people.math.carleton.capeople.math.carleton.ca/~robertb/thesis.pdf · 5 3.3 A worked example in GAP . . . . . . . . . . . . . . . . . . . . . 42 4 Base-transitive groups of rank 3

Permutation Groups,

Error-Correcting Codes

and Uncoverings

Robert Francis Bailey

Queen Mary, University of London

Thesis submitted to the University of London for the degree ofDoctor of Philosophy

2nd November 2005

2

Abstract

We replace the traditional setting for error-correcting codes (i.e. linear codes) withthat of permutation groups, with permutations in list form as the codewords. Weintroduce a decoding algorithm for these codes, which uses the following notion.A basefor a permutation group is a sequence of points whose stabiliser is trivial.An uncovering-by-bases(or UBB) is a set of bases such that any combination oferror positions is avoided by at least one base in the set. In the case of sharplyk-transitive groups, anyk-tuple of points forms a base, so a UBB can be formedfrom the complements of the blocks of a covering design. (In this case, we use thetermuncovering.)

A large part of the thesis (chapters 2 to 5) is concerned with constructingUBBs for groups which arebase-transitive, i.e. which act transitively on theirirredundant bases, which were classified by T. Maund. Various combinatorial,algebraic and number-theoretic techniques are employed in this. Other topics in-clude a case study of the Mathieu groupM12, where we investigate ways in whichwe can improve the performance of our algorithm, and how the group behavesas a code when number of errors received exceeds the correction capability. Wealso develop an alternative algorithm which works for the groupCmoSn, and com-pare its complexity with the original. There is also a chapter about what can besaid for more general permutation groups when regarded as codes in this manner,where we formulate a conjecture (the “single orbit conjecture”) about UBBs forpermutation groups in general, which we prove for some special cases.

For my grandparents

D. M. J. (25. 9. 1920–30. 5. 2003)

H. O. W. J. (13. 2. 1913–19. 9. 2003)

3

Contents

1 Introduction 10

1.1 The idea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.2 Sharplyk-transitive groups . . . . . . . . . . . . . . . . . . . . . 13

1.3 The algorithm: uncoverings . . . . . . . . . . . . . . . . . . . . . 14

1.4 Generalising . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.5 Why base-transitive groups? . . . . . . . . . . . . . . . . . . . . 20

1.6 Where next? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2 Infinite families of arbitrary rank 26

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.2 Wreath products of regular groups . . . . . . . . . . . . . . . . . 26

2.3 Zero-sum subgroups . . . . . . . . . . . . . . . . . . . . . . . . 28

2.4 GL(n,q) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.5 AGL(n,q) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

2.6 Comparison ofGL(n,q) andAGL(n,q) . . . . . . . . . . . . . . . 38

3 Base-transitive groups of rank 2 39

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.2 A general construction . . . . . . . . . . . . . . . . . . . . . . . 39

4

5

3.3 A worked example inGAP . . . . . . . . . . . . . . . . . . . . . 42

4 Base-transitive groups of rank 3 45

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4.2 An uncovering by triples . . . . . . . . . . . . . . . . . . . . . . 45

4.3 “Blow-ups” ofPGL(2,q) . . . . . . . . . . . . . . . . . . . . . . 48

4.4 Background material on finite fields . . . . . . . . . . . . . . . . 52

4.5 Uncoverings-by-bases forGL(3,q) . . . . . . . . . . . . . . . . . 55

4.6 Uncoverings-by-bases forAGL(2,q) . . . . . . . . . . . . . . . . 62

5 Exceptional base-transitive groups 69

5.1 The action ofA7 on 15 points . . . . . . . . . . . . . . . . . . . . 69

5.2 Two groups related toA7 . . . . . . . . . . . . . . . . . . . . . . 70

5.3 Two Mathieu groups,M11 andM12 . . . . . . . . . . . . . . . . . 71

6 Further topics: some case studies 74

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

6.2 Patterns of repeated symbols . . . . . . . . . . . . . . . . . . . . 74

6.3 Optimising the algorithm . . . . . . . . . . . . . . . . . . . . . . 77

6.3.1 A measure of optimality . . . . . . . . . . . . . . . . . . 77

6.3.2 Nested uncoverings . . . . . . . . . . . . . . . . . . . . . 81

6.3.3 Calculating the expected duration of the algorithm . . . . 83

6.3.4 Some concluding remarks . . . . . . . . . . . . . . . . . 89

6.4 M12: words with four errors . . . . . . . . . . . . . . . . . . . . . 89

6.5 M12: more than four errors . . . . . . . . . . . . . . . . . . . . . 93

6

6.6 Cm oSn: an alternative algorithm . . . . . . . . . . . . . . . . . . 95

6.7 Cm oSn: more thanr errors . . . . . . . . . . . . . . . . . . . . . 98

7 Other groups: the single orbit conjecture 103

7.1 The conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

7.2 Direct and wreath products . . . . . . . . . . . . . . . . . . . . . 104

7.3 Sm acting on 2-subsets . . . . . . . . . . . . . . . . . . . . . . . 106

7.4 Another family of groups . . . . . . . . . . . . . . . . . . . . . . 118

7.5 A stronger conjecture . . . . . . . . . . . . . . . . . . . . . . . . 119

8 Complexity issues 121

8.1 Complexity of the ‘uncovering-by-bases’ algorithm . . . . . . . . 121

8.2 Improvements . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

8.3 Complexity of the alternative algorithm . . . . . . . . . . . . . . 125

8.4 Comparison of the two algorithms . . . . . . . . . . . . . . . . . 127

A SomeGAP programs 129

A.1 The decoding algorithm for sharplyk-transitive groups . . . . . . 129

A.2 The “uncovering-by-bases” decoding algorithm . . . . . . . . . . 130

A.3 Programs for investigatingM12 . . . . . . . . . . . . . . . . . . . 130

A.4 The “alternative” decoding algorithm . . . . . . . . . . . . . . . . 133

A.5 Programs for verifying the single orbit property . . . . . . . . . . 135

Bibliography 137

Acknowledgements

I should begin for thanking my supervisor, Peter Cameron, for all his help andencouragement during the writing of this thesis, being an endless supply of ideas,coping with my inability to understand most of them, and tolerating me spendingthe first two years turning up to supervisions with blank sheets of paper.

This work would not have been possible without the CASE studentship pro-vided by the EPSRC and GCHQ. I should thank Cliff Cocks and Richard Pinchfrom GCHQ for their interest in this work, and also for not insisting that weworked on “Trelllises in Communication Theory” (the original project title) whichbears pretty much no resemblance whatsoever to the contents of the next 165pages.

I should also thank the members School of Mathematical Sciences at QueenMary, University of London, for making the place such an enjoyable, and stim-ulating, place to work. So many people there have shown an interest or helpedout that there are too many such people to all be listed individually, but some ofthem deserve a particular mention: Rosemary Bailey, my second supervisor (es-pecially for the apparently limitless confusion over the rightful ownership of thename ‘Bailey’), Leonard Soicher (who wouldn’t let me get away without includ-ing chapter 8), Bill Jackson, John Bray, Thomas Prellberg and many more.

I’m grateful to Sophie Huczynska at the University of St Andrews for takingan interest in what I’ve been doing, and in particular for a most useful discussionon finite fields. I’m also grateful to Jonathan Dixon and Mike Newman for helpingout with that thankless task, proofreading.

The past three-and-a-bit years would not have been the same without my fel-low occupants of rooms 201, 202 and 203. Again, they have been too numerousto all be listed, but again some are worthy of a particular mention: Jan Grabowskiand Jonathan Dixon for all the overly-detailed cricket/LATEX/London Transport-related conversations (which frequently would drive others away), John Arhin forpatiently putting up with my inability to useGAP, Cheng Yeaw Ku for his flipping

7

ACKNOWLEDGEMENTS 8

door, Matt Ollis for the alcohol supply, Becky Thorn for being utterly bonkers(but in a nice way), Debbie Lockett for inadvertantly stalking me, and RebeccaLodwick for helping to preserve my sanity in room 203, when all around weredescending into madness.

Of course, I can’t possibly forget those who convinced me that being a puremathematician was what I wanted to be, namely people at the University of Leedslike Dugald Macpherson, John McConnell, Chris Robson, John Truss and (mostof all) R. B. J. T. (Reg) Allenby.

Finally, I should thank my family for putting up with me whilst doing all ofthis (rather than making me get a proper job); none of them will understand aword of this, but hopefully will still be impressed by it all the same. While I’m atit, I should probably also thank my hairdresser, my therapist,. . . . . .

Robert BaileyLondon, 2nd November 2005

The results in this thesis, other than those clearly marked, are the unaided work ofthe author.

Signed:

Date:

9

Chapter 1

Introduction

1.1 The idea

This thesis is concerned with error-correcting codes. However, the usual notion oflinear codes, which are vector spaces over finite fields, does not feature. Instead,we will use a group of permutations as the code, where the codewords are groupelements written inlist form, with the usual Hamming distance. We begin byclarifying what we mean by this.

In fairly broad generality, we have the following definition.

Definition 1.1.1. An (error-correcting) codeis a setC of strings of symbols,calledcodewords, chosen from some alphabet.

If the strings are all of the same length, then we can make the following defi-nition.

Definition 1.1.2. Let x = x1x2 · · ·xn andy = y1y2 · · ·yn be two codewords. ThentheHamming distanceis

dH(x,y) = #i : xi 6= yi ,

i.e. the number of co-ordinates in whichx andy differ.

This then leads naturally to the next definition.

Definition 1.1.3. Theminimum distanceof a codeC is

d(C) = minx,y∈Cx6=y

dH(x,y),

i.e. the least value ofdH over all pairs of words inC.

10

INTRODUCTION 11

This is important for error correction, as if a codeword is transmitted and thereceived word contains errors, then as long as there are at mostr = bd(C)−1

2 c errors,there will be a unique nearest neighbour inC and we can decode correctly. Wecall this parameterr thecorrection capabilityof C.

In our situation, our codewords are strings of lengthnchosen from the alphabet1, . . . ,n, i.e. permutations written in list form. For example,

231794685

is an element ofS9. Furthermore, in this context the Hamming distance has auseful alternative interpretation.

Proposition 1.1.4.For g,h∈ Sn,

dH(g,h) = n−∣∣Fix(gh−1)

∣∣whereFix(g) denotes the set of fixed points of g.

Proof. By applyingh−1 to bothg andh, dH(g,h) = dH(gh−1, Id) is the numberof placesgh−1 differs from the identity, which isn−

∣∣Fix(gh−1)∣∣.

Thus the minimum distance of a setC of permutations is equal to

ming,h∈Cg6=h

n−∣∣Fix(gh−1)

∣∣= n− maxg,h∈Cg6=h

∣∣Fix(gh−1)∣∣ .

When the set of permutations forms a groupG, this becomes

n−maxg∈Gg6=1

|Fix(g)| ,

which is known to group theorists as theminimum degreeof G. There is an anal-ogy here to the theory of linear codes, in that the minimum distance of a linearcode is equal to its minimum weight, i.e. the distance from the all-zero codeword,which plays the role of the identity permutation in that setting.

One way to calculate the minimum distance (i.e. minimum degree) is to usecharacter theory. For any groupG acting on a finite setΩ of sizen, we define thepermutation representationof G to be the map

ρ : G → GLn(C)g 7→ P(g)

INTRODUCTION 12

where

P(g)i j =

1 if g(i) = j0 otherwise.

This defines a representation in the usual sense (see, for example, [26] or chapter2 of [7]). Then the function

π : G → Cg 7→ Trace(P(g))

is called thepermutation characterof G. Soπ(g) is the number of ones on thediagonal of the matrixP(g), i.e. the number of points inΩ fixed byg. Thus whenwe consider the elements ofG to be permutations written in list form, we havethatdH(g, Id) = n−π(g). So therefore the minimum distance (minimum degree)is

n−maxg∈Gg6=1

π(g).

For any code, we can define a polynomial which “counts” the number of code-words at a each distance from a fixed codeword, which is as follows.

Definition 1.1.5. Let C be a code of lengthn and minimum distanced. Then fora fixed codewordc∈C we define thedistance enumerator,

∆c(x) = ∑w∈C

xdH(c,w).

Clearly, the coefficient ofxi gives the number of codewords at distancei fromc. In the case whereC is a permutation group, we obtain the same polynomialregardless of the choice ofc (because each codeword can be mapped to any otherby a permutation), so we can takec to be the identity permutation and call thepolynomial∆(x). In the case whereC is a linear code, we takec to be the zerovector and obtain theweight enumerator1. This gives us another analogy withlinear codes.

Also, we can rephrase the distance enumerator in terms of the permutationcharacter. As is well-known (see for instance [26], proposition 13.5), charactersare constant on conjugacy classes. So using the notation above, the distance enu-merator becomes

∆(x) = ∑g∈R

|Class(g)|xn−π(g),

1Usually this is given as a two variable polynomial,W(x,y) = ∑w∈C xn−wt(w)ywt(w), wherewt(w) = dH(w,0): see [12], page 121.

INTRODUCTION 13

whereR is a set of conjugacy class representatives forG, and Class(g) denotes theconjugacy class containingg.

Example 1.1.6.Let G be the alternating groupA5. Using the character table intheATLAS [18], page 2, we obtain the polynomial

∆(x) = 1+20x3 +15x4 +24x5.

Note that by evaluating∆(1), we obtain|G|.

The idea of using permutation groups as codes in this way is not new; in fact itwas first decribed in 1974 by Blake [3], while Blake, Cohen and Deza developedthe idea further in their 1979 paper [4]. To begin with, we shall concentrate on thesame groups as [3], namely sharplyk-transitive groups.

1.2 Sharplyk-transitive groups

SupposeG is a permutation group acting onΩ = 1, . . . ,n (soG hasdegree n).Then we sayG is sharply k-transitiveif for any two orderedk-tuples of distinctelements ofΩ, there is a unique group element mapping the first to the second. Asharply 1-transitive group is aregular permutation group, so we restrict ourselvesto the casek ≥ 2. As Blake points out in [3], the minimum distance of such agroup is easy to calculate.

Proposition 1.2.1.Let G be a sharply k-transitive permutation group of degree n.Then the minimum distance of G is n−k+1.

Proof. By definition, only the identity element can fixk points, so the maximumnumber of fixed points of a non-identity element isk−1. (It is easy to show thatsuch an element exists, for instance mapping(x1, . . . ,xk−1,xk) to (x1, . . . ,xk−1,y).)Hence the minimum distance isn− (k−1) = n−k+1.

Example 1.2.2.The symmetric groupSn is both sharplyn-transitive and sharply(n−1)-transitive, while the alternating groupAn is sharply(n−2)-transitive. Thusthe minimum distance ofSn is n− (n−1)+1 = 2, and the minimum distance ofAn is n− (n−2)+1 = 3.

Thus, regarded as a code, the symmetric group is not much use, as its correc-tion capability isb2−1

2 c= 0. The correction capability of the alternating group is

INTRODUCTION 14

1. However, there are some more interesting examples. In fact, a complete clas-sification of sharplyk-transitive groups (fork ≥ 2) is available. Fork ≥ 4, it isa classical result of Jordan from 1873 [28], and the classification was completed(for k = 2 and 3) by Zassenhaus [39, 40] in 1936. A sharplyk-transitive groupmust be one of the following.

k≥ 6: Sk, Sk+1 andAk+2 only.

k = 5: S5, S6, A7 and the Mathieu groupM12.

k = 4: S4, S5, A6 and the Mathieu groupM11.

k = 3: The group

PGL(2,q) =

τ : x 7→ ax+bcx+d

∣∣∣∣ a,b,c,d ∈ Fq,ad−bc 6= 0

acting on the “projective line”Fq∪∞ (which includesS3

∼= PGL(2,2),S4∼= PGL(2,3), A5

∼= PGL(2,4)), plus an additional infinite family as de-scribed by Cameron [7], page 16.

k = 2: The group

AGL(1,F) = τ : x 7→ ax+b | a,b∈ F,a 6= 0

whereF is a finite near-field, acting onF . In the case whereF = Fq, weuse the notationAGL(1,q). Note thatAGL(1,2) ∼= S2, AGL(1,3) ∼= S3 andAGL(1,4)∼= A4.

A near-fieldis an object which satisfies all of the axioms of a field, with theexception of the commutativity of multiplication and a left distributive law. Soany field is a near-field, and there is also an additional infinite family of finitenear-fields plus seven “exceptional” examples. All have prime-power order, andare described by Cameron [7], page 16. A detailed account of the proof of thisclassification is given by Dixon and Mortimer [20], section 7.6. The MathieugroupsM11 andM12 are perhaps the most interesting examples of all, and will bediscussed in more detail in later chapters.

1.3 The algorithm: uncoverings

Regarded as a code, the correction capability of a sharplyk-transitive groupG ofdegreen is r = bd−1

2 c, whered = n−k+1. So our decoding method will assume

INTRODUCTION 15

that at mostr errors have occurred, i.e. that at leastn− r symbols are correct.Now, anyk-tuple of these correct symbols will uniquely identify the codeword.However, we don’t know in whichr positions the errors are, so we need a way ofchoosing a set ofk-tuples so that we can becertain that at least one contains noerrors. That is, we need a set ofk-subsets of1, . . . ,n such that anyr-subset ofX is disjoint from at least onek-set. More formally, we have the following.

Definition 1.3.1. A set U of k-subsets ofΩ = 1, . . . ,n is called an(n,k, r)-uncoveringif it has the property that for anyr-subsetR⊂ Ω, there existsS∈ Usuch thatR∩S= /0.

Provided thatk≤ n− r, an uncovering will always exist, by taking the set ofall k-subsets ofΩ. So the problem becomes one of finding the smallest possibleuncovering for a given set of parameters. We call an uncovering of least size aminimal(n,k, r)-uncovering.

The complementary problem of finding a set ofm-subsets ofΩ such that everyr-subset ofΩ (wherer < m) is containedin at least onem-set is of interest indesign theory. Such a set ofm-sets is referred to as an(n,m, r) covering design,and them-sets are known asblocks. So by taking the the complements of theblocks of an(n,m, r) covering design, one obtains an(n,n−m, r)-uncovering.For this reason, we use the termcoblocksfor the subsets in an uncovering.

Consequently, the extensive literature on covering designs is of use to us infinding uncoverings. For instance, there is a large internet database of cover-ing designs with small parameters maintained by Gordon, the “La Jolla CoveringRepository” [22]. Many of the constructions featured in the database are describedin the paper of Gordon, Kuperberg and Patashnik [23], while a more general sur-vey can be found in Mills and Mullin [33].

Example 1.3.2.For the sharply 3-transitive groupPGL(2,7), we haven = 8,k = 3, r = b8−3

2 c= 2, so need an(8,3,2)-uncovering, as shown below.

1 2 3

4 5 6

2 3 7

1 7 8

The above example was obtained from an(8,5,2) covering design in Gordon’sdatabase [22]. It is small enough for the “uncovering” property to be verifiedeasily by hand.

INTRODUCTION 16

For a given sharplyk-transitive groupG, once we have a suitable uncoveringwe can apply the following decoding procedure.

Algorithm 1.3.3. Suppose we have received the wordw = w1w2 · · ·wn. Take ak-setSand look at the entrieswi for i ∈ S. First check that there are no repeatedsymbols in these positions. If not, then becauseG is sharplyk-transitive, thereis a unique permutation mapping eachi to wi for i ∈ S; that is, there is a uniquecodeword with entrywi in positioni for i ∈ S. So we search through the codebook(i.e. the list of elements ofG) until we find it. We then check the distance betweenthis codeword and the received wordw. If it is within distancer then we concludethat this must be the transmitted word and stop the procedure. If not, we considerthe nextk-set of co-ordinate positions and repeat the procedure. (If at any stage ak-tuple of entries includes a repeated symbol, then we know that there must be anerror here and proceed immediately to the nextk-set.)

A diagram which describes this algorithm is given in figure 1.1. It has beenimplemented in the computer systemGAP [21], and is given in appendix A.1.

START

NO

Identify correspondingcodeword in codebook

Within

of received word?

STOP

YES

Choose first k−tuple

Choose next k−tuple

distance r

Figure 1.1: The decoding algorithm

Example 1.3.4.We continue with the example ofPGL(2,7), which is generatedby the permutations(3 8 7 6 5 4) and(1 2 6)(3 4 8) (in disjoint cycle form). Sup-

INTRODUCTION 17

pose we transmit the permutation

g = 1 2 3 4 5 6 7 8

and that the following word is received:

w = 4 2 3 6 5 6 7 8.

(This has two errors, in positions 1 and 4.) Using the uncovering in example1.3.2, we first identify the element ofPGL(2,7) which maps the 3-tuple(1,2,3)to (4,2,3), which is

4 2 3 6 8 7 5 1.

This is distance 4 fromw, so is rejected. Then we look at the next 3-tuple in theuncovering, which is(4,5,6). In w, these positions contain the symbols(6,5,6),so clearly no permutation can exist here. So we move to the next 3-tuple,(2,3,7);these positions contain the symbols(2,3,7). Thus we find the element whichmaps the first to the second, which is

1 2 3 4 5 6 7 8.

This is distance 2 fromw, so is accepted, and the algorithm terminates.

In this thesis, we find suitable uncoverings for all the sharplyk-transitivegroups, with the exception of the symmetric group in its natural action onnpoints. (Since the correction capability ofSn is zero, this is a degenerate case.)The alternating group forms a 1-error correcting code, so for eachn we require an(n,n−2,1)-uncovering.

Construction 1.3.5. We need a set of(n−2)-subsets of1, . . . ,n with the prop-erty that every point lies outside of one of the subsets. Alternatively, we needan (n,2,1) covering design. Ifn is even, then we can take12n disjoint pairs, andwe are done. Ifn is odd, then we take12(n−1) disjoint pairs, then a pair whichcontains the remaining point and any other point. Then to obtain an(n,n−2,1)-uncovering, we take the complement of each pair.

Example 1.3.6.For the alternating groupA7, we need a(7,5,1)-uncovering. Us-ing the construction above, we obtain:

3 4 5 6 7

1 2 5 6 7

1 2 3 4 7

2 3 4 5 6

INTRODUCTION 18

Uncoverings for the remaining sharplyk-transitive groups will be given inlater chapters.

We conclude this section by mentioning that the decoding method we give hereis not the already-known method ofpermutation decoding. This method, whichis attributed to F. J. MacWilliams, is a decoding method used for linear codes,involving finding a subset of the automorphism group of the code to move any setof errors out of the “information positions”. A full description of this method isgiven in the survey article by Huffman [25] in theHandbook of Coding Theory.While this method is also related to covering designs, it differs from our methodin that it involves finding a set of group elements, rather than a set of subsets ofthe domain of a permutation group.

1.4 Generalising

Algorithm 1.3.3, as stated in the previous section, works only for sharplyk-transitive groups. However, as the classification of these is quite restricted, thenwe would like to be able to utilise this algorithm for other groups. The followingdefinition, originally due to Sims [37], is fundamental here.

Definition 1.4.1. Let G be a group acting on a finite setΩ. A basefor G inthis action is a sequence of points(x1, . . . ,xb) from Ω such thatG(x1,...,xb) = 〈1〉,i.e. the pointwise stabiliser is the identity. Anirredundantbase is a base whereG(x1,...,xi ,xi+1) 6= G(x1,...,xi) for i = 1, . . . ,b−1.

For instance, in a sharplyk-transitive group, any sequence ofk points formsan irredundant base. The next example is also straightforward.

Example 1.4.2.Consider the general linear groupGL(n,q) acting on the non-zerovectors inFn

q. Then a basis for the vector spaceFnq forms an irredundant base for

GL(n,q).

We saw in the previous section that in a sharplyk-transitive group, the ac-tion of a group element on ak-tuple uniquely determines that element. The nextproposition shows that bases capture this property.

