fin de siecle Turin in the books by H. C. Kennedy [2],E. Marchisotto and J. T. Smith [3] and J. R. Goodstein [4].

Today, some traces of the past mathematical glories ofTurin are still visible. Since Turin is not a large city and itscentre is best explored on foot, we suggest taking a strollpast its mathematical points of interest. This is best done onweekdays, since some of the places are closed on week-ends. The main part of the walk is about 2 km long andtakes almost a full day to complete. If you wish, you can ofcourse divide this walk into several separate trips. Motorists

are warned that most of the city center is closed to trafficduring the day.

The walk starts at Porta Nuova railway station. As youstand in Piazza Carlo Felice, with the facade of the stationbehind you, walk under the arcades along the right side ofthe square. The first narrow road on the right leads to a smallsquare, the Piazzetta Lagrange. In the centre of the squarestands the monument to Lagrange (who else?). The monu-ment was conceived and sculpted by the Piedmontese artistGiovanni Albertoni in 1867. The statue shows Lagrange asa middle-aged man, standing upright, slightly stooping,wearing anold-fashionedwaistcoat.He is gazingdownward,apparently immersed in profound thoughts. His arms hangdown; he has a quill pen in his right hand and a manuscript inhis left. There are four books at his feet, perhaps representingthe treatises he published late in life. Although not a greatwork of art, the statue is a simple but effective depiction of aquiet and bookish man (Fig. 2).

Leave Piazzetta Lagrange at its upper right-hand (north-east) corner. Walk along Via Lagrange for a couple of blocks.At No. 29, on the so-called piano nobile, (i.e., the storeyimmediately above the ground floor) Lagrange was born onJanuary 25, 1736 (Fig. 3). A simple commemorative plaquerecalls the figure of the illustrious mathematician (Fig. 4).The facadeof thebuilding has been carefully restored, so thatit is not difficult to imagine how the place looked in theeighteenth century. Unfortunately, Lagrange’s apartment,now in private hands, has been remodelled since then.

.........................................................................

AU

TH

OR SANDRO CAPARRINI holds degrees in

Physics and in Mathematics and a Ph.D. in

Mathematics from the University of Turin.

However, all of his work has been in thehistory of mathematics. His research is mainly

focused on the interaction between math-

ematics and mechanics from 1750 onward.

In 2004 he was awarded the Slade Prize

from the British Society for the History ofScience.

Institute for the History and Philosophyof Science and Technology

Victoria College 316

91 Charles St. West

Toronto M5S 1K7

Canada

E-mail: [email protected]

Figure 2. The statue of Lagrange.


From Lagrange’s home, continue walking down ViaLagrange in the same direction until you reach the end of thestreet, at the intersection with Via Maria Vittoria. Pause for amoment to admire on the right the Baroque Chiesa di SanFilippo, where Lagrange was christened, then turn your atten-tion to the seventeenth-century building on your left, theCollegio dei Nobili. Any tour guide will tell you that this is thelocation of the Egyptian Museum and of the Galleria Sabauda,both requisite stops for tourists. For us, this building is impor-tant as the location of the Turin Academy of Science (Fig. 5).

The Academy was founded in 1757 by Lagrange togetherwith two friends, the physician and physicist GiovanniFrancesco Cigna (1734–1790) and Count Angelo Saluzzo diMonesiglio (1734–1810). A few kindred spirits joined thisinitial group in the years that followed. At first, this was aninformal institution, devoted essentially to discussing theworks and readings of its own members, and was thus calledSocieta privata (‘‘Private society’’). The members met at thehouse of the Count of Saluzzo. In 1759 the Society beganpublishing a scientific journal, originally entitled MiscellaneaPhilosophico-Mathematica Societatis Privatae Taurinensis.After 14 years, five volumes had been completed; these ini-tiated the long series of Melanges, Memoires and Atti thathave spread the fameof the earlier Society and later Academythroughout the world of science. Among the contributors tothese early volumeswere some of the leading scientists of thetime: Jean le Rond d’Alembert (1717–1783), Marie Jean

Antoine Nicolas de Caritat Marquis de Condorcet (1743–1794), Leonhard Euler (1707–1783), Albrecht Haller (1708–1777), Pierre Simon Laplace (1749–1827), Gaspard Monge(1746–1818) and, of course, Lagrange himself. Shortly afterthe publication of the Miscellanea, the Society obtained thepermission to add the adjective reale (‘‘royal’’) to its name.Finally, the creation of an Academy under royal patronagewas suggested to the King; it was formalized on 25 July 1783.Lagrange, then in Berlin, was elected honorary president.

Figure 4. Close-up of Lagrange’s plaque.Figure 3. The house where Lagrange was born, showing the

commemorating plaque.

Figure 5. The Collegio dei Nobili.


The Academy’s location also deserves attention. TheCollegio dei Nobili, a wonderful example of seventeenth-century Italian baroque architecture, was built between 1679and 1687 by the great architect Guarino Guarini (1624–1683)as a school for the sons of nobles. (In fact, behind this projectwas a Jesuit plan to infiltrate the centres of political power inPiedmont.) After the suppression of the Jesuits in 1773, thebuilding passed to the state, and in 1784 King Vittorio Ame-deo III designated it as the seat of the new Academy.

Now that you are at the Collegio dei Nobili, go to thepoint at which Via Maria Vittoria meets Piazza San Carlo.Here, a plaque on the wall indicates that ‘‘Giovanni Plana,while living in this building, composed the theory ofthe movement of the moon between 1807 and 1832.’’ Theastronomer and mathematician Giovanni Plana, author ofthe Theorie du mouvement de la lune (1832) was onceconsidered the most important Italian scientist of his time.The Theorie was an attempt to improve the approximationsLaplace had devised for the movements of the moon; itconsists of three massive volumes full of incredibly longand complicated formulae.

If you want to become better acquainted with Plana, enterthe Collegio through its main entrance in Via Accademiadelle Scienze, then go all the way to the back of the atriumand turn left along the corridor. After a few meters, you willfind on your left a slightly larger than life statue of Plana. Thefamous astronomer is shown in his old age, sitting in anarmchair, a book in his hand, a pensive look on his face(Fig. 6). The statue was made in 1870 by Giovanni Albertoni,

the same sculptor who made the monument to Lagrange.These two statues give you an idea of how mathematicianswere viewed at the end of the nineteenth century.

Immediately before the statue of Plana there is a bust ofAngelo Genocchi (1817–1889), professor of analysis at theUniversity of Turin from 1865 to 1884, now rememberedmainly for a polemic with Peano (Fig. 7). In 1883, Peano,still Genocchi’s assistant, was given the task of writing atextbookbasedon theprofessor’s lectures. But Peanodidnotlimit himself to merely transcribing what the professor hadsaid. With youthful enthusiasm, he added several pages ofendnotes full of important observations and ingeniouscounterexamples. Unfortunately, these remarks had thecollateral effect of undermining many of Genocchi’s proofs.Obviously Genocchi was not happy with the result. Infuri-ated, he sent a letter to several mathematical journalsrenouncing authorship of the final text. Today the Genocchi-Peano is considered one of the most significant textbooks onanalysis ever published. History has not been kind toGenocchi, who, while not on the same level as Peano, was ineffect a rigorous mathematician for his time.

To visit the Academy it is necessary to request permissiona couple of months in advance; send an email to [email protected]. The Academy is open to visitors from 9 amto 1 pm and from 3 p.m. to 5 p.m., Monday to Friday.

The entrance to the Academy is a small door on ViaMaria Vittoria 3. While the entire Academy extends overseveral floors, its core consists of three salons. The mainroom is called the Sala dei Mappamondi (Fig. 8). Imagine a

Figure 6. Plana’s monument in the Palazzo dei Nobili. Figure 7. The Genocchi bust.


large space having approximately the structure of a church,complete with apses, columns and frescoes. However thereis no altar, and the walls are almost completely coveredwith shelves of old books. This arrangement of the rooms,dating back to the foundation of the Academy, is the workof the architect Mario Ludovico Quarini. It is a remarkableexample of architecture in the service of science rather thanof religion. Also of interest are the frescoes, painted between1786 and 1787 by Giovannino Galliari, all depicting scientificsubjects. The mathematical tourist should look for the por-traits of Pythagoras and Euclid over the two entrances to thesecond room, each recognizable fromappropriate geometricsymbols.

Leaving the Collegio dei Nobili, follow Via Maria Vittoriaas far as the intersection with Via Carlo Alberto, then turnleft and go on until you reach number 10. This is themain entrance to Palazzo Campana, a seventeenth-centurybuilding which now houses the modern Department ofMathematics of the University of Turin. In October of 2008,the Department was renamed in memory of GiuseppePeano, who taught there for about 40 years. Like many oldbuildings, Palazzo Campana hides a few secrets. UnderFascism it became the Casa Littoria, the provincial head-quarters of the party; underneath it remain the hidingplaces in which the city leaders could take refuge in theevent of air strikes during the war.

Since you are interested in the history of mathematics,you will probably want to visit the Library of the Depart-ment, which houses many old books of great interest. Notfar from Palazzo Campana, on the right-hand side of PiazzaCarlo Alberto, there is also the Biblioteca Nazionale, a realtreasure-trove of rare and important texts. A description ofthe riches of Turin libraries can be found in the catalogue ofan exposition held in 1987 [5].

From Piazza Castello, take Via Po, keeping to the left.Turn left at the first intersection with Via Giovanni Virginio.A few steps later, you will arrive at Piazzetta Accade-mia Militare. Here you will find a line of columns, all thatremains after the bombing in the last war of the TurinMilitary Academy, where Lagrange taught analysis andmechanics from 1755 to 1766 (Fig. 9). A plaque on the wallsuccinctly recounts the history of the site.

The Military Academy was founded in 1678 primarily as aschool for the pages and nobles of the court. There were manychanges before it assumed its definitive form in 1815. The rulesof 1692 explain how, other than mathematics and design, stu-dents would learn how to ride horses, how to joust, how tohandle weapons, and how to dance. To become perfect gen-tlemen, the young men were also encouraged to participate incourt festivals. The rules of 1754 indicate that the Academyaccepted men and boys between 10 and 30 years of age, divi-ded into three groups. The first consisted of true cadets, thesecond,ofuniversity studentswho tookpart inonly someof theactivities of the Academy, and the third was made up ofyounger boys. The success of this institution may be judged bythe fact that many of the students came from abroad. There is avivid description of life in the Academy around 1760 in theautobiography of the dramatist Vittorio Alfieri (1749–1803).

Military academies were among the best places to learnmathematics during the eighteenth century. In the periodwhen Lagrange taught there, the Military Academy of Turinoffered courses in arithmetic, algebra, plane and solidgeometry, trigonometry, surveying, mechanics, hydrostat-ics and the elements of calculus. (Lagrange’s Turin lectureson calculus have been published in [6].) The founders ofthe renowned Ecole Polytechnique in Paris were inspiredby military academies, and this institution, in turn, becamethe model for West Point.

Lagrange was not the only mathematician of importance toteach at the Military Academy of Turin. After him came, amongothers, Plana, Menabrea, Peano, and Burali-Forti. Any univer-sity would be proud of such a faculty of professors.

The ruins of the Military Academy take only a few minutesto explore, but the next building on our tour would requirehours or days to be fully appreciated. Turning onto Via Po,turn right and walk toward Piazza Castello. On the far side ofthepiazza, just to the right ofViaPalazzodi Citta, youwill findthe Church of San Lorenzo, one of Guarini’s masterpieces.

We already encountered the Theatine priest Guarini whendiscussing the Turin Academy of Science. While he is justlyconsidered one of the major architects of the seventeenthcentury, it would not be a stretch at all to call him a mathema-tician turned architect. Had he not designed a few innovativebuildings, he would be remembered today as the author of

Figure 8. The Sala dei Mappamondi. (photo: Marco Saroldi)

Figure 9. The old Turin Military Academy, circa 1890.


several excellent texts on pure and applied mathematics. HisEuclides adauctus (‘‘Euclid Augmented,’’ 1671), summarizes agood part of the mathematical knowledge of the seventeenthcentury, while his Coelestis mathematica (‘‘Celestial Mathe-matics,’’ 1683) is a kind of astronomical encyclopedia. On aslightly different note, his Placita philosophica, physicis ratio-nibus, experientiis, mathematicisque ostensa (‘‘PhilosophicalThoughts Demonstrated by Means of Physical Reasoning,Experiments and Mathematics,’’ 1665) is mostly a reflection onscientific methodology. All these works demonstrate aremarkable knowledge of the mathematics of that time. TheEuclides adauctus, for example, contains a reference tothe then recent descriptive geometry of Desargues. In manyrespects, Guarini is comparable to his contemporary Christo-pher Wren, who, like him, was both a great architect and animportant scientist.

