Optimization and the Geometry of Numbers: Packing and Covering

Optimization and the Geometry of Numbers: Packing and CoveringAuthor(s): Thomas L. Saaty and Joyce M. AlexanderSource: SIAM Review, Vol. 17, No. 3 (Jul., 1975), pp. 475-519Published by: Society for Industrial and Applied MathematicsStable URL: http://www.jstor.org/stable/2028887 .

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SIAM REVIFW Vol. 17. No. 3, July 1975

OPTIMIZATION AND THE GEOMETRY OF NUMBERS: PACKING AND COVERING*

THOMAS L. SAATYt AND JOYCE M. ALEXANDER:

Abstract. There are a number of useful applications gradually arising out of the geometry of numbers. In this paper we give a brief survey of the fields of packing and covering, stating some results, and giving illustrations of methods of proof and approaches to the subject. We then mention and discuss a variety of applications, summarizing results and occasionally providing proofs. Our hope is to focus attention on optimization problems not in the main stream of current optimization theories.

1. Introduction. There is an area of mathematics with a large number of potential applications which somehow has gone essentially unnoticed within the structure of optimization theories. This is the field of geometric number theory, and, more specifically, its branches of packing (a set of bodies within a larger body) and covering (a body with other bodies). It is interesting to note that we shall have to call on the subject more and more often as the population of the world increases, crowding available space, and as technology looks for more efficient ways to save costs, thereby requiring better methods of packing and covering. This along with geometric number theory and discrete geometry, which are also inevitably touched on in this paper, are topics of mathematics that are in considerable use in crystallog- raphy and in related areas of physics and chemistry, but their utilization in other fields, particularly in operations research, has been negligible. It is helpful at the start to note that the results of these fields are usually attempts to answer questions of ''how many" objects may be grouped together to satisfy a stated problem and "how" in fact the construction may be carried out geometrically.

The purpose of this paper is a relatively modest one. It is to introduce the subject of packing and covering by giving examples of useful applications, to state some results and, as far as is possible, also to give elementary methods of proof. We hope that we can stimulate greater interest in the subject even though the scope of the paper does not permit full coverage of all important known results and applications. We also note that the subject itself is not sufficiently developed to permit a useful look at all its ramifications. Nevertheless, its techniques are non- trivial and as usual tend to build upon one another, essentially leading down narrow but deep paths.

There are many mathematicians who have made significant contributions to this field. Reluctantly (because of the sin of omission) and as a matter of interest, we mention a few to familiarize the reader with this emergent subject and hope that we may be forgiven for not being exhaustive. Here is a sample of twenty-two contributors: R. P. Bambah (Chandigarh, India), E. S. Barnes (Adelaide, Australia), H. F. Blickfeldt' (United States, 1873-1945), J. W. S. Cassels (Cambridge, England), H. S. M. Coxeter (Toronto, Canada), H. Davenport' (Cambridge, England), L. Few (London, England), L. Fejes Toth (Budapest, Hungary), A. Florian

* Received by the editors May 17, 1973, and in revised form October 25, 1974. t University of Pennsylvania, Wharton School, Vance Hall, Philadelphia, Pennsylvania 19174. 1 Immaculata College, Immaculata, Pennsylvania, 19345. ' Deceased.

475



476 THOMAS L. SAATY AND JOYCE M. ALEXANDER

(Salzburg, Austria), C. F. Gauss' (Gottingen, Germany, 1777-1855), E. Hlawka (Vienna, Austria), C. G. Lekkerkerker (Netherlands), M. Goldberg (Washington, D.C., United States), A. Heppes (Budapest, Hungary), J. L. Lagrange' (Paris, France, 1736-1813), H. Minkowski' (Gottingen, Germany, 1864-1909), J. Moln'ar (Budapest, Hungary), L. J. Mordell' (England), C. A. Rogers (London, England), C. L. Siegel (Gottingen, Germany), H. Steinhaus' (Wroclaw, Poland, 1887-1972) and A. Thue' (Sweden, 1863-1922).

David Gregory, an Oxford astronomer, recorded as early as 1694 Newton's conjecture that in a densest packing in 3-space, each sphere touches no more than 12 others [12].

In 1773, Lagrange showed that the density of the closest lattice packing of circles in the plane is 7r/(2,/3) = 0.9069... . In 1892, Thue gave the first mathematical proof that the densest packing of equal discs in the plane is obtained by arranging them in a hexagonal pattern with each disc touching six others.

Harris Hancock, in the introduction to his two-volume series Development of the Minkowski Geometry of Numbers, says, "In every subject that occupies the human mind, be it history, philosophy, law, medicine, science, music, etc., there arise outstanding men who evince an innate genius in their special fields, an innateness that seems as if it were of divine origin. Minkowski was one of the great mathematicians of all time. His grasp of geometrical concepts seemed almost superhuman... Minkowski came to his theorems through special intuitions." In a sense, then, the subject owes Minkowski the credit for its formal birth.

Among the known elementary references which include a discussion of packing and covering are the books by D. Hilbert and S. Cohn-Vossen [60], H. S. M. Coxeter [11], [14], H. Steinhaus [93], H. Meschkowski [73], A. Kitai- gorodsky [65], and W. R. Ball [1]. Several scholarly books given to the subject are those by H. Minkowski [74], H. Hancock [57], L. Fejes Toth [25], [27], J. W. S. Cassels [8], C. A. Rogers [80] and C. G. Lekkerkerker [71].

In order to introduce the formalism of the subject of packing and covering, we need to give a few (minimal) definitions. We shall then state some problems. In the following sections, we give a brief summary of some interesting results, followed by examples of elementary proofs. Finally, we go on to illustrate with some applications.

2. Definitions and examples. 2.1. Definitions [60], [80]. DEFINITION 1. A collection of sets {Si4, i = 1, 2, ,is said to form a packing

into the setSif(i)si n si = 0 for all i and j (i # j), and (ii) U Si c S. Thus no two sets have common elements and each element of each Si belongs to S.

DEFINITION 2. A collection of sets {Si}, i = 1, 2 ... , is said to cover the set S if U Si a S. Thus each element of S belongs to at least one Si.

These general definitions will only be used here in the context of Euclidean spaces.

A third concept closely related to packing and covering is tiling, given here for a Euclidean space.

DEFINITION 3. Tiling is a complete, simple covering (i.e., no multiple coverings occur) of a Euclidean space by congruent polytopes.



OPTIMIZATION AND GEOMETRY OF NUMBERS 477

The possible tiling patterns for a complete covering of the plane are not known. Examples are tiling with an ordinary triangular tile or with any quadrilateral. For higher dimensions the tiling problem is even more difficult.

DEFINITION 4. A convex body K in n-dimensional real space is a bounded, convex set of points (confining attention to convex bodies is convenient and excludes difficulties associated with arbitrary sets of points).

DEFINITION 5. For any point x of the space, the set K + x consists of points k + x, where k ranges over the points of K. It is the body obtained from K by translating it by the vector x. We say that the bodies K + x1, *- -, K + x, are packed if no two of them have a common interior point, i.e., if every point of the space belongs to either the interior of at most one of the bodies or to a point on the boundary of one or more of the bodies.

DEFINITION 6. A covering of a space by a convex body K is obtained by requiring that every point of the space should belong to at least one of the N bodies K + x1, --, K + XN.

DEFINITION 7. A general lattice in n-dimensional real space is the set of all points whose coordinates are given by n linear forms with real coefficients in n variables, where the variables can assume only integer values. Lattices play an essential role in packing and covering. For example, in the special case of n = 2, for the closest packing of equal circles in the plane, the centers of the circles form a regular hexagonal lattice whose points have the coordinates x = 2u + v, y = 3/ v, where u and v assume all integral values.

DEFINITION 8. The square lattice in the plane is the set of all points whose cartesian coordinates are integers.

The square lattice can be generated by translations of a square or a particular parallelogram using no points in its interior other than its vertices. The parallelogram must have the same area as the square.

DEFINITION 9. A unit lattice is a lattice which can be constructed from an arbitrary parallelogram of unit area.

A unit lattice can also be constructed from equilateral triangles. A pair of such triangles forms a generating parallelogram. It can be shown that the area of a large region in the plane is approximately equal to the number of lattice points multiplied by the area of the generating parallelogram and that for a given minimum distance between a pair of lattice points, the equilateral triangle lattice has the smallest generating parallelogram (for proof, see below). It follows from the preceding that such a lattice has the maximum number of points for a given (large) region.

The concept of lattice, also known as a lattice-point system, generalizes to a regular-point system.

DEFINITION 10. A regular-point system is an infinite collection of points, such that the number in any subset of the points contained in a circle or sphere goes to infinity as the radius goes to infinity; any finite region contains a finite number of points and each point of the system has the same position relative to the remaining points as any other point of the system.

It turns out that the only regular-point systems besides lattices are those corresponding to several lattices interlocking in parallel positions as in a diamond.

DEFINITION 1 1. The density of the closest packing i(K) for a convex body K is defined as follows (we start with a cube): Consider the open unit n-cube




C:0 < xi < 1, i = 1, , n, and the magnified cube AC :0 < xi < A. Referring to Definition 5, if all N bodies are contained in AC and if N is maximal with N = N(i), then

6(K) = lrm N(-A)V(K) -o n

obtained from the proportion of the cube A occupied by the bodies, where V(k) is the volume of K.

This limit can be shown to exist and satisfies 0 < 6(K) < 1. One can also show that the same limit is obtained if the cube is replaced by any convex body.

If, in the definition of closest packing, the translation vectors form a lattice, we obtain 5L(K), the density of closest lattice packing for K. Obviously

0 < 6L(K) < 6(K) < 1.