Proposition 1.4.3. For any group G, the action of an element g∈ G on a base(x1, . . . ,xb) uniquely determines that element; that is, if(x1, . . . ,xb)g =(x1, . . . ,xb)h,then g= h.

INTRODUCTION 19

Proof. Supposeg,h ∈ G, (x1, . . . ,xb) is a base forG and thatxgi = xh

i for each

i. Thenxgh−1

i = xi for eachi, that isgh−1 ∈ G(x1,...,xb) = 〈1〉. Hencegh−1 = 1,i.e.g = h.

So, if a groupG is to be used as a code, if the received word contains errors inpositions outside those labelled by a base, we can decode successfully. Thereforeto apply our algorithm we should replace “k-tuple” by “base”. Consequently, wereplace “uncovering” with the following.

Definition 1.4.4. SupposeG is a group acting onΩ, where|Ω|= n, with correc-tion capabilityr. Then anuncovering-by-basesfor G is a set of bases forG suchthat anyr-subset ofΩ is disjoint from at least one base.

The revised algorithm, which uses uncoverings-by-bases rather than uncover-ings, is essentially the same as algorithm 1.3.3. It has two subtle modifications:first, it allows for the possibility that the bases in the uncovering have differentsizes; second, because the existence of a group element agreeing with the receivedword in the positions labelled by a given base is no longer guaranteed, we includea check for non-existence. As it is of fundamental importance in this thesis, wedescribe it explicitly below.

Algorithm 1.4.5. Suppose we have a groupG, an uncovering-by-basesU andhave received the wordw = w1w2 · · ·wn. Take the first baseB∈U and look at thepositions inw indexed by the entries ofB. If they are all distinct, then we checkif there exists an elementg∈ G which agrees withw in those positions. If so, wethen check ifg is within distancer from w. If this happens, then we conclude thatg was the transmitted word and stop. Otherwise, we move to the next base inUand repeat the procedure.

This algorithm has been implemented inGAP and is given in appendix A.2.The procedure used to determine the element agreeing with the images of a baseis a standard technique in computational group theory, and is described in section8.1, where we determine the complexity of algorithm 1.4.5.

Clearly, in the case whereG is sharplyk-transitive, an uncovering-by-basesis just an(n,k, r)-uncovering, as defined above. Unlike that situation, in generalit is not immediately obvious that for an arbitrary group an uncovering-by-basesshould exist. However, we are saved by the next result.

Proposition 1.4.6.For any finite group G acting on a setΩ with |Ω| = n, therealways exists an uncovering-by-bases.

INTRODUCTION 20

Proof. Let d be the minimum distance ofG, so r = bd−12 c. We show that for

an arbitraryr-subset ofΩ, there exists a base forG disjoint from it, arguing bycontradiction.

Suppose there exists anr-subsetR⊆ Ω that meets every base forG. Then thepointwise stabiliser ofR= Ω\R is non-trivial, asRdoes not contain a base. There-fore there exists a non-identity elementg that fixesR pointwise, so|Fix(g)| ≥|R|= n− r. But the maximum number of fixed points of a non-identity element isn−d < n− r, giving a contradiction.

We remark that the definition of uncovering-by-bases, and indeed the proof ofproposition 1.4.6, is vacuous in the caser = 0. Although groups with zero correc-tion capability are, of course, useless as error-correcting codes, an uncovering-by-bases for such a group consists of a single base only.

Proposition 1.4.6 is, of course, only an existence result: it gives us no idea asto how to set about finding them. A sensible starting point would be groups whosebase structure is known.

1.5 Why base-transitive groups?

The following definition is due to Cameron and Fon-Der-Flaass [11].

Definition 1.5.1. A permutation group in a given action is called anIBIS groupifall irredundant bases have the same size.

The acronym stands for “irredundant bases of invariant size”. Examples ofIBIS groups include the sharplyk-transitive groups and general linear groups wementioned above. The easiest example of a group that isnot an IBIS group is thesymmetric group acting on 2-subsets; this will be decribed in chapter 7. In their1995 paper [11], Cameron and Fon-Der-Flaass proved the following theorem.

Theorem 1.5.2.The following statements are equivalent:

1. G is an IBIS group;

2. the irredundant bases of G are preserved by re-ordering;

3. the irredundant bases of G form the bases of a matroid.

INTRODUCTION 21

A matroid is a set together with a family of subsets, calledindependent sets,which satisfy certain axioms generalising the notion of linear independence ina vector space. (A full treatment is given in Oxley’s book [35].) The maximalindependent sets are calledbasesand the size of a base is therank of the matroid.(Bases must all have the same size: see [35], page 16.)

If G is an IBIS group, we call the size of an irredundant base therank of G, asthis is the rank of the corresponding matroid. It is the second property in theorem1.5.2 that is of use to us here: if we know that all the irredundant bases ofG havethe same size, then we can order them in any way we choose, and so can regard abase as a set rather than a sequence. This should hopefully ease the constructionof an uncovering-by-bases.

Unfortunately, a complete classification of IBIS groups seems unlikely. Thanksto a construction due to Cameron [8], any linear code gives rise to an IBIS group,so a classification of IBIS groups would include a classification of linear codes.However, in a more special situation, a complete classification is available.

Definition 1.5.3. If a group acts transitively on its irredundant bases, we say it isabase-transitivegroup.

Obviously, sharplyk-transitive groups are base-transitive, and base-transitivegroups are IBIS groups. In the literature, the rather ambiguous name ofgeomet-ric groupsis often used for base-transitive groups. This name is due to Cameronand Deza [9]; it arises from the fact that the associated matroids arepermuta-tion geometries. Base-transitive groups of rank 1 are regular permutation groups.However, for rankk ≥ 2, a complete classification of base-transitive groups isknown. Part of the classification involves determining thetype of each base-transitive group, which is defined as follows.

Definition 1.5.4. Let G be a base-transitive group of degreen and rankk, andlet (x1, . . . ,xk) be an irredundant base forG. Then thetype of G is the pair(l0, l1, . . . , lk−1,n), wherel0 is the number of fixed points ofG, and l i is thenumber of fixed points ofG(x1,...,xi) for 1≤ i ≤ k−1.

Sincelk−1 is the maximum number of fixed points of a non-identity element,we can therefore obtain the correction capability ofG from its type.

Essentially a base-transitive group falls into one of the following cases:

• “generic” examples for arbitrary rank;

INTRODUCTION 22

• infinite families for ranks 2 and 3;

• “sporadic” examples of rank≤ 5.

The “generic” examples are as follows:

(i) Sn, which has rankn−1 and type(0,1, . . . ,n−2,n);

(ii) An, rankn−2, type(0,1, . . . ,n−3,n);

(iii) G = H oSn, whereH is a regular group of degreem, soG has rankn andtype(0,m, . . . ,(n−1)m,nm);

(iv) the semidirect productG = X oSn, where

X = (a1,a2, . . . ,an) ∈ An | a1 +a2 + · · ·+an = 0

(A is an abelian, regular group of degreem, written additively), soG hasrankn−1 and type(0,m, . . . ,(n−2)m,nm);

(v) GL(n,q), rankn, type(0,q−1,q2−1, . . . ,qn−1−1,qn−1) (i.e. the gen-eral linear group);

(vi) the stabiliser ofd independent vectors inGL(n,q), where 0< d < n, whichhas rankn−d and type(0,(q−1)qd, . . . ,(qn−1)qd,(qn−1)qd);

(vii) AGL(n,q), rankn+ 1, type(0,q,q2, . . . ,qn−1,qn) (i.e. the affine generallinear group);

(viii) the groupV oH, whereV is the additive group ofFnq andH is the group in

(vi), this has rankn−d+1 and type(0,qd+1, . . . ,qn−1,qn).

Zil’ber [41] showed that for rank at least 7, there are only “generic” examples.The classification for ranks 2 to 6 is due to Maund [31]: we summarise her resultbelow. In addition to the appropriate “generic” examples, we have the followinggroups.

–Rank 2

• the sharply 2-transitive groups, which have type(0,1,n) (wheren is aprime power);

• C(q−1)/2×PSL(2,q) for q≡ 3 mod 4, of type(0, q−12 , (q2−1)

2 );

INTRODUCTION 23

• Cq−1×Sz(q), type(0,q−1,(q−1)(q2+1)) (whereq is an odd power of2);

• PSL(3,2), type(0,2,14);

• PSL(3,3), type(0,6,78).

In the last four cases, the base-transitive action is on the set of right cosets ofparticular subgroups.

–Rank 3

• PGL(2,q) and the other sharply 3-transitive groups, type(0,1,2,q+1);

• “blow-ups” of PGL(2,q), type(0,qd,2qd,qd(q+1)) (see below);

• A7 acting onF42\0, type(0,1,3,15);

• V o H, whereV is the additive group ofF42 andH is the point-stabiliser of

A7 in the above action, type(0,2,4,16).

–Ranks 4, 5 and 6

• V oA7 (whereV is as above), rank 4, type(0,1,2,4,16);

• M11, rank 4, type(0,1,2,3,11);

• M12, rank 5, type(0,1,2,3,4,12).

Unlike Zil’ber, Maund’s proof uses the classification of finite simple groups.It involves first classifying the base-transitive groups of rank 2 (by using the clas-sification of 2-transitive groups), and then using induction to analyse the higherranks. Unfortunately, neither Maund nor Zil’ber’s work has ever been publishedin a particularly accessible form: Cameron [6] contains a brief survey.

One important notion in the study of base-transitive groups is that of ablow-up. The construction was introduced by Cameron, Deza and Frankl [10] for per-mutation geometries. In the language of groups, it is as follows.

Definition 1.5.5. Let G be a base-transitive group of rankk and type(L,n). Thena blow-upof G is a base-transitive groupG of rankk and type(mL,mn) (wheremL= ml | l ∈ L), such that:

INTRODUCTION 24

(i) G has a normal subgroupN which hasn orbits of sizem,

(ii) G/N ∼= G,

(iii) G induces the given action ofG on theN-orbits.

From the third property, it follows that a base forG consists ofk points chosenfrom k “independent”N-orbits, i.e. from orbits which form a base for the actionof G induced on them.

The symmetric groupSn is something of an anomaly here, as although it hasrankn−1 and type(0,1, . . . ,n−2,n), when it comes to construct blow-ups, itbehaves as if it also has an action of rankn and type(0,1, . . . ,n−1,n). (This isbecause it is both sharply(n−1)-transitive and sharplyn-transitive.) We see thisin the following example.

Example 1.5.6.Consider the entry (iii) in the list of “generic” base-transitivegroups. Here we haveG = Sn andG = H oSn, whereH is a regular group of orderm. SinceH oSn meansHn oSn, we takeK = Hn (the direct product ofn copies ofH). SoK hasn orbits of sizem, G/K ∼= Sn, andG induces the usual action ofSn

on these orbits.

In the list of “generic” examples, entries (iv), (vi) and (vii) are all blow-ups, ofSn, GL(n−d,q) andAGL(n−d,q) respectively. There is also a family of blow-upsof PGL(2,q), but these are quite complicated to describe.

Most of this thesis (chapters 2, 3, 4 and 5) is devoted to the construction ofuncoverings-by-bases for base-transitive groups, employing a variety of methods.

1.6 Where next?

Although most of this thesis is devoted to base-transitive groups, and constructinguncoverings-by-bases for them, we also consider some other topics.

In chapter 7, we consider permutation groups in more generality. In partic-ular, we formulate a conjecture (the “single orbit conjecture”) which asserts thatany finite permutation permutation groupG not only has an uncovering-by-bases,but one which is contained in an orbit ofG on irredundant bases. While we do notprove this conjecture (which is probably quite hard in fullest generality), we prove

INTRODUCTION 25

some reduction theorems and also consider some special cases. One of these spe-cial cases (the action ofSm on 2-subsets of1, . . . ,m) uses techniques borrowedfrom the theory of graph decompositions. One of the questions left open at theend of this thesis is to prove this conjecture, and some stronger versions describedin section 7.5.

In chapter 6 we investigate some different aspects of this theory of using per-mutation groups as codes. For instance, we investigate methods of improving thedecoding algorithm, usually with the aid of the Mathieu groupM12 as a case study.Section 6.2 simply investigates how patterns of repeated symbols can be used toincrease the efficiency of the decoding algorithm.

Section 6.3 considers “optimality” in more detail, discussing exactly what“optimality” means here, ways to measure it, and what steps could be taken toimprove it. This in turn gives rise to some interesting questions about coveringdesigns in general. While it is clear that an(n,m, r) covering design is also an(n,m,s) covering design for alls∈ 1, . . . , r, an optimal(n,m, r) design is notan optimal(n,m, r −1) design, and may not even contain one. This then leads todifferent notions ofnestingcovering designs. There is plenty of scope for futurework in this area also.

Also in chapter 6, we also develop an alternative algorithm which works in aspecial case (the “generic” family of groupsCmoSn). We study how this algorithmcan be utilised to correct more thanr errors (wherer is the correction capabil-ity). We also investigate such a capability forM12 with our original algorithm.Investigating this for other groups is another possibility for future research.

In chapter 8, we determine the complexity of algorithm 1.4.5 and explain whythe “single orbit conjecture” from chapter 7 is beneficial with regard to this. Wealso determine the complexity of the alternative algorithm, and use this to comparethe performance of the two algorithms.

Chapter 2

Infinite families of arbitrary rank

2.1 Introduction

In this chapter we consider some of the “generic” base-transitive groups discussedin the previous chapter in more detail. We determine their parameters as error-correcting codes and construct uncoverings-by-bases for them. We have alreadyseen (in construction 1.3.5) how to find an uncovering-by-bases for the alternat-ing groupAn. Entries (iii) and (iv) on the list (i.e. “blow-ups” ofSn) are dealtwith completely in the following two sections. Finally, we consider the generallinear and affine general linear groups; however, as will be seen, these are onlypartial results, as the uncoverings we construct do not utilise the full correctioncapabilities of these groups.

2.2 Wreath products of regular groups

These are the third entry on the list of “generic” base-transitive groups, and havealready been mentioned in example 1.5.6, as an example of a “blow-up”. LetH beany regular permutation group of orderm. H could be any group of ordermactingon itself by right-multiplication, for instance the cyclic group of orderm. Thenconsider the permutation groupG = H oSn, that is the wreath product ofn copiesof H by the symmetric groupSn. The domain ofG, which we callΩ, consists ofn copies of a set of sizem, which we think of asn columns each ofm rows, asdepicted below.

· · · ...

......

...

26

INFINITE FAMILIES OF ARBITRARY RANK 27

G acts on the columns asSn, and the kernel of this action (i.e. the group fixing allthe columns) isHn.

To apply our decoding algorithm to useG as an error-correcting code, we needto know the structure of the bases ofG, and also the minimum degree ofG. Thesequestions are answered by the two results below. First, however, we need thefollowing definition.

Definition 2.2.1. A transversalof Ω is a set ofn points inΩ, formed of a singlepoint from each of then columns ofΩ.

For example, the shaded points in the diagram below form a transversal.

• • • · · · •...

......

...

Proposition 2.2.2.A transversal ofΩ forms a base for G.

Proof. Let x = (x1, . . . ,xn) be a transversal. To show thatx is a base, we haveto show that the stabiliser of the columns containing points fromx is trivial, andthen thatx is a base forHn.

The first part is clear, since as every column contains a point fromx then thestabiliser of these columns inSn must be the identity. Now suppose thath∈ Hn

lies in the stabiliser ofx. As the action ofHn is regular on each column, if onepoint in a column is fixed by an element, then all points in that column are fixed.It then follows thath must fix all points in all columns, so thereforeh must bethe identity element ofHn. Hencex is a base forHn, and so is also a base forG = H oSn.

It is easy to see thatG acts transitively on the set of all transversals; to mapone to another merely requires applying the appropriate cyclic permutation in eachcolumn. It is also quite easy to determine the minimum degree ofG.

Proposition 2.2.3.The maximum number of fixed points of a non-identity elementof G is(n−1)m, and so the minimum degree of G is m.


Proof. As mentioned in the proof of Proposition 2.2.2, if a point in a column ofΩ is fixed, then all points in that column must be fixed, so therefore fixed pointsoccur in multiples ofm. Since the action is faithful, no non-identity element ofG can fix allnmpoints ofΩ, so the most that can be fixed is(n−1)m. However,we need to show that such an element exists. Consider an element ofG where theelement fromSn permuting the columns is trivial (i.e. all columns stay where theyare), and where all columns except one are fixed. This element has the requirednumber of fixed points. Hence the minimum degree ofG is nm− (n−1)m= m.

Immediately, we have the usual corollary about the number of errors we cancorrect.

Corollary 2.2.4. When used as an error-correcting code, the correction capabilityof G is r= bm−1

2 c.

Thus it remains to construct an uncovering-by-bases for this group. However,this is surprisingly straightforward. Recall from Proposition 2.2.2 that a base forG consists of a transversal ofΩ.

Theorem 2.2.5.A set of r+1 disjoint transversals ofΩ forms an uncovering-by-bases for G.

Proof. For simplicity, we take the firstr +1 rows ofΩ to be our disjoint transver-sals. Now, if there arer errors, these could be spread across many columns, butcan appear in at mostr different rows. Thus at least one of ther +1 transversalsmust avoid the error positions.

A similar technique is required for the related family of groups described inthe next section.

2.3 Zero-sum subgroups

In this section, we consider entry (iv) on the list of “generic” groups. These arealso examples of “blow-ups” ofSn, in its base-transitive action of rankn−1 andtype (0,1, . . . ,n− 2,n). Let A be a finite abelian group of orderm, writtenadditively. DefineX ≤ An as follows:

X = (a1,a2, . . . ,an) | a1 +a2 + · · ·+an = 0.


The group we are interested in isG = X o Sn, where the action ofSn on X isby permuting the copies ofA. We refer toG as thezero-sum subgroupof A oSn.As a permutation group, acting on the copies ofA, this has degreemn, where anelement(a1,a2, . . . ,an) ∈ X acts by componentwise addition.

Proposition 2.3.1. A set n− 1 points, each chosen from a different copy of A,forms a base for G. We call such a base apartial transversalof An.

Proof. We use the same pattern of argument as we did with proposition 2.2.2.Let a = (a1, . . . ,an−1) be a partial transversal. First we consider the action ofSn

on An. As we haven−1 points chosen from distinct copies ofA, if an element ofSn fixes these copies it must also fix the remaining copy. Hence the stabiliser inSn of then−1 copies must be trivial.

Now we show thata is a base forX. Choose someg∈ X which stabilisesa.Sinceg∈ X, we have thatg = (g1,g2, . . . ,gn). By definition, we have

(a1,a2, . . . ,an−1)g = (a1 +g1,a2 +g2, . . . ,an−1 +gn−1)= (a1,a2, . . . ,an−1)

(sinceg lies in the stabiliser of(a1,a2, . . . ,an−1)). Henceg1 = g2 = · · ·= gn−1 =0A. But we then have

n

∑i=1

gi =n−1

∑i=1

gi +gn

= 0A +gn

= gn

(by the definition ofX), so thereforegn = 0A also. Consequently, we haveg =(0A,0A, . . . ,0A). Therefore the stabiliser ofa in X is trivial, soa forms a base forX and thus also does forG = X oSn.

Furthermore, we have this next result.

Theorem 2.3.2.G acts transitively on the set of all partial transversals, so there-fore is base-transitive.

Proof. Given two partial transversalsa = (a1, . . . ,an−1) andb = (b1, . . . ,bn−1),we need to demonstrate the existence of someg∈ G with ag = b.


SinceSn acts transitively on the labels, we can assume botha andb are bothinside the firstn− 1 copies ofA. Thus it also suffices to find someh ∈ X withah = b.

Supposeh = (h1, . . . ,hn), where∑ni=1hi = 0. Then we have

ah = (a1 +h1,a2 +h2, . . . ,an−1 +hn−1) = b = (b1,b2, . . . ,bn−1)

by choosingh1 = b1−a1, . . . ,hn−1 = bn−1−an−1. This subsequently determineshn, as we require the sum to be zero. Thus we have constructed a suitableh, andsoG acts transitively on partial transversals.

As usual, we also need to know the minimum degree ofG.

Proposition 2.3.3.The maximum number of fixed points of a non-identity elementof G is(n−2)m, and so the minimum degree of G is2m.

Proof. SinceA acts regularly, fixed points must occur in multiples ofm. How-ever, unlike the group in section 2, we can’t have(n−1)mfixed points here. Recallfrom the proof of Proposition 2.3.1 thatGhas an “Sn-action” and an “X-action”. IntheSn-action, a non-identity element ofSn has at mostn−2 fixed points, so if therewere an element ofG with (n−1)mfixed points, the “Sn-action” would have to betrivial. So we consider the “X-action”. Recall thatX ≤ An. If g = (g1,g2, . . . ,gn)were to fix(n−1)mpoints, it would have to look like(g1,0, . . . ,0), whereg1 6= 0,which clearly does not belong toX. Hence it is impossible for an element ofG tohave(n−1)m fixed points.

Now we must show that there do exist elements ofG with (n− 2)m fixedpoints. Again, suppose the “Sn-action” is trivial. There exist elements ofX with(n−2)m fixed points, for example(g1,−g1,0, . . . ,0), whereg1 6= 0. Hence theminimum degree ofG is nm− (n−2)m= 2m.

Once more, we have the usual corollary about the correction capability ofG.

Corollary 2.3.4. When used as an error-correcting code, G has correction capa-bility m−1.

Proof. We haver = b2m−12 c= m−1.

Finally, to enable us to use our decoding algorithm withG, we will need anuncovering-by-bases.


Theorem 2.3.5.A set of m disjoint partial transversals forms an uncovering-by-bases for G.

Proof. The proof is analogous to that of Theorem 2.2.5. For simplicity, we takethe firstn−1 entries of each of them rows ofΩ as our partial transversals. Now, ifthere arer = m−1 errors, then these can be spread across many columns, but canoccur in at mostm−1 rows. Hence there will be at least one row which containsno error positions, so there will be a partial transversal which avoids them all.

2.4 GL(n,q)

We consider the natural action ofGL(n,q) on the non-zero vectors of the spaceFnq.

SinceGL(1,q) is just the multiplicative group ofFq, which is cyclic, we considerthe cases wheren≥ 2. The degree of this action isqn− 1, and a basis for thevector space forms a base for this group in this action. Calculating the minimumdegree is equivalent to counting the maximum number of fixed points, so we havethe following lemma.

Lemma 2.4.1.Suppose M∈GL(n,q) fixesv1, . . . ,vk. Then M fixes the space thatthey span,〈v1, . . . ,vk〉.

Proof. Clearly Fix(M) is a subspace ofFnq, as forx,y ∈ Fix(M) andλ ∈ Fq, we

haveM(x+y) = Mx+My = x+y andM(λx) = λMx = λx. Now by assumption,we have

Mv1 = v1, . . . ,Mvk = vk.

Choose somew ∈ 〈v1, . . . ,vk〉. Then

w =k

∑i=1

λivi , whereλi ∈ Fq,

so therefore

Mw = M

(k

∑i=1

λivi

)=

k

∑i=1

λiMvi =k

∑i=1

λivi = w.

This immediately leads to this result.


Proposition 2.4.2.The minimum degree of GL(n,q) is qn−qn−1.

Proof. By lemma 2.4.1, the most that can be fixed by a non-identity element ofGL(n,q) is a subspace ofFn

q of codimension 1, which containsqn−1−1 non-zerovectors. Thus

maxg∈GL(n,q)

|Fix(g)|= qn−1−1,

so the minimum degree is

(qn−1)− (qn−1−1) = qn−qn−1.

Again, the correction capability is an immediate corollary.

Corollary 2.4.3. As an error-correcting code, the correction capability of GL(n,q)is r = 1

2(qn−qn−1)−1.

Proof. By the standard result thatr = bd−12 c (whered is the minimum degree)

and observing thatqn−qn−1 is always even forn≥ 2, the result follows.