Visiting one of Guarini’s buildings is like entering a giantthree-dimensional geometric construction. The plan of SanLorenzo is a curvilinear octagon formed from the intersectionof eight circles around a central area (Fig. 10). Looking up,you will see a cupola, parabolic in cross-section, criss-cros-sed by ribs that, seen from below, form abstract polygons. Asinteresting as this superimposition of geometrical forms may

be, it gives only a hint of just howcomplicated the structureofthe church really is. In fact, if you look harder, you will beginto see many unusual things (Fig. 11). You might notice, forexample, that the oval windows in the cupola are so largethat they compromise its stability. Lowering your gaze toground level, you will become aware that the columns aretoo slender to support the enormous weight of the massoverhead. And if this weren’t enough, the arches that appearto hold up the entire weight of the building are perforatedwith holes just where the keystones should be, that is, exactlywhere the weight of the arches should bear down on thecolumns. In essence, most of what you see is fake. The col-umns and arches do not support anything. The real weight-bearing structure is hidden inside the walls; even today,researchers are not sure exactly how it works. This doublestructure, one visible but false, and the other real but hidden,is typical of Guarini, as are the secret stairways, the hiddenfrescoes and the complicated system of passages in the spacebetween the internal and external cupolas. Within theChurch of San Lorenzo lurk enough eccentricities to delightthe readers of historical mysteries like Umberto Eco’s TheFoucault Pendulum.

As you leave San Lorenzo, look to the left. Above theroofs of the Palazzo Reale you will see the spiral steeple ofthe Church of the Holy Shroud (1694), another of Guarini’sworks. The Church of the Holy Shroud, one of the majormasterpieces of Baroque architecture, is mainly knownbecause it is home to the sheet in which, according totradition, the body of Christ was wrapped after theFigure 10. Ground plan of San Lorenzo.

Figure 11. Interior of San Lorenzo.


crucifixion. The structure of this church was even morecomplicated than that of San Lorenzo. What you see today,though, is only an empty shell: The inside was completelydestroyed by a fire in 1997.

From Piazza Castello, retrace your steps back to Via Po. Atnumber 17 you will find the entrance of the old Palazzodell’Universita. This is the place where distinguished math-ematicians, such as Cauchy, Volterra and Peano, gave theirlectures. ThePalazzowasbuilt between1712 and 1720underthe direction of the architect Michaelangelo Garove. KingVittorio Amedeo II decided to modernize university studiesin Piedmont, and the new edifice was to be the tangiblemarker of this reform. Thanks to teachers such as Beccariaand the Abbe Nollet, in a few years the University of Turinbecame an important centre for physics. This tradition con-tinued in the following century: Avogadro and Cauchytaught fisica sublime (‘‘sublime physics’’) in Turin, corre-sponding more or less with modern theoretical physics.

The courtyard of the Palazzo dell’Universita is anotherinteresting site for mathematical tourists. Starting around1860, a collection of busts of a number of the most celebratedprofessors of the University of Turin began to be assembled,following the example of the collections of statues of famousItalians displayed in the arcades of the Uffizi Museum inFlorence. Today, the names of these once famous professorsare little known, and one feels a twinge of sadness uponcontemplating these busts gathering dust. The mathemati-cians collected here are not among the most important,confirmation that fame can play strange tricks. There are theeffigies of Beccaria, Avogadro, Felice Chio (1813–1871),Tommaso Valperga Caluso (1737–1815), Carlo Ignazio Giu-lio (1803–1859), Plana and Genocchi. The busts are arrangedalong the side walls, on two floors. Entrance is free, thoughthe Palazzo dell’Universita is now home to the Rector’s offi-ces, restricting its hours to between 9 a.m. and 5 p.m.

Turn again toward Via Po and follow it in the direction ofthe river. On the other side of the street, just one block fromthe Palazzo dell’ Universita, you will spot the facade of theChurch of the San Franceso da Paola. Construction of thechurch was begun in 1632 and was completed by the end ofthe century. It is a fine specimen of Baroque architecture,but for us its real interest lies elsewhere: according toAntonioMaria Vassalli Eandi (1761-1825), an early biographer ofLagrange, this is the birthplace of the calculus of variations[7, p. 50]. Vassalli Eandi writes that in 1755 Lagrange, thenonly 19 years old, while assisting at Mass in the church, wasinspired by the music to create the delta algorithm. He wenthome immediately to write down the result, which he thensent to Euler on 12th August 1755. Today, anyone seriouslyinterested in mathematics will gladly spend a few minutes incontemplation of the church.

Continue along Via Po until you reach the intersectionwith Via Montebello. Turn left on Via Montebello and lookup: There before you stands the stately Mole Antonelliana,designed by Antonelli in 1862. Since the Mole is mentionedin every guidebook and in every tourist brochure aboutTurin, it hardly needs comment here. Suffice it to say that itis an absolute must for anyone visiting Turin, particularlynow that it houses the Cinema Museum. It is an excitingexperience to take the elevator with the glass floor all the

way up (about 170 m). On a fine day, the view from the topis impressive.

However, mathematical tourists will perhaps be moreinterested in another, more eye-catching feature of the Mole.Displayed on the south side of the dome is the Fibonaccisequence executed in red neon, each number about 2 mhigh (Fig. 12). This is the work of one of the most distin-guished Italian postwar artists, Mario Merz (1925–2003).Merz was fascinated, almost obsessed, by the Fibonaccisequence, in which he saw a ‘‘spiraliform mathematicalorganisation that differs from the Renaissance perspectiveand is organic’’ [7, p. 200]. He used the sequence in many ofhis works, beginning in the late 1960s, most notably on thechimney of the Turku Power Station in Turku, Finland, in1994 (see [8]) and outside the Zentrum fur InternationaleLichtkunst (International Center for Light Art) in Unna,Germany, in 2001. The story of the Turin installation beganin 1984, when Merz presented the Fibonacci sequence aspart of an art exhibition held at the Mole. When, in 2000,Merz was requested to contribute to the ‘‘Luci d’Artista,’’ anopen-air exhibition of large-scale light installations by Italianartists, held every year in Turin during the winter, he againsubmitted his old project.

The Fibonacci sequence on the Mole has alreadybecome a new symbol of the city. The sequence is invisiblein daylight; to see it, come here after dusk and approach

Figure 12. The Fibonacci numbers on the Mole.


the Mole from the south. However, you can get a betterview of the Fibonacci sequence from the hills surroundingTurin, especially from the Monte dei Cappuccini, on theother side of the River Po.

The next stop on our tour is at a rather demandingdistance to be reached by walking. Go back to Via Po andtake tram number 13 toward Piazza Castello. Stay on thetram until the stop just after Porta Susa railway station, inPiazza Statuto. A few metres away, in the middle of the littlegardens in the southern part of the piazza, you will find themonument to Giambattista Beccaria.

Aside from the aforementioned experiments on elec-tricity, from 1760 to 1764 Beccaria busied himself withmeasurements of an arc of meridian in Piedmont. Thesemeasurements were important not only for the productionof an accurate geographical map, but also to preciselyevaluate the flattening of the earth and, consequently, toverify Newton’s theories. Beccaria published his own resultsin the Gradus Taurinensis (1774), a book that was widelydiscussed throughout Europe. To have a base for his trian-gulations, Beccaria measured with extreme precision adistance of about 8 kilometers. The end points of this seg-ment were marked by signs on two slabs of marble. In 1808,two obelisks topped with armillary spheres were raised nearthese slabs in memory of Beccaria and his measurements(Fig. 13). They were one of the first examples of monu-ments dedicated to pure science. Oddly enough, to somepeople who dabble in esotericism, the obelisks are nowconsidered to possess magical significance.

From Beccaria’s obelisk it is only 10 minutes’ walk to thelast point of interest. Turn west and go to Via San Donato.At number 31 you will find the Institute of Faa di Bruno,built by Francesco Faa di Bruno (1825–1888).

Faa di Bruno was a remarkable man by any standard(Fig. 14). He came from an old Piedmontese family; afterreceiving his primary education at the Military Academy, hewent to the Sorbonne, where he graduated under Cauchy.Then he became in turn a soldier, a cartographer, a com-poser, a mathematician, an inventor, a social reformer, anarchitect, a publisher and, finally, a Catholic priest. Some-how he managed to be successful in all of his enterprises.He also created a religious order of nuns that still existstoday, the Suore Minime di Nostra Signora del Suffragio. In1988 the Church formally approved his veneration with hisbeatification by Pope John Paul II. While there are manybooks available about Faa di Bruno, there is no satisfactoryoverall treatment of his many achievements. The best singlesource available is a recent collection of essays [9].

Faa di Bruno’s work as a social reformer is extraordinary.He was particularly concerned with women’s problems: Forinstance, the welfare of teenaged mothers and of servantwomen fired when they became old. To help them, Faa diBruno created an institution dedicated to providing thesewomen with respectable work and housing. Bit by bit headded other activities: A retirement home for old washer-women, a school for female teachers, a high school. Faa diBruno had a modest apartment here, in which he kept hisbooks and his collection of scientific instruments.

When we look at Faa di Bruno’s variety of interests andduties, we cannot but wonder how he could find the timeto do serious work in mathematics. His favorite topic ofresearch was the algebra of invariants according to the

Figure 14. Francesco Faa di Bruno (1825–1888).

Figure 13. The obelisk commemorating Beccaria.


views of Cayley, Salmon and Sylvester. He wrote one of thebest textbooks on the subject, the Theorie des formesbinaires (1876), which was also translated into German(1881). Today, Faa di Bruno is best known in connectionwith a complicated formula which gives the n-th derivativeof composite functions (1857).

When Faa di Bruno died, the nuns did what they couldto keep his memory alive, preserving his home and per-sonal effects for future generations. His apartment is nowa fascinating little museum, one not ordinarily seen bytourists. Faa di Bruno’s rooms are almost like a time

capsule, preserved exactly as they were in the 1880s, withhis hat and his cane still lying on the table. (It is anunsettling experience to look at a photograph of Faa diBruno taken immediately after his death and, at the sametime, to stand in front of the very chair where his corpsewas laid out.) One entire wall of his living room consists ofbookshelves full of classics of mathematics (Fig. 15).Among the other exhibits on display is a fine collection ofteaching aids for physics, chemistry and mathematics, andeven some rare cameras from the 1860s. The Faa di BrunoMuseum can be visited only by appointment. Call or fax+39-011-489145, or send an e-mail to [email protected]. For more information, see the websitewww.faadibruno.com.

In the middle of the complex of buildings making up theIstituto stands the campanile (i.e., the bell tower), a 75-m-high edifice with yellow and white walls, designed by Faadi Bruno himself and erected in 1876 (Fig. 16). The cam-panile is made up of three different parts: A lower squaresection, a middle octagonal section, and, at the top, acircular steeple. With its vivid colors and unusual forms, itcontrasts strongly with the plain architecture of its sur-roundings. The architectural style of the campanile isdifficult to describe, since it shows several different influ-ences. Some of its design features were probably inspiredby the Gothic Revival style that was popular at the time itwas built. However, its overall appearance is quite differentfrom other buildings of the same period.

At first sight, the campanile appears much too tall andnarrow tobestable—thebase is a squareofonly5 mby5 m—especially considering that it was built at a time when rein-forced concrete had not yet been discovered. It is notsurprising that, originally, the project causedmanyobjections.However, the building turned out to be structurally sound.Indeed, it was so sturdy that the only damage it sufferedduring the bombings of World War II was the loss of the angeldecorating the top. The secret of this stability lies in the cam-panile’s innovative structure. Most notably, the belfry is notlocated at the topof the campanile, but 35 mup, or a little overhalfway to the top. It is composed of 32 cast iron columns.These columns are aparticularlyunusual feature, since theuseof cast iron in construction was almost unheard of when thetower was built. The belfry columns serve to join the upperand lower parts of the tower, making the campanile roughlyequivalent to two rigid bodies connected by a spring. Unlessan external force is periodic, with a period close to the naturalfrequency of the system, any dangerous oscillation will berapidly attenuated. This mechanism for ensuring stability wasthen completely new, a triumph of applied mathematics.

If you are still in the mood for mathematical memora-bilia, you might want to take a look at the monumentdedicated to the mathematical physicist Galileo Ferraris(1847–1897). Ferraris is mainly known for his work on thetechnical application of electricity, but he also wrote one ofthe first treatises on vector calculus (1895). The monumentwas made in 1903 by the sculptor Luigi Contratti. It is sit-uated at the intersection of Corso Montevecchio and CorsoTrieste, quite a long way from our main itinerary. On thepedestal there are two bas-reliefs, showing Ferraris meetingHelmholtz (1891) and Edison (1893). The bronze figure of a

Figure 15. Faa di Bruno’s apartment.