The density of the thinnest covering is denoted by 0(K), and that of the thinnest lattice covering by OL(K). If all N bodies cover AC, and if N is minimal with N = N(F), then

0(K) = lim N( nV(K) -o

n

and as above, if the translation vectors form a lattice, we obtain OL(K), the thinnest lattice covering. We have 1 < 0(K) < OL(K).

The idea of closest packing is so important that we illustrate it with an example that is intuitively appealing. For simplicity, we consider closest lattice packing in which we take unit circles in a square of side A.

Then N(F) V(K) N(L)n

in >2

We wish to evaluate N(F). We assume that the circles are packed so that one row touches a side of the square. The maximum number of circles in any row is A/2. The distance between rows is 3/, and thus the maximum number of rows is

-2)/V/3] + 1. Hence,

N(i) < 2[ -2)/,/] + 1I}

and

N(A)V(K) < (/2) [(A" - 2)/ 3] + 1}r A2 = ,2

Since the bound is approached as i becomes large,

Nl()V(K) = (/2){[( -2)/,/3] + 1}It lim.. = lim A i2 A-o i

-2,/3




THE FUNDAMENTAL PROBLEM OF THE GEOMETRY OF NUMBERS (Minkowski). The problem of closest lattice packing for K is equivalent to: Find the lattice with a maximum number of points which has no other point than the origin 0 in the given body 2K.

Proof. Suppose K is symmetric with respect to 0. A lattice provides a packing for K if and only if k + x = k' + x', where k and k' are any points in K, and x and x' are distinct points of the lattice. But this requirement is equivalent to the condition that the lattice has no point other than 0 in the body 2K, since k - k' is a point of 2K.

2.2. Number theory and sphere packing. Minkowski, in his collected works, attributes to the research of Dirichlet and Hermite on quadratic forms the starting point for his development of the theory of sphere packing and its connection with quadratic forms. It is known [94] that the minimum M(f) of a quadratic form f (x1 , xj) for a general choice of xi, i = 1, , n, with real coefficients has the crude estimate

4-[n(n- 1)2/2] Mn(f) I1 always, D(f) = 3n(n 1)/2 if f is not a zero form,

where D(f) is the determinant of the coefficients off. This bound is sharp for a particular class of quadratic forms defined below and known as extreme forms. The ratio on the left plays an important role in linking quadratic forms to sphere packing when the variables xi are restricted to integer values. Thus, for this case, we look for a new bound on the ratio on the left. In this procedure we assume that the forms involved have been reduced to a standard form for whose corresponding ratio we find a bound.

We now consider a positive definite quadratic form f which has been reduced to its Hermitian form, i.e., is such that the minimum is given by M(f) = a,1 and has been arranged as follows:

O < al _ a22_* < an

Let this Hermitian form f be further transformed to the form

f = Y 2 + Y 2 + + y 2

where the y's are linear forms in the x's. Let YI, Y2,' . , Yn be rectangular coordinates in Rn. Then f ? 1 is a sphere of radius 1 in Rn, whose volume Vn is given by

7rn/2 Vn =)'

and the volume of any sphere of radius r is r n. Let yi = bilxl + bi2X2 + + binXn i = 1, 2, , n. Then x1 = Ml,

x2 = M2, ...Xn = mn, where the m's are arbitrary integers, will generate a lattice in Rn. Note that since the m's are integers, the smallest possible change in xj is 1, leading to a corresponding distance between lattice points

(b1? + b22 + . . . 1/2




Since a11 = b 1, a22 = b22, , an = , and a1,

- 2 , the distance between lattice points is at least 11f = .M(f

The parallellopipeds

Mj- < X. < M. + 2,j = 1,2, ,n,

will fill the space Rn completely with no overlapping. The volume of each parallelopiped is the determinant IBI, B = {bjjl, and since the determinant of f is D(f) = IBBTI, the volume of the parallelopiped is D(f). (Here, BT is the transpose of B.)

Now we consider the following spheres about the individual lattice points in the space y2 + y ?2 + + y2 < r2 = M(f)/4.

Since the distance between the lattice points cannot be less than /Mf)/2, these spheres cannot overlap. Hence, the volume of each sphere is less than the volume of the enclosing parallelopiped. Thus

V( / }( <n

from which we have

M(f) 4 [Df]lln< ( V2)lln .

Now it can be shown that [57] D(f) ? )la 1a22 a ... (where An is independent of aij)

> )n(11 = in(allI

> [M(f)]n (since a11 = M(f), by reduction).

It follows that

M(f) []lln

[D(f)]Il n=l

A positive quadratic form in n variables is called an extreme form if, with infinitesimal variations of the form, the ratio M(f)/[D(.f)]IIn never increases. Thus a positive form f is extreme if by any infinitesimal variation of the form which leaves M(f) unchanged, D(f) never decreases.

Note, however, that if, in our consideration of the geometric part of the problem, we vary the coefficients aij off over a region in such a way as to leave M(f) unchanged, the spheres do not change in size and are still nonoverlapping, although the lattice of the centers is now different; hence the inequality M(f )/[D f )] lln < 4/( Vn2)1n still holds.

Thus consideration of extreme forms is essentially equivalent to the densest lattice packing of equal spheres.

2.3. Examples. 1. The parking problem. Given a relatively large piece of land to be used as a

parking lot, what is a good doctrine for allocating spaces to cars so that it has a




maximum number of parking spaces and cars can be manueuvered in and out by their own drivers without moving or touching other cars? (See [7], [21], [59].)

2. The orange packing problem. A citrus fruit company which exports its fruits in crates by ship wishes to minimize the number of crates of uniform size and shape and thereby maximize the use of limited ship space. The crates satisfy certain restrictions on their dimensions which normally arise from a weight, volume or size condition because of the opening through which they must pass. How best should they pack oranges (congruent spheres) in crates, and what is the optimum size of a crate?

3. The real estate problem. Given a large area of land and n congruent rectangular houses placed in it (each of sides a and b), how does one arrange the houses so that the distance between any two of them is maximized? Conversely, given an optimum desired distance d between houses, what is the maximum number of houses which can be packed in the area? (See [30].) This problem has been solved completely by G. Fejes Toth [22] for n = 10, 14 and is commented on by L. Fejes Toth in [33]. J. Moln'ar [75], [76] also considered placing houses as efficiently as possible so that each touches a circular park or heliport, obtaining a variety of interesting arrangements.

4. The fuel depot problem. On any spherical planet, it is desired to locate n = 9, 10, 11, . , fuel depots so that the distance of any point on the surface from the nearest depot is as small as possible. The problem of minimizing distances from railway stations to airports also has this form, as has the satellite tracking problem in which it is desired to cover the surface of the earth by n spherical caps (satellite tracking stations) [4], [66].

5. An economical distribution of points problem. Distribute n points on the sphere so that the least distance between any two of them attains its maximum. (This problem has been studied by L. Fejes Toth in the elliptic plane [25], [31]).

6. The mine problem (which is related to Problem 5) [88]. What is the smallest radius r of a sphere (a mine) on which 10 (or in general n) points (horns) can be placed at a minimum distance of 1 unit apart? This problem is sufficiently difficult that people have been willing to settle for an answer restricted to a few points. For example, for n = 24, the problem was recently solved by R. M. Robinson. There are also a number of good conjectured solutions for other values of n in the papers by M. Goldberg [46]-[48].

7. The packing of cans problem. How should cylindrical cans be packed most effectively in boxes of given shapes? (See [67].)

8. The bandaid problem. What is the smallest plane figure (e.g., a bandaid) with which every cluster of points (a wound) of unit diameter (greatest distance between two points) can be covered? The object may be to minimize waste by making one or several standard items. This problem in its most general form is essentially the unsolved problem of Lebesgue: What is the minimum area which a universal cover in R2 can have? (A compact convex set K c Rn is a universal cover if any Q c Rn having diameter ? 1 can be covered by a congruent copy of K.) Particular cases of this problem have been considered by Chakerian [9] and Grunbaum [55].

9. The blood particle problem. Is there an optimum number of blood particles of a given size, floating in a fluid pumped at a certain rate, in order to reach and




maintain healthy tissue? If this number is correctly calculated, it may turn out to coincide with the statistically observed answer. If they are crowded too much, they would not move easily and would be slow to get to the tissue. If they are separated too far apart, there would not be enough of them to do the job.

10. The chessboard packing problem. How many kings, queens, bishops, rooks or knights may be packed on an ordinary chessboard without any pair challenging each other? Solve the same problem on an m x m chessboard [87].

11. The circus car or phone booth problem. What is the maximum number n of people who could be packed, unharmed, in a circus car or phone booth? This interprets into a problem of packing circles and other ovals into a circle and has been solved for n < 11 by U. Pirl [78].

12. The tower of Babel problem. What is the maximum number of speakers, with the same level speaking voice, who can all talk in a hall at the same time such that no person in the room can clearly hear simultaneously more than two? This is related to the illumination problem discussed later in the paper.

13. The orchard tree problem. How should n trees in an orchard be planted so that the distance between neighboring trees be as great as possible? (See [45].) This problem has been solved for n = 8 for a square orchard, in a paper by A. Meir and J. Shaer [72]. They mention availability of results for n up to 9. For n = 8, their result for the minimum distance is (2 - 3/j)1/2.

14. In crystal structures, ions of the compound are packed into the crystal with some degree of efficiency. The packing of different ions together in a crystal may be regarded as a packing of unequal spheres. Some of the physical properties of the compound are determined by the nature of the packing; for example, there may be obvious lines along which breaks may occur. What is the underlying theory? There has been a good deal of work recently done on the packing of unequal spheres, in particular, that of A. Florian [41], [42] and the joint work of J. Molnair and L. Fejes Toth [35].

15. In the construction of their hives, the bees attempt to use a minimum amount of wax to produce cells of given volume. However, it turns out that they could do better. How? (See [26].)