However, finding infinite families of(qn−1,n, 12(qn−qn−1)−1) uncoverings-

by-bases for arbitraryn andq has proved extremely difficult. The only generalconstruction we have for arbitraryn only yields an uncovering-by-bases utilisingthe full correction capability forn= 2. (This is actually superfluous, as an alterna-tive construction is given in the next chapter which works forany rank 2 group.)For the casen = 3, a construction is given in chapter 4. What we present here isa method for finding a(qn−1,n,bqn−1

n c−1) uncovering-by-bases. First we showhow to construct a general(v,k,bv

kc−1)-uncovering.

Construction 2.4.4.Suppose we have a setX of sizev. Putt = bvkc−1. Now, the

largest number of disjointk-subsets ofX is bvkc= t +1. Theset +1 disjointk-sets

must uncover allt-subsets ofX, as it is not possible for for at-set to intersect witht +1 sets. So we have a(v,k,bv

kc−1)-uncovering.

So, to find the uncovering-by-bases we require, we must find a set ofbqn−1n c

disjoint bases for the vector spaceFnq. Recall that the extension fieldFqn can be

regarded as a vector space of dimensionn overFq. (The techniques we use here


and the terminology used is explained in more detail in chapter 4.) Using field-theoretic notation, we haveFqn = Fq(α), whereα is the root of some irreduciblepolynomial inFq[x] of degreen. ThenFq(α) has

B =

1,α,α2, . . . ,αn−1as a basis overFq.

Now we can suppose thatα is a primitive element ofFqn. That is,α is agenerator of the multiplicative group ofFqn, which is cyclic, so each non-zeroelement ofFqn can be written as a distinct power ofα. In particular, we observethat any ‘translate’ ofB (i.e. the set obtained fromB by multiplying each elementby some given power ofα) is also a basis ofFqn = Fq(α).

We obtain thebqn−1n c disjoint bases ofFn

q by first obtaining disjoint bases forFq(α), then using the isomorphism fromFq(α) to Fn

q. The disjoint bases forFq(α)are obtained by taking translates by powers ofαn of the standard basisB, e.g.

B =

1,α, . . . ,αn−1αnB =

αn,αn+1, . . . ,α2n−1

α2nB =

α2n,α2n+1, . . . ,α3n−1and so on. We can findbqn−1

n c disjoint translates ofB in this way (includingBitself).

Now, each of these powers ofα can be expressed in terms of elements ofB.Reading off the coefficients in order gives us an element ofFn

q. For each basis ofFq(α), thesen ‘coefficient vectors’ then give us a basis forFn

q. Consequently, we

now havebqn−1n c disjoint bases forFn

q.

Example 2.4.5.We consider the caseq = 3, n = 3. Here we haveF27 = F3(α),


whereα is a root ofx3−x+1, i.e.α3 = α−1.

Basis forF27 Corresponding basis forF33

1,α,α2 100,010,001

α3,α4,α5= −1+α,−α+α2,−1+α−α2 210,021,212

α6,α7,α8= 1+α+α2,−1−α+α2,−1−α2 111,221,202

α9,α10,α11= 1+α,α+α2,−1+α+α2 110,011,211

α12,α13,α14= −1+α2,−1,−α 201,200,020

α15,α16,α17= −α2,1−α,α−α2 002,120,012

α18,α19,α20= 1−α+α2,−1−α−α2 121,222,112

α21,α22,α23= 1+α2,−1−α,−α−α2 101,220,022

The remaining powers ofα are α24 = 1− α− α2 (corresponding to 122) andα25 = 1−α2 (102).

We can now incorporate these uncoverings into our decoding algorithm. How-ever, only in the casen = 2 do they give satisfactory performance. Whenn = 2we have ⌊

q2−12

⌋−1≥ q2−q

2−1

so this construction “uncovers” more points than needed. However, forn≥ 3 wehave ⌊

qn−1n

⌋−1 <

qn−qn−1

2−1

so we can decode rather fewer errors than the theoretical correction capability.

2.5 AGL(n,q)

The last family of “generic” base-transitive groups we consider are defined asfollows:

Definition 2.5.1. Let V = Fnq, and consider theaffine transformations

TA,b : V →V

x 7→ Ax+b.


Then the setTA,b |A∈ GL(n,q),b ∈V

is called theaffine general linear group, denotedAGL(n,q).

It can easily be verified that this is indeed a group. It is, in fact, a split extensionof the additive group of the vector spaceV by GL(n,q), so has orderqn |GL(n,q)|.It acts onall vectors inV (not just the non-zero ones), so to calculate the param-eters ofAGL(n,q) we regardV as an affine space. (For explanation of the termsinvolved, see chapter 12 of Neumann, Stoy and Thompson [34].)

Theaffine span(defined formally in [34], chapter 12) ofk+1 affine indepen-dentvectors is known as ak-dimensional affine subspaceor anaffine k-flat. Ifone of thesek+1 vectors is0, then we have ak-dimensionallinear subspace; theaffinek-flats are the cosets of these. For example, an affine 1-flat is a line throughtwo arbitrary points and an affine 2-flat is a plane through three arbitrary points;neither necessarily includes the origin. The theory forAGL(n,q) described belowis a direct analogue of what we described above forGL(n,q).

In order to distinguish between the linear span and the affine span of a set ofvectors, we use the notation〈〈v0, . . . ,vk〉〉 to denote the affine span ofv0, . . . ,vk.

Lemma 2.5.2. If TA,b ∈ AGL(n,q) fixes each of the vectorsv0, . . . ,vk, then it fixestheir affine span.

Proof. Consider〈〈v0, . . . ,vk〉〉 = λ0v0 + · · ·+ λkvk|λi ∈ Fq, ∑λi = 1. Now,for 0 ≤ i ≤ k, we haveTA,b(vi) = vi , i.e. Avi + b = vi . Choose some vectorw ∈ 〈〈v0, . . . ,vk〉〉, sow = µ0v0 + · · ·+µkvk, where∑µi = 1. Then we have

TA,b(w) = Aw+b

= A(µ0v0 + · · ·+µkvk)+b

= (µ0Av0 + · · ·+µkAvk)+b

= µ0(TA,b(v0)−b)+ · · ·+µk(TA,b(vk)−b)+b

= µ0TA,b(v0)+ · · ·+µkTA,b(vk)− (∑µi)b+b

= µ0v0 + · · ·+µkvk−b+b

= w.

As with theGL(n,q) case, the above result leads us directly to the following.


Theorem 2.5.3.The minimum degree of AGL(n,q) is qn−qn−1.

Proof. By lemma 2.5.2, the fixed-point sets are precisely the affinek-flats ofV,for 0≤ k≤ n−1. (Since the onlyn-flat isV itself, the only element ofAGL(n,q)that fixes every vector in here is the identity,TI ,0.) These are the cosets of thek-dimensional linear subspaces ofV, so containqk vectors. Thus

maxg∈AGL(n,q)

|Fix(g)|= qn−1,

so therefore the minimum degree isqn−qn−1.

Consequently, we now know the correction capability ofAGL(n,q) when viewedas a code.

Corollary 2.5.4. The correction capability of AGL(n,q) is r = 12(qn−qn−1)−1.

Proof. Exactly the same as the proof of corollary 2.4.3.

In order to find an uncovering-by-bases forAGL(n,q), we use the same tech-nique as we used in section 2.4; that is, we construct a set ofbv

kc disjoint bases

for AGL(n,q). Here we havev = qn andk = n+ 1, so we wantb qn

n+1c− 1 dis-joint affine bases forV. We note that the affine independence of then+1 vectorsv0,v1, . . . ,vn is equivalent to the linear independence of then vectorsv1−v0, . . . ,vn−v0 (see [34], page 128).

Once again we use the trick of identifying the vector spaceFnq with the ex-

tension fieldFqn. Again, we suppose thatα ∈ Fqn is a primitive element ofFqn,and thatFqn = Fq(α). As we observed earlier, the set1,α,α2, . . . ,αn−1 is lin-early independent inFq(α). Now we apply the linear transformation given by thematrix

1 1 · · · · · · 1

0 1...

... 0...

......

... 1 10 · · · · · · 0 1

to obtain the linearly independent set

1,α+1,α2 +α+1, . . . ,αn−1 +αn−2 + · · ·+α+1.


Multiplying by a scalar does not affect linear independence, so multiplying by(α−1) gives another linearly independent set,α−1,α2−1,α3−1, . . . ,αn−1.But this is equivalent to the setA = 1,α,α2,α3, . . . ,αn being affine indepen-dent.

We then obtain our set ofb qn

n+1c−1 disjoint affine bases by obtaining trans-lates, that is multiplying each element ofA by powers ofαn+1. This time weobtainbqn−1

n+1 c disjoint translates ofA . However, we can improve this by startingwith the obvious affine basis0,1,α, . . . ,αn−1, then taking the translate(αn)A ,then translates ofthisby powers ofαn+1. This gives usb qn

n+1c disjoint affine bases.Then, just as we did for theGL(n,q) case, we then convert these affine bases forFq(α) into bases forFn

q, which gives us the uncovering-by-bases we require.

Example 2.5.5.Again we consider the caseq = 3, n = 3, whereα3 = α−1 (cf.example 2.4.5).

Affine basis forF27 Corresponding basis forF33

0,1,α,α2 000,100,010,001

α3,α4,α5,α6 210,021,212,111

= −1+α,−α+α2,−1+α−α2,

α7,α8,α9,α10 221,202,110,011

= −1−α+α2,−1−α2,1+α,α+α2

α11,α12,α13,α14 211,201,200,020

= −1+α+α2,−1+α2,−1,−α

α15,α16,α17,α18 002,120,012,121

= −α2,1−α,α−α2,1−α+α2

α19,α20,α21,α22 222,112,102,220

= −1−α−α2,1+α−α2,1+α2,−1−α

α23,α24 andα25 remain unused.

As with GL(n,q), we remark that this uncovering does not achieve the fullcorrection capability ofAGL(n,q).


2.6 Comparison ofGL(n,q) and AGL(n,q)

We conclude with a comparison of these two groups as codes.

GL(n,q) AGL(n,q)

Length of codewords qn−1 qn

Number of codewords ∏n−1i=0 (qn−qi) qn∏n−1

i=0 (qn−qi)

Minimum distance qn−qn−1 qn−qn−1

Errors correctible(in theory)

12(qn−qn−1)−1 1

2(qn−qn−1)−1

Errors correctible(in practice)

bqn−1n c−1 b qn

n+1c−1

The principal observation we make about the above table is that, for a smallincrease in the length of the codewords, usingAGL(n,q) gives us a much largernumber of codewords thanGL(n,q) but without reducing the (theoretical) cor-rection capability. So in a non-technical sense,AGL(n,q) could be said to be a“better” code thatGL(n,q).

Chapter 3

Base-transitive groups of rank 2

3.1 Introduction

In this chapter we consider base-transitive groups of rank 2. We are fortunate herein that we are able give a construction of an uncovering-by-bases that works foran arbitrary rank 2 group, rather than having to consider each group (or family ofgroups) individually. The final section contains an example of how to apply ourgeneral construction inGAP.

3.2 A general construction

We begin with a lemma about permutation groups in general. Throughout,Gdenotes a permutation group acting onΩ. Fora∈ Ω, let Ga be the stabiliser inGof a.

Lemma 3.2.1. Let G be a permutation group acting on a setΩ. Then for anyg∈ G and a∈ Ω, Gag = g−1Gag.

Proof. First, pick someh∈Ga. Then we have(ag)g−1hg = ahg = ag, so it followsthatg−1Gag⊆ Gag. The reverse inclusion is similar. Now pick somek∈ Gag, sothat (ag)k = ag. Thenagkg−1

= (agk)g−1= agg−1

= a, so we havegGagg−1 ⊆ Ga.ConsequentlyGag ⊆ g−1Gag, so the two sets are equal.

Now we progress to the main theorem of this section, which is solely aboutfinite base-transitive groups. Henceforth, we make the assumption that there areno points fixed by the whole group.

Theorem 3.2.2.Let G be a finite base-transitive group of rank 2 acting on a setΩ. Then there is a block system onΩ such that two points drawn from distinctblocks form a base for G.

39

BASE-TRANSITIVE GROUPS OF RANK2 40

Proof. We define aG-invariant equivalence relation onΩ and show that the ensu-ing block system has the property we require. Define∼ onΩ by a∼ b⇔Ga = Gb.Clearly this is an equivalence relation. We need to check that it isG-invariant.Supposea∼ b. By lemma 3.2.1, we haveGag = g−1Gag = g−1Gbg = Gbg, i.e.thatag ∼ bg.

The next step is to determine the equivalence classes. Let[a] = b : b∼ a,and let Fix(H) denote the set of points fixed by all elements ofH ≤ G. We claimthat[a] = Fix(Ga).

First, supposeb∈ [a], i.e. thatGb = Ga. By definition,b∈Fix(Gb) = Fix(Ga).Conversely, supposeb 6∈ [a], i.e. thatGa 6= Gb. Now, sinceG is base-transitive andtherefore transitive,|Ga| = |Gb|, so there exists someg∈ Ga with bg 6= b. Thusb 6∈ Fix(Ga), and the claim is proved.

Finally, we explain the base structure ofG. SinceG is base-transitive of rank2, a base consists of a pair of points whose pointwise stabiliser is trivial. Thechoice of the first point is arbitrary, so let this be some pointa. Now, for a,bto be a base forG, we require thatb is not fixed byGa, and so must therefore lieoutside the block[a] = Fix(Ga). Consequently, we can chooseb to be any pointfrom the remaining blocks.

We remark that this block system is trivial (i.e. the blocks all have size 1) if andonly if G is sharply 2-transitive: this follows easily from the proof above. IfG issharply 2-transitive, thenGa 6= Gb for a 6= b, soa∼ b meansa= b. Conversely, ifthe equivalence relation∼ is equality, becauseG is base-transitiveGa is transitiveon Ω \Fix(Ga) = Ω \ [a] = Ω \ a. ThusG is 2-transitive, and must thereforebe sharply 2-transitive. This corresponds to our knowledge that any pair of pointsforms a base for a sharply 2-transitive group.

Another useful fact is a general formula for the maximum number of fixedpoints of a non-identity element.

Proposition 3.2.3.Let G acting onΩ be a finite base-transitive group of rank 2.Then the number of fixed points of a non-identity element of G is either0 or m,where m= |Fix(Ga)| for some a∈ Ω.

Proof. We need to show that ifg 6= 1 has a fixed point, saya, then it has exactlym= |Fix(Ga)| fixed points. (We note that the size of Fix(Ga) is not dependent onthe choice ofa.)


SinceG is base-transitive of rank 2,Ga acts regularly onΩ\Fix(Ga) (becauseany point in here, along witha, forms a base forG, by theorem 3.2.2). Conse-quently, if an element ofGa fixes a point inΩ\Fix(Ga), it fixes all these points,and therefore must be the identity.

Supposeg fixesa. Clearly,g belongs toGa, sog (when acting onΩ) fixes allpoints in Fix(Ga) and no others. Henceg has|Fix(Ga)| fixed points.

Thinking ofG as an error-correcting code, we can now calculate the minimumdistance ofG. From a remark in section 1.1, this is given by

|Ω|− maxg∈ Gg 6= 1

|Fix(g)| .

Here,Ω is partitioned into blocks of the form Fix(Ga), each of which has sizem. Thus |Ω| = km for somek. By proposition 3.2.3, the maximum number offixed points of a non-identity element ism. Hence the minimum distance ofG iskm−m= (k−1)m. Consequently, we have this result.

Corollary 3.2.4. With notation as above, the correction capability of a base-transitive group of rank 2 is

r =⌊

(k−1)m−12

⌋.

We now give a construction for a general(v,2, t)-uncovering, which we willthen try to convert into an uncovering-by-bases forG. If we taket + 1 disjointpairs of elements from ourv-set, then this forms an uncovering, as clearly not-setcan coincide with all pairs. Clearly, this will only work if 2(t + 1) ≤ v, i.e. ift ≤ 1

2v−1. We need to show that this holds for the case we are interested in.

Here,v = kmandt = r (wherer is as in corollary 3.2.4). We have

r =⌊

(k−1)m−12

⌋≤ (k−1)m−1

2.

Now,(k−1)m−1

2≤ 1

2km−1

⇔ (k−1)m−1 ≤ km−2⇔ km−m ≤ km−1


which is clearly true, sincem≥ 1. So this method of constructing an uncoveringwill always work, provided we can find a way of allocating the disjoint pairs sothat each forms a base forG (i.e. that each pair is chosen from distinct blocks).Recall that there arek blocks, and that we will need at leastr +1 disjoint pairs.

If k is even, then we can partition the set of blocks into12k pairs of blocks. Let

A,B be a pair of blocks, whereA = a1, . . . ,am andB = b1, . . . ,bm. Thenwe can havem disjoint bases of the formai ,bi (for i = 1, . . . ,m). Doing thisfor each pair of blocks gives12kmbases in total. As shown above, we know thatr +1≤ 1

2km, so this is sufficient.

If k is odd, we carry out the same trick tok− 1 of the blocks, producing12(k−1)m disjoint bases. However, sincek is odd, we know that

r =⌊

(k−1)m−12

⌋=

12(k−1)m−1

so thereforer +1 = 12(k−1)m. Hence we have exactly the right number of bases.

It is worth remarking that ifG is sharply 2-transitive, the blocks all have size1, so therefore this construction reduces to merely taking disjoint pairs of pointsto form an uncovering.

Although we have already constructed an uncovering-by-bases forGL(2,q) inchapter 2, we use this example again to demonstrate this construction.

Example 3.2.5.Let G be the general linear groupGL(2,q) acting onF2q \ 0.

Now, by lemma 2.4.1, an element fixingv fixes〈v〉, i.e. a line through the origin.So each line (with 0 removed) forms an equivalence class of sizeq−1, and thereareq+1 classes. The minimum distance is(q2−1)− (q−1) = q2−q. To con-struct an uncovering, we arrange the lines into pairs of lines, and takeq−1 pairsof points to giveq−1 bases. This gives us a total ofbq+1

2 c(q−1) disjoint bases.

3.3 A worked example inGAP

We conclude this chapter by constructing inGAP one of the “sporadic” rank 2groups, and constructing an uncovering-by-bases for it. The group we choose isPSL(3,3), in an action on 78 points. This action is described in example 5.12 ofCameron and Deza [9] (and also by Cameron in [6]). LetG = PSL(3,3). In theaction ofG on the projective plane of order 3, the stabiliser of a line contains anormal subgroupK which induces the Klein groupV4 on that line.K has index 78


in G, so the action ofG on the right cosets ofK has degree 78. Also,K has index6 in its normaliserNG(K), so elements ofK fix 6 cosets in this action. It can beshown that this action is base-transitive of rank 2, and since non-identity elementshave either 0 or 6 fixed points, the type is(0,6,78).

There are two ways we can construct this action inGAP. First, we need theusual action ofG on the projective plane, and the stabiliser of a line:

gap> G:=PSL(3,3);Group( [ (5,8,11)(6,9,12)(7,10,13),(1,2,5)(3,8,7)(4,11,6)(9,10,13) ])

gap> H:=Stabilizer(G,[1,2,3,4],OnSets);Group([ (6,7)(8,11)(9,13)(10,12), (5,7,6)(8,10,9)(11,13,12),(5,8,11)(6,9,12)(7,10,13), (3,4)(6,7)(9,10)(12,13),(2,3)(6,7)(8,9)(11,13), (1,2,4)(6,11,10)(7,8,12) ])

By producing a list of normal subgroups ofH and observing that only the thirdhas the correct order, we choose this asK:

gap> list:=NormalSubgroups(H);;gap> K:=list[3];Group([ (1,2)(3,4)(6,11,7,8)(9,12,13,10),(1,4)(2,3)(6,12,7,10)(8,13,11,9), (6,7)(8,11)(9,13)(10,12),(5,12,10)(6,13,8)(7,11,9), (5,7,6)(8,10,9)(11,13,12) ])

Finally, we can construct the action ofG on the right cosets ofK:

gap> G1:=Action(G,RightCosets(G,K),OnRight);<permutation group with 2 generators>

Alternatively, we can obtainK as the second derived subgroup ofS, whereSis the stabiliser of a point inG, and construct that as follows:

gap> S:=Stabilizer(G,1);Group([ (5,8,11)(6,9,12)(7,10,13),(2,7,12,4,6,11)(3,5,13)(8,10), (6,7)(8,11)(9,13)(10,12) ])

gap> SS:=DerivedSubgroup(S);;


gap> K:=DerivedSubgroup(SS);Group([ (2,7)(3,5,4,6)(8,11,9,12)(10,13),(2,6,4,5)(3,7)(8,12)(9,13,10,11),(2,13)(3,12,4,11)(5,9,6,8)(7,10) ])

gap> G2:=Action(G,RightCosets(G,K),OnRight);<permutation group with 2 generators>

By using a short program to determine the sets of fixed points for non-identityelements, we have our equivalence classes. (This used the second construction ofthe action.)

[ [ 1, 2, 3, 4, 5, 6 ], [ 7, 8, 9, 10, 11, 12 ],[ 13, 14, 15, 16, 17, 18 ], [ 19, 20, 21, 22, 23, 24 ],[ 25, 26, 27, 28, 29, 30 ], [ 31, 32, 33, 34, 35, 36 ],[ 37, 38, 39, 40, 41, 42 ], [ 43, 44, 45, 46, 47, 48 ],[ 49, 50, 51, 52, 53, 54 ], [ 55, 56, 57, 58, 59, 60 ],[ 61, 62, 63, 64, 65, 66 ], [ 67, 68, 69, 70, 71, 72 ],[ 73, 74, 75, 76, 77, 78 ] ]

Then we can easily construct the uncovering-by-bases we require from this, givingthe following:

[ [ 1, 7 ], [ 2, 8 ], [ 3, 9 ], [ 4, 10 ], [ 5, 11 ],[ 6, 12 ], [ 13, 19 ], [ 14, 20 ], [ 15, 21 ], [ 16, 22 ],[ 17, 23 ], [ 18, 24 ], [ 25, 31 ], [ 26, 32 ], [ 27, 33 ],[ 28, 34 ], [ 29, 35 ], [ 30, 36 ], [ 37, 43 ], [ 38, 44 ],[ 39, 45 ], [ 40, 46 ], [ 41, 47 ], [ 42, 48 ], [ 49, 55 ],[ 50, 56 ], [ 51, 57 ], [ 52, 58 ], [ 53, 59 ], [ 54, 60 ],[ 61, 67 ], [ 62, 68 ], [ 63, 69 ], [ 64, 70 ], [ 65, 71 ],[ 66, 72 ] ]

Note that the correction capability ofG is b (78−6)−12 c = 35, and that the un-

covering above contains 36 disjoint pairs, as required.

Chapter 4

Base-transitive groups of rank 3

4.1 Introduction

In this chapter we consider base-transitive groups of rank 3, other than the twoexceptional examples which are dealt with in the next chapter. First, we give con-structions of uncoverings in which the coblocks have size three. This will then beable to be used immediately for the sharply 3-transitive groups. We then considerthree infinite families of rank 3 groups, namely the “blow-ups” ofPGL(2,q), thegeneral linear groupsGL(3,q) and the affine general linear groupsAGL(2,q), andshow how we can convert our uncovering into a suitable uncovering-by-bases. Themethods forGL(3,q) andAGL(2,q) make considerable use of the theory of finitefields, so we include a section which contains the relevant background material.

4.2 An uncovering by triples

The construction we present first gives a(2m,3,m−1)-uncovering, so this can beused when the number of points is even. As the coblocks have size three, we referto them astriples.

Theorem 4.2.1.The set of all2m triples of the formi−1, i, i +m, for i ∈ Z2m

and with addition modulo2m, forms a(2m,3,m−1)-uncovering.