Figure 16. The campanile.


http://www.faadibruno.com

naked woman on the front side, representing the ElectricScience, was considered quite risque in its time.

We are now at the end of our walk. This guide to his-torical-mathematical Turin was written in the hope of beingas informative as possible. But, for a few hours, why not gooff the beaten track? With its many cultural attractions,Turin will amply repay even a casual exploration.

ACKNOWLEDGMENTS

I gratefully acknowledge the information about Faa di Bruno

supplied by the Suore Minime di Nostra Signora del Suffragio.

For her assistance with the illustrations, my thanks go to Cri-

stina Palermo. I also wish to thank Prof. Livia Giacardi

(University of Turin) for her support.

REFERENCES

[1] Coolidge, Julian Lowell. 1904. The Opportunities for Mathematical

Study in Italy. Pp. 9–17 in Bulletin of the American Mathematical

Society 11.

[2] Kennedy, Hubert Collings. 1980. Peano: Life and Works of

Giuseppe Peano. Dordrecht: Reidel.

[3] Marchisotto, Elena Ann and James T. Smith. 2007. The Legacy of

Mario Pieri in Geometry and Arithmetic. Dordrecht: Reidel.

[4] Goodstein, Judith R. 2007. The Volterra Chronicles: The Life and

Times of an Extraordinary Mathematician, 1860–1940, Providence,

RI: American Mathematical Society; London: London Mathematical

Society.

[5] Giacardi, Livia and Clara Silvia Roero, eds. 1987. Biblioteca

mathematica: documenti per la storia della matematica nelle

biblioteche torinesi. Turin: Allemandi.

[6] Borgato, Maria Teresa and Luigi Pepe. 1987. Lagrange a Torino e

le sue lezioni inedite nelle R. Scuole di Artiglieria. Pp. 3–200 in

Bollettino di Storia delle Scienze Matematiche 7.

[7] Eccher, Danilo, ed. 1995. Mario Merz. Turin: Hopefulmonster.

[8] Gyllenberg, Mats and Karl Sigmund. 2000. The Fibonacci Chim-

ney. P. 46 in The Mathematical Intelligencer 22, 4 (December,

2000).

[9] Giacardi,Livia,ed.2004.FrancescoFaadiBruno: ricercascientifica, inse-

gnamento e divulgazione. Turin: Deputazione subalpina di storia patria.


Reviews Osmo Pekonen, Editor

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The Man Who Flattenedthe Earthby Mary Terrall

THE UNIVERSITY OF CHICAGO PRESS, 2002, 408 PAGES, ISBN 0-226-

79360-5 (CLOTH) $48

REVIEWED BY ANDREW J. SIMOSON

AAshort quiz: Name a half-dozen mathematicians of

the eighteenth century. To make the quiz a littlemore difficult, we limit the list to include only those

who did most of their work in the eighteenth century. Bythis rule we disallow both Gottfried Leibniz and IsaacNewton, who belong mostly to the seventeenth century,and mathematicians such as Pierre-Simon Laplace andJoseph Fourier, who belong more to the nineteenth century.

Okay, time’s up. Who’s on your list?If your list is like most, it may include one of the Ber-

noullis, perhaps Johann or Daniel, but not Jacob, as he’s tooold. Of course, almost everyone’s list will feature LeonhardEuler. A few might include Alexis Claude de Clairaut, as,among other things, he gave conditions for the equivalenceof mixed partial derivatives and for the existence of inte-grating factors for solving first-order differential equations.

One important mathematician of this period might fail toappear on almost anyone’s list. Who is that?

Aha! As you are reading this review, you might guess theman who flattened the Earth, the title of a recent book byMary Terrall. Pierre-Louis Moreau de Maupertuis (1698–1759) was one of the more colorful and influential mathe-maticians of his time. As the reader learns in this book,Maupertuis was mentored by Johann Bernoulli at the startof his career and died in the home of Johann Bernoulli II,Johann’s son. In mid-career, Clairaut and Maupertuis wereteam members on a grand French Academy of Sciencesexpedition that captivated the imagination of the readingpublic of their time. And at the end of his career, as pres-ident of the Berlin Academy of Sciences, Maupertuis’s chieflieutenant was Euler.

Mary Terrall’s overarching thesis is that Maupertuis, as aman of science, was motivated by ‘‘personal honor andambition.’’ But how does one get inside another’s head?Despite his voluminous writings—Terrall catalogs 58 worksin her book’s appendices—she observes that Maupertuis‘‘was not an introspective person.’’ Therefore, Terrall arguesindirectly by summarizing Maupertuis’s scientific arguments,by copiously translating passages into English (each ofwhich is footnoted in the original), and by analyzing hisactions.


As the reader learns, many of Maupertuis’s enterpriseshad minimal a priori chances for success. For example, hismentor, Johann Bernoulli, was notorious for ‘‘being easilyroused to anger’’ and for having a ‘‘vituperative’’ and a‘‘possessive jealousy’’ with respect to his methods. Whychoose to start one’s career with a veritable volcano? Yetafter nine months, Bernoulli says their time together wasone of ‘‘revealing to [Maupertuis] the deepest parts of mysmall stock of wisdom, without hiding anything from him.’’

Secondly, in the French Academy of Sciences, Mau-pertuis championed the idea of geodesic expeditions, ofmeasuring degrees of arc along the Earth’s surface to helpdetermine the shape of the Earth. He then captained ageodesic expedition to Lapland, the region now dividedbetween Finland and Sweden, putting his reputation liter-ally on the line. Voltaire called Maupertuis’s subsequentaccount of the expedition ‘‘a story and piece of physicsmore interesting than any novel.’’ Incidentally, Voltairewent on to write a romance, called Micromegas, in whichan extraterrestrial giant stumbles upon Maupertuis and histeam in the Baltic who then converse at length about howthey know what they know (a new annotated translationappears in [4]).

Finally, Maupertuis, an extremely successful member ofthe Paris Academy, chose to be president of a fledglingPrussian Academy of Science. Such a career move was a riskto the nth degree. For in the aftermath of this gamble, in themidst of the Seven Years’ War between Prussia and France,Maupertuis was branded a traitor to France. Therefore,Maupertuis did not pursue ‘‘personal honor and ambition’’out of a sense of entitlement, but out of risk-taking and thespirit of enlightenment, or as Terrall puts it, Maupertuisthrived on being part of a ‘‘circle of like-minded friends,where amusement and science came together in sociable andwitty terms,’’ which in turn ‘‘pushed each other to take moredecisive [and insightful] steps than they might have takenotherwise.’’

Since it will illustrate the idea of the book’s title, let’s seewhy Maupertuis championed a French expedition to Lap-land, or to a region as far north as was then possible. In thePrincipia, Isaac Newton proposed a theory of gravitationand concluded that our rotating Earth must be flattened atits poles. He estimated [2] the difference Dr in equatorialradius q and polar radius R, to be 17.1 miles (where Dr =

q - R). This calculation expressly contradicted the reigningview on the continent that the Earth was bulging at thepoles, a view championed by the father and son team ofastronomers Giovanni Domenico Cassini (1625–1712) andJacques Cassini (1677–1756). Newton gave some data: In1635, Richard Norwood measured 1 degree of arc along ameridian near London to be 57,300 toises, while the Cas-sinis at the turn of the century measured 1 degree of arcnear Paris to be 57,061 toises. To convert to miles, take 1toise as 1.949 meters and 1 mile as 1.609 km. Thus, theNorwood measure is 69.39 miles between, let us say, 51�and 52� N, and the Cassini measure is 69.10 miles between,let us say, 48.5� and 49.5� N.

Based upon these data points, what is Dr ? The Cassinisand Newton agreed that the difference between q and Rwas relatively small and that the profile of the Earth wasmore or less elliptical. Had they chosen to do so, with a lotof work Maupertuis and his associates could have deter-mined that the best radii guesses for q and R usingNewton’s two old data points are 4000 and 3872 miles,giving Dr & 128 miles. Such a result means that Newton’stwo old measurements are inconsistent. Of course, thesemeasurements were taken years apart using differentinstruments and were very close in latitude, which wasprecisely Maupertuis’s argument for launching twin geo-desic expeditions: One to the equator and one to Lapland.Technology had improved, and now was the time to settlethis 40-year-old dispute. Contrasting Maupertuis’s leader-ship—of emceeing the measurement of one degree of arcalong a meridian (1736–1737) in Lapland—versus the con-fusion of operations for the expedition sent to Peru (to theregion now called Ecuador) lasting nine years (1735–1744),suggests that Maupertuis was a skilled administrator. Fur-thermore, after his return to Paris, he defended the team’sresults decisively. Despite heated objections by JacquesCassini andhis allies about issues of accuracy,Maupertuis gotalong ratherwellwithCassini deThury, the second sonof theoutraged astronomer. Thury subsequently remeasuredParisian arclength and ultimately vindicated the Lapland

Figure 1. Engraved portrait of Maupertuis by Jean Daulle of

Tournieres, courtesy of the Owen Gingerich Collection.

THE MATHEMATICAL INTELLIGENCER70

team’s findings, thereby augmenting Maupertuis’s prestige.Of this accomplishment, Voltaire said that ‘‘Maupertuis hadflattened the Earth and the Cassinis too.’’ To commemoratethis event, Maupertuis commissioned a self-portrait, anengraved version of which is Figure 1.

To demonstrate the improvement in the French 1735–1744 geodesic measurements over Newton’s two oldmeasurements, Maupertuis’s team measured 1 degree of arcbetween 66� and 67� N as 57,395 toises (69.52 miles), andthe equatorial team—one of whose leaders was CharlesMarie de La Condamine (1701–1774) who in returninghome to Paris from Peru began by going down the Amazonto the Atlantic, collecting cinchona bark and seedlingsamples, which in turn led to a very effective kind of qui-nine—measured 1 degree of arc between 0.5� S and 0.5� Nas 56,768 toises (68.75 miles) [1]. Using these two newarclength values gives q& 3974.0 and R & 3956.5 miles, fora difference of Dr & 17.5 miles—very close to Newton’soriginal estimate of 17.1 miles. (Earth’s actual measure-ments are q & 3964.1, R & 3950.8 and Dr & 13.3 miles.)

Terrall’s secondary thesis is ‘‘that Maupertuis made astrategic move by writing in [a] hybrid genre,’’ namely,writing scientific ideas for the reading public, most rep-resentatively, Venus physique. As one critic observed,‘‘Our ladies have abandoned their novels to read it.’’Other critics decried this unprofessional behavior as one‘‘seeking fame and reputation, for being fashionable.’’ Butsuch damnation has, over the years, transformed intoapprobation—and just as Voltaire is ‘‘a poet who writesgeometry’’ and La Mettrie (Frederick the Great’s courtphysician) is ‘‘a doctor who writes about the soul,’’ soMaupertuis is a mathematician who writes about pleasure.That is, in Maupertuis’s words, ‘‘In spite of a thousandobstacles to the union of two hearts and a thousand tor-ments that are bound to follow, pleasure directs the loversto the goal nature intended.’’

Maupertuis’s personality fitted him well for this venture,for the public enjoyed following the literary exploits of aneccentric yet important savant. For example, Maupertuis‘‘had a reputation as a libertine man-about-town, equallyhappy to consort with duchesses and their maids.’’ Of Mau-pertuis’s life-force experiments: ‘‘He threw salamanders intothe fire to show that they burn, and allowed scorpions to bitedogs to test the effect of their venom; he enclosed scorpionswith spiders to watch their battles.’’ He traced the genealo-gies of six-fingered men, conducted breeding experimentswith his pet dogs, and maintained a houseful of exotic ani-mals. Maupertuis describes this managerie, ‘‘You would notbelieve the multiplication of animals of all species I have atmy home. When one has lived like this one finds almost asmuch stimulation from them as from people.’’

Any account of Maupertuis’s accomplishments is sure toinclude two items: The Lapland expedition, which we havealready summarized, and the physical principle of leastaction compounded by the Konig affair. Maupertuis con-sidered his work on the principle of least action to be hisfinest. Intuitively, the principle says ‘‘that nature acts assimply as possible,’’ and formally, that nature acts on matterso as to minimize a product-like combination of its velocityand position. Terrall amplifies these ideas at chapter-

length, giving, among other things, the example of howlight in following Snell’s law follows the path of least timerather than least distance. However, Samuel Konig, a long-time friend of Maupertuis, accused Maupertuis of plagia-rizing these ideas from Leibniz. As Terrall points out,‘‘Following the dispute meant following a complicated trailof print, often mediated by journal articles and letters, asauthors and publishers printed a bewildering array of oldand new texts.’’ Terrall guides the reader through thislabyrinth for 18 pages of spirited give and take.