16. The factory location problem. Given n producers, each of whom supplies a commodity to those demand points (in a uniformly populated domain) that are nearer to it than to any other producer, how should the producers be located to minimize the total cost of transportation given that the transportation cost is an increasing function of the distance. G. Fejes Toth [23] has supplied the result which proves a standing conjecture that the best arrangement is given by the vertices of an equilateral triangular lattice.

3. Some interesting results. 3.1. Circles and spheres-no constraints. Small cables are frequently packed

inside a larger cable which contains them, cans are packed in boxes and regular cylindrical objects are packed inside a container of regular constant cross-section. All of these may be regarded as two-dimensional packing problems.

Let us consider the packing of equal circles in the plane. Two obvious configurations spring to mind. The first is to surround the circle with a square which




is tangent to the circle, i.e., the length of the side of the square is equal to the diameter of the circle. These squares may be fitted together to form a lattice which covers the plane (Fig. 1).

FIG. 1

It is obvious here, however, that there is a considerable amount of space wasted. If the density of packing is defined as the ratio of the area covered by circles to the area of the entire plane, then this ratio is the same as the ratio of the area of the circle to the area of the containing square, which is

7ir2 71

(2r)2 = 1 = .785

However, a second configuration is possible, for one may take one row of circles and "slide" them down so as to utilize some of the indentations between the circles. This leads to a packing as in Fig. 2. Here each circle is touched by six others, and from symmetry, each circle is enclosed in a regular hexagon whose sides are tangent to the circle.

FIG. 2

A regular packing of circles may be obtained by drawing a circle with center at each lattice point and radius equal to half the minimum distance between a lattice pair of points. Clearly there is no overlap of circles. A packing of circles is said to be closer than another if it has more circles than the other. It would appear that the equilateral triangle packing has the closest packing of circles.

The minimum distance between two lattice points of a unit lattice can be arbitrarily small as in a lattice generated by a rectangle of sides c and 1/c. However, in order to have a unit lattice, c cannot be arbitrarily large. The bound on c is obtained as follows: consider a unit lattice and any pair of its points at a minimum distance c. The line through these two points has an infinite number of points




spaced by intervals of length c. Parallel to this line at a distance 1/c is another line with the same property. The strip between the two lines cannot have lattice points because we have a unit lattice. If each lattice point of one line is used as center of a circle of radius c, every interior point of the strip (bounded by circular arcs) generated by the line and the new line passing through the intersection points of the circles is less than the distance c from the same lattice point and hence cannot itself be a lattice point. Thus 1/c ? the distance between the boundary of the strip and the line itself. This distance is the altitude of an equilateral triangle of side c. Thus 1/c _ 3/i c/2, from which the desired bound on c is given by c ? (2/ /3)1/2. This bound is attained by a lattice generated by a parallelogram consisting of two equilateral triangles [60].

In general, if a2 is the area of the generating parallelogram, the minimum distance between two lattice points is <a (2/ /3)1/2 with equality holding in the case of equilateral triangles. Thus we have proved the following theorem.

THEOREM 1. For a given minimum distance, the lattice generated by a parallelogram composed of two equilateral triangles has the smallest generating parallelogram.

By analogy with the use of parallelograms to construct plane lattices, parallelepipeds of different forms may be used to construct a space lattice, but they must all have the same volume. We have a unit lattice if the volume is equal to one. A generating parallelepiped must have the lattice points as its vertices, with none on the surface or in the interior. By analogy with the previous proof in the plane, we obtain a parallelepiped composed of two regular tetrahedra and one regular octahedron, leading to what is known as the rhombohedron, and, corre- spondingly, we obtain the closest lattice packing of spheres in a rhombohedral pattern.

In higher dimensions, the property of going from triangular to rhombohedral patterns does not generalize to give the closest lattice packing.

Let N(R) be the number of unit circles packed wholly in a larger circle of radius R. The density 5 of the packing in this case is given by

N(R)7r N(R) 3=lm 2 = lm R -oo 71R2 R +oo R2

It is known that

7t < = .9069....

The density of covering the plane with a system of unit circles such that every circle with a center in the plane is covered by a finite number of circles of the system is given by 0 > 27r/(3,3_) = 1.209 ... (see [63]). The equalities hold when the centers of the circles are the vertices of a lattice formed by equilateral triangles.

A set of open discs forms a Minkowskian arrangement if none of the discs contains the center of another one. What is the densest Minkowskian circle arrangement? For congruent circles of radius r in the plane, this is equivalent to the problem of the densest packing of the plane by equal circles. To see this, replace each circle of the arrangement by a concentric circle of radius r/2, thus obtaining a circle packing whose density (already given) is < 7/ 1 and is 1/4 the density of




the original set, i.e., the density of the original set is ? 27// 3 = 3.627 ... . General- izing to 3-space, one obtains 87/ /18 as an upper bound for the density of a Minkowskian set of balls [6], [28], [29], [77].

Analogous to the packing of equal circles in the plane by means of the incircles of regular polygons which fill the plane without overlapping, one may pack spheres in three-dimensional space by using the inspheres of regular polyhedrons.

One obvious such polyhedron is the cube whose side is equal to the diameter of the sphere, and all of whose faces are tangent planes to the sphere.

The density of packing in this case is

(4/3)mrr3 -7_

(2r)3 6 = .5236...,

which is not very efficient. An alternative is to use the midspheres (spheres through the midpoints of

each edge of a cube) of alternate cubic cells as the spheres. Imagine the cells of a regular cubic honeycomb to be colored alternately black and white, so that a black cube is touched along each of its six faces by a white cube (a kind of three- dimensional chessboard). We may divide each of the white cubes into six square pyramids by planes joining pairs of opposite edges (the apex of each pyramid is at the center of the cube). Each black cube may now be considered as having six white pyramids attached to it to form a dodecahedron whose faces have the edges of the black cube as shorter diagonals.

This configuration can also be obtained by considering a cubic lattice, taking a cube and replacing each face by the pyramid associated with it, the apex of which is the center of the adjoining cube. The result is a figure known as the rhombic dodecahedron. The rhombic faces are a result of certain adjacent pairs of triangles falling on a plane and forming rhombi; in all there are twelve such rhombi. The insphere of this dodecahedron is the circumsphere of the black cube, and the volume of the dodecahedron is twice that of the cube (this scheme is due to Kepler and to the team of Hilbert and Cohn-Vossen). Each sphere touches twelve others in this arrangement. The density of the packing, known as cubic close packing, is

(4/3)m(,/2 1)3 7 Z = .74048 2(21)3 32/2

Another alternative form of packing is hexagonal close packing. Here one fits together a group of spheres in a plane according to the hexagonal packing of circles, i.e., each sphere will touch six others in the same plane.

Then three spheres may be placed over the shaded "gaps" as shown in Fig. 3. This is repeated over the whole plane, so that a further hexagonal packing is created. Here again each sphere will touch twelve others, but the configuration is different. The third "layer" is placed vertically above the first.

The density of hexagonal close packing (Barlow, 1883) is the same as that for cubic close packing, i.e., .74048... (see [14]).

An alternative description of cubic close packing may be given as follows. Here the first and second layers are placed as in the above case, but the packing of the third layer is different: instead of being placed vertically above the




FIG. 3

(a) (b) (c) FIG. 4. (a) Cubic close packing, (b) hexagonal close packing, (c) cubic close packing

first layer, the spheres of this layer lie above the unshaded "gaps" of Fig. 3 (see Fig. 4(c)).

The problem of finding the densest distribution of nonoverlapping unit spheres in space is still unsolved. However, in 1874, R. Hoppe gave the answer for the densest lattice packing. If we assume that the centers of the spheres form a space lattice, then the lattice must be composed of rhombic dodecahedra or the centers are the vertices of a face-centered cubic lattice. The density is ( = 2 7r/6 =m/ 1 8 % 0.7448 ... . As before, this density also is believed to be the greatest without lattice restriction. It is known that an arbitrary packing satisfies ( < .7797 ... and a nonrigorous proof says that it is ( < .7545 ... . In general, the density associated with a single sphere, in an equal sphere packing of n-space, is the ratio of its volume to the volume of the interior of the polyhedron Ti, composed of those points which are nearer to the center Pi of the sphere than to any other center. This polyhedron is known as a "Voronoi polyhedron" or a "Dirichlet region". Such regions, each surrounding a sphere, fit together to fill the whole space. The density of the distribution is then taken as the limit of the average of the densities associated with those unit spheres which lie within a large cube Wwhen the volume of W becomes infinite. It is known that this distribution density cannot exceed 0.7545 ... , which is the density associated with a single sphere when the polyhedron Ti is the circumscribed regular dodecahedron. But, since space cannot be filled completely by regular dodecahedra, the average distribution density for the most dense space distribution must be less than this value. The thinnest covering density in 3-space is 5312 7/24 = 1.464.. . , attained when the centers of the spheres form a body-centered cubic lattice [38].




The Voronoi polyhedron idea is used by Rogers [80a] to show that the closest possible packing of n + 1 spheres of radius 1 in n dimensions is attained when their centers are at the vertices of a regular simplex of edge 2, whose volume is 2("12)n(n + 1)12/n! Note that this is not the same thing as the general packing of spheres in the whole space. Roger's result is given later on.