Proof. Let X denote an arbitrary(m− 1)-subset ofZ2m, and letY denote itscomplement, so|Y| = m+1. We assign colours to the 2m points as follows: redif the point lies inX, green if it lies inY.

We partitionZ2m as the union of them “antipodal” pairs, i.e. the set of pairsof the formi, i +m. Now, with the colouring described above, the pairs are asfollows:

45


• greenpairs (i.e. both points are coloured green);

• red pairs (both points red);

• bichromaticpairs (one point of each colour).

Let G denote the set of green pairs,R the set of red pairs andB the set of bichro-matic pairs.

We need to show that there exists a green pair that can be extended to a tripleof the formi−1, i, i +m that is disjoint fromX, i.e. all three points are colouredgreen. This will only fail if every green pair is preceded by a red pair (i.e. bothi−1 andi +m−1 are coloured red), which would require|G | ≤ |R |.

However, counting the total number of red and green points, we have that

|X|= m−1 = 2|R |+ |B|and

|Y|= m+1 = 2|G |+ |B|.This in turn gives

2|G |+ |B|= m+1 = (m−1)+2 = 2|R |+ |B|+2,

so therefore|G |= |R |+1 and we cannot possibly fail.

Example 4.2.2.Consider the casem= 5. In each row, the framed elements forma coblock in an uncovering, the remaining elements form a block in the corre-sponding covering design.

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10


Note: Examples of these uncoverings (or, more specifically, the correspondingcovering designs) appear in [22], as “cyclic coverings found by search program”.No reference for a general construction is given.

We remark that this construction is within a constant factor of the least pos-sible size. Thanks to a result of W. H. Mills (which is too complicated to statein full generality here, but can be found in [32], theorem 2.3), a(2m,3,m−1)-uncovering (or, equivalently, a(2m,2m− 3,m− 1) covering design) must havesize at least53m. Our construction gives an uncovering of size 2m, so this is withina factor of at most65 of the optimal size.

From the construction in theorem 4.2.1, we can obtain uncoverings for wherethere are an odd number of points, thanks to the following lemma. This is aspecial case, rephrased in terms of uncoverings rather than covering designs, of the“induced construction” due to Gordon, Kuperberg and Patashnik [23], in section4 of their paper.

Lemma 4.2.3.LetU be a(v,k, t)-uncovering with point set1, . . . ,v, and letWbe the subset ofU obtained by removing all coblocks containing the point v. ThenW is a (v−1,k, t−1)-uncovering.

Proof. Let E be an(t − 1)-subset of1,2, . . . ,v− 1. SinceU is a (v,k, t)-uncovering, there exists a coblockT ∈ U disjoint from thet-setE∪v. Now,clearlyT cannot be one of them-sets containingv. ThusT ∈ W , and is disjointfrom E. HenceW is a(v−1,k, t−1)-uncovering.

Thus we can apply lemma 4.2.3 to the construction from theorem 4.2.1 toobtain a(2m−1,3,m−2)-uncovering.

Corollary 4.2.4. Let U be a (2m,3,m− 1)-uncovering ofZ2m as described intheorem 4.2.1 above, and letW be the subset ofU obtained by removing alltriples containing the point2m. ThenW is a (2m−1,3,m−2)-uncovering, ofsize2m−3.

Example 4.2.5.By applying corollary 4.2.4 to the(10,3,4)-uncovering in exam-ple 4.2.2, we obtain the(9,3,3)-uncovering below:


1 2 7

2 3 8

3 4 9

1 5 6

2 6 7

3 7 8

4 8 9

As mentioned in section 4.1, the above constructions can immediately be ap-plied to sharply 3-transitive groups. Recall from section 1.2 that such a group ofdegreen has minimum distancen−2 and correction capabilityr = bn−3

2 c. If n iseven, sayn = 2m, we haver = m−2; if n is odd, sayn = 2m−1, again we haver = m−2. So in fact whenn is odd, we actually have a better uncovering than weneed.

Recall thatPGL(2,q) is sharply 3-transitive of degreeq+1, and exists for allprime powersq. So, for instance, the uncovering in example 4.2.2 can be used forPGL(2,9) and that in example 4.2.5 used forPGL(2,8).

4.3 “Blow-ups” of PGL(2,q)

The family of groups described by Maund [31] as “blow-ups” ofPGL(2,q) arequite hard to define from scratch. However, as they are “blow-ups” (see definition1.5.5) their base structure is relatively uncomplicated. The following theorem istaken from Maund’s D.Phil. thesis [31].

Theorem 4.3.1.Let G be a blow-up of PGL(2,q). Then G has degree qd(q+1)(for some d≥ 1), where these points can be thought of as being arranged in(q+1)columns, each containing qd points. A base for G consists of three points, eachdrawn from different columns.

The base structure follows from the fact that the action on theq+1 columnsis the usual action ofPGL(2,q) on q+1 points, which is sharply 3-transitive, soa base forG consists of three points, each drawn from different columns.


Proposition 4.3.2.Let G be a blow-up of PGL(2,q). Then, when considering Gas an error-correcting code, its correction capability is

r =qd(q−1)

2−1.

Proof. According to Maund [31], the maximum number of fixed points of anon-identity element is 2qd. Hence the minimum distance isqd(q+ 1)−2qd =qd(q−1) and so the correction capability is

r =⌊

qd(q−1)−12

⌋=

qd(q−1)2

−1.

Observe thatqd(q+ 1) is always even, so we only need the(2m,3,m− 1)-uncovering in theorem 4.2.1, and do not need to resort to any induced construc-tions. However, we notice that this is actually better than we really need, since thecorrection capabilityr is actually less thanm−1 = 1

2qd(q+1)−1.

What we need to do to convert our construction of an uncovering (from section4.2) into an uncovering-by-bases forG is to ensure that each triple forms a base.So we arrange the 2m= qd(q+1) points into a rectangle ofqd rows of lengthq+1,such that in each triple in the uncovering the three points come from differentcolumns. As will become evident, we must considerq odd andq even separately.

Construction 4.3.3. Suppose that the set of points is1,2, . . . ,2m. The easiestway to arrange them into a rectangle would be to put 1,2, . . . ,q+ 1 into the firstrow, thenq+ 2,q+ 3, . . . ,2(q+ 1) into the second row, and so on. Thus thej th

column contains all points congruent toj (modq+ 1). For example, withq = 3andd = 1 we have 12 points arranged as follows.

1 2 3 4

5 6 7 8

9 10 11 12

Theorem 4.3.4.Let G be a blow-up of PGL(2,q), where q is odd. Then applyingtheorem 4.2.1 to construction 4.3.3 yields an uncovering-by-bases for G.

Proof. Recall that the uncovering in theorem 4.2.1 contains all triples of the formi−1, i, i +m, where 1≤ i ≤ 2m and addition is modulo 2m. (In particular, this


means thati +m= i−m.) We wish to avoid situations where any two of the threepoints lie in the same column. Asq is a prime power we haveq+1≥ 3, so clearlythe pointsi − 1, i are always in different columns. The other two possibilitiesrequire more thought.

First, consider the pairi, i +m. We have:

i, i +m in same column

⇔ i +m≡ i modq+1

⇔ m≡ 0 modq+1

⇔ (q+1) | m

(recalling thatm= 12qd(q+1))

⇔ (q+1) | 12

qd(q+1)

⇔ 2(q+1) | qd(q+1)⇔ 2 | qd

i.e. if and only if q is even. Butq is odd, so this is impossible. Secondly, weconsider the pairi−1, i +m. We have:

i−1, i +m in same column

⇔ i +m≡ i−1 modq+1

⇔ m+1≡ 0 modq+1

⇔ (q+1) | m+1

⇔ (q+1) | 12

qd(q+1)+1

⇔(

qd

2+

1q+1

)∈ Z.

Sinceq is odd, thenqd

2 = s+ 12 for some integers, so we have(

qd

2+

1q+1

)∈ Z⇔ 1

q+1=

12,

i.e. q = 1. Butq is a prime power, so this is impossible.

So construction 4.3.3 definitely works whenq is odd. For example, withq= 3andd = 1 we have the 12 triples listed below.


Example 4.3.5.By comparing these with the arrangement above, we see that eachof the 12 triples is spread across three columns.

12,1,7 6,7,1

1,2,8 7,8,2

2,3,9 8,9,3

3,4,10 9,10,4

4,5,11 10,11,5

5,6,12 11,12,6

However, the case whereq is even is still to be resolved. As the proof oftheorem 4.3.4 shows, the same method won’t work, so we need to modify it. Firstwe observe that sinceq is even, we now have an even number of rows, so we

can split these into a “top half” (firstqd

2 rows) and a “bottom half” (lastqd

2 rows).Furthermore, we observe that, in construction 4.3.3 above, wheni−1 andi lie inthe top half,i +mmust lie in the bottom half (with the exception of the “boundarycases”, wheni = 1 or i = m+1). So we perform the following “trick”.

Construction 4.3.6.We arrange the points1, . . . ,2m into a rectangle as in con-struction 4.3.3, except that all the rows in the bottom half are “cycled” two placesto the right. For instance, withq = 4, d = 1 we have the following:

1 2 3 4 5

6 7 8 9 10

14 15 11 12 13

19 20 16 17 18

We need to demonstrate that this trick works in general.

Theorem 4.3.7.Let G be a blow-up of PGL(2,q), where q is even and q> 2.Then applying theorem 4.2.1 to construction 4.3.6 yields an uncovering-by-basesfor G.

Proof. Recall that our triples are of the formi−1, i, i +m. First, we deal withthe case wheni −1 andi are both in the top half, and lie in columnsj −1 and


j (modq+1). Now, before the shift is applied, by the same arguments as in theprevious proof,i +m will also lie in column j. But after the shift, it will now liein column j +2.

Second, we deal with the case when bothi −1 andi are both in the bottomhalf. Because of the shift applied, these will lie in columnsj +1 and j +2 whilei +m, which will be in the top half, lies in columnj.

Then there are the two “boundary” cases. First, ifi = 1, then clearlyi lies inthe column 1. However, before the shift,i−1 will be in the last column of the lastrow (and thus is in the bottom half), so after the shift will lie in column 2. Also,i +m will be in column 1 before the shift, and so is in column 3 afterwards.

Second, ifi = m+1, theni will be in the first row of the bottom half, and in thefirst column before the shift. Thus after the shift it will lie in column 3.i−1 willbe in the last column of the last row of the top half, i.e. in columnq+1. Finally,i +m will be in the first row and the first column, i.e. column 1.

We see that in all of these cases the three entries are in distinct columns.

However, this does not work ifq = 2, as we will have three columns and, forinstance,j −1≡ j + 2 (mod 3). Thus forq = 2, an entirely different approachis required. But sincePGL(2,2) ∼= S3, we have already dealt with “blow-ups” ofthis group (in section 2.2).

4.4 Background material on finite fields

In order to give general methods for constructing uncoverings-by-bases for thegeneral linear groupsGL(3,q), rather than work in the vector spaceF3

q, we movethe problem into the extension fieldFq3 and use primitive roots to obtain solutions.We then use a similar method for the affine general linear groupsAGL(2,q).

Most of the following background material can be found in theEncyclopædiaof Mathematicsvolume on finite fields by Lidl and Niederreiter [30]. LetFq bethe finite field of orderq andFqn be its degreen extension field. The followingtwo facts are well-known.

Fact 1. Fqn is a vector space overFq of degree n.

A basis for this vector space is referred to as a basis forFqn overFq.


Fact 2. The multiplicative group of the non-zero elements ofFq, F∗q, is cyclic oforder q−1.

A generator of this cyclic group is known as aprimitive elementof Fq, or as aprimitive rootof q. Thus if α is a primitive element, the non-zero field elementscan be written as

1,α,α2, . . . ,αq−2.

If αr1,αr2, . . . ,αrn (wherer1 > r2 > · · · > rn is a basis forFqn over Fq, thenthis is equivalent to saying thatα satisfies no polynomial of the form

f (x) = ar1xr1 +ar2x

r2 +arnxrn

(where eachai ∈ Fq and are not all zero). Clearly, we can divide through byxrn toobtain a polynomial of smaller degree, and also divide by the leading coefficientto obtain a monic polynomial.

Definition 4.4.1. Supposeα ∈ F = Fqn. Then the elementsα,αq,αq2, . . . ,αqn−1

are called the(algebraic) conjugatesof α with respect toK = Fq.

Let f (x) denote the minimal polynomial ofα overK (i.e. the unique monic,irreducible polynomial overK of least degree withα as a root). Thenf (x)has degreed say, whered divides n. Then the conjugates ofα are the rootsα,α2, . . . ,αqd−1

of f (x), each repeatedn/d times. In particular, iff (x) has degreen, then the conjugates are all distinct. A related definition is the following.

Definition 4.4.2. For α ∈ F , thetraceof α overK, TrF/K(α), is defined as

TrF/K(α) = α+αq +αq2+ · · ·+αqn−1

,

i.e. the sum of the conjugates ofα.

Properties of the trace are explored in detail in [30], section 2.3. The followingproperty is the only one we will require. Again, letf (x) be the minimal polyno-mial of α and suppose it has degreed. Then the polynomialg(x) = f (x)n/d isknown as thecharacteristic polynomialof α. Now, the roots ofg(x) are preciselythe conjugates ofα with respect toK. That is,

g(x) = xn +an−1xn−1 + · · ·+a1x+a0

factorises asg(x) = (x−α)(x−αq) · · ·(x−αqn−1

).

We observe that, when multiplying out the second form ofg(x), the coefficient ofxn−1 is−TrF/K(α), i.e. we have:


Proposition 4.4.3.TrF/K(α) =−an−1.

Returning to the subject of bases, the next definition is related to the abovediscussion.

Definition 4.4.4. A normal basisfor F over K is a basis consisting of the con-jugatesα,αq, . . . ,αqn−1 of some elementα ∈ F . Such an elementα is called afree element.

Free elements have the following useful property.

Lemma 4.4.5.A free elementα ∈ F satisfiesTrF/K(α) 6= 0.

Proof. Suppose not. Then we would have

α+αq +αq2+ · · ·+αqn−1

= 0,

i.e. the conjugates would be linearly dependent and do not form a basis, i.e.α isnot free.

The Normal Basis Theorem(desribed in [30], theorem 2.35) states that forany prime powerq andn > 1, there exists a free elementα, i.e. that there exists anormal basis. A more specific type of basis is aprimitive normal basis, which is anormal basis but with the extra condition that the free elementα is also a primitiveelement. The following theorem is important here.

Theorem 4.4.6.The Primitive Normal Basis Theorem (Carlitz; Davenport; Lenstraand Schoof; Cohen and Huczynska). For any prime power q and n> 1, there ex-ists a primitive normal basis ofFqn overFq.

The theorem was originally proved by Carlitz [13, 14] in 1952 for “sufficientlylarge” q andn and in 1968 by Davenport [19] for anyn in the case whereq isprime. Later (1987), Lenstra and Schoof [29] completed the proof, but with theaid of a computer; more recently (2003) Cohen and Huczynska [17] gave a non-computational proof. The result we will need in the next section is the followingcorollary.

Corollary 4.4.7. For any prime power q and n> 1, there exists a primitive elementα ∈ F = Fqn such thatTrF/K(α) 6= 0 (where K= Fq).


Proof. By the primitive normal basis theorem, a primitive, free elementα exists.By lemma 4.4.5, free elements have non-zero trace.

We remark that the Primitive Normal Basis Theorem is a much stronger resultthan we actually need, and that there are other results that we could use in its place.For instance, a result of Cohen [16] shows that with one non-trivial exception,Fqn

contains a primitive element of arbitrary trace overFq.

4.5 Uncoverings-by-bases forGL(3,q)

In this section, we construct an uncovering-by-bases for the groupGL(3,q) actingonF3

q\0. However, to do this we will not work in the vector space, but insteadwe move the problem into the extension fieldFq3, since as vector spaces overFq

they are isomorphic. This enables us to make use of the material on finite fieldsin the section above. Thus we obtain a collection of bases forFq3, and then applythe isomorphism to obtain bases forF3

q.

As was the case with the groups in section 4.3, the uncoverings we constructedin section 4.2 are actually capable of uncovering more points than the correctioncapabilityr. From corollary 2.4.3, we haver = 1

2(q3−q2)−1. Whenq is oddwe have 2m= q3−1, som−1 = 1

2(q3−1)−1, while whenq is even we have2m−1 = q3−1, som−2 = 1

2q3−2. In both cases we have more thanr points.

For converting our uncoverings into uncoverings-by-bases, we consider thecases whereq is odd and even separately. First, we supposeq is odd, so thereforeq3− 1 is even, sayq3− 1 = 2m. If α is a primitive element ofFq3, then thenon-zero elements ofFq3 are the powers ofα, i.e.

1,α,α2, . . . ,α2m−1.

Note thatα2m = 1, and also that(αm)2 = α2m = 1, soαm =−1.

Suppose we take our “cyclic triples” construction of an uncovering, from the-orem 4.2.1, and attempt to convert it into an uncovering-by-bases, using the bijec-tion

ψ : Z2m → F∗q3

i 7→ αi .


Unfortunately, this will not work, as each triple of field elements will have theform αi−1,αi ,αi+m. This will be a basis if and only if1,α,αm+1 is a basis.However, sinceαm = −1, we have the triple1,α,−α, which is clearly not abasis.

All is not lost, however. Instead of using the ordering above, we modify it asfollows. Define the bijection

ϕ : Z2m → F∗q3

wherei 7→ ϕ(i) as follows:

i 0 1 2 · · · m−1 m m+1 · · · 2m−3 2m−2 2m−1

ϕ(i) 1 α α2 · · · αm−1 αm+2 αm+3 · · · α2m−1 αm αm+1

That is, the first half remain as they were underψ, and the second half are movedtwo places to the left, with the remaining two elements replaced at the end.

Theorem 4.5.1.Suppose q is odd. Then the set of triples of the form

ϕ(i−1),ϕ(i),ϕ(i +m) ,

for i ∈ Z2m forms an uncovering-by-bases for GL(3,q) acting onFq3.

Proof. We know from theorem 4.2.1 that these triples form an uncovering, so wenow need to verify that each tripleϕ(i−1),ϕ(i),ϕ(i +m) is a basis forFq3 overFq. There are a number of cases to consider.

Case 1. 1,α,αm+3 independent⇔ 1,α,−α3 independent⇔ 1,α,α3 independent (?)

This also covers all cases up toαm−4,αm−3,α2m−1, as they are all scalar multi-ples of the triple in case 1, and so are equivalent.

Case 2. αm−3,αm−2,αm independent⇔ 1,α,α3 independent (?)

Note that the next triple,αm−2,αm−1,αm+1, is equivalent to case 2.


Case 3. αm−1,αm+2,1 independent⇔ α,αm,αm+3 independent⇔ α,−1,−α3 independent⇔ 1,α,α3 independent (?)

Case 4. αm+2,αm+3,α independent⇔ −α2,−α3,α independent⇔ 1,α,α2 independent (†)

This covers all cases up toα2m−2,α2m−1,αm−3, as they are equivalent underscalar multiplication to case 4.

Case 5. α2m−1,αm,αm−2 independent⇔ αm+1,α2,1 independent⇔ −α,α2,1 independent⇔ 1,α,α2 independent (†)

Case 6. αm,αm+1,αm−1 independent⇔ 1,α,α2 independent (†)

Case 7. αm+1,1,αm+2 independent⇔ 1,−α,−α2 independent⇔ 1,α,α2 independent (†)

In the cases labelled (†), the requirement that each triple is a basis is satisfied, as1,α,α2 is the natural basis ofFq3 = Fq(α). In the cases labelled (?), we have tobe more careful. Now, the triple1,α,α3 forms a basis if and only ifα satisfiesno polynomial of the formx3 + ax+ b = 0 (with a,b,∈ Fq). Since the minimalpolynomial ofα, mα(x), is a cubic, we require thatmα(x) is not of this form.

In other words, by proposition 4.4.3, we require that Tr(α) 6= 0. But by corol-lary 4.4.7, we have that such a primitive elementα always exists. Hence if weensure thatFq3 = Fq(α) for such a specificα, then we are done and these triplesdo indeed form bases forFq3 overFq.

The theorem above gives us an uncovering-by-bases for the action ofGL(3,q)onFq3; however, we would like an uncovering-by-bases for the action onF3

q. Foreachq, we need to exhibit an isomorphism betweenF3

q and the extension fieldFq3

as vector spaces. The smallest example for this is still quite large:GL(3,3) actingon the 26 non-zero vectors inF3

3.


Example 4.5.2.In a similar manner to example 2.4.5, in order to determine theisomorphism between the vector spaceF3

3 and the extension fieldF27, we need toknow the minimum polynomial of our primitive elementα. By Table F in Lidl andNiederreiter [30], the polynomialx3 + 2x2 + 1 is irreducible and has a primitiveelement as a root. Clearly this polynomial has non-zero trace, so it satisfies ourrequirements, although it can be verified that the primitive element obtained isalso a free element. By using the equationα3 = α2 + 2, we can now write eachpower ofα as a linear combination of 1,α,α2, and use the triples of coefficientsas corresponding elements of the vector space. So we have the following.

Field element Vector1 001α 010α2 100α3 = α2 +2 102α4 = α2 +2α+2 122α5 = 2α+2 022α6 = 2α2 +2α 220α7 = α2 +1 101α8 = α2 +α+2 112α9 = 2α2 +2α+2 222α10 = α2 +2α+1 121α11 = α+2 012α12 = α2 +2α 120

Field element Vectorα13 = 2 002α14 = 2α 020α15 = 2α2 200α16 = 2α2 +1 201α17 = 2α2 +α+1 211α18 = α+1 011α19 = α2 +α 110α20 = 2α2 +2 202α21 = 2α2 +2α+1 221α22 = α2 +α+1 111α23 = 2α2 +α+2 212α24 = 2α+1 021α25 = 2α2 +α 210

From this information, we can apply our construction above to obtain 26 triples ofvectors which form bases for the vector spaceF3

3, and thus for the groupGL(3,3),as given below.


Basis forF27 Basis forF33

1,α,α16 001,010,201

α,α2,α17 010,100,211

α2,α3,α18 100,102,011

α3,α4,α19 102,122,110

α4,α5,α20 122,022,202

α5,α6,α21 022,220,221

α6,α7,α22 220,101,111

α7,α8,α23 101,112,212

α8,α9,α24 112,222,021

α9,α10,α25 222,121,210

α10,α11,α13 121,012,002

α11,α12,α14 012,120,020

α12,α15,1 120,200,001

α15,α16,α 200,201,010

α16,α17,α2 201,211,100

α17,α18,α3 211,011,102

α18,α19,α4 011,110,122

α19,α20,α5 110,202,022

α20,α21,α6 202,221,220

α21,α22,α7 221,111,101

α22,α23,α8 111,212,112

α23,α24,α9 212,021,222

α24,α25,α10 021,210,121

α25,α13,α11 210,002,012

α13,α14,α12 002,020,120

α14,1,α15 020,001,200


The lines are to indicate cases 1 to 7 from above. It can be verified that each ofthe triples of vectors is indeed a basis.

We now deal with the case whereq is even, sayq= 2s. ConsiderFq⊂ Fq3, i.e.F2s ⊂ F23s = F . This time, as there are an odd number of elements inF∗, we wantto use the induced construction of an uncovering in corollary 4.2.4. To do so weintroduce an additional element,∞, construct an uncovering ofS= F∗ ∪∞,then remove the triples containing∞. Let α be a primitive element ofF . We have

S= 1,α,α2, . . . ,α23s−2,∞

so|S|= 23s = 2m, say. We construct a bijection

χ : Z2m → Si 7→ αi for 0≤ i ≤ 2m−2i 7→ ∞ for i = 2m−1.

Theorem 4.5.3.Suppose q is even. Then the set of triples of the form

χ(i−1),χ(i),χ(i +m) ,

for i ∈ Z2m, excluding those containing∞, forms an uncovering-by-bases forGL(3,q) acting onFq3.