In the midst of the Konig affair, Maupertuis wrote Lettresur le progres des sciences in 1752. As president of the BerlinAcademy, in this open letter he proposes a number of pro-jects for the scientific community to consider. His list is grandand sweeping, not unlike David Hilbert’s 1900 list of 23problems with which the mathematical community mightwrestle into the next century or two. Sprinkled amidst mostlysound proposals are these: Test new medical procedures oncriminals, such as removing the kidney as a treatment forkidney stones; use opium to explore the mind; and raise agroup of children in isolation from adults to determine thelanguage they would develop. This last item jumps off thepage. Was Maupertuis serious?

Of course, one of the reasons that the Konig affairmushroomed beyond anyone’s expectations is that Voltaireentered the debate in defense of Konig. As Terrall pointsout, Maupertuis and Voltaire had been long-time friends.Both Maupertuis and Voltaire had had an affair with thesame woman, Emilie du Chatelet (1706–1749), whose lifework was the translation of Newton’s Principia from Latininto French. Voltaire lived with Emilie for her last 15 years.Maupertuis had tutored both Emilie and Voltaire in math-ematics. After Maupertuis’s return from Lapland, Voltairerecommended Maupertuis to Frederick the Great as can-didate for leading the Berlin Academy, saying, ‘‘A man like[Maupertuis] would establish in Berlin an academy of sci-ence that would outdo the Parisian one.’’ After acceptingthe presidency of the academy, Maupertuis returned thefavor to Voltaire, arranging a private first meeting betweenFrederick and Voltaire. But, a mutual friend to both Mau-pertuis and Voltaire predicted, ‘‘Maupertuis and Voltaire arenot made to live together in the same room.’’

Terrall explains Voltaire’s ‘‘perverse’’ entrance into theKonig affair: ‘‘[Voltaire] was motivated by personal animos-ity, [cloaking his actions as] a self-styled enemy of tyrannyand defender of press freedom. He concentrated on ridi-culing Maupertuis as a tyrant and a buffoon, and on makingfun of the more speculative parts of his works [such asMaupertuis’s Lettre as described above].’’ But to be fair toVoltaire, there’s more. Konig had been a two-year houseguest of Emilie andVoltairewhile serving as tutor helpingherto understand the Principia. Furthermore, Voltaire had ahabit of championing the underdog, even if the underdogadvocated ideas contrary to his own. He also thrived oncrossing verbal swords with almost anybody. As one exam-ple among many, of Jean-Baptiste Rousseau’s poem Ode toPosterity, Voltaire said that ‘‘it was unlikely to reach its des-tination.’’ Indeed, one of the reasons Emilie invited Voltaireto a long-term stay in the country was to protect Voltaire fromhimself, that is, from his almost uncontrollable wit. So in the


midst of the controversy at the Prussian Academy of Science,when ‘‘[Maupertuis’s] more outlandish suggestions hadbecome stock jokes at court’’ [3], Voltaire yielded todefending the underdog once again. To wrap up this Konigaffair, Terrall concludes that ‘‘it did not incapacitate [Mau-pertuis],’’ as he continued to work on ideas of heredity andthe nature of matter until his death.

Mary Terrall’s book is the fruit of 20 years of work onshowcasing a quasi-forgotten, yet prominent, member ofthe scientific community of the enlightenment. The manwho flattened the Earth, who was called the first FrenchNewtonian, who popularized science for the masses, andwhose somewhat endearing nickname was the Flea—Terrall tells a fascinating story backed by interesting detail,careful citation and enlightening insight.

REFERENCES

[1] Hoare, Michael Rand (2005). The Quest for the True Figure of the

Earth. Surrey: Ashgate.

[2] Newton, Isaac (1999). The Principia, translated by I.B. Cohen and

A. Whitman. Berkeley: University of California Press.

[3] Pearson, Roger (2005). Voltaire Almighty New York: Bloomsbury

Press.

[4] Simoson, Andrew J. (2010). Voltaire’s Riddle: Micromegas and the

Measure of All Things. Washington, DC: The Mathematical Associ-

ation of America.

Professor of Mathematics

King College

Bristol, TN 37620, USA


THE MATHEMATICAL INTELLIGENCER72

Mathematics and Music(Mathematical World, Vol. 28)by David Wright

PROVIDENCE, RI: AMERICAN MATHEMATICAL SOCIETY, 2009, 161 PP.,

US$35.00, ISBN-10: 0-8218-4873-9; ISBN-13: 978-0-8218-4873-9

REVIEWED BY EHRHARD BEHRENDS

IIt is a commonplace that there are links between theworld of mathematics and the world of music. But inthe literature on these connections, the two areas play

asymmetric roles. The reader is usually assumed to havesome mathematical background: Mathematical terms andtheories are used with little explanation. These investiga-tions are hardly accessible to nonspecialists.

Mathematics and Music is written in a different spirit. Itreviews some basic concepts in both mathematics and musicfrom the very beginning, presuming no background in eitherof these fields. It’s addressed to students of all fields who areinterested in both subjects. The 12 chapters cover a widevariety of mathematical and musical themes. Chapter 1 isdevoted to ‘‘basic concepts.’’ Here, the various sets of num-bers are introduced (N, Q, etc.), and one also learns, forexample, that the integers are well ordered, how to visualizefunctions by their graphs, and how an equivalence relation isdefined. Basics for the musical counterpart include thetranslation of pitches to notes by the treble and bass clefs,musical intervals (for example, the fifth or the octave), andthe use of accidentals. At the end of this chapter, cyclic per-mutations are introduced to explain how the different modes(Ionian, Dorian, etc.) can be derived from a single scale.

Chapter 2 is concerned with ‘‘horizontal structures.’’How are whole notes, half notes, and so on written, whichsymbols are used for rests, and how do dots change thelength of a note? I never realized before that the length d ofa note increases to d(2 - 1/2m) if the note is m-dotted, afact proved here by geometric series. It is also explainedthat translation (resp. transposition, resp. retrogression) ofpatterns corresponds to replacing f(x) by f(x - c) (resp.f(x) + c, resp. -f(x)) for functions f.

Let’s turn to Chapter 3: Harmony and Related Numer-ology. The mathematics starts with the algebraic structureof Z12. In this setting, a major chord is just the sequence(4, 3, 5) of modular intervals. Similarly, diminished chordsand many others are described and correctly translated tomusical notation. (That is, one must write E # and not F inthe major chord of C #.)

Chapter 4 introduces ratios as equivalence classes whichone can hear as pitches: The octave, for example, corre-sponds to 2 : 1. Clearly, the number

ffiffiffi212p

plays an importantrole here: This is the ratio associated with a semitone. Onealso learns how intervals can be converted to cents and viceversa.

Logarithms and the exponential function are introducedat the beginning of Chapter 5. The graphs are sketched and

the basic properties summarized. This is used to transformthe description of intervals from ratios to a measurement insemitones (‘‘the interval ratio r is measured in semitones by12 log2r.’’)

In Chapter 6, it is noted that the partition of the octaveinto 12 semitones is rather arbitrary: For every n 2 N onecould consider the n-chromatic scale based on the intervalratio

ffiffiffi2np

. To be able to discuss this scale in more detail,Euler’s /-function is introduced (/(n) is the number ofintegers k \ n which are relatively prime to n): There are/(n) ways to generate the n-chromatic scale by consideringk, 2k, 3k, ... (modulo n).

In Chapter 7, the modular arithmetic continues, startingwith the properties of N and ending with a number of basicalgebraic notions: monoids, groups, homomorphisms, andso on. It is shown how modular arithmetic can be used togenerate a 12-tone row in 12-tone music.

Algebraic investigations are also central to Chapter 8;they culminate in the proof of the fact that Z is a principalideal domain. It is noted in passing what prime numbersare and how one can find them with the sieve of Eratos-thenes. It is then easy to describe how musical passages arebuilt where patterns of m notes stand against patterns ofn notes (with m^n ¼ 1).

n-chromatic scales are studied in more detail in Chapter9. For example, the 19-chromatic scale is appropriate if onewants to generate a scale by an interval of ratio 3.

Calculus is introduced in Chapter 10. The e-d definitionof continuity is given, and piecewise smooth and periodicfunctions are defined. As the main result, one learns thatsuch functions have a Fourier expansion. With this back-ground, it is easy to explain the importance of formants forthe sound of instruments and the human voice.

Chapter 11 starts with the old observation that two pit-ches played simultaneously sound ‘‘harmonious’’ when theratio of their frequencies is rational with small numeratorand denominator. It is shown how a scale constructed byusing only a just fifth necessarily has a small imperfection,the comma of Pythagoras. And it is proved that it is impos-sible to avoid irrational numbers in the n-chromatic scales.

Problems concerning tuning are investigated further inChapter 12. For many centuries, various scales have beenproposed: The problem is to have as many justly tunedintervals as possible, and, at the same time, intervals whichsound ugly have to be avoided. Advantages and disad-vantages of the Pythagorean scale, the mean tone scale andthe equal temperament are discussed in some detail in thisfinal chapter.

Each chapter is complemented by (mostly mathemati-cal) exercises of various difficulty, and understanding isfacilitated by many graphics and musical scores.

Wright has packed an ambitious overview into 150printed pages. He had to make choices, and it is legitimatethat he followed his own preferences. And, of course, it is amatter of taste which of the many aspects of the theme‘‘mathematics and music’’ a reader will consider more fas-cinating, or less.

The author writes that ‘‘the treatise is intended to serveas a text for a freshman college course.’’ This purpose iscompletely achieved. This book can be an inspiring basis


for lectures presented to all students. For readers with amathematical background who are interested in music, thebook will be a valuable resource. And even those who havestudied music for many years may learn many new facts(formants, just fifths, equal temperament, to mention afew).

I also recommend these books on the same subject:

• G. Assayag, H. G. Feichtinger, J. F. Rodrigues (eds.):‘‘Mathematics and Music: A Diderot Mathematical forum,Lisbon, Paris and Vienna, December 3–4, 1999’’ Berlin,Springer, 2002.

• J. Fauvel, R. Floyd, R. Wilson, (eds.): ‘‘Music andMathematics: From Pythagoras to Fractals,’’ OxfordUniversity Press, Oxford, 2003.

• L. Harkleroad: ‘‘The Math behind the Music,’’ CambridgeUniversity Press, Cambridge and the MathematicalAssociation of America, Washington D.C., 2007.

Free University of Berlin

Arnimallee 2-6

Berlin, D-14195

Germany



Mathematicians Fleeingfrom Nazi Germany: IndividualFates and Global Impactby Reinhard Siegmund-Schultze

PRINCETON AND OXFORD: PRINCETON UNIVERSITY PRESS, 2009, 472 PP.

US $90.00, ISBN: 978-0-691-12593-0 (CLOTH); US $38.22,

ISBN 978-0-691-14041-4 (PAPERBACK)

REVIEWED BY G. L. ALEXANDERSON

TThis is an important book for mathematicians andother scientists, for those in the field of intellectualhistory, and for general readers interested in our

not-too-distant past. I got so caught up in it I could scarcelyput it down. It is well written and meticulously researchedand documented. For many, there will be connections topersonal experience. Anyone who spent the 1950s or evenlater in the vicinity of Stanford University will have encoun-tered many in its cast of characters: Faculty at Stanford(S. Bergman(n), C. Loewner,G. Polya,G. Szeg}o,H. Samelsonand M. Schiffer) and visitors (R. Courant, H. Lewy,O. Neugebauer, I. Schoenberg and S. Warshawski, amongothers). These people shaped the professional and personallives of many Stanford students. Similar stories could be toldof research universities throughout the United States at thattime.

This book follows by a few years another fascinatingwork, Mathematicians Under the Nazis, by Sanford Segal(Princeton, 2003), which covered those ‘‘German speaking’’mathematicians who remained in Germany and the occu-pied countries during World War II. Siegmund-Schultze’sbook covers, in a sense, the complement. Up until now, theliterature on the emigres has not been large, due to a varietyof reasons: Archives that remained closed to scholars,unwillingness of some to speak on the subject because ofpolitical sensitivities, and the possibility that the reminis-cences of the emigres were sometimes unreliable due to thepassage of time. For those interested in the subject, one of thebest sources has been Max Pinl’s series of articles (somewritten with A. Dick) ‘‘Kollegen in einer dunklen Zeit,’’ thatappeared in the Jahresbericht der Deutschen Mathematiker-Vereinigung in the 1960s and 1970s. Siegmund-Schultzepoints out, however, that Pinl is incomplete and, on occa-sion, can be misleading, but he has high praise for ConstanceReid’s biography of Courant and gives credit for much of hiswork to Courant’s papers at New York University.