TABLE 1

DimensnNumber of Minimum distance Dimension Numberes o between the points Density of the packinig Reference spheres in a lattice packing

2 6 (2/ ,/)1 2 = 1.075 ... r/'(2 /3) = 0.9060. Lagrange, 1773;

Gauss, 1831 3 12 2 = 1.122 ... 7r/(3/2) = 0.7404 ... Gauss, 1831 4 24 = 1.189... 72/16 = 0.6168 ... Korkine and

Zolotareff, 1872 5 40 1 = 1.074 ... 7r2/(15X/2) = 0.4652 ... Korkine and

Zolotareff, 1877 6 72 73/(48,/3) = 0.3729 ... Blichfeldt, 1925 7 126 7m3/105 = 0.2952... Blichfeldt, 1926 8 240 74/384 = 0.2536... Blichfeldt, 1934 9 > 274/(945,/2) = 0.1457 ... Chaundy, 1946 10 > 7t5/(1920 /I) = 0.0920 ... Chaundy, 1946 11 _ 647t5/(197,110X/3) = 0.0604 ... Barnes, 1959 12 > 7r6/19,440 = 0.0494 ... Coxeter and

Todd, 1951

Table I gives the number of spheres touching another in a densest lattice packing in n-space. The solution is not yet exactly known for n _ 9 For n = 2 the density is for a general packing [60], [80] It is known that triangular and rhombohedral lattices do not generate the closest packing of spheres for n _ 3

Coxeter [14] observes that experiments on sphere packing have a long history. In 1727, Stephen Hales noted that when he compressed peas with forces of up to 1,600 pounds, they took up regular shapes; he described these as regular dodecahedra, although since the dihedral angle of the regular dodecahedron is less than 1200, the shapes could not all have been completely regular. Much later, in 1939, this experiment was repeated by Marvin and Matzke, two botanists who used uniform lead shot. They increased the pressure to 40,000 pounds, so as to eliminate all interstices. When the shot were cannonball-stacked as in Fig. 4(a), they became almost perfect rhombic dodecahedra. However, if the shot were stacked randomly, irregular 14-faced bodies were formed, almost all of those faces were quadrilaterals, pentagons, or hexagons. They also examined a froth of bubbles, and found that for 600 central bubbles, the commonest shape had 13 faces: 1 quadrilateral, 10 pentagons, and 2 hexagons.

Later experiments include those of Bernal, who in 1959 took equal balls of modeling clay, rolled them in powdered chalk, packed them irregularly and pressed them together. The resulting shapes had an average of 13.3 faces. Scott in 1960 tried to find a random packing between .7405 and .7797, but was not successful (see [12]).

One may note also [14] the Hindu fakir's trick in which a table knife is plunged repeatedly into a jar of uncooked rice, so that eventually the knife will bind and one can support the jar of rice by the knife handle.




We may now proceed to the consideration of the densest packing of unit circles in a given polygon. If the required circles do not have a unit radius, the problem may be scaled down. In general, the density of the pack;ng is the ratio of the area of the circle to the area of the polygon in which it is inscribed, provided that these polygons cover the plane. If the polygon is an n-gon of side 2b, its area is nbr, where r is the radius of the circle (see Fig. 5).

b b

FIG. 5

It may be seen that r = b cot (7t/n), and the density of packing is

7ir2 7rr 71 71 7/n =-=-cot-= nbr nb n n tan (i/n)

Note that this increases with n and tends to 1 as n tends to infinity. However, since the polygons must cover the plane, n = 3, 4 or 6 and thus 6 is the "best" value. Hence the packing by fitting together regular hexagons is the most efficient form of regular polygon packing of equal circles in the plane.

The density of such a packing is

- cot - = = .9069 ... 6 6 2,/3

(of course in a general packing this need not be true). It is worth mentioning here that one of the oldest geometrical problems is to determine those plane figures which, like the square and regular hexagon, by regular repetitions, can be used to cover the plane without gaps or overlaps. This is the problem of tiling, plane paving, or tesselation. It is easy to see, using Euler's formula, that no convex polygon with more than six sides can be used to pave the plane. Kershner [64] has shown that there are exactly three types of hexagons and eight types of pentagons which can do so and it is readily seen that all triangles and quadrilaterals do so. Later on, we shall mention paving with polyominoes.

If A is the area of a finite convex region whose perimeter is L, then the region can be covered with, at most,

-2 2 - -A + -L + I

unit circles. The square brackets indicate the greatest integer function. A result for the maximum number of unit circles which can be packed in the region is also known [25].




Let K be a convex n-gon of area A and perimeter L and let r be its inradius and R its circumradius; then nr2 tan (7n/n) ? A < 1nR2 sin (27/n), 2nr tan (7r/n) < L < 2nR sin (7r/n). Equalities hold if the polygon is regular. Thus among all convex n-gons inscribed in (circumscribed about) a given circle, the regular n-gon has maximum (minimum) area and perimeter.

We now consider briefly the problem of the most efficient packing of circles of radius r in a larger circle of radius R. The essential parameter is R/r [67].

Two main questions arise. (a) Given R/r, how large may one make n, the number of small circles to be

packed and (b) Given r, n, how large must R be? There are virtually no theoretical results for this problem. A few cases have

been considered, but, in general, the problem is approached in what is essentially an empirical manner. If R/r is given, one possible approach is to create an annulus of circles which touch each other and also the larger circle (it may not be possible to complete the ring). We then fit another annulus inside this, and proceed until no further circles may be packed (see Fig. 6). If N circles are packed into an annular ring as shown in the figure, then

2 Rr =

1- tan2 (7r/4 - 7r/2N)

If N as given by this formula is not an integer, the greatest integer function [N] is used. This indicates that there is a gap in the ring, as in the figure. If, on the other hand, n and r are given, one may build up a nest of circles in a roughly circular

FIG. 6

pattern, maintaining as much symmetry as possible in order to make R as small as possible. Neither of these methods is particularly satisfactory, since there is no general procedure to be followed.

If A is the area of a convex region and An that of an inscribed regular n-gon of maximum area, then [25]

A >A n

sin 2-. n= 27t nl




Equality holds only for an ellipse, and hence the approximation by an inscribed polygon is worse when the region is an ellipse. The problem is unsolved for circumscribed polygons, but it is known that for large N the external curve approxi- mates a circle. To approximate the boundary of a convex region by inscribed or circumscribed n-gons of perimeters Ln and in respectively, it is known that such n-gons exist for which

n n < 2 sin2-. n 2n

If n congruent convex regions are packed into a convex hexagon of area A, then n < A/a, where a is the area of the smallest hexagon circumscribed about any one of the packed regions.

Let R be the diameter of the smallest sphere on which N nonoverlapping circles of unit (straight) diameter can be drawn. For large N, the circles tend to be inscribed in n regular hexagons of area 3/2 covering a sphere of area 7R2, where asymptotically R2 - 3'12N/27c. This result has been refined by L. Fejes Toth [25] to

R2>4 (4-csc2NN26)

Equality holds when the circles are inscribed in the faces of a regular N-hedron {P, 3}. (A regular polyhedron {P, q} has a regular polygon of P sides as its face, with q faces surrounding each vertex.) We have N = 12/(6 - P), R2 = 4/(4 - CSC2 7r/P), P = 2,3,4,5, N = 3,4,6, 12. The case N = 5 is the same as N = 6, i.e., the best one can do is to inscribe the circles in five of the six faces of the unit cube. For N = 8, the centers of the circles are the vertices of an antiprism. For N = 9, the figure has the symmetry of a triangular prism.

If d is the density of packing N > 3 congruent spherical caps on the surface of a sphere (where the radius of the cap sphere is small in relation to the radius of the covered sphere), then

N I N 7c\ d < 1- -csc 2\

This result solves the problem of how to place N points on the surface of a unit sphere so that the minimum distance between the -N(N - 1) pairs of points is maximum. If the caps were to cover the sphere instead with a covering density D, then

Nl I N 71

-2 _, /3 N -2 6.

In both cases, equality occurs only when N = 3, 4, 6, 12, with the cap centers being the vertices of an equilateral triangle (inscribed in a great circle) a regular tetrahedron, octahedron or icosahedron.

If V is the volume of a polyhedron of e vertices, f faces and k edges which contains the unit sphere, then

V > ? sin 7

tan 2 tan2 2- . - 3 k 2k 2k /




Equality holds only if the polyhedron is regular and circumscribes the sphere. We find M. Goldberg's following isoperimetric theorem useful. If a convex

polyhedron of n faces has volume V and surface area F, then [811

>3 > 54(n - 2)tan ( -26) L4sin2 ( 26) -]

If K is a symmetrical convex body in n-dimensions whose in-radius (the radius of a maximal inscribed sphere) is greater than 1, then packing into K and covering K with unit spheres give

6(K) > 1/2 -' and 6(K) > 0(K)/2n. The following results are in Rogers [80] and apply to an n-dimensional sphere

S or convex body K as indicated:

6(K) > 2(n !)2/(2n!),

(5(S) _< an-~ (nle) ( 11 ,) n

0(K) < n log n + n log log n + 5n, n > 2,

2n n/2 n 0(S)> an+1 e e

0(K) < OL(K) < n lO2fn+logelog,en n > 3,

where un is the ratio of the volume of that part of a regular (n + 1)-simplex, (whose edge is of length 2) covered by n + 1 unit spheres (whose centers are at the vertices of the simplex) to the volume of the whole simplex.

The following gives a lower bound on the density of covering a region in n-space with n + 1 equal spheres. Again, we have unit spheres whose centers lie on the vertices of a regular (n + 1)-simplex of edge /2(n + I1)/n. The spheres just cover the entire simplex (with overlap). The volume of this simplex is

(n + 1)1/2 (2(n + 1)\n/2

2 n12n ! n

The following two results have already been mentioned above: 1. If-Cn is the sum of the volume of the portions ("sectors") of the spheres lying

in the simplex to the volume of the simplex, then a lower bound on the density of covering the space with unit spheres is given by

2n n/2

tn=n + 1I S

It was Rogers [80a], using an ingenious argument suggested by H. E. Daniels, who derived the asymptotic formula for an:

n an 2n/2e0v

which is an improvement on the bound (n + 2)/(/2)n +2 on 6(S), given by Blichfeldt [5a].