Proof. Using the induced construction from corollary 4.2.4, we know that thesetriples will form an uncovering, so we need to verify that each triple forms a basisfor Fq3 overFq. The triples we have are as follows.

1,α,αm+1= 1,α,α23s−1+1

α,α2,αm+2...

αm−3,αm−2,α2m−2

We note that these are all equivalent under scalar multiplication.

αm−2,αm−1,∞

This triple is excluded, as it contains∞.


αm−1,αm,1= 1,α23s−1−1,α23s−1

αm,αm+1,α...

α2m−3,α2m−2,αm−2

These are also all equivalent.

α2m−2,∞,αm−1

∞,1,αm−1

Again, these are both excluded as they contain∞. So in fact there are just twocases to consider.

(i) 1,α,α23s−1+1

(ii) 1,α23s−1−1,α23s−1

Thusα cannot be a root of a polynomial of the forms

(i) x23s−1+1 +ax+b = f (x)

(ii) x23s−1+cx23s−1−1 +d = g(x)

(wherea,b,c,d ∈ F2s).

In case (i),α is a root of f (x) if and only if it is a root of f (x)2. Now,

f (x)2 = (x23s−1+1 +ax+b)2

= (x23s−1+1)2 +a2x2 +b2 (sinceF has characteristic 2)

= x23s+2 +a2x2 +b2.

Note thatx23s= x for all elements ofF , so

f (x)2 = x3 +a2x2 +b2.


In case (ii),α is a root ofg(x) if and only if it is a root ofxg(x)2. This time, wehave

xg(x)2 = x(x23s−1+cx23s−1−1 +d)2

= x(x23s+c2x23s−2 +d2)

= x23s+1 +c2x23s−1 +d2x

= x2 +d2x+c2.

Sinceα cannot satisfy any quadratic overF2s, we conclude that the triple in case(ii) must be a basis. So case (i) remains.

If α is a root of f (x)2, thenα−1 is a root ofh(y) = f (y−1)2 (wherey = x−1).Soh(y) = y−3 +a2y−2 +b2, and multiplying through byy3 gives the cubic poly-nomial 1+a2y+b2y3. Dividing through byb2 to obtain a monic polynomial, wehave

y3 +a2

b2y+1b2 .

Thusα is a root of f (x) if and only if α−1 is a root of a cubic polynomial of theform y3 + λy+µ, i.e. Tr(α−1) 6= 0. Hence if we takeα−1 to be a free, primitiveelement, then we are done. By corollary 4.4.7, such an element will always exist.Thus by a suitable choice ofα we can ensure that each of these triples are bases,and we do indeed have an uncovering-by-bases.

4.6 Uncoverings-by-bases forAGL(2,q)

As in the previous section, we convert our constructions of uncoverings from sec-tion 4.2 into uncoverings-by-bases. Here we apply similar techniques to those weused forGL(3,q), except now we are working in a 2-dimensional vector space,and need to consider affine independence rather than linear independence. Weremark that althoughAGL(2,q) acts on all the vectors in the space (including thezero vector), we don’t necessarily need 0 to appear in our uncovering, so we canstill use the method of working in the corresponding extension field, which in thiscase isFq2.

Once more, we remark that the uncoverings constructed are more powerfulthan we need. By corollary 2.5.4, we know the correction capability ofAGL(2,q)to be r = 1

2(q2− q)− 1. For q odd, we have an even number of non-zero fieldelements, so use the(2m,3,m− 1)-uncovering with 2m = q2− 1, so m− 1 =


12(q2−1)−1. Forq even, we use the(2m−1,3,m−2)-uncovering with 2m−1=q2−1, som−2 = 1

2q2−2. In both cases, we have something larger thanr.

As with GL(3,q), to construct our uncovering-by-bases we considerq odd andeven separately. In the case whereq is odd, we have an even number of non-zerofield elements,

1,α,α2, . . . ,α2m−1

(where 2m= q2−1). This case is relatively straightforward.

Theorem 4.6.1.Suppose q is odd. Then the set of translates (by powers ofα)of the triple

1,α,αm+1

forms an uncovering-by-bases for AGL(2,q) acting on

Fq2.

Proof. By theorem 4.2.1, we know that this set is an uncovering, so thereforewe need to verify that each triple is a base forAGL(2,q). It suffices to show thatthe triple1,α,αm+1 (and hence each translate of it by powers ofα) is a base,i.e. that it is affine-independent. This is equivalent to the pairα−1,αm+1−1being linearly independent. Sinceαm =−1, this pair is actuallyα−1,−α−1.Now, 1,α forms a basis forFq2 over Fq. Then we apply the following lineartransformation: (

1 −1−1 −1

)(α1

)=(

α−1−α−1

).

This matrix has determinant−2 6= 0 (sinceq is odd), so we have that this lin-ear transformation is invertible. Thusα− 1,−α− 1 is also a basis, and so1,α,αm+1 is affine-independent. It follows that each translate is also affine-independent, so therefore the set of translates forms an uncovering-by-bases.

The case whereq is even is considerably more complicated. First we mentionthat we do not need to consider the caseq= 2, sinceAGL(2,2)∼= S4 has correctioncapability 0. We require some preliminary lemmata, the first two of which arenumber-theoretic in nature. Henceforth,φ denotes Euler’s totient function.

Lemma 4.6.2.For any positive integer n,

φ(n)≥ nlog2(n)

.

Proof. Let p1, p2, . . . , pr be the prime divisors ofn. Son≥ p1p2 · · · pr ≥ 2r , and


thereforer ≤ log2(n). Also,

φ(n)n

=(

1− 1p1

)(1− 1

p2

)· · ·(

1− 1pr

)≥

(1− 1

2

)(1− 1

3

)· · ·(

1− 1r

)=

1r

≥ 1log2(n)

and the result follows.

Lemma 4.6.3.Where q= 2s (for s> 1), φ(q2−1) > 2(q−1).

Proof. Suppose not, i.e. supposeφ(q2−1) ≤ 2(q−1). By lemma 4.6.2 above,we have

φ(q2−1) ≥ q2−1log2(q2−1)

>q2−1

log2(q2)

=q2−1

2s.

By assumption, we haveq2−1

2s< 2(q−1)

i.e. 2s+1 < 4s, which fails fors≥ 4.

Checking small cases by hand, we have:

s= 1 : φ(3) = 2, 2(q−1) = 2s= 2 : φ(15) = 8, 2(q−1) = 6s= 3 : φ(63) = 36, 2(q−1) = 14

so the proposition holds fors> 1.

As we will see shortly, we will be restricted in our choice of primitive elementin this case. The next lemma guarantees the existence of the specific kind ofprimitive element we will require.


Lemma 4.6.4.Let q= 2s (for s> 1) and considerFq ⊂ Fq2. Then there exists aprimitive elementα of Fq2 which satisfies no polynomial of the form x2 + x+ a,where a∈ Fq and a6= 0.

Proof. There are exactlyq−1 such polynomials, which have a total of at most2(q−1) roots, which are not necessarily primitive. So there are at most 2(q−1)primitive elements which satisfy such polynomials. However, there are a total ofφ(q2−1) primitive elements altogether, and by lemma 4.6.3 above,φ(q2−1) >2(q−1). Therefore a primitive element satisfying no such polynomial must exist.

In this case, we will use triples of the formi, i + 1, i + m rather than whatwe have used previously. (The set of all such triples for 0≤ i ≤ 2m− 1 formsa (2m,3,m−1)-uncovering, simply by replacingi−1 with i + 1 in the proof oftheorem 4.2.1.)

But as we have an odd number of points here (q2− 1 points), we do as wedid in theorem 4.5.3, namely introducing an additional point∞, applying theconstruction from theorem 4.2.1, then throwing away each triple which contains∞. That is, we first produce an uncovering on the set

S= 1,α,α2, . . . ,α2m−2,∞

and then remove all triples containing∞. To do this, we define a bijection

θ : Z2m → Si 7→ αi for 0≤ i ≤ 2m−2i 7→ ∞ for i = 2m−1.

Note that 2m= q2 here.

Theorem 4.6.5.Suppose q is even and q> 2. Then the set of triples of the form

θ(i),θ(i +1),θ(i +m) ,

for i ∈ Z2m, excluding those containing∞, forms an uncovering-by-bases forAGL(2,q) acting onFq2.

Proof. By corollary 4.2.4, we know that this set of triples is an uncovering, sowe must show that each triple is a base forAGL(2,q), by showing that it is affine-independent overFq. Underθ, the triples we have are as follows.


1,α,αm

α,α2,αm+1...

αm−2,αm−1,α2m−2

These are all equivalent, under multiplication by powers ofα.

αm−1,αm,∞

This triple contains∞ and is excluded.

αm,αm+1,1

αm+1,αm+2,α...

α2m−1,α2m−2,αm−1

Again, these are all equivalent.

α2m−2,∞,αm

∞,1,αm+1

These last two triples are both excluded. Thus we have two cases to consider:

(i) 1,α,αm

(ii) 1,αm,αm+1

We require that these two triples are affine-independent overFq. In case (i),we note that the triple1,α,αm is affine-independent if and only if the pairα+1,αm+1 is linearly independent. (Note that the characteristic of this fieldis 2, so addition and subtraction are the same.) Now, again because the charac-teristic is 2 but also becausem is even, we have thatαm+ 1 = (α + 1)m, so in


particularα+1 dividesαm+1, with quotient(α+1)m−1. Thusα+1,αm+1 willbe linearly independent if and only if(α+1)m−1 6∈ F∗q.

So suppose not, i.e. suppose(α+1)m−1∈ F∗q, and consider the quotient group

H = F∗q2/F∗q

which is cyclic and has orderq+1. Also, consider the natural projection map

π : F∗q2 H.

Let β = π(α +1), so thereforeβm−1 = 1. Thus the order ofβ divides bothm−1andq+1, so divides

GCD(q+1,12

q2−1) = GCD(q+1,q2−2)

= GCD(q+1,(q+1)(q−1)−1)= 1.

Henceβ has order 1, i.e.β = 1. But then this meansα+1∈ Ker(π) = F∗q, whichis impossible sinceα is a primitive element ofFq2. Therefore(α + 1)m−1 6∈ F∗q,so we have thatα + 1,αm + 1 is linearly independent, and that1,α,αm isaffine-independent.

It is case (ii) that requires the lemmata we proved above. Here we want to showthat1,αm,αm+1 is affine-independent. At this point we impose the restrictionmentioned in lemma 4.6.4, i.e. thatα is chosen so that it satisfies no polynomialof the formx2 +x+a (?) (for a∈ F∗q).

By multiplying each element byαm−1 the triple1,αm,αm+1 is equivalent toαm−1,α2m−1 = 1,α2m = α. Thus it suffices to show that1,α,αm−1 is affine-independent, i.e. thatα + 1,αm−1 + 1 is linearly independent. Now,α + 1dividesαm−1 +1, with quotient

C = 1+α+α2 + · · ·+αm−2.

If α+1,αm−1 +1 is to be linearly independent, then we needC 6∈ F∗q.

So we suppose not. Since 1+α+α2+ · · ·+α2m−2 = 0, we have the equation

1+Cα+Cαm = 0.

Squaring both sides (and remembering that we have characteristic 2) gives

1+C2α2 +C2α2m = 1+C2α+C2α2 = 0


so therefore

α2 +α+1

C2 = 0.

But thenα satisfies a polynomial of the form (?), giving a contradiction. SoC 6∈ F∗q, soα + 1,αm−1 + 1 is linearly independent and thus1,αm,αm+1 isaffine-independent.

Consequently, by a judicious choice ofα, it follows that each triple is affine-independent, and so we have an uncovering-by-bases.

Chapter 5

Exceptional base-transitive groups

5.1 The action ofA7 on 15 points

Because of an exceptional isomorphism (see theATLAS [18], page 22), the al-ternating groupA8 is isomorphic toGL(4,2). Consequently, the alternating groupA7 also has an action onF4

2\0. The irredundant bases forA7 in this action arethe linearly independent triples of vectors inF4

2, andA7 acts transitively on these.Thus, in this action,A7 is a base-transitive group of rank 3.

To obtain this action on 15 points, we useGAP. While it is possible to constructit directly (by finding the maximal subgroups ofA8

∼= GL(4,2) and so on), itis easier to use the in-built library of transitive groups. By using the commandAllTransitiveGroups and inputting the required degree (15) and order (2520),it returns the unique such group.

With the knowledge that this action is base-transitive, we can then producethe set of all bases ofA7 in this action on 15 points, by taking one base andconstructing the orbit ofA7 on it. We also know from [31] that the maximumnumber of fixed points of a non-identity element of this group is 3, so thereforethe minimum distance is 12 and correction capability is 5.

A (15,3,5)-uncovering can be obtained from the(15,12,5) covering design inGordon’s database [22]. By comparing this with the list of bases forA7 obtainedabove, it happens (by sheer good fortune!) that all the triples in this uncoveringare bases, so no “fiddling” is required here.

We summarise theGAP computations below.

# Construct action of A_7 on 15 pointsgap> L:=AllTransitiveGroups(NrMovedPoints,15,Size,2520);[ A_7(15) ]gap> G:=L[1];;

69

EXCEPTIONAL BASE-TRANSITIVE GROUPS 70

# Construct orbit of A_7 on irredundant basesgap> b:=BaseOfGroup(G);;gap> B:=Orbit(G,b,OnSets);;

# (15,3,5)-uncovering found in La Jolla repositorygap> U:=[ [ 13, 14, 15 ], [ 10, 11, 12 ], [ 9, 12, 15 ],[ 9, 13, 14 ], [ 10, 11, 15 ], [ 6, 7, 8 ],[ 3, 4, 5 ], [ 1, 2, 8 ], [ 2, 6, 7 ] ];;

# Test if this is contained in the set of all basesgap> IsSubset(B,U);true

5.2 Two groups related toA7

There are two other groups related toA7 which are base-transitive. The first is the“affine extension” of the action ofA7 onF4

2\0. This is constructed in the sameway thatAGL(n,q) is constructed fromGL(n,q), giving us the groupG = V oA7

(whereV is the additive group ofF42). This action ofG has degree 16 and is

base-transitive of rank 4; the bases are affine-independent 4-tuples of vectors inF4

2.

As with A7, we can construct this action inGAP. Again, we find it in thelibrary. This time we obtain a list of three candidate groups. However, only thethird of these has base size 4, so this must be the group we require.

In order to useG as a code, we need to know its minimum distance. From [31],we know that the maximum number of fixed points of an non-identity element ofG is 4, so the minimum distance is 12, and thus the correction capability is 5. Thuswe require a(16,4,5)-uncovering, where each 4-set is a base forG.

Once more, we can obtain an uncovering from [22]. Unfortunately, as initiallyobtained, it is not an uncovering-by-bases. However, by relabelling the points itis possible to convert it into one. This has been tested usingGAP, in a similarmanner to the above, and the uncovering-by-bases obtained is given below.


[ [ 1, 3, 9, 16 ], [ 1, 6, 9, 12 ], [ 1, 8, 9, 14 ],[ 2, 4, 13, 15 ], [ 2, 5, 10, 15 ], [ 2, 7, 11, 15 ],[ 3, 6, 12, 16 ], [ 3, 8, 14, 16 ], [ 4, 5, 10, 13 ],[ 4, 7, 11, 13 ], [ 5, 7, 10, 11 ], [ 6, 8, 12, 14 ] ]

The second group related toA7 is constructed similarly. LetK denote thestabiliser of a point in the action ofA7 on 15 points described above. Then we canconstruct the affine extensionH of this group,H = V oK, whereV is as above.

This group also has degree 16, and is base-transitive of rank 3; the bases hereare the affine-independent triples of vectors inF4

2. Again, we use theGAP libraryto find this group. From [31], the maximum number of fixed points of a non-identity element ofH is 4, so the minimum distance is 12 and therefore this codecan also correct 5 errors.

As before, we consult the database [22] to find a suitable uncovering; herethe required parameters are(16,3,5). Once again, we need to relabel the pointsbefore it becomes an uncovering-by-bases. Having done this, we find that theuncovering given below forms an uncovering-by-bases.

[ [ 1, 13, 15 ], [ 2, 5, 8 ], [ 3, 6, 7 ], [ 4, 9, 12 ],[ 4, 11, 16 ], [ 9, 12, 16 ], [ 10, 11, 14 ], [ 10, 14, 16 ] ]

5.3 Two Mathieu groups,M11 and M12

Two of the Mathieu groups are the final two exceptional base-transitive groups.M12 is sharply 5-transitive and has degree 12;M11 is the stabiliser of a pointin M12, so is sharply 4-transitive and has degree 11. They are two of the 26sporadic simple groups decribed in theATLAS [18]. There are various ways ofconstructing them, but the following (forM12) is probably the easiest to remember.Consider the following diagram:

First, take the permutation given by the 2-cycles of the labels on each edge,(1 2)(3 4)(5 6)(7 8)(9 10)(11 12). Second, at each vertex, form cycles by readingthe labels anti-clockwise around each one (ignoring the trivial cycles), which gives(1 3 2)(4 7 5)(8 9 11). Then these two permutations generateM12, which is asimple group of degree 12 and order 95040.M11 is defined as the stabiliser of apoint in M12. The origins of the diagram above are in the Groethendieck theoryof “dessins d’enfants”, as decribed by Jones [27].


&%'$6 6

?

6*

?

6

BBBN

BBB

BBBMs s

s

ss s

1 2

34

56

7

8

9

10

11

12

Figure 5.1: Generators forM12

SinceM12 is sharply 5-transitive and has degree 12, we have that its minimumdistance is 12−5+1 = 8, while the minimum distance ofM11 is 11−4+1 = 8.Hence, when used as error-correcting codes, both have correction capability 3.

As both groups are sharplyk-transitive (for appropriatek), we require onlyuncoverings, as anyk-tuple forms a base. Specifically, we require an(11,4,3)-uncovering forM11 and a(12,5,3)-uncovering forM12. The following uncover-ings were obtained from the corresponding covering designs in the database [22].


1 2 3 4

1 2 5 6

3 4 5 6

7 8 9 10

7 8 9 11

7 8 10 11

7 9 10 11

8 9 10 11

Figure 5.2:(11,4,3)-uncovering forM11

1 2 3 4 5

1 2 6 11 12

1 3 7 8 9

1 4 6 7 10

1 5 8 9 11

2 4 8 9 12

2 5 7 10 11

3 4 7 11 12

3 5 6 10 12

3 6 8 9 11

6 7 8 9 10

Figure 5.3:(12,5,3)-uncovering forM12

Chapter 6

Further topics: some case studies

6.1 Introduction

In this chapter we investigate some further topics related to our use of permutationgroups as error-correcting codes. For each topic, we have one of two case studiesin mind, these being the Mathieu groupM12 and the wreath productCm oSn.

First, we investigate ways in which we can improve the performance of ouralgorithm, either by using patterns of repeated symbols or by ordering uncover-ings. We then move on to investigating the perfomance of our codes for when thenumber of errors received exceeds the correction capability. We also introduce anew algorithm which can be used for a special case.

6.2 Patterns of repeated symbols

Suppose we are using a group as a code and have transmitted a permutation. If thereceived word contains errors, then it may not necessarily be a permutation anymore, i.e. it contains one or more repeated symbols. In this section, we investigatehow looking at patterns of repeated symbols can help us with decoding. We willbe using the Mathieu groupM12 as a case study.

Recall thatM12 has degree 12 and has correction capability 3. There are sev-eral different possibilities for how the (assumed) maximum of three errors couldoccur.

If the received word is a permutation, then we have no choice other than toproceed as normal, and apply algorithm 1.4.5. However, if it is not a permutation,we can take advantage of the repeated symbols to improve the algorithm. Forexample, if the 3 errors are 3 repeats of the same symbol (so that symbol nowappears 4 times), then we know that the remaining 8 symbols must be correct and

74

FURTHER TOPICS: SOME CASE STUDIES 75

we can immediately find the unique codeword that corresponds to any 5-tuple ofthese. Thus we have decoded the received word in one step only, a considerableimprovement.

As we are consideringM12, there are a sufficiently small number of possibili-ties for us to be able to determine them all explicitly. In each case, we can isolatesome of the digits where weknowthere to be an error, so we apply a similar algo-rithm to the remaining digits to locate a smaller number of errors, which shouldrequire a smaller number of steps.

If the received word is not a permutation, then there are several possibilitiesfor this, as listed below. Each example is based on the transmitted word being theidentity word, 1 2 3 4 5 6 7 8 9 10 11 12.

–Three errors

• 1 repeated symbol, 2 symbols movede.g. 1 1 4 3 5 6 7 8 9 10 11 12Here, we need a(10,5,2)-uncovering, as we know that at least one of thetwo 1s is an error.

• 2 different symbols repeated, 1 symbol movede.g. 1 1 3 3 2 6 7 8 9 10 11 12Here, we need an(8,5,1)-uncovering.

• 3 different symbols repeated (∗)e.g. 1 1 3 3 5 5 7 8 9 10 11 12Here, we know that there are three errors in the first six places, so can ex-amine any five out of the last six and find the unique permutation inM12

that agrees in those places.

• 1 symbol repeated thrice (∗)e.g. 1 1 1 1 5 6 7 8 9 10 11 12Here, we can choose any five of the last eight places and proceed as above.

• 1 symbol repeated twice, another symbol repeated (∗)e.g. 1 1 1 4 4 6 7 8 9 10 11 12Here, we can choose any five of the last seven places and proceed as above.

• 1 symbol repeated twice, another symbol movede.g. 1 1 1 2 5 6 7 8 9 10 11 12Here, we need a(9,5,1)-uncovering.


• 1 symbol repeated twice, both occurrences incorrectly placed, another sym-bol movede.g. 3 3 1 4 5 6 7 8 9 10 11 12Here, we need a(10,5,2)-uncovering and apply to places 3 to 12.

–Two errors

• 2 symbols repeatede.g. 1 1 3 3 5 6 7 8 9 10 11 12Here, we need an(8,5,1)-uncovering. (Note that although only two errorshaveactuallyoccurred, we have to continue our assumption that there maybe three.)

• 1 symbol repeated twicee.g. 1 1 1 4 5 6 7 8 9 10 11 12Here, we need a(9,5,1)-uncovering.

–One error

• 1 symbol repeatede.g. 1 1 3 4 5 6 7 8 9 10 11 12Here, we need a(10,5,2)-uncovering.

Using Gordon’s database of covering designs [22] we can obtain some of theuncoverings we require. A(12,5,3)-uncovering was given earlier (in figure 5.3),and a(10,5,2)-uncovering (complement of a(10,5,2)-covering design) is givenbelow.

1 2 3 4 5

1 2 7 8 10

1 5 6 7 9

2 3 6 8 9

3 4 7 9 10

4 5 6 8 10


Constructing(n,k,1)-uncoverings is straightforward enough for the correspond-ing covering designs not to be included in the database. For completeness, weinclude the ones we need below.

• A (9,5,1)-uncovering:

1 2 3 4 5

1 6 7 8 9

5 6 7 8 9

• An (8,5,1)-uncovering

1 2 3 4 5

1 2 6 7 8

4 5 6 7 8

Note that depending on where the repeated symbols occur in a received word, arelabelling will almost certainly be required. Consequently, we will now need tostore several uncoverings (including the relabelled ones) instead of just one.

6.3 Optimising the algorithm

Up to now, when discussing uncoverings-by-bases we have referred to them as setsof bases, which are unordered. However, when applying our decoding algorithm,the order of the bases is important, as it will affect performance. Although wehave usually presented uncoverings in lexicographic order, this is only for easeof reading and there is no sensible reason why this ordering should be used inpractice. In this section we discuss how to order an uncovering-by-bases in an“optimal” way.