The author is extremely conscientious in defining thewords he uses. For example, he does not use ‘‘NationalSocialism’’ because it was neither socialism nor national incharacter, nor does he use ‘‘Aryan,’’ ‘‘Third Reich,’’ and othersuch words, because after the war they carried too muchadditional baggage. He prefers ‘‘Nazis,’’ though sometimeshe refers to ‘‘Hitler’s regime.’’ He also makes careful dis-tinctions between those who left Germany or Austria before

1933 or after, and between those who left ‘‘voluntarily’’ orthose who were ‘‘forced.’’ The author decided to concentrateon ‘‘forced emigration’’ after 1933. He also relies most onemigres fleeing racial persecution, thoughother groupswerealso leaving central Europe—pacifists, some Catholics andhomosexuals.

Some who left were able to reach ‘‘safe’’ countries likeSweden, Switzerland and England, though many camedirectly to the United States. Some were unable to leave orchose to stay, believing the situation would improve.Notable among the latter was F. Hausdorff who stayed, buteventually, with his wife and sister-in-law, committed sui-cide rather than face the death camps. These personalstories are heart-wrenching. There are many personal sto-ries of death or hardship such as the grim task of travellingacross Siberia to end up eventually on the West Coast of theUnited States (Max Dehn and Kurt Godel, for example).

We also read here of the difficult questions of just howmuch the state of mathematics changed as a result of thismass movement of some of the most brilliant mathemati-cians of the time from one continent to another. The authorwarns of the post hoc, ergo propter hoc phenomenon: Thewidespread assumption that mathematics prospered in theUnited States as never before because of the infusion of allthat talent. Though probably true, we have no proof thatAmerican mathematics might not have shown remarkablegrowth in any case. The war itself created jobs, particularlyin applied mathematics, and this provided work for manyAmerican scientists. At the same time, to say that Europeanmathematics declined in prominence only because of theemigration may be simplistic. These are provocative ideasand will surely be discussed for years to come. The authorstates that he is quite aware that his will not be the lastword on the subject. In particular, he points out that thequestions raised about G. D. Birkhoff’s alleged anti-Semi-tism could only be treated by a much larger biographicalstudy of Birkhoff, well beyond the scope of this volume. Hecorrectly argues, however, that good work is more easilydone in a community of scholars: It is best to be able tocommunicate one’s ideas directly with colleagues ratherthan relying solely on reading published work. In this waythe United States, and other countries, obviously benefitedsignificantly from the emigration.

Chapter One covers questions of terminology. ChapterTwo is devoted to the extent to which the emigrationaffected mathematics more than some of the other sciences.The author observes that the United States accepted moreemigres by far than other countries, and a disproportionatenumber of these were mathematicians. One side-effect ofthe migration was to make English the lingua franca ofmathematics, finishing off German as the internationallanguage of science. A complication in compiling statisticsabout the emigres was that some mathematicians (notablyE. Artin, K. Friedrichs, E. Kamke and H. Weyl) were notJewish but were forced to leave their positions becausetheir wives were. And we note, too, that age made a dif-ference: The oldest of those who came to the United States(F. Bernstein, M. Dehn, H. Hamburger, E. Hellinger andA. Rosenthal) failed to get regular appointments. Amongthose who were eventually successful in locating positions


appropriate to their ability were those at the Institute forAdvanced Study (IAS) in Princeton, or at Stanford, Berkeleyor NYU. But with very few exceptions, mainly due tospecialized fields of expertise like applied mathematics orhistory of mathematics, emigres did not get regular posi-tions at the leading departments at Harvard, Princeton,Yale, Chicago, Brown, MIT or Caltech.

Chapter Three largely deals with emigres to the UnitedStates prior to 1933, mainly motivated by the economicconditions in Europe. Some of these had illustrious careersin America: E. Hille, E. Hopf, T. Rado, D. Struik, J. vonNeumann and A. Wintner. Others from non-Germanspeaking countries also came to the United States at thattime: C. Lanczos, I. Sokolnikov, J. Tamarkin, S. Timo-shenko, and T. von Karman, and, in the critical year 1933,one of the most illustrious additions to the IAS faculty,H. Weyl.

A striking table in Chapter Four shows that 90 of the 145emigres, and 130 of the 234 persecuted (including nonemi-grants and those killed) came from only four of the 42 citiescovered (from Berlin, 41 faculty members out of 62, and fromGottingen 24 out of 28). The Hitler regime was remarkablyeffective in clearing out the best and the brightest.

In this and the next chapter, we read letters and docu-ments pertaining to those who succeeded in their efforts toemigrate as well as those who waited too long or were justplain unlucky. Among the latter were: O. Blumenthal andA. Tauber, who both died at Theresienstadt; Hausdorff,who was mentioned earlier; and F. Noether, who made themistake of going to Russia, where he was executed by theSoviets.

Chapter Six is devoted to those who emigrated to ‘‘safe’’European countries, theMiddle East, Australia or India.Manyof these were eminent mathematicians, but the numberswere comparatively small, and some were also in transit toother destinations.

In Chapter Seven the author addresses the attitudes ofthe emigres following their move to the United States.Curiously enough, though they were grateful for havingbeen saved from almost certain death in Germany, oftenthey still held out hope that they could at some time returnto Germany and the colleagues and institutions that hadbeen hospitable to them early in their careers. Since Gauss,the German mathematical community had been extremelystrong, with support from outside the universities by thegovernment and publishers such as Springer, for example.Many emigres retained their concern for the health ofGerman science and culture. Germany was, after all, thecountry of Heine, Schiller and Goethe, Bach and Beetho-ven. A few mathematicians even returned (notablyEberhard Hopf and, at least temporarily, Carl Ludwig Sie-gel). Some who could have left Germany did not, for avariety of reasons. A prime example was the Dutch alge-braist, B. L. van der Waerden, prompting Courant to writeto him in 1945, ‘‘Your friends in America, for example,could not understand why you as a Dutchman chose to staywith the Nazis.’’ This criticism followed van der Waerdenthrough the remainder of his long career. Some who leftopenly expressed their regret over leaving behind Germanculture—von Neumann and Feller, for example. Even

Courant found it hard to give up his loyalty to Springer andadvised Szeg}o to publish his Orthogonal Polynomials withthat eminent publisher. (But Szeg}o did not agree, and itwas published instead in 1939 by the American Mathe-matical Society.) There are many well-known storieshere—Neugebauer’s eventually establishing MathematicalReviews to substitute for the largely unavailable Zentralblattduring the war, and many not so well known

The author explores at some length the reactions in theUnited States to the crisis in Europe (Chapter Eight) and thevarious committees and organizations set up to expeditethe granting of visas and making the necessary arrange-ments for appointments, even when only temporary. Hecites faculty at three institutions who stood out for theirefforts: (1) The Institute for Advanced Study; (2) Thegraduate school at NYU under Courant; and (3) The grad-uate school at Brown under R. G. D. Richardson.

On the other hand, there were those like G. D. Birkhoffwho questioned the wisdom of hiring the emigres duringthe Great Depression when native-born Americans werehaving such difficulties in getting jobs, a view expressed inBirkhoff’s well-known statement that American mathema-ticians would be reduced to being ‘‘hewers of wood anddrawers of water,’’ on the occasion of the semicentennial ofthe American Mathematical Society in 1938. Some viewedthe remark as clearly anti-Semitic. The American govern-ment under Roosevelt could have done more to speed upthe process of getting the emigres into the United States,but it was politically difficult because of the Depression anda strong wave of isolationism in the country. Siegmund-Schultze, however, makes it clear that ‘‘it is imperative tostress that this kind of anti-Semitism cannot be compared,let alone put on an equal level, with the criminal, institu-tionally legalized and incited anti-Semitism in Germanyafter 1933.’’

These observations are supported by a large number ofcitations of documents and letters and are followed by anassessment of the effect of the immigration on mathematicsin the United States (Chapter Ten) and an Epilogue. Muchof this is concerned with the question of how well emigresadjusted to American life. With so much attention paid toundergraduates in American universities, the Europeanprofessors were disappointed in American students whoneeded background in mathematics that would have beencovered in the gymnasium in Europe. Further, in Germany,professors had traditionally held a higher social positionthan was common in America. These conditions madeadjustment difficult. The author quotes L. Coser: ‘‘Theintimacy of the coffee house had to give way to the dis-tance and strangeness of the American lifestyle, and so theywere for the most part happy but not glucklich.’’ Further, ina quotation from M. R. Davis, we read that, ‘‘Another barbetween the foreign professor and his students was thedifference in attitude which characterized the European asdistinguished from the American professor. The former haddeveloped to a fine art the technique of social distancefrom his students.’’ C. L. Siegel wrote, ‘‘I no longer have thehope, which led me to America four years ago, of finding atolerable position abroad…. I can no longer adapt, I am toomuch of a Prussian.’’ He also wrote to Courant in 1935, ‘‘It


would be meaningless to escape the sadism of Goring’sonly to get under the yoke of Mrs. Eisenhart’s notion ofmorality…. Please do not be offended that I do not likeyour America.’’ (Luther Eisenhart was dean at Princeton.)These seem to be the exceptions, however. Most adjustedas best they could. Richardson wrote that Loewner, ‘‘themost distinguished mathematician in Kentucky,’’ held aposition at the University of Louisville, teaching manysections of trigonometry each term. He eventually movedon to Brown and Stanford. Many had little choice. They hadto adjust to whatever positions were open to them.

There is no shortage of heroes in these pages, but one,for me, stands out: Charles Loewner, who, before he emi-grated, travelled back and forth from Germany to Praguewhere he had received his Ph.D., because it was easier toget news out of Prague to American friends, reporting onthe state of colleagues and asking for assistance in locatingpositions for them. In a letter here to L. Silverman atDartmouth, quoted in its entirety, he makes the case for I.Schur, G. Szeg}o and S. Cohn-Vossen. The tone of the letteris consistent with my own impressions of Loewner, who

was one of the kindest people I have ever met and one ofthe most popular mentors and teachers at Stanford, wherehe advised the Ph.D. dissertations of 16 students. But hewas quiet and self-effacing and never got the credit hedeserved for his work for the emigres and his veryimportant mathematical contributions to the solution of theBieberbach conjecture, for example.

Almost as interesting as the text itself are the numerousappendices—lists of those who escaped, those killed, andthose persecuted in various ways, along with many lettersand documents. All around, this is a rewarding andimpressive piece of scholarship, a story that is at once grimbut also uplifting, since for many of those who escaped,there was a happy ending.

Department of Mathematics and Computer Science

Santa Clara University

500 El Camino Real

Santa Clara, CA 95053-0290

USA



Perfect Rigorby Masha Gessen

BOSTON, NEW YORK: HOUGHTON MIFFLIN HARCOURT, 2009, 256 PP.,

US $26.00, ISBN: 978-0-15-101406-4

REVIEWED BY REUBEN HERSH

TThis book is a biography of Grigori (Grisha) Perelman,the Russian mathematician who is now famous forproving Thurston’s classification of 3-manifolds. As a

corollary, he proved the Poincare conjecture—one of theoutstanding open problems in mathematics.

Thurston had conjectured, and proved in importantspecial cases, that all 3-dimensional manifolds can beclassified into combinations of 8 basic types, each of whichcan be represented geometrically using 3-dimensional non-Euclidean geometry. The simplest of these cases would justbe the 3-sphere, which is the subject of Poincare’s century-old conjecture.

In the course of telling about Perelman, Gessen tells muchelse that is of great interest. She leads us into the hidden innerlife of ‘‘under cover’’ mathematics in the Soviet Union, includ-ing ‘‘special schools,’’ ‘‘math circles’’, and ‘‘math clubs’’. There,dedication to truth itself remained possible, for years on end,right under the noses of the Party and the KGB. All this wasclosely connected with the beneficent influence and inspira-tion of one man—Andrei Kolmogorov. He was, of course, agreat international pioneer and researcher in many differentfields of mathematics. But he was also the energizer andinspirer of a whole special Russian system of mathematicaleducation and indoctrination for talented young people. Ges-sen paints an amazing portrait of him, hitherto quite unknowntome, includinghis long-time intimate friendshipwith thegreattopologist Pavel Sergeevich Alexandrov, and his dedication toan all-round life devoted to beauty and refinement, both cul-tural and physical.

Masha Gessen has never met her subject, Grigori Per-elman. Indeed, it seems that for a while now nobody at allhas met him—except for his mother, Lyubov, who sharestheir modest apartment on the outskirts of St. Petersburg(formerly Leningrad). Gessen thinks that her never havingmet Perelman may have been an advantage in writing thebook. She certainly seems to have met and thoroughlyinterviewed every major friend, acquaintance, and influ-ence in Perelman’s life (except for his mother and hissister). As a result, she has been able to paint a convincingand fascinating psychological portrait of him that makescredible and understandable his refusal of the Fields Medaland the Clay Prize, and even his present total withdrawal,not only from the mathematics community of Russia and ofthe world, but even from almost all human contact. This lifestory raises deep, disturbing questions about the stressesand the values of a life entirely devoted to mathematics,especially in the world as it is today.