2. In general, if 6L denotes the densest lattice packing and 6 is the densest packing, it is known that for a certain constant C,

nC 2n -L <6 < n

where 6 can be evaluated in theory for all n and has been for n = 2, 3, 4. The upper bound <_ unis the best known for every n. Bounds better than

6L > nC/2n

are known for 9 ? n < 12, and for some larger values of n.

3.2. Mixed packing of spheres. A variation of the sphere packing problem would be the packing of unequal spheres in a region. One simple form of this problem would be: What is the densest packing of spheres of two different radii R1, R2 if we wish to pack approximately equal numbers of each in a given region?

Without any loss of generality we may let R 1 = 1 and suppose that R2 < R1. Our first thought would be to pack the unit spheres in cubic close packing and try to fit smaller spheres inside the "gaps". The centers of the "gaps" are at a distance 2/ from the centers of the nearest unit spheres.

Thus, if R2 + 1 <_ 2/, the smaller spheres may be packed inside the empty spaces.

The density of this packing would be

4/37(/ 2)3 + 4/37rR' - 1{2 2 + R7}

2(2)3 ~122

If R2 = 2 - 1, the density becomes (77c/12)( -2 1)- .76. If R2 > 2 -1, this packing is no longer possible. If 2 < (R2 + 1)2 < 3, i.e., /2 - 1 < R2

3_ - 1, the unit spheres must be moved apart sufficiently to permit lattice packing of the smaller spheres. If 3 < (R2 + 1)2 < 4, i.e., /3 - 1 < R2 < 1 we may use the lattice for unit spheres but replace alternate unit spheres by the spheres of radius R2 (for details, see [37]).

This type of mixed packing is exhibited in the common salt molecule, where the sodium and chlorine ions may be regarded as spheres of different sizes which have to be packed together.

4. Examples of methods of proof. Let us now look at a few more examples in 2 dimensions. In the discussion of the next section, a number of ideas relating to higher-dimensional problems will be briefly examined.

THEOREM 2. There are at most three points on or inside the unit circle such that the distance between any pair is greater than 2/ (see [15]).

Proof. Let P1, ..., Pn be n points inside the unit circle, with center 0. Then OPi.< 1, and /2 < PLPj. The cosine law gives

2 < opi + OP2 -2 OPi OPj cos PiOPj. This means that 2 OP. OP . cos PiOPi < 0, from which we have that (the smaller) angle PiOP is obtuse for any pair of points, which is impossible for n > 3.

Alternative proof. This result may be proved as a particular illustration of a stronger theorem.




THEOREM 3. If n ? 2 points lie in a unit circle, then some pair of these points is separated by a distance of no more than max (1, 2 sin (7/n)).

Proof. If a1,a2, , an lie in a unit circle, and x lies in the convex hull (a 1, a2, , an), then d(x, aj) < 1 for some j. This is obvious for n = 2. For n > 3, we may choose three of the ai so that T, the triangle they form, contains x. But no point inside this triangle can be more than a unit distance from some vertex of T (the circumcircle of T has radius < 1).

Now assume that no a' lies in the convex hull of the remaining aj (if it does, it may be ignored as its sole effect would be to decrease distances). Any ai which is not on the circle may be moved out to the circle in such a way as not to decrease the distances between ai and the remaining points. If now all ai lie on the circle, the circumference would be divided into n arcs. Consequently, some arc must have length < 27t/n, and hence some pair of points ai, aj is such that d(ai, aj) < 2 sin (7n/n).

Thus,

d(ai, aj) < max 1, 2 sin-), for some i and j.

Note that if n ? 4, d(ai, ai) < 2, for some i and j. THEOREM 4 [18]. The number of circles of radius r packed so that each is tangent

to six others and contained partly or entirely in a circle of radius R > r that is concentric with one of the circles is given by

N = I + 6 n3 _s ,,k 2] n ( = -

(square brackets indicate the largest integer function). Proof. For simplicity, we assume first that the distance between the centers

of two adjacent circles is unity. Because of the triangular nature of the array of centers, we shall compute the number of centers in one of six similar sectors of the large circle. Let C be the center of this circle. Then C belongs to no sector (see Fig. 7). We need to determine the number of centers along an arbitrary vertical line in the sector and scan over this number from 0 to n, where n is the number of centers along the radius R.

R

60 0

C \ FIG. 7

Let k be the number of centers up to an arbitrarily chosen vertical line, whose number of centers x will now be computed. We have

(x + h1 + k sin ) + (kcos )= (n + h2)2 0 < h1, h2 < 1.




Therefore k

x= (n+h2)2-3/4k2---hl, 2 i.e.,

X=L [Xn2 _ 3/4k2k]

and the total number of centers N is obtained by summing over k, multiplying by six (sectors), and adding one for C.

Therefore

N= 1 + 6 n2 _ 3/4k2k-]

If the centers are 2r units apart, then a circle of radius R = 2rn + h3, 0 < h3 < 2r, will contain N centers, and a circle of radius R = 2rn + h3 + r will contain N circles.

Therefore R -r -h3FI(R1 n= 2r L2\ r/I

which completes the proof. We now consider the number of points with integer coordinates in a circle

or ellipse. For example, we may wish to determine the number of integral solutions of X2 + y2 < n, where n is a given integer. This estimate is easy to compute for small values of n. In fact, one can list all the solutions. There is one for n = 0, five for n = 1, nine for n = 2, etc., and they are easily enumerated. Points with integral coordinates are uniformly distributed in the plane. A unit square corresponds to exactly one such point. The area of a circle is approximately equal to the number of unit squares it contains and, hence, also to the number of solutions of the inequality, because they lie on the vertices of the squares.

An order of magnitude estimate in general may be given by the following theorem [44].

THEOREM 5. The number of integral solutions N of X2 + y2 < n, with n an integer, may be approximated as N - n7r, with error IN - cnl < 27r(/2n + 1).

Proof. Consider all points of the Cartesian plane with integer coordinates. These points comprise the vertices of unit squares. Associate with each unit square the point on its upper right-hand vertex. Let Cn be a circle, with center at the origin, which contains N integral points. Let Sn be the area of the corresponding squares. Let C and C2 be two circles, concentric with C, with radii /n - 2/ and ,n + 2, respectively. Sn lies completely within C2 and contains in its interior C'. Since the area of Sn is N, we have

nr(-_ 2)2 < N < 7r(,/; + 2/5)2.

This gives N ; n7 and IN-inl < 27n(,/2 + 1)

and completes the proof.




By a geometric argument like the one given above, it can be shown that the number of integral solutions of X2 + y2 + Z2 < n is approximately equal to 47i;n312/3.

THEOREM 6. At most A//12 unit circles may be placed in a convex polygon of area A, so that no two of the circles overlap. The angles of the given polygon must not be greater than 27/3. (A consequence of this restriction is that since the sum of the angles of an n-gon is (n - 2)r, at least one angle is >(n - 2)ir/n, and hence 1 - 2/n < 2/3 or n < 6 (see [89]).)

Proof. We first define the Voronoi polygon of this packing. We let P, P1, ,P2 , P, be a set E of points in the plane such that

PPi > 2, PiP >2,

for all i and j, and let C(P), C(P1), - be unit circles of centers P, P1, P2, , respectively. Then no two of these circles can overlap.

Now let K, K' be concentric circles of radius r, r + 1, respectively. If p("), p(2), ... , p(l) are points of E in K, then C(P(l)), C(p2)), C(P(1)) must be in K', and their total area 1lr is less than the area of K'. Thus

1 < (r + 1)2.

Hence every circle in the plane contains at most a finite number of points Pi of E. For all points Pi in E, let S(Pi) be the set of points Q in the plane for which

PiQ <- PkQ, k k i.

Thus {C(Pj)} c {S(Pj)}. The points Q for which PQ= PkQ lie on the perpendicular bisector Ak of

PPk, and the points Q for which PQ < PkQ lie on a closed half-plane to one side of Ak or on Ak.

Thus S(P) is a closed convex region. Let S(P) be of finite area A(P), and Q # P be a point in S(P). Let T1, T2 be the endpoints of the diameter of C(P) which is perpendicular to PQ (see Fig. 8). Since Q, T1, T2 are in S(P), and S(P)

T - x

2

FIG. 8

is convex, the area of AQT1 T2 ? A(P). But area AQT1 T2 = PQ. Therefore

PQ < A(P),

for all Q in S(P). Thus, if P' lies on Ak, PP' < A(P), and it follows that

PPk 2A(P).

Since there are only a finite number of points Pk inside a circle of center P and radius 2A(P), there are only a finite number of lines Ak on the boundary of S(P). Consequently, S(P) is the Voronoi polygon of P.




Let S(P) have n sides A1, A2, , An and n vertices R1, R2, Rn in order. (Rk, Rk+1 are on Ak, k = 1, 2, n - 1, R, R1 are on An).

We divide S(P) into n triangles by joining P to R1, R2, , R, and let ARkPRk+ 1 be of area ak, LRkPRk+ 1 = ?(k'

Then we must show that ak _ /3k/2T and A(P) ? 12, equality holding only for a regular hexagon touching C(P), because

n n

I ak= A(P), I Zk-= 22. R =1 R= 1

We may now divide any polygon of area A, in which 1 unit circles are packed, into 1 Voronoi polygons each of area at least /12 (we need angles < 27/3). Thus

1,i 2 2A and I A/ 2.