6.3.1 A measure of optimality

Suppose we are using a groupG of degreen and correction capabilityr. Thensuppose we have an uncovering-by-basesU of size u, which we have ordered


(B1,B2, . . . ,Bu). For a given set of error positionsE (where|E| ≤ r), we definethe number ofsearchesto be the numberi, which is the label of the first basewhich is disjoint fromE. (So this is the number of times the “reconstruction”algorithm has to be performed.) Then we can calculate the average number ofsearches, which will be dependent on the ordering.

To count the average number of searches required forj ≤ r errors, we de-fine the sequencesj = (s1 j ,s2 j , . . . ,su j), wheresi j is the number ofj-subsetsof 1, . . . ,n avoided byBi but not byB1, . . . ,Bi−1. So the average number ofsearches will be

a j =∑u

i=0 isi j(nj

) .

Example 6.3.1.Consider the Mathieu groupM12, which has degree 12 and cor-rection capability 3, and the uncovering in figure 5.3 in lexicographic order. Sup-posing that there are three errors, and usingGAP to perform the calculations, weobtain the sequence

(35,31,30,23,17,21,20,15,17,8,3).

Using the formula above, we find that the average number of searches isa3 = 1016220 .

One way to “optimise” an ordering would be to order the uncovering in sucha way to minimise the quantityar . In general this is not an easy problem to solve.However, there is a relatively easy way to improve the performance of a givenordering.

Algorithm 6.3.2. Suppose that we are given the ordering(B1, . . . ,Bu), whichgives rise to the sequences= (s1,s2, . . . ,su) (we assumej = r and drop that in-dex for clarity). Then suppose thats1 ≥ s2 ≥ ·· · ≥ sl and thatsl < sl+1 (?), i.e.this is the first place where the sequence increases. We obtain a new ordering(B′1, . . . ,B

′u) by interchangingBl andBl+1 and leaving all others alone. So we

haveB′1 = B1, . . . ,B

′l−1 = Bl−1,

B′l = Bl+1,B′l+1 = Bl ,

B′l+2 = Bl+2, . . . ,B′u = Bu.

This new ordering gives rise to the sequences′ = (s′1, . . . ,s′u). We make the

following observations about this:

• s′1 = s1, . . . ,s′l−1 = sl−1 ands′l+2 = sl+2, . . . ,s′u = su (??);


• s′l ≥ sl+1 ands′l+1 ≤ sl (???);

• s′l +s′l+1 = sl +sl+1 (†).

That this new ordering is indeed an improvement is shown by the following propo-sition.

Proposition 6.3.3.With the notation as above,

u

∑i=1

is′i <u

∑i=1

isi .

Proof. By (??), it is sufficient to show

ls′l +(l +1)s′l+1 < lsl +(l +1)sl+1.

By inequalities (?) and (???), we havesl < sl+1 ≤ s′l , so therefores′l = sl +d forsomed > 0. Rearranging equation (†), we obtain

s′l −sl = sl+1−s′l+1 = d.

Hence we have

l(s′l −sl ) = l(sl+1−s′l+1) < (l +1)(sl+1−s′l+1).

Rearranging this inequality we obtain

ls′l +(l +1)s′l+1 < lsl +(l +1)sl+1

as required.

So algorithm 6.3.2 does indeed reduce the average number of searches. We canthen iterate it until we obtain a decreasing sequence.

Example 6.3.4.Consider the uncovering in example 6.3.1. We see that the fifthterm in the sequence obtained is less than the sixth, so we change the order of the


uncovering by interchanging the fifth and sixth bases, as follows:

1 2 3 4 5

1 2 6 11 12

1 3 7 8 9

1 4 6 7 10

2 4 8 9 12

1 5 8 9 11

2 5 7 10 11

3 4 7 11 12

3 5 6 10 12

3 6 8 9 11

6 7 8 9 10

This new ordering gives rise to the following sequence:

(35,31,30,23,23,15,20,15,17,8,3).

We then iterate algorithm 6.3.2 until we obtain the following ordering:

1 2 3 4 5

1 2 6 11 12

1 3 7 8 9

1 4 6 7 10

2 4 8 9 12

2 5 7 10 11

3 5 6 10 12

3 4 7 11 12

1 5 8 9 11

3 6 8 9 11

6 7 8 9 10


which gives rise to the following decreasing sequence:

(35,31,30,23,23,23,20,13,11,8,3)

The average number of searches (for 3 errors) for this ordering is now988220.

We remark that while this clearly does not guarantee a best possible order-ing, it does tell us that a best possible ordering must give rise to a decreasingsequence, which could reduce the amount of searching required to locate a bestpossible ordering. An exhaustive search usingGAP (by testing all possible or-derings) revealed that, for theM12 example, the minimum value for the averagenumber of searches is959

220, and that there are 288 of the possible 11! orderingswhich give rise to this value.

6.3.2 Nested uncoverings

Whilst optimising forr errors may be desirable, in some ways it makes sense toconsider optimising forj ≤ r errors. One way to do this is the idea ofnestinguncoverings. Clearly, an(n,k, r)-uncovering is also an(n,k, r − 1) uncovering(and so on), but is not one of best possible size. There are two possible strategieshere: a “top-down” strategy and a “bottom-up” strategy.

By “top-down” we mean the following: starting from an(n,k, r)-uncoveringU, find the smallest(n,k, r − 1)-uncovering insideU (this may not be unique)and place it at the beginning. Then inside that place the smallest(n,k, r − 2)-uncovering at the beginning, and so on.

Example 6.3.5.Again, we consider the(12,5,3)-uncovering in figure 5.3 and wedenote it byU. According to Gordon’s database [22], the best known(12,5,2)-uncovering has size five, but a computer search reveals that no such(12,5,2)-uncovering is contained inU. It does, however, contain several (32) of size six,of which we choose one. There is no clear way to choose which of these to use,so we take one which (when in lexicographic order) gives the least value ofa2.


There are two which havea2 = 163, one of which is:

1 4 6 7 10

1 5 8 9 11

2 4 8 9 12

3 4 7 11 12

3 5 6 10 12

3 6 8 9 11

This in turn will contain some(12,5,1)-uncoverings. The smallest such uncover-ing possible has size two (consisting of two disjoint 5-tuples). Indeed,U containssuch, but the list above does not. It does, however, contain several of size three.As before, we take one which gives rise to the least value ofa1; this time, wea1 ≥ 18, and an example satisfyinga1 = 18 is:

1 4 6 7 10

2 4 8 9 12

3 5 6 10 12

Overall, the “nested”(12,5,3)-uncovering looks like this:

(12,5,1)-uncovering

1 4 6 7 10

2 4 8 9 12

3 5 6 10 12

Rest of(12,5,2)-uncovering

1 5 8 9 11

3 4 7 11 12

3 6 8 9 11

Rest of(12,5,3)-uncovering

1 2 3 4 5

1 2 6 11 12

1 3 7 8 9

2 5 7 10 11

6 7 8 9 10


The “bottom-up” strategy would work as follows. Start with a best-possible(n,k,1)-uncovering, add as few extrak-tuples as possible to it in order to producean (n,k,2)-uncovering, then iterate this procedure until an(n,k, r)-uncoveringis obtained. Of course, there is no guarantee that a best-possible(n,k, j + 1)-uncovering will be obtained by expanding the(n,k, j)-uncovering we have at stagej, so it is possible that the(n,k, r)-uncovering obtained could be much larger thannecessary.

6.3.3 Calculating the expected duration of the algorithm

What would be even more desirable would be a way to determine the expectednumber of searches required, based on a fixed probability of there being an error.Specifically, we suppose that an error in a single position occurs with probabilityp, and that these are independent and identically distributed (i.i.d.), so that theprobability thatj errors occur is

(nj

)p j(1− p)n− j . Hence we have

P(A) = ∑j≤r

(nj

)p j(1− p)n− j .

Let X denote the random variable corresponding to the number of searches.We want to determineE(X | A), that is the expected value ofX conditional on theeventA, whereA is the event that the number of errors is at mostr. For a givenordering of an uncovering, we letsi j denote the number of patterns ofj errorsthat requirei searches (as above), and letTj = ∑i isi j , soTj is the total number ofsearches over allj-error patterns.

Theorem 6.3.6.The expected number of searches, given a maximum of r errors,is given by

E(X | A) =∑ j≤r Tj p j(1− p)n− j

∑ j≤r(n

j

)p j(1− p)n− j

.

Proof. Defineqi j to be the probability of havingi searches forj errors, soqi j =p j(1− p)n− jsi j . Now (see [24], page 11), we have P(X andA) = P(X | A)P(A),


so we have

E(X | A)P(A) = ∑i

i P(X = i | A)P(A)

= ∑i

i P(X = i andA)

= ∑i

i ∑j≤r

qi j

= ∑i

i ∑j≤r

p j(1− p)n− jsi j

= ∑j≤r

p j(1− p)n− j ∑i

isi j

= ∑j≤r

p j(1− p)n− jTj .

So therefore

E(X | A) =∑ j≤r p j(1− p)n− jTj

P(A)

=∑ j≤r p j(1− p)n− jTj

∑ j≤r(n

j

)p j(1− p)n− j

.

From now on, we denote the functionE(X | A) by F(p) for a given ordering.

Example 6.3.7.Suppose we consider the Mathieu groupM12, and are using theuncovering in figure 5.3 in lexicographic order. Then we haven = 12, r = 3and, from earlier calculations,T0 = 1, T1 = 22, T2 = 189 andT3 = 1016. So theexpected number of searches is

F(p) =(1− p)3 +22p(1− p)2 +189p2(1− p)+1016p3

(1− p)3 +12p(1− p)2 +66p2(1− p)+220p3 .

(Note that we were able to cancel a factor of(1− p)9 from both the numeratorand denominator.) Using MAPLE we can plot this function as shown in figure 6.1.We can also evaluate its Taylor series about 0, obtaining the following:

1+10p+13p2 +116p3 +O(p4).

We need to explain what is meant by “optimality” when discussingF(p). Anordering of an uncovering could be considered optimal if its functionF(p) attains


1

0.60

0.40.20

p

5

1

4

3

0.8

2

Figure 6.1: Expected number of searches forM12, with lexicographic order

a global minimum (for all values ofp). However, it is not clear that such a min-imum should exist! We illustrate how different ordering can affect that functionF(p) with an example.

Example 6.3.8.Again, we considerM12, with our uncovering in various orders.

Ordering Colour T0 T1 T2 T3

Lexicographic Red 1 22 189 1016

“Improved” lexicograhic (Example 6.3.4)Green 1 21 180 988

“Nested” (Example 6.3.5) Yellow 1 18 161 976

Best possible forT3 (†) Blue 1 18 162 959

(The ordering marked (†) is an arbitrarily-chosen example of those with the least


value ofT3 found by exhaustive search.) The functionF(p) is plotted for thesefour orderings below.

1

0.60

0.40.20

p

5

1

4

3

0.8

2

Figure 6.2:F(p) for various orderings, I

This illustrates the effect that the “improvement” algorithm (algorithm 6.3.2)and the “nesting” procedure can have on the order of an uncovering. However, itcan be seen that while they reduce the values taken byF(p), they don’t achievewhat is best possible (although “nesting” comes close for small values ofp). Also,the next example demonstrates that the effect of algorithm 6.3.2 is not uniform.

Example 6.3.9.Once more, we consider our uncovering forM12. This time, wecompare lexicographic order with a randomly-chosen order and the effect that


algorithm 6.3.2 has on each of them. The random order used is as follows:

6 7 8 9 10

3 5 6 10 12

3 4 7 11 12

2 4 8 9 12

1 5 8 9 11

1 3 7 8 9

1 2 3 4 5

2 5 7 10 11

3 6 8 9 11

1 2 6 11 12

1 4 6 7 10

After the usual calculations, we have the following:

Ordering Colour T0 T1 T2 T3

Lexicographic Red 1 22 189 1016

“Improved” lexicograhic (Example 6.3.4)Green 1 21 180 988

Random Yellow 1 19 177 1022

“Improved” random Blue 1 19 173 959

The functionF(p) for these four orderings is plotted in figure 6.3 below.


1

0.60

0.40.20

p

5

1

4

3

0.8

2

Figure 6.3:F(p) for various orderings, II

As can be seen, for small values ofp the random ordering works faster thanthe lexicographic ordering, but whenp is very large lexicographic order is better.However, after applying algorithm 6.3.2, we see that the improved version of therandom order is faster than the improved lexicographic order for all values ofp. So it appears that the algorithm is more effective when applied to the secondordering.


6.3.4 Some concluding remarks

We conclude this discussion with a few remarks. First, we observe that if theuncovering consists of disjoint bases (such as in sections 2.2 and 3.2), then anyorder is automatically optimal, as all bases are the same with respect to each other.

Secondly, ordering of block designs for some kinds of optimality criteria havebeen studied previously; for instance, Cohen and Colbourn [15] consider certainoptimal orderings of Steiner triple systems. It is likely that some of the issuesraised and techniques discussed here (such as nesting) could have applications tocovering designs in general.

6.4 M12: words with four errors

Suppose that we transmit a permutationg∈ M12 and receive a wordw with fourerrors. Since there are elements ofM12 at distance 8 fromg, it is possible thatg isnot unique as a codeword at distance 4 fromw. For example, ifh∈ M12 satisfiesd(g,h) = 8, then we can construct a word halfway between them (i.e. halfwayalong ag−h path of length 8 in the Hamming graphH(12,12); Hamming graphsare described in [5], section 9.2.) The question which then arises is, “How manywords with four errors do not have a unique nearest neighbour?”. It is this whichwe will address in this section.

We begin by calculating the distance enumerator ofM12. Recall definition1.1.5:

∆(x) = ∑g∈R

|Class(g)|xn−π(g).

Using the character table in theATLAS [18], page 32, and following a few boringcalculations, we then find that

∆(x) = 1+3465x8 +1760x9 +21384x10+33120x11+35310x12.

The most important observation at this stage should be thatthe proportion ofwords at distance 8 is very small(in fact, about 3.5%).

Lemma 6.4.1. If w is at distance 4 from two codewords g,h, then d(g,h) = 8.

Proof. d is a well-defined metric, so satisfies the triangle inequality. Thus

d(g,h)≤ d(g,w)+d(w,h) = 4+4 = 8.


But the minimum distance of the code is 8, sod(g,h)≥ 8. Henced(g,h) = 8.

Lemma 6.4.2.Fix g∈ M12. Then there are exactly7(12

4

)elements h∈ M12 satis-

fying d(g,h) = 8.

Proof. Supposed(g,h) = 8. Theng andh agree in exactly four positions; thereare(12

4

)choices for these. Suppose without loss of generality that we have chosen

positions 1, 2, 3, 4. BecauseM12 is sharply 5-transitive, the entry in position5 of h will then determine all of the remaining entries. There are 7 choices ofsymbol to put there (anything from positions 6, . . . ,12 of g). Hence there are7(12

4

)codewords at distance 8 fromg.

Note that 7(12

4

)= 3465, agreeing with the distance enumerator above.

Lemma 6.4.3.For g,h∈ M12 satisfying d(g,h) = 8, there are exactly(8

4

)words

w satisfying d(g,w) = d(h,w) = 4.

Proof. If d(g,h) = 8, theng andh agree in exactly 4 positions. Sincew lieson a shortestg− h path, w must also agree withg and h in these 4 positions.Additionally, w must agree withg in 4 of the 8 positionsg andh differ (for whichthere are

(84

)= 70 such choices), and agree withh in the remaining four places.

Lemma 6.4.4.Suppose we have g,h,w as in Lemma 6.4.3 above. Then there isatmost onefurther codeword k∈ M12 satisfying d(k,w) = 4.

Proof. Suppose without loss of generality that:

g,h,w all agree in positions 1, 2, 3, 4g,w additionally agree in positions 5, 6, 7, 8h,w additionally agree in positions 9, 10, 11, 12.

Now, if k ∈ M12, thend(g,k) = d(h,k) = 8 (by Lemma 6.4.1). Sok must agreewith g in 4 positions. Sincew lies on a shortestg− k path,w must also agreein these 4 positions, so thereforek agrees withg (and w) in positions 5, 6, 7,8. Similarly, k agrees withh (and w) in positions 9, 10, 11, 12. Now, sinced(k,w) = 4, k andw agree in exactly 8 positions, so therefore they must differ inpositions 1, 2, 3, 4.


But how cank be a permutation? So far, we know that

k = [ ? ? ? ? g(5) g(6) g(7) g(8) h(9) h(10) h(11) h(12) ].

First, we needg(5), . . . ,g(8),h(9), . . . ,h(12) to all be distinct. If this happens,then choose any five of them (without loss of generality, we can choose these tobeg(5), . . . ,g(8),h(9)). SinceM12 is sharply 5-transitive, exactly one permutationk∗ ∈ M12 agrees in these points. (Note that becauseg(9) 6= h(9), the permutationk∗ cannot be equal tog.) These five positions uniquely determine all entries ofk∗.If it happens thatk∗(10) = h(10), k∗(11) = h(11), k∗(12) = h(12), then we cantake this unique permutationk∗ to be the required permutationk.

So we now know that if we have a word with four errors, then we have one ofthe following situations:

(a) s s4g w (b) s s s4 4

g w h (c)J

JJ

s s

s

s4

4

4g w

h

k

Supposeg is fixed. Letyi be the number of pairs(g,w) (with d(g,w) = 4) suchthat there exist exactlyi codewordsh (whereh 6= g) with d(h,w) = 4. By lemma4, we know thati = 0,1,2 only. Soy0+y1+y2 is the number of configurations oftype (a), 2y2 is the number of configurations of type (c) (since we will be countingeach possibility twice, withh andk interchanged) andy1 + 2y2 is the number ofconfigurations of type (b) (as each occurence of type (c) contains two type (b)configurations inside it).

Clearly, the number of configurations of type (a) is(12

4

).114, as there are

(124

)choices for the 4 positions where the errors lie, and 11 possibilities for each error.Combining the results of lemmas 2 and 3, we see that there are 7.

(124

)(84

)config-

urations of type (b). So it remains to determine the number of configurations oftype (c), which we shall do by computer search.

Without loss of generality, we may assume thatg is the identity permutation,and thatg andh agree in the first four positions. By lemma 2, there are 7 possibil-ities forh, and then

(84

)choices forw. For each of these, we then test if the unique

permutation satisfies our requirements.


We use theGAP [21] programs described in appendix A.3. As can be seenfrom the results (also in appendix A.3), for a given codewordg and codewordshagreeing withg in a given four positions, there are 18 configurations of type (c).So we now have three equations to solve:

y0 +y1 +y2 =(

124

).114

y1 +2y2 =(

124

).7.

(84

)2y2 =

(124

).18

which gives

y1 +y2 =(

124

).481.

That is, for a given codeword, there arey1+y2 configurations which do not decodeuniquely. Or, the probability of not being able to decode uniquely is

y1 +y2

y0 +y1 +y2=

48114641

' 0.0329.

We conclude that in the majority of cases, we can decode correctly, soM12 can beused to correct four errors with a high probability(0.9671).

To decode four errors, we will use our exisiting algorithm but with slight mod-ification. First, we now need a(12,5,4)-uncovering. From the database [22] weused before, we obtain the complement of the(12,7,4)-covering design (in thiscase, a “Nurmela andOstergard Simulated Annealing Cover”), and relabel this toobtain the following:


1 2 3 4 5

1 2 3 7 9

1 2 4 8 12

1 2 5 7 10

1 2 6 11 12

1 3 5 6 11

1 3 7 8 10

1 4 5 8 12

1 4 6 9 10

1 8 9 10 11

2 3 4 7 10

2 3 5 6 10

2 3 5 8 12

2 4 6 8 11

2 5 7 9 11

2 9 10 11 12

3 4 5 9 11

3 4 6 11 12

3 7 9 11 12

4 5 7 10 12

4 6 7 8 9

5 6 7 9 12

6 7 8 10 11

6 8 9 10 12

The other necessary modification comes from the fact that in some cases ofwords with four errors there are two or three possible nearest neighbours. In thiscase, we need the algorithm to continue until all three are found. Because in themajority of cases there will only be one answer, three will not be found, so thiswill result in the algorithm checkingall 24 5-tuples. For this reason, there is nopoint in calculating the expected number of ‘searches’ as we did earlier.

6.5 M12: more than four errors

The obvious next step is to think about what happens if more than four errorsoccur. However, there are issues to consider here that did not arise in the earliercases. In all possibilities where there were three errors or less, and in the majorityof possibilities where there were four errors, we would recover a unique codeword,and be certain that we have decoded correctly. In the remaining cases with fourerrors, the decoding algorithm returns two or three possible codewords, one ofwhich we know to be the transmitted word. However, whenw is a received wordwith five errors, there are some cases where the nearest neighbour tow in M12 isnot the transmitted word.


Example 6.5.1.Suppose we transmitg, where

g = 1 2 3 4 5 6 7 8 9 10 11 12

and receivew, where

w = 4 1 2 3 6 6 7 8 9 10 11 12.

As we can see,w contains five errors, sod(g,w) = 5. However, the algorithm willdecodew ash, where

h = 4 1 2 3 6 7 8 5 9 10 11 12

andd(h,w) = 3. In this example, we have decoded incorrectly.

To help visualise this problem, we can allocate a colour to each possible re-ceived word to fit into one of these three types:

• Greenwords are those which will definitely decode correctly and uniquely;

• Yellowwords are those where the algorithm returns several possibilities forthe transmitted word, one of which is correct;

• Redwords are those which will definitely be incorrectly decoded.

So, for example, if there are three errors or less, all words are green, or if thereare four errors, 96.7% of words are green and the remaining 3.3% are yellow. Ifthere are more than four errors, determining all the configurations that give rise toyellow or red words (in the manner of section 2 above) is not straightforward.

However, if we assume that the transmitted word is the identity and that theerrors occur in the first five positions, we can use aGAP program to determinethe colour of each possible received word and to count the number of each colour.Modifying this program lets us count the the number of words of each colour withsix errors. The results of this are shown below.


No. of errors % Green % Yellow % Red

0 100 0 0

1 100 0 0

2 100 0 0

3 100 0 0

4 96.7 3.3 0

5 18.8 73.3 7.9

6 0.5 19.2 80.3

(See also figure 6.4 below.)

We conclude by observing that there will be some words with seven errors thatare “green” (uniquely decodable), such as 999999989101112. However, thenumber of these will surely be very small indeed, and as our method of checkingall possible error patterns will leave the computer with 117 patterns to check, thiswould not be worth doing!

6.6 Cm oSn: an alternative algorithm

In this section we describe an alternative algorithm for decodingCm oSn, whichdoes not use uncoverings. AsCm is a regular permutation group,Cm oSn is anexample of a “generic” family of base-transitive groups, as listed in section 1.5.Our new algorithm makes use of their relatively simple combinatorial structure.This is most easily demonstrated with an example. The following permutation isan element ofC5 oS4.

7 8 9 10 6 15 11 12 13 14 20 16 17 18 19 5 1 2 3 4

As can be seen, the permutation splits into four “blocks” of length five, containing1, . . . ,5, 6, . . . ,10, etc. Within each block, a cyclic permutation has been applied,and then some permutation is then applied to the blocks. In general, we havenblocks of lengthm. Now, recall from corollary 2.2.4 that the minimum distanceof Cm oSn is m, so the correction capability isr = bm−1

2 c. Consequently, if weassume there to be a maximum ofr errors, there will be a majority of positions ineach block which contain the correct symbol. The decoding algorithm works asfollows.


100%

90%

80%

70%

60%

50%

40%

30%

20%

10%

0%0 1 2 3 4 5 6

Number of errors

Perc

enta

ge o

f wor

ds o

f eac

h co

lour

Figure 6.4: Percentage of words of each colour

Algorithm 6.6.1. First, we divide the received word into then blocks of lengthm. Each entryi is first written in the form

i = qm+s

then converted into a pair(b,c), as follows:

b :=

q if s 6= 0q−1 if s= 0

c := s− j modm

where j is the position in the current block. Henceb corresponds to the blocklabel andc to the cyclic shift in that block. So in this block we determine the mostfrequently occurring values ofb andc (given the assumed maximum number of


errors, unique maximum frequencies will exist). So we now know what the shiftis in that block. Having performed this procedure on alln blocks, the majorityvalues ofb in each block will give us a permutation of0,1, . . . ,n− 1, whichtells us the correct order of the blocks. This gives us the information we requireto reconstruct the transmitted permutation.