Grisha’s mother Lyubov herself is mathematically gifted.In fact, she declined the offer of a position as a graduate

student of mathematics in Leningrad in order to give birth toandnurtureher sonGrigori.WhenGrigoriwas10yearsold, shewentback tohermentorProfessorNatanson, to tell him thatherson was mathematically talented. Natanson sent Lyubov andGrisha to Sergey Rukshin, a famous coach of mathematicalproblem-solving teams, and boss of a math club in St. Peters-burg. It seems that Rukshin is more than just a famous mathcoach; he is the greatest math coach in the world. He has sentmany contestants to the International Mathematics Olympiad.Rukshin and Grisha became inseparable companions. UnderRukshin’s coaching, Grisha actually did become one of thebest, maybe the very best mathematical problem-solver in theworld. First in Rukshin’s math club, and then in national andinternational competitions, Grisha seems almost never to havefoundaproblemhecouldn’t solve. In sessionsof themath club,he sat quietly in the back. He was often the last to speak, for hissolutionsusuallywere clearly optimal.Nothing left out, nothingunnecessary, nothing open to challenge. While working on theproblem, he might rub his leg, hum softly, or toss a ping pongball back and forth. Not only did he solve the hardest problems,he then explained his solutions in perfectly clear, concise lan-guage to anyone who asked. His only difficulty seemed to behow to help anyone who failed to understand his clearexplanation. In such a case, he seemed to have no recourse butto simply repeat the same explanation.

As a boy, Grisha was reasonably fit physically. At somemeetings of Olympiad contestants, he played volley ball withthe others. But his mental energy seems to have been totallyfocused on mathematics, from early childhood until matu-rity. Another geometer, who was reported to have beenPerelman’s friend during his stay in the United States, toldGessen that they often had conversations, and that the con-versations were never on any topic except mathematics.

Perelman did have one setback. The first time he com-peted in the all-Russian mathematics Olympiad, he came insecond. This was a very severe shock and disappointment.Gessen writes that Grisha decided that he hadn’t workedhard enough in preparation. He resolved never again toallow such a mishap to occur. In fact, it never did. Healways came in first, before and after that one ‘‘failure.’’

Like many other male mathematicians of relatively youngyears, Perelman gave little attention to matters of physicalappearance. He always wore the same brown corduroyjacket. He did not waste time or effort about cutting his hairor his fingernails. With food he also preferred simplicity. Itseems that while in the United States he rarely ate anythingbut bread and cheese. He did prefer one particular variety ofblack bread which he procured, while living and working inNew York, at a bakery on the far south side of Brooklyn, atBrooklyn Beach. He would walk there after each day’s workat the Courant Institute in Manhattan.

Gessen’s book gives a rather brief treatment of the Poin-care conjecture itself. Many readers of this journal will knowthat the strongest attack on it had been made by RichardHamilton of Columbia University. Hamilton used what hecalled ‘‘Ricci flow,’’ a nonlinear parabolic partial-differentialequation satisfied by a certain geometric quantity associatedto a 3-dimensional manifold. The time-evolution of thesolution to the equation decribes a smoothing of an arbitrary3-manifold. The smoothing action eventually would bring an


arbitrary manifold to a form recognizable according to Thur-ston’s classification. However, before reaching that stage, theevolution could get stuck by encountering a geometrical sin-gularity, one of several possible kinds of singularity. To getpast such a singularity, it was necessary to perform whattopologists like to call ‘‘surgery’’—that is, a cutting and pastingoperation which removes the singularity and renders theevolving manifold again sufficiently regular. Hamilton wasunable to show that such surgery was always possible. Per-elman succeeded in doing so. Complex, detailed geometricaland analytical reasoning permitted Perelman to provide thenecessary surgery instructions to complete Hamilton’s Ricciflow program, thereby proving both the Thurston Classifica-tion and the Poincare Conjecture.

Perelman never submitted his solution for publication in ajournal. He posted three announcements on a well-knownsite intended for such early warnings of new results. Henever even announced that he had proved Thurston orPoincare, merely that he had obtained certain technicalresults about the Ricci flow. Those who are qualified to readhis abstracts would understand their significance. Those whoare not so qualified need not attempt to read them.

Once the word got around to the ‘‘Ricci-flow commu-nity’’ and other interested topologists, they had to decidewhether Perelman really had solved those problems. Thiswas not very quick or easy, for his abstracts were concise,even in certain places perhaps a bit obscure. It took a yearand a half for several teams of topologists to renderthe verdict—yes, he did it! During this process, Perelmanspent time traveling the U.S., giving talks and answeringquestions. People found him well-prepared, patient andforthcoming.

It was always clear that this work was very likely goingto win a Fields Medal and a Clay Prize. As Perelman trav-eled, giving talks at elite math departments, he received joboffers, some very favorable. However, he expressed verylittle interest in any of them. It is now clear that rather thanbeing excited and flattered by this experience, Grisha wasdisappointed, repelled, perhaps even disgusted. This wasnot what he had expected, not what he was looking for.

Hamilton did not seek him out, did not express greatenthusiasm or gratitude to him. Others who wanted to talkto him about job offers at high salaries for little work didnot seem to have even studied or understood his mathe-matical work. In fact, Grisha was becoming a celebrity,something it seems he had never sought, expected, desired,understood or valued. His celebrity status, even within the

academic community, seemed to outweigh and overbal-ance the actual content of his mathematical achievement.To Grisha, this was unattractive, unpleasant, even immoral.He practiced mathematics only for its own sake, hebelieved in mathematics only for its own sake. Mathematicsfor the sake of fame, money or power were alien to him,perhaps even incomprehensible. Certainly alien, repellent.Unclean. Degenerate.

In Russia also there were unpleasant incidents involvingmoney, and horrible surrounding incidents by the Russianpress media. Grisha quit his position at the Steklov Insti-tute. There was a kind of embarrassment—I wouldn’t say ascandal—when Shing-Tung Yau, one of the greatest livinggeometers, seemed to try to squeeze some of the credit forthe proof of the Poincare conjecture from Grisha towardtwo of his proteges—possibly for the sake of political cloutin the People’s Republic of China. Then Sylvia Nasar andDavid Gruber managed to get Grisha to spend time withthem in St. Petersburg, and published a somewhat sensa-tional article in The New Yorker. Of course Grisha refusedthe Fields Medal, refused to attend the International Con-gress of Mathematicians, and finally refused one milliondollars from the Clay Institute. In her book, Masha Gessenreports that Grisha has now broken off from his lifelongfriend and mentor Rukshin. He has told people he islooking for something new to do instead of mathematics.He continues to live in their apartment with his mother.

Masha Gessen devotes one chapter of her book to thetopic of Asperger’s Disorder, a form of autism dispropor-tionately found among mathematicians. She never actuallysuggests that Grisha Perelman suffers from Asperger’s.Whether he does or not is a medical question. But there is anissue here of good taste and good manners. People maywonder about such things and talk about them privately.Decent consideration for the feelings of the subject of herbook would have suggested abstaining from publishing sucha chapter.

Much more important is the cultural and moral questionwhich this story forces one to ask. Does today’s world haveroom for a mathematician who practices mathematics for itsown sake, and only for its own sake?

1000 Camino Rancheros

Santa Fe, NM 87505

USA

e-mail: [email protected]; [email protected]


Mythematics: Solving theTwelve Labors of Herculesby Michael Huber

PRINCETON, OXFORD: PRINCETON UNIVERSITY PRESS, 2009,

XIX + 183 PP., US $24.95, ISBN: 978-0-691-13575-5

REVIEWED BY JOHN J. WATKINS

MMichael Huber is not the first author to havebeen inspired by the enduring myth of the twelvelabors of Hercules. In 1947, Agatha Christie

published The Labors of Hercules, a novel in which herbrilliant Belgian detective Hercule Poirot decides he willease his way into retirement by solving precisely twelvefinal cases: cases selected only with reference to the‘‘twelve labors of ancient Hercules.’’ In the hands of DameChristie the foes of the mighty Hercules are wonderfullytransformed. The fearsome Nemean lion becomes forHercule Poirot a small Pekinese dog; the awe inspiringflock of Stymphalian birds becomes two ominous womenwith long curved noses dressed in cloaks walking by a lakeat a European resort; the filth of the Augean stablesbecomes instead a political scandal at the very highest levelof government, a mess which Poirot is called upon to cleanup; and appropriately in his final ‘‘labor’’ Poirot is forced, aswas Hercules, to deal with an all-too-real Cerberus guard-ing the gates of Hell.

Each chapter in Mythematics: Solving the Twelve Laborsof Hercules, by Michael Huber, is also based on one oftwelve tasks imposed upon Hercules by Eurystheus. Her-cules was born the son of the god Zeus and the mortalwoman Alcmena. From infancy, the jealous wife of Zeus,Hera, had but one goal, the destruction of Hercules, andshe almost succeeded. Hera was able to eventually driveHercules mad and he murdered his own three sons. Herculeswas thus forced into exile to serve Eurystheus and performtwelve labors. Upon the completion of these labors, hewould become immortal.

Each chapter of the book follows the same generalformat and begins with a quote from Apollodorus, the mostreliable author of ancient times who wrote about Herculesand his labors, describing the particular task assigned toHercules. Huber then uses this task as a springboard fromwhich to pose three or four mathematical problems for thereader to attempt. Next, he provides detailed solutions forthese problems and also—in passages that are by far themost entertaining sections of the book—elaborates furtheron the characters and stories from Greek mythology.

There is much to be admired in this book. MichaelHuber, who teaches mathematics at Muhlenberg College inPennsylvania, has a real passion for Greek mythology and acreative flair for making connections with a wide range ofmathematical topics. This book could be used in manyways. Its most obvious use will be as a source of livelyversions of familiar problems that can be used in fresh new

ways in courses. Or, more ambitiously, I can imagine usingthis book as the main text in an interdisciplinary course thatis co-taught by a mathematician and a classicist where thegoal is to introduce students simultaneously to the ancientGreek world and also many of the varied fields of mathe-matics. This is a course I would truly love to teach.

Hercules’ first task is to bring back the skin of theNemean lion, and he attempts to shoot the lion with anarrow. Huber uses this episode to pose a pair of routinequestions: What is the speed at which an arrow strikes thelion at a distance of 200 meters given a launch angle of20 degrees, and how long does it take the arrow to travelthe distance from the bow of Hercules to the invulnerablelion? Huber does ‘‘solve’’ this problem in that he correctlyfinds the speed at which the arrow leaves Hercules’ bow(about 200 kilometers per hour) and also the time of travel,but he never gets around to saying how fast the arrow isgoing when it strikes the lion. Of course, the answer is‘‘about 200 kilometers per hour’’ (here I would invokeconservation of energy, but one could also plug the time oftravel into the velocity function to compute this speed).Unfortunately, all Huber says on the matter is ‘‘the speed ofthe arrow remains constant in flight’’, which of course isutter nonsense. So, while this book is both entertaining andat times inspired, it does need to be used with some care.

Hercules’ third labor deals with capturing the Cerynitiandeer. Huber turns the first part of this tale into a familiarproblem in optimization. The deer, in trying to escape fromHercules, must swim across the Ladon River (which is250 meters wide) and reach shelter in a forest 1600 metersalong the shore on the other side. Of course, the deer runsfaster than she swims (8 meters per second versus 5).Where should she land in order to reach the forest asquickly as possible? The artificiality of this particular prob-lem reminded me of a similar problem I came across a fewyears ago in a new calculus book touting applicationsto biology and one ‘‘applied’’ problem involved a duckwishing to get from point A to point B as quickly as pos-sible. This mathematically inclined duck could fly at acertain speed over land but could fly faster over water dueto an often observed phenomenon whereby water birds flyextremely close to the surface of the water in order toincrease efficiency. I was also somewhat bothered inHuber’s version of this problem by his unrealistic assump-tion that the deer maintained a constant swimming speed of5 meters per second independent of the angle at which shewas swimming relative to the river’s current.