Therefore, at most, A/ 12 unit circles may be packed in a polygon of area A. There are exactly A/i 2 circles if and only if the polygon is the sum of a finite

number of regular hexagons of side 2. Otherwise, the inequality is strict. We are particularly interested in illustrating a proof of a useful result in

n-dimensional space. We follow Rogers [80, Chap. 2], a well-known reference in this field. Let /(K) be the Lebesgue measure of a set K in Euclidean space En, and let DK be the difference set of all points of the form x - y, with x and y in K.

THEOREM 7. If K is an open convex body in En, then

()>2(n !)2 =(K) (2n)!

Proof. The proof is developed in two stages.

(i) (K) ? 4(K)

(ii) p(K < n ().

Combining these two results will yield the answer. LEMMA 1. If K is a bounded open set in En, then

6(K) > ? (K)

Proof. Let K contain the origin 0 of the coordinate system, and let s(K) be a number such that K is contained in a cube of side s(K) whose edges are parallel to the coordinate axes. Let C be another cube, of side s(C) > 2s(K).

Let a be a set of points such that K + a c C. Choose

a - {(a('i), a , aU)}, i = 1, 2, 3, , from a so that

(K + ai) n (K + aj) =0 for j < i, and so that ai') is as small as possible. Let the sequence terminate at an.




Also, let y = (Yi, Y2, ... , yj) lie in cube C', of side s(C) - 2s(K), whose center coincides with that of C. We have a(1' I <y1. Let r be the largest integer < N such that a7r < Yi Then either a7(r) is not defined, or a7(r+ > yl. Therefore either

(i) K + y is not contained in C, or (ii) K+y has a point in common with one of the sets{K + a},j = 1,2, r.

Since y is in C', (i) is impossible, and hence (ii) holds. Thus there are points k1, k2 in K such that k1 + ai = k2 + y, i.e., y = k1 -k2+ ai or y c DK + ai. Since Yi _ (), y C HDK + a (HDK is the set of points x of DK such that x1 ? 0). Thus HDK + ai, i = 1, 2, N, is a covering of the cube C'. It follows that

K> y_ (? )_O(HDK). - f1(HDK)

Since DK is symmetrical about 0, p(HDK) = (DK), and, further, O(HDK) ? 1. Thus

6 (K) ? 21i(K) K)_ DK)

LEMMA 2. If K is a closed convex set, then

n

Proof Let

1 if xe K, %X) = o0 if xo K.

Then integration over the entire space gives

jj Z(y - x)Z(y) dy dx = { %(y) Z(y - x) dx dy}

Z { (y)M(K) dy= =M(K)}2.

Now, for each x,

X(Y - x)Z(y) dy = 0,

unless y is such that y, y - x, belong to K, i.e., x = y - (y - x) e DK. Thus

{XK { x(y - x)Z(y) dy dx = { (K)}2 DK

Let A = A(x), O < A < 1, for each x(# O) of DK, such that A x lies on the boundary of DK. Let z = z(x) = A-1lx. Then 3a c K,b c K, such that z = b - a, (1 - A)K + Ab c K, (1 - A)K + Aa + x c K + x. But

b - (a + x) = z - x = 0,




so that the two sets on the left coincide, and hence each is in K n (K + x). Thus

J (y - x)(y) dy = 1(K n [K + x]) ? ,u([I - ]K)

= (1 - A)n (K)

from which we have

{DJ {1 - i(x)} 4(K) dx ? [1i(K)]2 DK

or

{D -A(x)}ndx < 1l(K). JK Now

{ 1 - ~(X)}~ = J n(I - t)n 1 dt,

so that

TDJ LLxJ n(1 t)n- dt] dx ? 1(K). DK A(X) Therefore

{JL{[JDK n(1-t)n- dx] dt <? (K). Oxc-DK A.(x) ?t

Since {A(x)}- lx is on the boundary of DK, x is on the boundary of A(x)DK, and, if A(x) < t, x e tDK. We now have:

J1: L T xeDn(1 - t)n- dx] dt = { n(1 -t)n- '1(tDK)dt O X -DKO A(X) < t

= P(DK) tn .n(1 - t)'n - 1dt

= f(DK) (n!)2 (2n)!'

It follows that

p(DK). 2 ) < M(K), (2n)! =

14DK) ? (2n)M(K).




Combining Lemmas 1 and 2, we have

5(K) > ( ? )2 - (2n)!

5. Applications. 5.1. Signal transmission. In the transmission of signals from an origin we

may regard the point at which the signal is received as lying in a sphere of given radius about the signal position in n-dimensional space. The point representing the signal position must be in or on a larger sphere of given radius with center at the origin of the signals (the radius depends on the power of the signal).

The spheres about different signals must not overlap, since this would cause interference. Consequently, this becomes a problem of packing spheres inside a larger sphere.

(a) Spheres in a sphere. Let U be the unit sphere with center 0 in an n-dimensional Euclidean space, and let M(r) denote the maximum number of open spheres of radius r with centers in or on U that do not overlap. M(r) is, of course, the greatest number of spheres of radius r that can be packed into a sphere of radius 1 + r. We have

M(r)= I when r > 1 and

M(r) =2 when X < r ? . 2

Blachman [3] has shown that

N(arc sin r), r >

2r21 n +

L2r 2 _ij' >

2n M(r) = n

2n, r = 2

where N(x) is the greatest number of open spherical caps of angular radius x that can be placed without overlapping on the n-dimensional sphere. (To avoid con- fusion, note that N(arc sin r) is, in fact, equal to the given expressions over the ranges indicated beside each.)

He has further shown that

(N(arcsinr) < M(r) ? N(arcsinr) + M*(1 r)

for r _ 2, where M*(r) is the maximum number of nonoverlapping spheres of radius r with centers interior to U.

Finally,

M(r ? 2rn3(1 - 2r2) + 0(Xn) e(T2r)n-

forr< r.




(b) Packing spherical caps in n dimensions. If the caps are such that their angular radius is not too much greater than 450, they may be arranged with their centers at the vertices of a regular simplex concentric with the sphere; no more than n + 1 caps may be so fitted. If the angular radius is 450, 2n caps may be fitted, each centered over the face of a hypercube inscribed in the sphere. If the angular radius is less than 450, only divergent bounds are known. The expression due to Rankin [79]

/(n3 cos 20)

(A/2 sin n0) n

gives an upper bound. This has been improved by Coxeter [16]. Another problem related to transmitting information is the coding of messages

in such a way as to eliminate errors. Considerable work on this problem has been done by Golomb [50], [51], who also relates it to sphere and polyomino packing. Polyominoes are discussed later on.

5.2. Covering: Radar, lighting and sonar buoys. It is possible to site radar stations in such a way that the ground area over which signals may be picked up is completely covered, i.e., the circles within which any individual station may oper- ate effectively provide a covering for the ground. This is of little use as far as tracking is concerned, since one wishes to be aware of an unusual object above ground level. Thus the objective is to ensure that all points up to a certain height above the ground area are within the coverage hemisphere of some station. Two other problems are concerned with placing lights in such a way that the illumination at all points does not fall below a certain level, and with locating sonar buoys so that a signal may be obtained at all necessary points. Note that these are small scale problems, i.e., we ignore the curvature of the earth. They are all problems in covering: all points in a given domain must be inside at least one circle or sphere.

g' hh

A B C FIG. 9

(a) Radar stations. The radar stations must be located to pick up a clear signal from a distance r or closer. It is required to provide coverage to a height h, i.e., the sections of the hemispheres in any direction must cut at a height of at least h. It may be seen that no point on the ground may be at a distance greater than (r2 - h2)1"2 from a station. Thus we require a minimum covering with circles of radius (r2 - h2)1/2 (see Fig. 9).

As in packing, the centers form a regular hexagonal lattice to both pack and cover the plane by the hexagons inscribed in the circles [64] (see Fig. 10).

The distance between the centers of any two intersecting circles is 2 (/3/2) 2 r2 -h2 = 3(r2 - h2), and the density of the covering is

7 -r h2) _27t _________________ _ = = 1.209....

6 (1/2) (v 3/2)(r2-h2)3 3X/




FIG. 1 0

We may mention here the works of L. Few and L. Fejes Toth, included in the references.

(b) Illumination. It is required to provide an acceptable level of lighting over a given area, up to a specified height. Since the intensity of illumination varies inversely with the distance from the source, we have as in the previous problem hemispheres of influence, although we are usually considering a region below the level of the lamps (see Fig. 11).

A B C D

T ) r .s T 1 '. so T T; ~Lamp Level

< , . , ,- ~~~~~~~~Ground Level

FIG. 11

This is essentially the same problem as before, and the same lattice provides the solution.

An interesting illumination problem was posed by L. Fejes Toth [32]. Con- sider an infinite straight road along which identical lamps are erected with a given average number per mile; how should they be placed so as to maximize the illumination of the road at the worst illuminated point? Prove or disprove that the best solution (regardless of the illumination function or density) is a distribution of the lamps consisting of the union of a finite number of congruent sets of lattice points along the line. B. R. Henry [58] recently completed a disproof of this conjecture by showing that it did not hold for a given illumination function.

(c) Sonar buoys. The problem of the sonar buoy placement is effectively a combination of the previous two problems, since coverage may be required above and below the level of the buoy. The solution is, however, as before, with the qualification that if either r or h (or both) are different above and below the level of the buoy, the smaller value of r2 - h2 is to be used in determining the lattice.

5.3. Crystals. A characteristic property of many crystals is that of cleavage. The cleavage planes in a crystal are those across which the forces between the atoms are weakest. For example, in sodium chloride, the cleavage planes follow the cube, since the even numbers of positive and negative ions ensure that the




overall charge is zero. From symmetry, there are three directions of cleavage. Other materials show different cleavage planes. A cleavage plane does not refer to any given plane in the crystal, but to a direction, for the crystal may be cleaved along any one of a family of planes in that direction.

The most striking example of cleavage planes are those of the diamond, which is held together by chemical bonds between its carbon atoms and has octahedral cleavage planes.