Example 6.6.2.Suppose we transmit the following element ofC5 oS4:

7 8 9 10 6 15 11 12 13 14 20 16 17 18 19 5 1 2 3 4.

Then suppose we receive the following word:

17 1 9 10 6 15 11 12 13 14 20 16 17 18 19 5 1 2 3 4.

This clearly has errors in positions 1 and 2. Having split this into four blocks oflength five, we obtain the following information for each block:

[ [ 3, 1 ], [ 0, 4 ], [ 1, 1 ], [ 1, 1 ], [ 1, 1 ] ][ [ 2, 4 ], [ 2, 4 ], [ 2, 4 ], [ 2, 4 ], [ 2, 4 ] ][ [ 3, 4 ], [ 3, 4 ], [ 3, 4 ], [ 3, 4 ], [ 3, 4 ] ][ [ 0, 4 ], [ 0, 4 ], [ 0, 4 ], [ 0, 4 ], [ 0, 4 ] ]

Taking the most frequently occurring values of the first and second entries ofeach list, we obtain the permutationσ =[ 1, 2, 3, 0 ] of the blocks and shiftsof [ 1, 4, 4, 4 ]. Thus we have the information we need to reconstruct thetransmitted permutation: in positionj of block i, we put the entry

iσm+ j +ci

whereci is the ith entry in the list of shifts. (This is subject to a correction ifj +ci > m.) For instance in position(1,1), we putm+1+1, in position(1,2) weput m+2+1, in position(2,1) we put(2m+1+4) (which gets corrected), andso on, until we recover

7 8 9 10 6 15 11 12 13 14 20 16 17 18 19 5 1 2 3 4.

Algorithm 6.6.1 has been implemented inGAP and the code is given in ap-pendix A.4. We return to this algorithm in chapter 8, where we determine itscomplexity and compare its performance with that of our original algorithm.


6.7 Cm oSn: more than r errors

We observe that the algorithm described in the previous section only requires amajority of correct elementsin each block. Consequently, there will be patternsof up tonr errors that can successfully be corrected, as long as there are no morethanr errors in each block. In this section, we investigate how many such patternsthere will be.

For positive integersk, n andr, definePn,r(k) to be the set of all partitions ofthe integerk into at mostn parts, and where each part has size at mostr. We writea partitionπ of k in the following form:

π =

(i, fi)

∣∣∣∣∣∑i≥1i f i = k

.

So if π ∈ Pn,r(k), we have thati ≤ r and∑ fi ≤ n. We also define a quantityci tobe

ci =i−1

∑j=1

f j ,

that is, the number of parts inπ of size strictly less thani.

Theorem 6.7.1.For a word in Cm oSn, and for k≤ nr, then the following numberof patterns of k errors can be corrected:

En,r(k) = ∑π∈Pn,r (k)

r

∏i=1

(n−ci

fi

)(mi

) fi

.

Proof. For a pattern ofk errors to be corrected, they can be spread across upto n blocks, as long as there are no more thanr errors in each block. So for agiven partitionπ ∈ Pn,r(k), for each part ofπ we have to choose (i) which blockcontains that many errors and (ii) where in that block they lie. Working throughi in increasing order, for eachi there aren− ci blocks remaining, of which wechoosefi (corresponding to thefi parts of sizei). Then in each of thefi blockswe have chosen, we choosei error positions from them available.

Note that ifk > nr, the setPn,r(k) is empty, soEn,r(k) = 0.

While this is a tidy combinatorial expression for the desired quantityEn,r(k),its behaviour cannot easily be seen, especially as we wish to compare it with thetotal number of patterns ofk errors,

(mnk

). A first step would be to find a recurrence

relation.


Proposition 6.7.2.The numbers En,r(k) satisfy the recurrence relation

En,r(k) =r

∑l=0

(ml

)En−1,r(k− l).

Proof. Suppose there arel errors in thenth block; there are(m

l

)ways of arranging

these. Then there arek− l errors in the remainingn− 1 blocks, so there areEn−1,r(k− l) ways of arranging these. Summing over all possible values ofl ≤ r,we obtain the required relation.

This recurrence relation assists us in studying the generating function forEn,r(k). Let En(x) denote this function, that is

En(x) = ∑k≥0

En,r(k)xk.

(The indexr is dropped for clarity.)

Proposition 6.7.3.The generating functionEn(x) can be rewritten as

En(x) =

(r

∑l=0

(ml

)xl

)n

.

Proof. First, we defineEn,r(k) = 0 for k < 0. So we have

En(x) = ∑k≥0

En,r(k)xk

= ∑k≥0

r

∑l=0

(ml

)En−1,r(k− l)xk−l xl

=r

∑l=0

(ml

)xl

(∑k≥0

En−1,r(k− l)xk−l

)

=r

∑l=0

(ml

)xl

(∑

k−l≥0

En−1,r(k− l)xk−l

)

=

(r

∑l=0

(ml

)xl

)En−1(x).

By iterating this, and with the observation thatE0(x) = 1, we have

En(x) =

(r

∑l=0

(ml

)xl

)n


as required.

It has been suggested [36] that methods from complex analysis such as saddle-point asymptotics could be used to determine the behaviour of the coefficients inthis generating function, and of the comparison with the total number ofk-errorpatterns,

(mnk

). We will not pursue this here, however, we will consider the special

casem= 3, which is much easier.

For m= 3, we haver = 1 so therefore

En(x) =

(1

∑l=0

(3l

)xl

)n

= (1+3x)n

=n

∑k=0

(nk

)3kxk

by the Binomial Theorem. Consequently the number of correctiblek-error pat-terns is

(nk

)3k, which we want to compare with the total number ofk-error patterns(3n

k

). To do this we use the standard asymptotic result (known as Stirling’s for-

mula) thatlogn! ∼ nlogn−n.

From this, we deduce the following lemma.

Lemma 6.7.4.For 0 < α < 1, we have

log

(αnβn

)∼ n(α logα−β logβ− (α−β) log(α−β)) .

Proof. We have

log

(αnβn

)= log

αn!(βn)!((α−β)n)!

= log(αn)!− log(βn)!− log((α−β)n)!∼ (αnlogαn−αn)− (βnlogβn−βn)−

− ((α−β)nlog(α−β)n− (α−β)n)= αnlogαn−βn(logβ+ logn)− (α−β)n(log(α−β)+ logn)= n(α logα−β logβ− (α−β) log(α−β)) .


We can then use lemma 6.7.4 to approximate the logarithm of the ratio(nk

)3k(3n

k

) .

We have

log

(nk

)3k(3n

k

) = log

(nk

)+ log3k− log

(3nk

)so by settingk = βn and usingα = 1 andα = 3 in lemma 6.7.4 we obtain:

log

(n

βn

)+βnlog3− log

(3nβn

)∼ n(−β logβ− (1−β) log(1−β))+βnlog3

−n(3log3−β logβ− (3−β) log(3−β))

= n(−(1−β) log(1−β)− (3−β) log3

+(3−β) log(3−β)) .

Let f (n,β) denote this approximation function. In order to see how good an ap-proximation this is, we consider an example.

Example 6.7.5.Considern = 100 andβ = k100 for 1≤ k≤ 100. Using MAPLE

we can plot exp f (100, k100). On the same axes we plot the ratios(100

k

)3k(300

k

)calculated fork = 1, . . . ,100. This plot is shown in figure 6.5.

This approximation function also allows us, for a given example, to estimatenumerically the number of error patterns that can be corrected with a given prob-ability. For instance, if we taken = 100 again, we can determine the value ofkwhich gives (100

k

)3k(300

k

) ∼ 0.9.

Using MAPLE for the calculations, we find thatk = 5.5519 is the solution to thatequation. Thus we can conclude that with probability 0.9 we can succesfully de-code up to 5 errors, which is an improvement on the original correction capabilityof r = 1.


60

0.4

0.2

0800 20 100

1

k

0.6

40

0.8

Figure 6.5: Plot of asymptotics forn = 100

Chapter 7

Other groups: the single orbit conjecture

7.1 The conjecture

We have already seen (in proposition 1.4.6) that, for any permutation groupG ofdegreen and minimum distance (i.e. minimum degree)d and correction capabilityr, there exists an uncovering-by-bases forG. Up until now, the only groups wehave been considering have acted transitively on their irredundant bases, so itfollows that an uncovering-by-bases contains irredundant bases from one orbitonly. It is this particular property that interests us in this chapter.

Definition 7.1.1. We say that a permutation groupG has thesingle orbit propertyif there exists an orbit on irredundant bases forG that contains an uncovering-by-bases.

Furthermore, we make the following conjecture.

Conjecture 7.1.2. (The single orbit conjecture.)Any permutation group has thesingle orbit property.

While we do not offer a proof of this conjecture, we have various pieces ofevidence that it should be true, such as the following.

• It holds trivially for base-transitive groups (as there is only one orbit) andfor groups withr = 0 (as we only need one base).

• The single orbit property is preserved by taking direct and wreath products:see section 7.2.

• The conjecture holds for the action ofSm on 2-subsets (section 7.3) and forsome further examples of groups (section 7.4).

• Computer searches (see below) show that the conjecture holds for transitivegroups of degree at most 19, and primitive groups of degree at most 30.

103

OTHER GROUPS: THE SINGLE ORBIT CONJECTURE 104

Using theGAP libraries of transitive and primitive groups, it is relatively easyto verify the conjecture for groups of low degree. For each group, one has todetermine the correction capability, then for a given base construct the orbit ofthe group on that base, and then check that this orbit forms an uncovering.GAPprograms such as those given in appendix A.5 were used for transitive groups ofdegree at most 19, and for primitive groups of degrees 20 to 30, and did not findany counterexample.

The single orbit conjecture is not entirely unmotivated; it has implications forthe space complexity of the decoding algorithm, as discussed in section 8.2.

7.2 Direct and wreath products

In this section, we show that the single orbit property is preserved by taking directproducts and wreath products.

Theorem 7.2.1.Suppose that G acting onΩ and H acting on∆ have the singleorbit property. Then G×H acting onΩ∪∆ also has this property.

Proof. We take suitable orbits forG andH, then use them to form an orbit forG×H, and show that this satisfies our requirements. SupposeG acting onΩhas degreen, minimum distanced and correction capabilityr = bd−1

2 c, and thatx = (x1, . . . ,xk) is a base forG in this action, such that the orbitxG forms anuncovering-by-bases. Suppose also thatH acting on∆ has degreem, minimumdistancee and correction capabilitys, and thaty = (y1, . . . ,yl ) is a base such thatyH forms an uncovering-by-bases.

Now consider the action ofG×H acting onΩ∪∆. Clearly |Ω∪∆| = n+ m.First, we show that the minimum distance ofG×H in this action is mind,e. Asa permutation in list form, an element(g,h) ∈ G×H is simply the concatenationof g andh. ThusdH((g1,h1),(g2,h2)) = dH(g1,g2)+dH(h1,h2). Suppose withoutloss of generality thatd≤ e. Then the Hamming distance is smallest whenh1 = h2

andd(g1,g2) = d, so clearly the minimum distance ofG×H is the minimum ofd ande. Consequently, the correction capability ofG×H is minr,s.

Next, we construct a base forG×H. Definez = (x1, . . . ,xk,y1, . . . ,yl ). SinceG acts onΩ only andH acts on∆ only, we have

z(g,h) = (x(g,h)1 , . . . ,x(g,h)

k ,y(g,h)1 , . . . ,y(g,h)

l ) = (xg1, . . . ,x

gk,y

h1, . . . ,y

hl )


for any(g,h) ∈ G×H. In particular, ifz(g,h) = z (i.e. if (g,h) ∈ StabG×H(z)), wehave thatxg = x andyh = y, so becausex andy are bases we haveg = 1G andh = 1H , so(g,h) = 1G×H . Hence the pointwise stabiliser ofz is trivial, soz formsa base forG×H.

Now, we have thatzG×H = (xg1, . . . ,x

gk,y

h1, . . . ,y

hl ) g∈ G,h∈ H. We show

that zG×H forms an uncovering-by-bases forG×H. Suppose without loss ofgenerality thatr ≤ s. We need to show that given an arbitraryr-subsetR⊂ Ω∪∆,there exists an element of the orbitzG×H that is disjoint fromR. Suppose thatR = A∪B, whereA ⊂ Ω, B ⊂ ∆ and |A|+ |B| = r (note that one ofA,B maybe empty). Since|A| ≤ r, there exists a base(xg

1, . . . ,xgk) for G disjoint from A.

Similarly, since|B| ≤ r ≤ s, there exists a base(yh1, . . . ,y

hl ) for H disjoint fromB.

Thus the base(xg1, . . . ,x

gk,y

h1, . . . ,y

hl ) ∈ zG×H is disjoint fromR= A∪B, so we are

done.

Theorem 7.2.2.Suppose G acting onΩ has the single orbit property, and that His an arbitrary permutation group of degree m. Then GoH acting on m copies ofΩ also has this property.

Proof. As in theorem 7.2.1 above, we take a suitable orbit forG and use it to finda suitable orbit forGoH. SupposeG acting onΩ has degreen, minimum distanced and correction capabilityr. Recall thatGoH is defined to beGmoH, whereGm

is the direct product ofm copies ofG.

Now, a non-identity element ofG oH with the maximum number of fixedpoints will be an element ofGm, as any non-trivial action ofH will reduce thenumber of fixed points. But a non-identity element ofGm with the maximumnumber of fixed points will fixm− 1 copies ofΩ, and haven− d fixed pointsin the remaining copy. Hence it has(n−1)m+(n−d) fixed points in total, andso the minimum distance ofG oH is d. Consequently the correction capability ofGoH is r.

Next, we need a base forG oH. Supposex is an irredundant base forG suchthatxG forms an uncovering-by-bases. Now, by the proof of theorem 7.2.1 above(and a straightforward induction),mcopies ofx (one from each copy ofΩ) formsa base forGm. We denote this base bymx. Clearlymx is also a base forG oH, asthe only elements ofGoH that fix each block block-wise are elements ofGm.

Finally, we observe that the orbit(mx)Gmis contained in the orbit(mx)GoH . By

theorem 7.2.1 above,(mx)Gmforms an uncovering-by-bases, so therefore(mx)GoH

also does.


7.3 Sm acting on 2-subsets

Consider the symmetric groupSm acting on the 2-subsets of1, . . . ,m. Now,m= 1 means nothing here,m= 2 gives an unfaithful action, and the action ofS3

on 2-subsets gives the usual action ofS3, so throughout this section we assumem≥ 4. We begin by determining the minimum degree and correction capabilityof this group in this action.

Proposition 7.3.1.The minimum degree of Sm acting on 2-subsets is2(m−2) andthe correction capability is r= m−3.

Proof. From section 1.1, we have that the minimum degree is equal to

n−maxg6=1

|Fix(g)| .

Now, the maximum number of fixed points ofSm in this action is(m−2

2

)+ 1,

corresponding, for example, to a transposition(1 2), which will fix the remaining2-sets chosen from3, . . . ,m and also the pair1,2. Thus the minimum degreeis (

m2

)−(

m−22

)−1 = 2(m−2).

Consequently the correction capability isr = b2(m−2)−12 c= m−3.

The 2-subsets of1, . . . ,m can be thought of as the edge set of the completegraphKm. Thinking in this way enables us to use graph-theoretic methods, whichwe shall exploit to construct uncoverings-by-bases. For instance, a base forSm inthis action will consist of a subset of these edges.

Lemma 7.3.2. The edges of a spanning subgraph of Km which have(i) at mostone isolated vertex and(ii) no isolated edges form a base for Sm acting on theedge set of Km.

Proof. Let Γ denote such a spanning subgraph. We have to show that the edge-wise stabiliser,G(EΓ), of Γ in G = Sm is trivial. First we suppose thatΓ containsno isolated vertex. Lete1 = i, j be an edge inΓ. Sincee1 is not isolated, thereexists another edgee2 that is incident withe1. We can suppose without loss ofgenerality thate2 = j,k. Now choose someg∈ G(EΓ), sog fixese1 ande2, i.e.

i, jg = i, j and j,kg = j,k.


The only way this can happen is ifg fixes the vertexj, which then forcesg tofix both i andk as well. But sinceΓ is a spanning subgraph, and there are noisolated vertices, every vertex must lie in such a configuration, so must be fixedby g. Hence allm vertices are fixed byg, and sog = 1.

If there is a single isolated vertex, we have by the same argument as abovethat the remainingm−1 vertices are fixed, which forces the remaining vertex tobe fixed. So in this case we also haveg = 1.

A minimal base forSm in this action obtained from the bases described abovewill be such a graph with the least number of edges. Thus a graph of one of thethree forms shown in figure 7.1 will form a minimal base for this action (accordingto congruence classes modulo 3).

m≡ 0 mod 3 ss@

@s ss@

@s · · · ss@

@sm≡ 1 mod 3 s

s@

@s ss@

@s · · · ss@

@s sm≡ 2 mod 3 s

s@

@s ss@

@s · · · ss@

@s s sFigure 7.1: Minimal bases forSm acting on the edges ofKm

We call a base of this form aV-graph. Clearly such a base is irredundant, asremoving any edge will leave a graph with an isolated edge or two isolated ver-tices, which will have non-trivial edge stabiliser. We note thatSm acts transitivelyon V-graphs, so they form a single orbit. There are, however, other graphs whichform irredundant bases, such as the following.

HHH

HHH

s ssss s s·· ·

Figure 7.2: Irredundant bases of differing size


The V-graphs are irredundant bases of size∼ 23m, but the “star” bases are

irredundant bases of sizem−2. Therefore,Sm in this action is not an IBIS group.

The following definition is well-known in graph theory.

Definition 7.3.3. A Hamilton circuit in a graphΓ is a circuit inΓ containing eachvertex exactly once.Γ is said to beHamiltonianif it contains a Hamilton circuit.

We observe that any V-graph is contained inside a Hamilton circuit ofKm, asis shown in figure 7.3. From a V-graph, by adding extra edges (shown ass s )we can obtain a Hamilton circuit; conversely, if we have a Hamilton cycle we canremove those edges to obtain a V-graph.

m≡ 0 mod 3 ss@

@s ss@

@s · · · ss@

@s m≡ 1 mod 3 s

s@

@s ss@

@s · · · ss@

@s s m≡ 2 mod 3 s

s@

@s ss@

@s · · · ss@

@s s s Figure 7.3: Hamilton circuits from V-graphs

Now, an uncovering-by-bases forSm in this action consists of a set of basessuch that anyr-set is disjoint from at least one base. Now, by proposition 1.4.6we already know that one exists. However, we would like to show that this grouphas the single orbit property, where the orbit on bases is specified to be that onV-graphs (see figure 7.1 above). That is, given an arbitraryr-set of edges ofKm,we need to show that there exists a V-graph that is disjoint from it.

Recall that a V-graph is contained in a Hamilton circuit inKm. Thus for anuncovering-by-bases to exist, it suffices to show that for a givenr-set of edges ofKm there exists a Hamilton circuit which avoids those edges. That is, we requirethat the graphΓ obtained fromKm by deleting somer-set of edges is a Hamiltoniangraph. Although there is no known characterisation of Hamiltonian graphs, thereare a number of partial results, some of which a described by Wilson [38]. Theresult that is of use to us here is known asOre’s Theorem.


Theorem 7.3.4.(Ore’s Theorem, 1960.)If Γ is a simple graph with m≥ 3 ver-tices, and wheredeg(v)+deg(w) ≥ m for all pairs of non-adjacent vertices v,w,thenΓ is Hamiltonian.

Proof. See Wilson [38], theorem 7.1.

We apply this in the following lemma.

Lemma 7.3.5.The graphΓ obtained by deleting r edges from Km is Hamiltonian.

Proof. Choose two non-adjacent verticesv,w in Km. In Km, they both have de-greem−1. LetRbe the set ofr edges that is to be removed to formΓ. Supposesedges ofRare incident withv andt are incident withw, with 0≤ s+ t ≤ r. (Sincev andw are not adjacent, there is no edge incident with both.) Then inΓ, we have

deg(v)+deg(w) = (m−1−s)+(m−1− t)≥ 2(m−1)− r

= 2(m−1)− (m−3)= m+1,

so by Ore’s theorem (7.3.4),Γ is Hamiltonian.

Thus it follows thatΓ contains aV-graph, inside a Hamilton circuit. Conse-quently, we have proved the following theorem.

Theorem 7.3.6.There exists an uncovering-by-bases for the action of Sm on 2-subsets, contained in a single orbit on irredundant bases.

In other words, this group has the single orbit property. However, this is onlyan existence proof: it doesn’t give us a way of constructing an uncovering-by-bases, or give us an idea of its size. In order to do this, we need some more graphtheory.

Definition 7.3.7. Let Γ = (V,E) be a graph. Adecompositionof Γ is a partitionof the edge-setE of Γ. A k-factor of Γ is ak-regular spanning subgraph ofΓ; ak-factorisationis a decomposition ofΓ where each part forms ak-factor.


k-factors are usually only studied fork = 1 ork = 2 (which is all we will needhere). Clearly, ifΓ has ak-factorisation then it must itself be regular, andk mustdivide the valency ofΓ. In particular, ifΓ has a 2-factorisation, then it must haveeven degree. A 1-factor is also referred to in the literature as aperfect matching. AHamilton circuit is a special case of a 2-factor, leading to the following definition.

Definition 7.3.8. A Hamiltonian decompositionof a graphΓ is a 2-factorisationof Γ where each 2-factor is a Hamilton circuit.

In this section we will give constructions for uncoverings-by-bases for theaction ofSm on 2-subsets, using graph decompositions of the complete graphKm

into Hamilton circuits. The following theorem is a consequence of a more generalresult due to Alspachet al [1].

Theorem 7.3.9.When m is odd, the complete graph Km has a Hamiltonian de-composition.

However, ifm is evenKm has odd valency so cannot have a 2-factoristaion.It does, however, have a decomposition into Hamilton circuits and a 1-factor, aswe explain below. It uses the construction of the 1-factorisation known asGK2n

(for m = 2n) described by Andersen [2], example VI.4.33. The vertices ofK2n

are labelled asZ2n−1∪∞, and the 1-factors are the translates of the so-calledstarter,

F0 = g,−g | g∈ Z2n−1\0∪0,∞.

By translates, we mean the sets

Fi = g+ i,−g+ i | g∈ Z2n−1\0∪i,∞

for 1≤ i ≤ 2n−2, giving us a set of 2n−1 1-factors. An example of a starter isF0 for GK8, shown in figure 7.4 below.

s ss

ssss s

0

1−1

2−2

3−2

∞

Figure 7.4: A starter forGK8


Lemma 7.3.10.Two consecutive 1-factors in GK2n form a Hamilton circuit. Thusby taking consecutive pairs, we will obtain12(m−2) disjoint Hamilton circuits,with a single 1-factor remaining.

Proof. First, we show thatF0∪F1 forms a Hamilton circuit. Recall that we have

F0 = g,−g | g∈ Z2n−1\0∪0,∞, and

F1 = g+1,−g+1 | g∈ Z2n−1\0∪1,∞.

Clearly, as these are both 1-factors, every vertex appears exactly once in each ofF0 and F1. We claim that the union ofF0 and F1 is a connected graph, whichgives us a Hamilton circuit. We do this by induction, showing that every vertex isconnected to 0.