Once Hercules captures the deer (presumably by antic-ipating its landing point) he must carry the deer back toEurystheus in Mycenae. Huber asks the reader to determinethe work needed to carry the deer a distance of 80 kilome-ters given that the mass of the deer is 125 kilograms. Hecomputes the animal’s weight (a vertical force) and multi-plies this force by 80 kilometers (a horizontal distance) to geta completely meaningless answer of 98,000,000 newton-meters (this is in fact the amount of work it would take tohaul this deer to the top of a tower 80,000 meters high!).Huber makes a similar blunder about work later in the bookwhen, having just computed the mass of the earth, he asks,‘‘How much work does Hercules do in placing the earth on


his massive shoulders?’’ Leaving the reader to answer thisquestion, he quips: ‘‘No wonder Atlas was tired of holdingup the earth.’’ But, of course, it takes no work at all for Atlasto hold the earth in one place, even if its mass is 6 9

1024 kilograms.In a good problem that is typical of the sort of modeling

problems that Huber favors, Hercules has shot his friendChiron in the knee with a poison arrow. However, Herculescan administer a protective antidote at 5 minute intervals.Huber models Chiron’s immune system by p(t + 1) =

.75p(t) + .1 (that is, his immunity is breaking down continu-ously by 25% every 5 minutes but also the medicationprovides an instantaneous boost). At this point, Huber asks thequestion: How long before Chiron’s immune system fallsbelow the .5 level? (This is where he will constantly be in greatpain.)Now, sincep(0) = 1 is the starting level forhis immunity,the most natural thing for students to do is a few iterations ofthis function and they quickly discover that p(6) = .5068 andp(7) = .4801. So, just prior to 30 minutes his immunity fallsbrieflybelow .5 (before the antidote again brings it back above.5), but after 35 minutes it will remain below the .5 threshold.Note that as long as p(t)[ .4 this function will decrease.

But, instead, without ever saying what it is that hesolving, Huber says the ‘‘solution’’ is p(t) = c(.75)t + d. Ofcourse, what he intends by this solution is a continuousfunction that agrees with the original discrete function p(t)modeling Chiron’s immune system for t = 1, 2, 3, . . . . Yethe never explains this strategy nor how the form of thisparticular continuous function is arrived at. He merelysolves for c and d and checks that this function then agreeswith the values p(1) and p(2). (Of course, countless othercontinuous functions also agree with p(1) and p(2) withoutnecessarily agreeing with the other values p(3), p(4), . . . .)He also provides a nice graph purporting to representChiron’s immune system protection level, but since thisgraph exhibits no step-function behavior, it is instead agraph of his continuous approximation p(t) = .6(.75)t + .4.

Huber uses the 2,000-year-old The Greek Anthology as asource for several of his problems. Here is a nice combina-torial problem he adapts slightly to suit his needs. Herculescalls for wine and the centaur Pholus poses the followingproblem. Five centaurs have 45 jars of wine, of which 9 eachare full, three-quarters full, half-full, one-quarter full, andempty. The centaurs want to divide the wine and the jarswithout transferring the wine from jar to jar in such a waythat each centaur receives the same amount of wine and thesame number of jars, and so that each centaur also receives atleast one of each kind of jar and no two of them receivethe same number of every kind of jar. Can the wine be sodivided?

Another problem taken from The Greek Anthology isrelated to the Labor of the Augean Stables and asks us tofind how many herds of cattle Augeas, the king of Elis, had.Hercules the mighty was questioning Augeas, seeking tolearn the number of herds, and Augeas replied: ‘‘About thestreams of Alpheius, my friend, are half of them; the eighthpart pasture around the hill of Cronos, the twelfth part faraway by the precinct of Taraxippus; the twentieth part feedin holy Elis, and I left the thirtieth part in Arcadia; but hereyou see the remaining fifty herds.’’ Huber, as is common,

tends to treat these as simple problems in algebra, but Iprefer to use basic ideas about numbers that were certainlywell known at the time The Greek Anthology was written. Ifeel this is more in the spirit in which these problems wereintended. We are told that the number in question isdivisible by 2, 8, 12, 20, and 30; hence, by 8, 3, and 5.Therefore, the number of herds is either 120, or a multipleof 120. A simple check shows that 120 is not the number(since 60 + 15 + 10 + 6 + 4 + 50 = 120) but that 240 isthe number of herds (since 120 + 30 + 20 + 12 + 8 + 50 =

240). Huber also treats another problem from The GreekAnthology as an algebra problem that I suspect may wellhave originally been an Egyptian problem about unitfractions. In this problem, three Hesperides pour water intoa tank at varying rates, and he asks how long it will take thethree together to fill the tank. The rates at which the threewomen are pouring water are, respectively, 1

2 ;14 ; and 1

6 (of atank per hour) and the final answer is that it takes them 1

11

of a day (one day = 12 hours) to fill the tank; all of thesefractions are unit fractions.

Huber takes the story of how Hercules made the riverStrymon unnavigable by filling it with rocks and turns it intoa nice mathematical problem about the relative degree towhich various lattice structures fill space. He chooses tocompare three cubic lattices: simple cubic, face-centeredcubic, and body-centered cubic. In each case, Herculesstacks equal-sized spherical boulders in the appropriatelattice pattern. Then Huber compares the packing factor ineach case. This is extremely well done, and I can just imaginethe mighty Hercules sitting in the middle of the river Strymonlike a child playing in a sandbox, stacking massive spheres ofstone and creating beautiful cubic lattices.

Huber can be very inventive, not so much in terms ofcreating original problems, but by the way in which he canassociate standard mathematical problems with the Herculesmyth in quite surprising ways. Most ancient sculptures ofHercules show him holding three apples in his left hand.These are the golden apples he was required to fetch fromthe Hesperides in his eleventh labor. To get the apples,Hercules first relieves Atlas of his burden of holding up theearth and sends Atlas to fetch the apples; then, when Atlasreturns with the three apples, Hercules tricks Atlas into onceagain accepting the responsibility for holding up the earthwhile Hercules departs with the apples. This provides Huberwith a convenient jumping off point for a problem in basicphysics: What is the mass of the Earth, given the followingthree pieces of information: i) the radius of the earth is 6378kilometers; ii) the acceleration due to gravity is 9.8 metersper second squared (a critical point Huber neglects to makeis that this is valid only at the surface of the earth); iii) theuniversal gravitational constant is 6.67 9 10-11 meters perkilogram per second squared (this quantity, whose valueeven at this level of precision is still in doubt, was firstdetermined by Henry Cavendish in 1798; unfortunatelyHuber twice refers to this universal constant as ‘‘earth’sgravitational constant’’).

At the end of every fourth chapter, Huber offers readers areward for our own labors in the form of sudoku puzzlesthematically related to a task just completed by Hercules.The first such puzzle follows the Labor of the Erymanthian


Boar and replaces the numbers 1-9 with the lettersA,E,H,I,M,N,R,T,Y (and the solution contains a very pleasantsurprise). His second sudoku puzzle follows the eighthlabor, which called upon Hercules to bring the four mares ofDiomedes to Mycenae. (Agatha Christie turned these savagemares into four high-spirited young women in an Englishcountry house.) Huber presents us with a clever variation ona standard sudoku puzzle: a ‘‘knight’s puzzle’’, in which asingle number is placed in the center, then sixteen morenumbers are symmetrically arranged in groups of four, eachgroup itself forming a knight’s move in chess. The additionalcondition in this delightful puzzle is that no two squares aknight’s move apart can contain the same number. I mustconfess that I then skipped ahead to the last sudoku puzzlelong before I had completed the twelfth labor. The specialwrinkle in this puzzle was appropriately inspired by thethree heads of Cerberus, and so uses each of the numbers1–6 exactly once, but the number 7 three times in each row,column, and 3 9 3 block.

The twelfth labor imposed on Hercules was to bringCerberus from Hades. Huber characterizes Hercules’ des-cent into the Underworld using a multivariable functionf(x, y) = -2x + y2 - 2xy to represent the terrain and asksat various points along the way whether Hercules isascending or descending and in which direction he shouldtravel to descend as quickly as possible. Then, since Her-cules is required to capture Cerberus with his bare hands hedecides to strangle the three-headed beast (in another Greekmyth, Orpheus chooses a method I much prefer: he lulls himto sleep with music from his lute). Huber models Hercules’attempted strangulation by assuming there are 6 milliliters

per second of blood flowing in each of the dog’s three brainsto begin with and that, when Hercules is strangling onehead, the blood flow is reduced by 7 percent per second.This yields a differential equation db/dt = -.07b. Herculeshangs on until the blood is less than 2 milliliters and thengrabs another head to strangle, at which point of coursesome blood begins to return to the head just released. Thedramatic tension Huber creates is whether Hercules cansucceed in getting the blood flow in all three brains belowthe critical threshold of 2 milliliters per second, thus sub-duing the beast and completing his final task.

Well, perhaps not his final task after all. As Huberexplains in an appendix about the authors of the Herculesmyth, the twelve labors sometimes vary in order and detail,and even the stories may not be the same from author toauthor. A version describing the deeds of Hercules in TheGreek Anthology adds a thirteenth ‘‘labor’’ which neitherAgatha Christie nor Michael Huber found suitable to includeamong their own collections of ‘‘the labors of Hercules’’:

First, in Nemea he slew the mighty lion. Secondly, inLerna he destroyed the many-necked hydra. Thirdly,after this . . . . Twelfthly, he brought to Greece thegolden apples. in the thirteenth place he had thisterrible labour: In one night he lay with fifty maidens.

Department of Mathematics and Computer Science

Colorado College

Colorado Springs, CO 80903

USA



Life After Geniusby M. Ann Jacoby

NEW YORK/BOSTON: GRAND CENTRAL PUBLISHING, 2008, 400 PP.,

US $24.99, ISBN: 9780446199711

Monster’s Proofby Richard Lewis

NEW YORK: SIMON & SCHUSTER, 2009, 288 PP., US $15.99,

ISBN-10: 1416935916, ISBN-13: 9781416935919

REVIEWED BY ALEX KASMAN

AAbrief plot summary of M. Ann Jacoby’s Life AfterGenius may sound quite familiar: A young manescapes from a small town and the family business

but then returns to both when he fails to achieve his dreamsof success in the big city. However, there are a few unusualtwists in the story of Theodore Mead Fegley. For instance,the Fegley family business is a combination of a furniturestore and a funeral home, with many aspects of the latterdescribed in gory detail. It is also unusual that, rather thanbeing a businessman or athlete, the protagonist of this bookis a prodigy who will be graduating from a major universityat the age of 18 and presenting his research on the RiemannHypothesis to an audience of prestigious mathematicians.

Another unusual feature of this book is that the reader isfocused on understanding past events rather than on seeingwhat will happen in the future. Since the chapters arepresented out of chronological order, we know from thestart that Mead quit school just days before his graduationand research presentation, but not why he would throwaway years of work and tuition in this way. Similarly, formuch of the book we know of the death of his cousin, butnot how he died nor why Mead feels responsible. Eventhough there is no murder and no detective, these puzzlesgive the book the feeling of a mystery novel.

There is not much more I can say about these plots andsubplots without damaging the effectiveness of this cleverliterary technique. Instead, we will be concerned here withanalyzing what this book says about mathematics andabout the nature of genius.

The Riemann Hypothesis is currently the most famousopen problem in mathematics. With any even, negativeinteger as input, the zeta function outputs the value zero,and so these input values are ‘‘roots’’ of the function. It wasconjectured by the great Bernhard Riemann that, aside fromthese, all of the roots of the analytic continuation of the zetafunction are complex numbers with real part equal to 1/2.Since the zeta function can be written as an infinite productinvolving the prime integers, a proof of the conjecturewould have implications for the distribution of primes. Onthe other hand, finding even one root that does not havethe conjectured property would disprove it. And so, muchwork has gone into finding the roots of zeta. As Jacoby

mentions in the novel, Alan Turing was one person whoworked on automating their computation, finding morethan a thousand of them. By the late 1960s, the number ofcomputed roots reached into the millions, and by now thenumber computed is much higher even than that. Needlessto say, all of the roots found so far support the conjecture.Aside from this ‘‘statistical evidence,’’ there are some nota-ble theoretical results relating to the Riemann Hypothesis.Hugh Montgomery studied the distribution of zeta rootsin the 1970s, and physicist Freeman Dyson pointed outan unexpected connection between Montgomery’s workand distributions of energy levels in mathematical physics.More recently, Michael Berry built upon this connection tophysics by showing a more specific relationship betweenthe distribution of zeta roots and the transition to chaoticdynamics.

Rather than creating completely fictional mathematicalresults for Life After Genius (a task which would requiregreat imagination as well as expertise in mathematics), theauthor ascribes some of these real mathematical resultsto Mead Fegley. In particular, he is shown traveling toPrinceton to use a supercomputer where, supposedly, he isthe first person to compute a large number of zeta functionroots. Later, by chance, he stumbles upon a physics papercontaining familiar formulas and recognizes the connectionbetween the Riemann Hypothesis and chaos. Among thepeople who are said to be attending the planned presen-tation of his discoveries are Hugh Montgomery and MichaelBerry. Perhaps the inclusion of the names of these peoplewho really contributed to the field is a sort of apology/acknowledgement from the author. (This would explainthe anachronism of including Berry’s name among theexperts on the Riemann Hypothesis, since the book takesplace around the year 1980 and Berry had not yet pub-lished any work on this topic.)