FIG. 12. 'Face' centered structurefor both sodium (white) and chloride ions [61], [65]

FIG. 13. Rhombohedral unit cell




The same ordered packing which we have discussed here is exhibited to a marked degree in nature and is closely related to the physical and chemical characteristics of the materials used (see Figs. 12 and 13). The honeycomb of the bees represent a form of packing which is related to the present case, even though the shapes of the cells are convex polyhedra [26].

5.4. Polyominoes [49], [92]. A polyomino is a plane figure made up from squares which are joined together with at least one other square along an edge. A domino is a simple polyomino consisting of two squares joined together. Many popular games have been based on these figures. Consider the pentomino: five squares may be joined together in twelve possible shapes which may be fitted together in various ways, or some of which may be placed on a chessboard so as to exclude other pentominoes. One problem with dominoes has applications to construction: is it possible to cover a rectangle with two or more dominoes so that every grid line of the rectangle intersects at least one domino? (Grid lines are the horizontal and vertical lines spaced at the width of one domino which are perpendicular to parallel edges of the rectangle). If a grid line does not intersect any dominoes, this may be regarded as a structural weakness and a possible fault line (the dominoes may be regarded as bricks). We thus identify those rectangles which it is possible to cover with dominoes without fault lines (see Fig. 14). We may

Fault Line

FIG. 14

rule out those rectangles both of whose sides cover an odd number of squares, since then the total number of squares would be odd. It is also obvious that rectangles with a side of two, three or four units must have a fault line, and we find that a 5 x 6 rectangle is the smallest which may be covered in a fault-free manner. Two different coverings are possible (see Fig. 15).

FIG. 15




One can prove that there are no fault-free 6 x 6 squares. Consider a 6 x 6 square completely covered by dominoes. There will be 18 dominoes and 5 horizontal, 5 vertical grid lines.

(i) Each grid line cuts an even number of dominoes, for each grid line has an even number of squares to one or to the other side. Since the complete dominoes cover an even number of squares, the cut dominoes must cover an even number of squares, and thus the number of cut dominoes is even. In fault-free cases, therefore, each grid line must cut two or a larger even number of dominoes.

(ii) The 10 grid lines must cut 20 or more dominoes if the covering is fault-free. Since there are only 18 dominoes, at least one grid line must be a fault line.

This reasoning may be extended to other configurations, and more compli- cated polyominoes may also be used.

The study of polyominoes yields some interesting proofs. One such is as follows and is due to Spira [92].

Replicas of the 3 x 4 L-hexomino cannot be fitted to form a rectangle. An L-hexomino is formed by joining in order the following points: (0 0), (3, 0), (3, 1), (1, 1), (1, 4), (0, 4), (0, 0). The L-hexomino may take eight positions, as in Fig. 16, and

=fl m cEr --f L St~ E 1 2 3 4 5 6 7 8

FIG. 16

we label the positions 1 through 8. Now let the positive x-axis and the negative y-axis of a coordinate system be edges of the rectangle which we are trying to cover. We label the unit squares in this quadrant by the row and column in which they lie, i.e., square (i,j) has corners

(j - 1, i - 1) (j, i - 1) (j,i) (j - 1, i).

We may now attempt to construct a covering, and the tree diagram in Fig. 17 shows that every possibility ends with a square which cannot be filled. The diagram is to be read as follows: the number to the left of each node of the tree indicates the position number of the hexomino (see Fig. 16). This hexomino is to be so placed as to cover the unit square given by the numbers to the right of the node above.

2 122 3 21 422

4 51 3 132 6 1 23 8 F14 7- 32 2 1 23

6 033 262 3 33 7 26 23 44142614 71138 14

8I14T7162 6134 5 5 72 234451343 145335233 6 4243 25615 744

5 3 2 152 4 51 21353 43 51 43 7 35

6- 14 8114 70 -43 5133 645 2 52 2 52 3 51

7 1 26 5 -33 6 45 6 45 26 62 7 55 7 5F

FIG. 17




At a terminal node, the number to the right describes a square which is impossible to fill, either immediately or soon afterwards. As an example, we may cover square (1, 1) with a hexomino in position 3. We would cover (2, 1) with a position 2 hexomino, but this will leave (2, 2) in an "uncoverable" position. By following through the tree diagram, we see that each possibility terminates with a square which cannot be filled, and that at no stage can we complete a rectangle [92].

A region R of the plane can be tiled with a finite set S of polyomino shapes if replicas of the members of S can be used to tile R. These members must have the same size unit square. The operations allowed are arbitrary rotation, reflection and translation of the replicas and the use of arbitrary quantities of each shape in S (excluding certain shapes if necessary). Golomb [52], [53], gives a hierarchy of tiling capabilities of polyominoes.

5.5. Housing sites [30]. This example is taken from a work of L. Fejes Toth, the well-known Hungarian geometer. The problem is to build as many houses as possible on a large area with the restriction that the houses (which have rectangular bases) must not be closer than a given minimum distance.

This minimum distance may be taken to be two units. We take a rectangle of sides a, b, a < b, and create what is known as a parallel domain, i.e., the locus of all points at a given distance from the given domain [27], by first drawing lines parallel to the sides, at a distance one unit away from the sides, and of the same length. We then draw parts of unit circles with centers at each corner of the rectangle. The boundary of the site so created contains all points which are at a distance 1 from the sides of the base (see Fig. 18). If two sites are placed together in any nonoverlapping manner, the rectangles cannot be closer than two units. The problem is then converted to one of packing the sites most efficiently in the given area. The centrosymmetric hexagons h of least area which contain the sites S are used for packing. (To fill the plane, we need triangles, squares or hexagons. Here we see that

FIG. 18




lattice packing of hexagons will be most efficient. The hexagons must be centrosymmetric for the packing to be also a covering) (see Fig. 19).

The axis symmetry of S parallel to the large side of the rectangle is called the axis of S. If a = b, one of the axes of symmetry is chosen as the axis of S.

D

FIG. 19

The centrosymmetric hexagon is circumscribed about S so that (i) it has a pair of sides parallel to the axis of S; (ii) all of its midpoints lie on the boundary of S (this may be seen to be a

necessary condition for a minimal hexagon, by a symmetry argument. These hexagons fall into three main classes. I. The hexagon has bilateral symmetry about the axis of S (see Fig. 20).

FIG. 20

II. The hexagon has no bilateral symmetry about the axis of S, and no side perpendicular to the axis of S (see Fig. 21).




FIG. 21

III. The hexagon has two sides perpendicular to the axis of S (see Fig. 22).

FIG. 22

THEOREM 8. Let S be the parallel domain at unit distancefrom a rectangle whose shorter side has length a. Then the centrosymmetric hexagon of least area containing S is of Type I, II, or III, according as:

a 4- 12,

4- /2 < a < 2- ,

a ? 2- 2/,

respectively. Proof. The result is proved initially with the rectangle a square, of side length

a = 2x, and then the results derived in this case are shown to apply for the more general rectangle.

(i) Consider a minimal centrosymmetric hexagon H with no side parallel to a side of S (see Fig. 23).




FIG. 23

The half-area of the hexagon H is given by

fo(x)= (1 + /2x) (2 1 - (1+ 2 )2 + 2/x)

= (1 + /ix) 3- 2/x- 2x2+ 2x).

For this case to be realized, 2(1 + 2x < 2\/2. The midpoints of the sides of the hexagon lie on site S, i.e., 2x < 2 - 2.

(ii) H has exactly one pair of sides parallel to a side of S. There are two cases. (a) H has an axis of symmetry (see Fig. 24). The half-area of the hexagon H

is given by

fi1(x) = (1 + x)IABI

(1 + x)(2.+1-(1 )+ 2x)

= (1+ x)(X/3 + 2x -x2 + 2x).

Note that in this case the following must hold:

1+x< or x<1. 2

(b) H has no axis of symmetry. Let AL = AM < BM = BN (see Fig. 25). The distance of M from TU is given both by AM sin 3a and by BM sin 3/B, i.e., by sin 3a/cos a and sin 3,B/cos /B. Let Z(4) = sin 30/cos 0. Then Z(a) - Z(,B). Also, projecting AM + MB perpendicularly to line AB gives

cos 3a cos 3/3 cos a cosf,




A

(lix

B

FIG. 24

M

U. T

FIG. 25

But

= 4 cos2o -3 = 1-4 sin2 oc. cos oc

Therefore

x = 2(sin2 oc + sin2 B)-1.

Now sin 3oc/cos oc > 1, sin 3f,/cos , > 1, from the geometry of the figure. Thus 3oc > 7r/2 - o, and oc > t/8. Also, since 2/B < 7r/2, / < t/4. Further, oc < ,B. Com- bining all of these gives

8 4 < < 4




In (t/8, t/4), the function Z(0) has a unique maximum given by

3 cos 3) cos ) + sin 4 sin 3) 0 '7PJ cos 0,

or 3cos34)cos 4 + sin 4 sin34 = 0, i.e., 3cos 4 (4cos34) - 3cos4)) + sin 4 *(3cos2 4 sin ) - sin3 4) = 0, or 8 cos44) - 4cos2 - 1 0 O. If one root is ', cos2y (1 + 3/)/4 (the alternative root for cos2 y = (1- 3)/4 will not give real cosy).

Now, in (t/8, y), Z increases, and in (y, t/4), Z decreases. If t/8 < oc < y, the equation Z(oc) = Z(/B) associates with a given value of oc a unique /3 such that ;' < /B < t/4. Thus x may be regarded as a function of oc only. It may be shown numerically that x strictly decreases as oc increases from t/8 to y.

When o /3 = x 4sin2y 1 3- 4cos2y= 2- /3.

In other limiting cases, oc = t/8, /3 =t/4, and

2x = 4 (sin2 + sin2 -2

=2(1 - /2/2)+2-2 = 2 - 2/2.

These limiting cases are strictly excluded and, thus,

4 - 2 /3 < 2x < 2 - /2.

The half-area of hexagon H is given by

fl2(x) = 8(sin3 ocos o + sin3 fBcosf) + 2x(1 + x).

(iii) H has two pairs of sides parallel to the sides of S (see Fig. 26). The half- area of the hexagon is given by

f2(x) = 1(2 + 2x)2 - 2(2)(F2 - 1)2

= 2 + 4x + 2x2 - 3 + 2 /2

= 2x2 + 4x + 2,/2-1.

For this case to be realized, 2(1 + 2/ix) > 2 /2 (again, because the midpoints of perpendicular sides must lie on straight edges of S, i.e., 2x > 2 - /). The next stage is to select from fo , f 1, f 2, f2, the least one for given values of x. We have

fo(x) = (1 + 2/2x)( 3- 2x -2x2?+ 2x), x <1- 2/ /2,

f11(x)=(1 +x)(X/3+2x-x2+2x), x<1,

f12(x) = 8(sin3 oc cos o + sin 3 ,cos /) + 2x(1 + x),

2- /3<x<1- 1 2-/2,

f2(x) = 2x2 + 4x + 8/ - 1, x >1- 2//2.




/~~

/

FIG. 26

If O < x < 2 - /, compare fo and f 1

f(O() 3, f11(O) 3,

f(0) =2 >6/ 3 + 2 /2 4.46...ff'(0) =4 /33 + 2 -4.31 ....

Thus, near 0, fo(x) > f1 1(x), and it may be checked that this remains true up to 2 - 3/3.

For 2 - / < x < 1 - //2, comparefo, f 1 andfl 2 .Checkin numerically shows that fl2 is the smallest over the whole interval. For 1 - 2/2 < x < 1 compare f1 1 and f2 . Again, numerical checks show that f2 < fi 1 throughout. For x > 1, f2 is the only possibility.

We conclude that the centrosymmetric hexagon of minimal area for (a) x < 2- 3is of type I, (b) 2-/3 < x < 2->/2/2 is of type II, (c) x ? 1 - //2 is of type III.

This the assertion that the required haxagons are of types I, II, or III, according as

a < 4- /,

4- /2 < a < 2- 2,

a ? 2- o2,

respectively, is proven for a square of side a = 2x. We now wish to establish the result in the general case. We transform a

convex polygon P into a new one as follows: Let a and b be two vertical "supporting" lines of P, and A and B be vertices of P lying on a and b respectively. Then A and B divide the boundary of P into an upper and a lower polygonal line, each having A and B as endpoints (see Fig. 27).

Then P may be regarded as the intersection of two convex point sets P1 and P2 bounded by a, AU . r B, b and b, B'loY Y A, a respectively. Translate P2 upwards through a distance d intersecting P1 to obtain a new polygon P'. This is called




b b' a} a

Pi

A

B

WI?

FIG. 27

telescoping. If a', b' are vertical supporting lines of P', and w' is the distance between a' and b', we have:

t = t' + w'd + t",

where t = area of P,

t' = area of P',

ti = area of part of P outside a', b'.

Now let S be a site with vertical axis, created from the rectangle with a < b, and let h be an arbitrary centrosymmetric hexagon enclosing S. Telescope h through distance b - a, obtaining hexagon h' of area t'. Note that this will change the rectangle into a square of side a. Then

t > t' + w'(b - a) > t' + (2 + a)(b - a).

Then certainly t > V + (2 + a)(b - a), where V is the area of the smallest centrosymmetric heagon h containing the new site. But V + (2 + a)(b - a) is the area of the hexagon obtained by a telescopic elongation through the distance b - a, since h has a pair of vertical sides. Thus this hexagon is the minimal hexagon of all centrosymmetric hexagons containing the site, and the results hold for general rectangular bases.

5.6. Parking. The arrangement of parking spaces for automobiles to make optimum use of space is a major problem in the design of a parking lot. One approach to a theoretical study of the design of a parking area borrows its tools by approximation from the geometry of numbers. The problem here is to find the upper density of packing of a set of equal circles which can be packed in a given domain without blocking each other's way out of the domain.

Heppes [59] considered this problem by finding the density of an "approachable" system of unit circules, i.e., each circle may be approached and touched by a circle of the same size sliding in from outside the convex hull of the domain. The packing which he considered was an "r-approachable" packing, in which each circle may be approached by a circle of radius r < 1. This is a weaker condition but provides an upper bound to the density of packing in the previous case.




It may be shown that an upper bound for the density of packing when r = 1 is

3_ - .56518... 2? + 6X/2

The proof involves consideration of a parallel domain (also called outer parallel domain because it is constructed outside the initial domain) as in the previous section and uses a similar argument, as we show below.

THEOREM 9. Let D be a simply connected bounded domain containing an r- approachable packing of n unit circles, where r > 2//3 - 1. The area A(Dr) of the outer parallel domain Dr of radius r of the domain D satisfies the inequality

A(Dr)> nA + 27t 2 - 3)'

where

Ar= w 2 - 3}t + 3(1 + i) + (r + 2r)

and

w = 2 arc sin

(Ar is the area of the domain shown in Fig. 28).

X ~~~

FIG. 28

DEFINITION. The angular area at 0 of a triangle AOB is the quotient of its area and its angle at 0.

We first need the following two lemmas: LEMMA 1. Let AOB be a triangle with thefollowing properties:

(a) it contains the sector AOB of the unit circle C1 of center 0; (b) neither A nor B lies in the interior of the circle C2 of radius 2//i3,

concentric with C1 (see Fig. 29). Then the angular area of AOB with respect to its vertex 0 is > /i/t. Equality

holds only if the triangle is regular with side 2//i3. We may suppose that AB meets C2, for otherwise the statement is trivially

true (by comparison of the sector of circle C2 and triangle AOB).




A B

0 C

C2

FIG. 29

LEMMA 2. Let C1, C2, C3 be three concentric circles of radii 1, 2//i3 and 1 + r > 2/3/3 respectively, with common center 0.

Let A1, Al, , Ak, At, k > 1, be points on C3 in cyclic order, such that the angle A'k OA1 > (2n)/3 - w, where w is the central angle of a chord of length 2 of C3 (see Fig. 30, and note that this implies that w = 2 arc sin (1/(1 + r)).

A2

FIG. 30

Then if P is a convex polygon with the properties (a') P contains C1, (b') P contains the arcs AiA o 3i ,2 ... ,k, and (c') all vertices of P lie outside or on the boundary of C2,

the area of P, A(P) satisfies

A(P) ? - r

}2 + 3(1 + + (r 2 + 2r)'1/2

where oc denotes the sum of the angles Ai OA', i I ,.. k. To prove this lemma, we split up the polygon P into three different types

of triangles and show that the sum of the lower bounds of the areas of the triangles gives the desired result.

Now for the proof of the theorem.




Proof. Let D be a bounded domain containing an r-approachable packing of n unit circles. The circles of radius 1 + r concentric to the circles of the packing lie in the domain Dr, the outer parallel domain of radius r, of D. These circles will be called "great circles" of the packing. The union of the great circles consists of one or more "islands" lying in Dr such that every great circle adjoins the "ocean" along one or more "shore arcs". These shore arcs can never consist of a single point, since no two unit circles may be inside any great circle (the packing is r-approachable).

To prove the theorem we need consider only a single island containing m circles (see Fig. 31) and show that the area 1(m) of this island satisfies the inequality

1(m) > mA + 2r(( 2 r)2 -/i3

This assertion is proved by first obtaining a lower bound on the angle of intersection of the shore arcs; this lower bound is w.

FIG. 31

We now consider the Dirichlet cells of the unit circles. (The Dirichlet cell of a circle consists of those points of the island which are nearer to the circle in question than to any other circle. These cells will thus be the intersections of the Voronoi polygons of the circles with the island.) A Dirichlet cell is said to be "good" if on the great circle there exists an arc of central angle 2t/3 - w which has no common point with the shore arcs. If no such arc exists, the cell is said to be "bad".

If the cells are all good, we apply Lemma 2 to each cell and add to obtain

1(m)> m{ { - ]w + 3(1 + -)+ (r2 + 2r)1/2}

+ 2 {(1 ) r

3} .




Repeating for each island and using

Ar = L 2 - w + /3(1 +-) + (r2 + 2r)'12,

we obtain the required result and prove the theorem. If some of the cells are bad, we perform a reduction which allows us to consider

modified cells to which Lemma 2 will apply. The theorem is then proved in all cases. If we take the special case r = 1, when w now becomes 7r/3, we have

A1 = (2/3)ir + 23/3 and D1 > n[(2/3)n + 2X/3] + 4i - 2,/3. Since the density of the packing in the plane must satisfy br < it/Ar, we have

< = .56518 ... 2? + 6 /

If one compares this figure with that obtained in the "best expected arrangement", i.e., double rows of touching circles divided by narrow roads, an interesting result is obtained. In the expected arrangement the density of packing is (n/,/12) (5 - 1)/2 = .56050..., while we have shown that under somewhat weaker conditions an upper bound to the packing density is .56518 ... . This suggests that the traditional parking arrangement is in fact the best one.

6. Conclusion. To our knowledge, this is the first such general and practically oriented paper on geometric number theory and the geometry of numbers (although we also make frequent use of results in discrete geometry). The story is by no means complete, but we hope to have demonstrated that results from these subjects provide useful techniques for optimization problems, and that we have stimulated the reader to think about a subject generally regarded as highly esoteric and theoretical.

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Optimization and the Geometry of Numbers: Packing and Covering