The first few times mathematics was discussed in LifeAfter Genius, it was done so smoothly I was certain that theauthor had advanced mathematical training herself. How-ever, the discussion of infinite series between Mead and hisfirst major advisor at college convinced me otherwise. Otherrevealing errors include the suggestion that Number Theory(a very large and ancient branch of mathematics) grew outof the Riemann Hypothesis and is nothing other than theattempt to prove it, reference to the ‘‘function plane,’’ andthe common mistake of describing the Riemann Hypothesisas ‘‘an equation’’ which needs to be solved. Also, the authorrefers several times to the periodicals that Mead consults inpreparation for his presentation: The Mathematical Intelli-gencer and the American Mathematical Monthly. I hopeI do not offend these fine publications, when I claim thatthey are not the right sources to utilize in preparing aresearch talk on the Riemann Hypothesis to an audience ofexperts on that subject!

But why am I being so critical? Overall, Jacoby demon-strates a reasonable understanding of the basic concepts.She recognizes that the key point is to determine the loca-tion of the roots of a certain function, and that a singlecounterexample would disprove the conjecture while areasoned argument would be required to prove it. Dwellingon the mathematical errors, irrelevant to the plot or literary


quality of the work, is precisely what Mead Fegley woulddo! Mead is especially insecure, competitive, confronta-tional, and cynical. When dealing with a kind offer fromothers, he is immediately suspicious of their motives. Whenother people make a mistake in logic or grammar, he is sureto point it out, as if their failure improves his own stature incomparison.

The book does implicitly offer some explanation forthese behaviors. The other children in his town are oftenparticularly mean to him. His mother is overbearing. Hecannot hope to share in the popularity of his charismatic andathletic cousin. Because he has skipped years in school, heis always younger than his classmates and misses outon some socialization. Of course, simply being rude andunpleasant alone are not enough to qualify a person as a‘‘genius.’’ The title and book jacket certainly suggest that thisbook is a character study of a genius; this is the word peoplein Mead’s hometown use to describe him. What does thismean?

It is interesting to note that Mead never does anythingparticularly brilliant in Life After Genius. Starting college atage 15 and graduating at 18 is impressive, but here itappears to be a consequence of the fact that his socialfailure left him more time for school work. Using a formulagiven to him by his professor to compute roots of the zetafunction by hand or with the aid of a computer demon-strate hard work, but require no insight. Stumbling upon amisshelved physics paper that makes use of a distributionfunction he recognizes from a different context is pureluck. Is this a deliberate statement on the part of the author,arguing against the idea that a genius is a person whosemind functions in a fundamentally different way, or doesshe not realize how unimpressive Mead’s achievementsseem? Mead does make fun of the concept of ‘‘genius’’when he teases the naive people in his small town, tellingthem that geniuses have a body temperature one degreelower than that of ‘‘ordinary people.’’ In fact, aside frombeing rather anti-social and having a lot of time to devote tohis studies, Mead seems essentially to be an ordinaryperson.

If Jacoby does indeed intend to offer Mead as a coun-terexample to the common view that geniuses are sodifferent from ordinary people as to almost be a differentspecies, then I would certainly be sympathetic. I believethat exaggerated anecdotes about geniuses, and a bias forpeople with notable ‘‘quirks’’ to be described as geniuseswhen seemingly ordinary people with equally impressiveintellectual qualifications are not, result in a somewhatunrealistic image of what the word really means.

However, there is one thing that leads me to suspect thatthe reader is supposed to view Mead as being a ‘‘truegenius.’’ Like John Nash in the film A Beautiful Mind andCatherine in the Pulitzer Prize-winning play Proof, Meadhas discussions with people who are not really there andcannot easily distinguish when this is taking place. (Heseems to realize that his discussions with Bernhard Rie-mann are figments of his imagination, but is less sure aboutsimilar conversations with classmates, professors, and rel-atives.) So far as I know, no real famous mathematician hashad this problem. (Even John Nash, who truly does suffer

from schizophrenia, did not have delusions of such con-versations. This was a clever and effective plot deviceinvented specifically for the film.) I have certainly neverheard anyone outside of this book suggest that AlbertEinstein, the twentieth century’s canonical genius, had suchdelusions. Yet, one of the men Mead meets in Princetontells him that as a young boy delivering newspapers, hesaw Einstein having a conversation with someone who wasnot really there. Since she gives both Mead and AlbertEinstein this trait shared by other fictional mathematicalgeniuses, Jacoby probably does intend readers to put Meadin this category.

In the book, there are professors and other studentswho try to benefit from Mead’s discoveries even thoughthey did not really contribute to the work. Probably, we aresupposed to see them as not merely unscrupulous, but alsoas being unable to have done what Mead did on their ownsince they are not geniuses like him. If so, her inability toconvey Mead’s exceptional mathematical brilliance is a flawin an otherwise enjoyable and thought-provoking book.

Like Life After Genius, Monster’s Proof by Richard Lewisfeatures a mathematical prodigy who is teased and abusedby his classmates, as well as detailed discussions of theRiemann Hypothesis. Yet, despite these similarities, thebooks could hardly be more different. In contrast to therealism and adult themes in Life After Genius, Monster’sProof is a book for young adults making use of elements ofboth fantasy and science fiction.

The general plot outline here is also likely to soundfamiliar, this time in the Frankenstein tradition. By the veryact of proving a conjecture, young prodigy Darby Ell releasesa powerful being of pure mathematics on an unsuspectingworld. This entity, who goes by the name of ‘‘Bob’’ and sayshe comes from Hilbert Space, seems benevolent at first but issoon recognized as being a significant danger. The source oftension here is not what has happened but whether Darby(with help from an angel, a demon, and his grandmother)will be able to save the world from his own creation.

Unlike Mead Fegley, Darby is a trusting and kind boy. IfMead’s upbringing is supposed to be responsible for hisantisocial tendencies, then perhaps Darby’s more pleasantdemeanor is a reasonable consequence of one major dif-ference. Although Darby is also subjected to humiliation byhis classmates and also faces competition from unprinci-pled academic competitors, Darby is only one of a family ofgeniuses. His grandmother, father, and mother are also allbrilliant scientists and mathematicians. (Only his teenagesister is ‘‘normal.’’) Consequently, he would have emotionalsupport that Mead lacked.

While working for the United States government’s topsecret nuclear weapons program, Darby’s grandmothermade a conjecture about an unusual Hilbert Space, whichshe called the ‘‘thingamabob’’ conjecture because she didnot quite understand what it was. Recognizing the potentialrisks that it entailed, she left it unproved. But, at age 10,Darby Ell is able to prove this conjecture that had stymiedteams of top mathematicians at the National SecurityAgency. This demonstrates more than just academic excel-lence and hard work. Even though he had some help from‘‘Bob’’ (who recognized and encouraged the boy’s brilliance


in the hopes that he could prove the conjecture and releasehim in this universe), Darby Ell seems to better capture whatI would mean by ‘‘genius’’ than does Mead Fegley.

A much broader range of mathematical topics are tou-ched upon in Monster’s Proof than in Life After Genius. Inaddition to the Riemann Hypothesis, we see a continuedfraction expansion of 4/p, the Mandelbrot set, a Pythago-rean cult, operator theory, a classroom discussion of whythe product of negative numbers is positive (which theteacher handles very poorly), and many popular mathe-matical jokes and anecdotes. I believe this reflects not justthe greater freedom of the genre, but also the author’sgreater familiarity with the subject. Aside from odddescriptions of the Thingamabob Conjecture itself, whichcan be forgiven as being entertainingly cute even if math-ematically nonsensical, all of the mathematics in Monster’sProof is essentially correct.

For a cheerleader who is failing algebra, Darby’s ‘‘ordin-ary’’ sister gives a surprisingly nice summary of the RiemannHypothesis: ‘‘The Hypothesis was this incredibly excitingidea that all the zeros of something called the zeta functionwere on a straight line. Well, excuse me, she thought, thenontrivial zeros.’’ In a subplot, ‘‘Bob’’ and Darby’s fatherwork together on the Riemann Hypothesis. When theyeventually disprove it, the father seems to be crushed. Thismay be based on the common misconception that it wouldsomehow be a horrible thing for mathematics if thehypothesis were false. In any case, ‘‘Bob’s’’ reaction is to saythat within mathematics, beauty is truth, and so if they haveshown that there is a nontrivial root of the zeta function off ofthe critical line, then this is a beautiful thing and should beappreciated as such.

The more important mathematical subplot in Monster’sProof is the Thingamabob Conjecture itself, which involves‘‘the Hilbert Space of all Hilbert Spaces’’. This is a cute idea,

reminiscent of the notion of ‘‘the set of all sets,’’ which seemsreasonable until one considers Russell’s Paradox. However,a Hilbert Space is required to have algebraic and topologicalproperties that the set of all Hilbert Spaces is not likely tohave. In a fantasy/science fiction story like this, such tech-nical details should not be a problem, and once one ignoresthis concern (and other similarly cute definitions, such as theoperator on the Hilbert Space whose action is defined inanalogy to the behavior of sharks), the mathematical workshown in this book is probably more realistic than that inLife After Genius. In particular, the mechanics of workingtowards a proof used by the characters here, not by rotecomputation or finding formulas in previously publishedpapers, but by building up a sequence of logical arguments,is what research mathematicians really do.

Here we have considered two relatively recent additionsto the library of ‘‘mathematical fiction.’’ That the author ofLife After Genius lacks the mathematical background toget all of the technical details correctly and misses anopportunity to explore what is ‘‘genius level’’ mathematicalresearch does not detract from the book’s thought-pro-voking analysis of the affect of the environment on thepersonality of a young mathematician (and vice versa).Monster’s Proof, which devotes a lot of pages to theromance between a cheerleader and a demon, does notaddress such sober topics, but it is a fun book and actuallydoes a better job of conveying mathematical ideas to thereader. I recommend both books.


College of Charleston

Robert Scott Small Building, Room 339

Charleston, SC 29424

USA



Stamp Corner Robin Wilson

RecentMathematicalStamps: 2005

Avicenna (980–1037)Avicenna, also known as ibn Sinah, was the most celebratedof Persian philosopher-scientists, best known for his trea-tises on medicine. He contributed to arithmetic and numbertheory, produced a celebrated Arabic summary of Euclid’sElements, and applied his mathematical knowledge toproblems from physics and astronomy.

Albert Einstein (1879–1955)In 1905, Einstein published his ‘special theory of relativity’,asserting that the basic laws of motion (including Maxwell’sequations) are the same for all observers in uniform motionrelative to one another. He thereby extended Newton’s ideason mechanics to include electromagnetism and Maxwell’sresults. A consequence is that mass is a form of energy, andthat the energy E and mass m are related by the well-knownequation E = mc2, where c is the speed of light.

GAMM 2005In 2005 the Gesellschaft fur Angewandte Mathematik undMechanik (Society of Applied Mathematics and Mechanics)organized the 76th International Congress of Applied

Mathematics and Mechanics in Luxembourg. The com-memorative stamp illustrates the calculation of the airstreamof a turbine in a hydroelectric power station.

Josiah Willard Gibbs (1839–1903)Gibbs was an American physicist and mathematician whospent his working life as professor of mathematical physics atYale. In mathematics he combined Grassmann’s ideas onexterior algebra with Hamilton’s quaternions, applying hisconclusions in vector analysis to areas of mathematicalphysics. He also contributed to statistical mechanics, helpingto provide a mathematical framework for quantum theory.

Edmond Halley (1656–1742)While still an Oxford University student, Halley sailed toSt Helena to prepare the first accurate catalogue of the starsin the southern sky. In 1684 he persuaded Isaac Newton topublish his ideas on gravitation in the Principia Mathem-atica. In 1704 Halley became professor of geometry atOxford, where he prepared a definitive edition of Apollo-nius’s Conics. He is primarily remembered for the comet,named after him, whose return he predicted.

William Rowan Hamilton (1805–1865)Hamilton was a child prodigy who mastered several lan-guages at an early age, discovered an error in Laplace’streatise on celestial mechanics while still a teenager, andbecame Astronomer Royal of Ireland while an undergradu-ate. He made several important advances in mechanics, anddiscovered the noncommutative ‘quaternions’ of the form.

aþ bi þ cj þ dk; where i2 ¼ j2 ¼ k2 ¼ ijk ¼ �1:

Halley

Hamilton� Please send all submissions to the Stamp Corner Editor,

Robin Wilson, Faculty of Mathematics,

Computing and Technology,

The Open University, Milton Keynes, MK7 6AA, England


Avicenna

Einstein GAMM 2005

Gibbs


Documents

The Mathematical Intelligencer volume 32 issue 4