As base cases, we have that 0 is connected to 1, along the edges0,∞ and∞,1 (so∞ is also connected to 0). Also, 0 is connected to 2, by puttingg = 1in F1. The induction hypothesis is this: suppose 0 is connected to allk where2≤ k < x. Now, x is connected to−x+ 2 (put g = x− 1 in F1), and−x+ 2 isconnected tox−2 (putg = −x+2 in F0). By the induction hypothesis, becausex−2 < x, it is connected to 0. Hencex is also connected to 0.

By the rotational symmetry ofGK2n, it follows that Fi ∪Fi+1 is a Hamiltoncircuit. Consequently, we have a set

F0∪F1,F2∪F3, . . . ,F2n−4∪F2n−3

of n−1 Hamilton circuits, andF2n−2 as a remaining 1-factor.

Our construction for an uncovering-by-bases uses these decompositions ofKm,as follows.

Construction 7.3.11.Formodd, take a Hamiltonian decomposition ofKm. Thenin each Hamilton circuit, take the set of all V-graphs contained in it. Form even,we take the Hamilton circuts in the decomposition described in lemma 7.3.10, andfor each of those take the set of all V-graphs contained in that.

In order to prove that these constructions do indeed yield uncoverings-by-bases, we need to show that for an arbitrary setR of r edges, there is a V-graphdisjoint fromRwhich is contained in one of the specified Hamilton circuits. Firstwe need two lemmata.


Lemma 7.3.12.Let π be a partition of the integer n into k parts. Then if n< 2k,π contains a part of size 1.

Proof. Suppose not, i.e. suppose thatπ hask parts, each of which has size atleast 2. Then clearlyn≥ 2k.

Lemma 7.3.13.Suppose that R contains a single edge e within a Hamilton circuitC. Then there is a V-graph contained in C which avoids e.

Proof. Recall from figure 7.3 how a V-graph, sayB, is obtained from a Hamiltoncircuit. Then by choosingB such thate is one of the edges inC\B, we have thatB avoidse.

We can now proceed with the proof. Not surprisingly, we consider the casesm odd andm even separately.

Theorem 7.3.14.Let H denote a Hamiltonian decomposition of Km, for m odd.Let U denote the set of V-graphs contained in the Hamilton circuits ofH . ThenU avoids any r-edge subset of the edges of Km.

Proof. Recall thatr = m−3 from proposition 7.3.1. Also note thatH containsc = 1

2(m−1) Hamilton circuits.

Let R denote an arbitrary set ofr edges ofKm. This may contain edges frommany Hamilton circuits. However, if it meets strictly less thatc of them, we aredone, as we can choose a V-graph from one of the remaining circuits. So weassume thatRmeets every circuit inH .

If this is so, then there exists a circuit which contains only one edge ofR for thefollowing reason. We haver edges, partitioned intoc parts (one for each circuit).Now, we haver = m−3 andc = 1

2(m−1), so thereforer = 2c−2. By lemma7.3.12, a partition of 2c−2 into c parts must have a part of size 1, so there mustbe a circuit containing just one edge ofR. LetC denote such a circuit, containinga single edgee∈ R. Then by lemma 7.3.13, there exists a V-graph contained inCwhich avoidseand therefore avoids the rest ofR.

The case wherem is even works similarly.


Theorem 7.3.15.Let F denote a decomposition of Km, m even, into Hamiltoncircuits and a 1-factor. LetU denote the set of V-graphs formed from the Hamiltoncircuits in F . ThenU avoids any r-edge subset of the edges of Km.

Proof. Again, we recall note thatr = m−3. We regard the parts of the decompo-sition F as “colour classes”; we havec = 1

2(m−2) “Hamiltonian” colour classes(corresponding to the Hamilton circuits) and a single distinguished colour class(corresponding to the 1-factor), which we label as “black”.

Let Rdenote an arbitrary set ofr edges ofKm. If it contains edges from strictlyless thatc of the Hamiltonian colour classes, then we are done, as we can choosea V-graph from inside the the remaining classes. So we assume thatRmeets everyHamiltonian colour class.

Suppose that this happens, and thatR also containsb≥ 0 “black” edges. Ig-noring theseb edges, we have a partition ofr−b edges intoc colour classes. Now,sincer = m−3 andc= 1

2(m−2), we have thatr = 2c−1, sor−b= 2c−b−1<2c. By lemma 7.3.12, this partition must contain a part of size 1. Hence thereis a Hamiltonian colour class (i.e. Hamilton circuit) which contains a single edgee∈ R. Then by lemma 7.3.13, there exists a V-graph contained inC which avoidseand therefore avoids the rest ofR.

The last step is to determine the sizes of these uncoverings. Before doingthis, we note that it is not always necessary to takeall V-graphs from inside eachHamilton circuit. Recall from lemma 7.3.13 that for each edgee in a given Hamil-ton circuitC, we need to provide a V-graph contained inC that is disjoint frome.The next result tells us how many V-graphs we need inside each circuit.

Lemma 7.3.16.Let m≥ 6. Then the number of V-graphs inside a Hamilton circuitof length m needed to avoid any single edge is3 when m≡ 0 mod 3or m≡ 1mod 3, or 4 when m≡ 2 mod 3.

Proof. Whenm≡ 0 mod 3, a V-graph consists of the circuit with every thirdedge removed, so there are only three V-graphs in there. Form≡ 1 mod 3 (withm≥ 7), again we have a circuit with every third edge removed, except at one pointwhere we remove two edges and leave an isolated vertex (see figure 7.3). Now,we can writem= 3s+ 7. Over the 3s edges, we arrange three V-graphs missingevery third edge, so that all these edges are avoided by one of these. This leaves uswith seven edges remaining, over which we arrange our three V-graphs as shownin figure 7.5 below.


s s s s s s s s· · · · · · (the circuit)

s s s s s s s ss s s s s s s ss s s s s s s s

Figure 7.5: Arranging V-graphs form≡ 1 mod 3

As can be seen from this, all seven edges are avoided by at least one of the threeV-graphs.

Now we considerm≡ 2 mod 3 (wherem≥ 8). This time, the V-graphs havethe repeating pattern of every third edge omitted, except at the end, where thereis a path of length 3 then two missing edges (see figure 7.3). This time, we canwrite m= 3s+8. Over the 3s edges of the circuit, we arrange three V-graphs asbefore, so that all these edges are avoided. We then add a fourth V-graph which isa duplicate of one of the first three. Over the remaining eight edges, we arrangethe V-graphs as shown in figure 7.6 below.

s s s s s s s s s· · · · · · (the circuit)

s s s s s s s s ss s s s s s s s ss s s s s s s s ss s s s s s s s s

Figure 7.6: Arranging V-graphs form≡ 2 mod 3

As can be seen, each of the eight edges is avoided by at least one V-graph.

This proof leaves us with the casesm= 4 andm= 5 outstanding, which canboth be handled easily.K4 can be decomposed into one Hamilton circuit and a1-factor, while a V-graph is two adjacent edges, and we haver = 1. So we splitthe Hamilton circuit into two V-graphs, and we are done.K5 has a Hamiltoniandecomposition into two Hamilton circuits, a V-graph is a path of length three andr = 2. In each of the two circuits, we arrange three V-graphs as shown in figure7.7, giving us a total of six V-graphs.

We can now determine the sizes of these uncoverings. Observing that we areusing 1

2(m−1) Hamilton circuits form odd, and12(m−2) Hamilton circuits for


s sss

s

BB

BB

s sss

s

BB

BB

ZZ

ZZ

s sss

s

ZZ

ZZ

Figure 7.7: Arranging V-graphs form= 5

m even, and combining this with the result of lemma 7.3.16 above, we obtain thefollowing result. As the size is dependent on congruence classes modulo 3andon whetherm is odd or even, we can phrase this in terms of congruence classesmodulo 6.

Theorem 7.3.17.Let m≥ 6. Then the sizes of an uncovering-by-bases for theaction of Sm on 2-subsets, as described above, are as follows:

m≡ 0 mod 6: 32(m−2)

m≡ 1 mod 6: 32(m−1)

m≡ 2 mod 6: 2(m−2)m≡ 3 mod 6: 3

2(m−1)m≡ 4 mod 6: 3

2(m−2)m≡ 5 mod 6: 2(m−1)

We now demonstrate our constructions with two examples.

Example 7.3.18.Consider the symmetric groupS7 acting on the edges of the com-plete graphK7. This graph has a Hamiltonian decomposition into three Hamiltoncircuits, as shown in figure 7.8. Applying our construction to these three Hamiltoncircuits, we obtain the nine V-graphs in figure 7.9.

Example 7.3.19.Now we consider the symmetric groupS6 acting on the edges ofthe complete graphK6. This graph has a decomposition into two Hamilton circuitsand a 1-factor, as shown in figure 7.10. Applying our construction to these twoHamilton circuits, we obtain the six V-graphs in figure 7.11.


Figure 7.8: A Hamiltonian decomposition ofK7

Figure 7.9: V-graphs forK7


Figure 7.10: A decomposition ofK6 into Hamilton circuits and a 1-factor

Figure 7.11: V-graphs forK6


7.4 Another family of groups

Let G be a finite permutation group with a transitive, abelian normal subgroupAand an irredundant base of size 2. For instance,G could be a transitive subgroup ofa sharply 2-transitive group, or a dihedral group. Now, a transitive, abelian groupis regular (see [7], exercise 1.5), soG is acting onA and therefore has degree|A|= n. As A is abelian we write it additively.

Suppose0,a is a base forG, wherea∈ A. We have the following theorem.

Theorem 7.4.1.For G as above, the orbit of G on0,a contains an uncovering-by-bases for G, and thus G has the single orbit property.

Proof. The setU = x,x+a | x∈ A

is theA-orbit on0,a, so therefore is contained in theG-orbit on0,a. Supposethata has orderm. We have two cases to consider.

First, supposem= 2. Thenx,x+a = x+a,x+2a, soU consists of12ndisjoint bases, which cover alln points. Since the correction capability ofG isr < 1

2n, we have thatU forms an uncovering-by-bases.

Second, supposem > 2. ThenU is formed ofk disjoint m-cycles (wheren = km), such as

x,x+a,x+a,x+2a, . . . ,x+(m−1)a,x.

So |U| = n, and alln points are covered twice. Now chooseR to be an arbitraryr-subset of then points. These will be contained in at most 2r bases. But 2r < n,so there exists a base inU disjoint fromR, soU forms an uncovering-by-bases.

In both cases,U is contained in a single orbit ofG on irredundant bases, soGhas the single orbit property.

In the second case of the above proof, whenm is even we can take every otherbase in eachm-cycle to form an uncovering, as this gives us1

2n disjoint baseswhich is quite sufficient.

Example 7.4.2.Let G be the dihedral group of order 12, acting on six points. Thiscontains a cyclic group of order 6, which we will regard as(Z6,+). Then0,1


is a base forG, and as 1 has order 6 inZ6, we can take every second translate of0,1. Thus

0,1,2,3,4,5

forms an uncovering-by-bases forG.

7.5 A stronger conjecture

The single orbit conjecture (7.1.2), as was the case with proposition 1.4.6, merelyasserts the existence of a suitable uncovering-by-bases, and does not attempt toshow how large it should be. In practice, this is an important thing to know: its fullimportance is explained in chapter 8, where we see how it affects the complexityof the decoding algorithm.

Conjecture 7.5.1. Let G be a permutation group of degree n. The G has thesingle orbit property, and furthermore, there exists such an uncovering-by-baseswith size polynomial in n.

This conjecture is likely to be much harder to prove than the single orbit con-jecture. However, the evidence we have suggests that it should be true. Con-sider table 7.1 below, and in particular compare the degrees with the sizes of theuncoverings-by-bases (UBBs).

In each case, we observe that the size of the uncovering-by-bases is boundedabove by the degree of the group. This leads us to the next conjecture, possiblythe most optimistic so far.

Conjecture 7.5.2. Let G be a permutation group of degree n. The G has thesingle orbit property, and furthermore, the uncovering-by-bases obtained has sizebounded above by n.

A computer search has demonstrated that conjecture 7.5.2 holds for all tran-sitive groups of degree up to 15. This computer search involves, for each group,randomly constructing uncoverings-by-bases until one of size less than the degreeis found. (For most groups, only one attempt was necessary.)


Group Degree Correctioncapability

Size of UBB

An n 1 dn3e

H oSn (H regular,|H|= m) mn bm−12 c bm−1

2 c+1

X oSn (“Zero-sum subgroups”) mn m−1 m

Rank 2 n = km b (k−1)m−12 c b (k−1)m−1

2 c+1

Sharply 3-transitive n bn−32 c n or n−3

Blow-ups ofPGL(2,q) qd(q+1) qd(q−1)2 −1 qd(q+1)

GL(3,q) q3−1 q3−q2

2 −1 q3−1 orq3−4

AGL(2,q) q2 q2−q2 −1 q2 or q2−3

A7 on 15 points 15 5 9

V oA7 16 5 12

V oK 16 5 8

M11 11 3 8

M12 12 3 11

Sm on 2-subsets(m

2

)m−3 Between3

2(m−2)and 2(m−1)

“Dihedral-like” (section 7.4) n bn−22 c n or n

2

Table 7.1: Parameters of various groups when viewed as codes

Chapter 8

Complexity issues

8.1 Complexity of the ‘uncovering-by-bases’ algorithm

In our earlier discussions of the ‘uncovering-by-bases’ decoding algorithm, wehave skipped over what it required to recover a group element from its action on abase. In order to determine the complexity of this decoding algorithm, we need torectify this situation. Suppose we have a groupG acting on a setΩ of sizen, andsuppose that(x1, . . . ,xb) is a base forG in this action. LetGi denote the pointwisestabiliser inG of (x1, . . . ,xi), with the convention thatG0 = G. The followingdefinition is due to Sims [37].

Definition 8.1.1. The setS= S1∪S2∪ . . .∪Sb, whereSi is a set of coset represen-tatives forGi in Gi−1, is called astrong generating setfor G.

This set is indeed a set of generators forG; furthermore, any element ofG canbe written uniquely as a productsbsb−1 · · ·s1, where eachsi ∈Si (see Cameron [7],section 1.13). This enables us to use the following “reconstruction” algorithm.

Suppose that we have a groupG which we are using as an error-correctingcode. Suppose thatB = (x1, . . . ,xb) is a base forG and thatS is a correspond-ing strong generating set, and that we have a received wordw which has symbols(y1, . . . ,yb) in the positions labelled byB. Then we want an answer to the follow-ing question:

Does there existg∈ G with xgi = yi for all i, and if yes, what is it?

At the first stage, we have that the setxs1 | s∈ S1 is theG0-orbit onx1, and see

if y1 appears in it. If not, then no suchg can exist, and the algorithm stops. If itdoes appear, we lets1 be the element that mapsx1 to y1, then replace(y1, . . . ,yb)

with (ys−11

1 , . . . ,ys−11

b ), and iterate as follows.

121

COMPLEXITY ISSUES 122

At stepi, we check if there exists somesi ∈ Si such thatxsii = y

s−1i−1···s

−11

i . If not,

then the algorithm stops; if somesi does exist, we replace(ys−1i−1···s

−11

1 , . . . ,ys−1i−1···s

−11

b )

with (ys−1i ···s−1

11 , . . . ,y

s−1i ···s−1

1b ), then repeat the iteration.

When we reach stepb, if we succeed we take the elementsbsb−1 · · ·s1 to beour required elementg. This works because for eachi we have

xgi = xsb···s1

i

= xsi ···s1i (sincesb,sb−1, . . . ,si+1 all lie in Gi)

= (yis−11 ···s−1

i−1)si−1···s1

= yi .

Now that we know what the procedure is, we are able to answer the questions,“how long does it take?” and “how much space is required?”. The latter ques-tion has two parts, as there are two kinds of space needed: storage space for anylook-up tables (ROM), and space needed for performing the actual computation(RAM). For the sake of simplicity, we make the following assumptions:

• finding the image of a point under a permutation takes one unit of time;

• the composition of two permutations of lengthn takesn units of time;

• the storage of a single symbol requires one unit of space.

We also use the convention whereg(n) = O( f (n)) means thatg(n) is boundedabove by some constant multiple off (n).

Lemma 8.1.2. The time required by the “reconstruct an element” algorithm isO(bn).

Proof. At step i we look throughSi to see ifyis−11 ···s−1

i−1 appears there. Since|Si | = |Gi−1 : Gi | ≤ n, there are at mostn operations to be made here. If wesucceed here, we replaceb symbols with their images unders−1

i , and as acting ona point by a permutation requires one operation, this givesb operations, so thereare at mostn+ b operations per step. As there are at mostb steps, this gives amaximum ofb(n+b).

If all b steps are completed successfully, we then have to composeb permu-tations, which requires(b− 1)n operations. Overall, the maximum number ofoperations required will beb(n+b)+(b−1)n, so this is O(bn).


Lemma 8.1.3.The storage space required by the “reconstruct an element” algo-rithm is O(bn2), and the space required to perform the algorithm isO(n).

Proof. With the convention that a single symbol requires one unit of space, a per-mutation needsn units. Our look-up table comprises a strong generating set andthe corresponding set of inverses. Now, a strong generating set has size boundedby bn, as eachSi has size at mostn. As we also need to store the inverse of eachelement, the overall number of storage units required is at most 2bn2, which isO(bn2).

When performing the algorithm, at each stagei we need to store the positionin the look-up table of the elementsi , which requires one unit of space, giving usa total ofb units. Also, when performing the composition of permutations at theend, we need a furthern units for this. So we have a total ofn+b units, which isO(n).

Recall that the “uncovering-by-bases” decoding algorithm works by workingthrough a set of bases and applying the “reconstruction” algorithm repeatedly untilthe correct permutation is obtained. LetU be the uncovering-by-bases being used.Then we have the following results.

Theorem 8.1.4.The time required by the “uncovering-by-bases” algorithm 1.4.5is |U|O(bn).

Proof. We apply the “reconstruction” algorithm, which by lemma 8.1.2 needsat mostb(n+ b) + bn time units. After this, we check the Hamming distancebetween the reconstructed permutation and the received word to see if it is withinthe correction capability, so there aren checks here. We then may have to repeatthis procedure until it has been carried out|U| times, so the total number of stepsis bounded by|U|(b(n+b)+bn+n), which is|U|O(bn).

Theorem 8.1.5.The storage space required by the “uncovering-by-bases” al-gorithm 1.4.5 is|U|O(bn2), and the space required to perform the algorithm isO(n).

Proof. For the look-up table, each base inU requires its own strong generatingset, so by lemma 8.1.3, we will need 2|U|bn2 storage units, which is|U|O(bn2).To perform the algorithm, whilst applying the “reconstruction” algorithm we need


b+n units of space. The samen units are then used for the comparison with thereceived word, meaning that the space required here is still O(n).

So, ultimately, both the time complexity and the amount of storage space re-quired are dependent on the size ofU.

8.2 Improvements

In this section we suggest some improvements which will reduce the complexityas described in the section above. It is here that the “single orbit” conjecture fromchapter 7 comes into its own.

The benefit gained from having an uncovering-by-bases contained within asingle orbit is that the space complexity is reduced. Instead of storing a list of allthe bases and, more importantly, a list of several strong generating sets, we onlyneed to store one base and one strong generating set, along with a list of groupelements that map the first base to each of the others, and to the correspondingstrong generating set. As such, we have the following.

Theorem 8.2.1.Assuming the truth of conjecture 7.1.2, then the storage spacerequired by algorithm 1.4.5 isO(bn2)+ |U|O(n).

Proof. Recall from theorem 8.1.5 that without the “single orbit” property, thespace required is|U|O(bn2). This was because each strong generating set re-quired O(bn2), and we had to store|U| of them. Now, if we assume conjecture7.1.2, we only need to store one, so the factor of|U| can be removed. How-ever, we now need to store a list of|U| permutations, which map the first baseto each of the others. As a permutation requiresn units, and we have|U| per-mutations, this gives us|U|O(n). So altogether we have a space complexity ofO(bn2)+ |U|O(n).

This is a reduction in complexity, although by how much is dependent on thesize of |U| when compared with O(bn). However, conjectures 7.5.1 and 7.5.2assist with this, as they assert the existence of bounds on|U|.

Theorem 8.2.2.Assuming the truth of conjecture 7.5.1, then the storage spacerequired by algorithm 1.4.5 isO(bn2)+O(nk+1). Assuming the truth of conjecture7.5.2, then the required space isO(bn2).


Proof. Conjecture 7.5.1 asserts that|U|= O(nk) for somek, while the strongerconjecture 7.5.2 asserts that|U|= O(n). Substituting these results into 8.2.1 com-pletes the proof.

Note that if the size of the uncovering-by-bases is O(n2), then this gives thesame space complexity as having an uncovering-by-bases of size O(n).

In its weakest form, the “single orbit” conjecture does not affect the timecomplexity. Recall that the amount of time needed to perform algorithm 1.4.5is bounded by|U|O(bn). Now, by using the modified version described above(which assumes conjecture 7.5.1), at the final step we must compose the recon-struct group element with the ‘base change’ element; this is another composi-tion of permutations, so requires anothern steps. This then needs a total timeof |U|(O(bn)+ n) = |U|O(bn), so there is no change. However, if the strongerconjectures are true, we have the following.

Theorem 8.2.3.Assuming the truth of conjecture 7.5.1, then the time requiredby algorithm 1.4.5 isO(bnk+1). Assuming the truth of conjecture 7.5.2, then therequired time isO(bn2).

Proof. By the above discussion, the time required is|U|O(bn). Conjectures7.5.1 and 7.5.2 give sizes for|U|; substituting these sizes gives the required re-sults.

So, as far as time is concerned, the single orbit property on its own does notaffect the time complexity, but the stronger versions which include a bound on thesize of the uncovering-by-bases reduces it.

8.3 Complexity of the alternative algorithm

Recall the alternative decoding algorithm (6.6.1) that can be used for the wreathproductCm oSn, described in section 6.6. Now, recall that that algorithm has threeparts:

• dividing the received word into blocks, which involves integer arithmetic;

• determining the most frequently occurring position label and shift withineach block;


• reconstructing the decoded word, which involves more integer arithmetic.

In order to determine the complexity of this algorithm, there are some assumptionswe need to make first, akin to those in section 8.1:

• integer arithmetic can be done via a look-up table, in constant time;

• comparing the sizes of two integers can be done in constant time;

• finding positioni in a list of lengthk takes O(logk) time.

However, the second step is more complicated and requires the following lemma.

Lemma 8.3.1.Let L be a list of length m with symbols chosen from S= 1, . . . ,k.Suppose L has a unique most frequently occurring element, x∈ S. Then the timetaken to determine x isO(k+mlogk).

Proof. We begin by producing an auxilliary listK of lengthk, initially set to[0,0, . . . ,0]. This takesk units of time. We then work through each of the positionsof L: in each position, we do as follows:

• read the symbol,i, (taking one unit of time);

• find positioni in K (taking O(logk) time);

• increment that entry by 1 (taking one unit of time).

This turnsK into a list of the frequencies of each symbol inS in the listL. Doingthis for each of them entries ofL requires a total of O(mlogk) time units. Wethen work throughK to find the position of the maximum element. This willrequire O(k) comparisons. Combining this, we have O(k)+O(mlogk)+O(k) =O(k+mlogk) as required.

We observe that the method described above is not necessarily the best possi-ble; other methods may be faster, and which method is the best may depend onfactors such as the relative sizes ofm andk. Anyway, we now use it to determinethe time complexity of this decoding algorithm.

Theorem 8.3.2.The time required to perform algorithm 6.6.1 isO(mnlogm) (ifm≥ n) or O(n2 +mnlogn) (if m≤ n).


Proof. The first stage is the decomposition into blocks, and converting each sym-bol into a pair corresponding to the block label and the shift within that block.There aremnsuch calculations to perform, and we have assumed that each takesa constant amount of time, requiring a total of O(mn) time units.

The next stage is, in each block, to determine the most frequently occurringposition label and most frequently occurring shift. That is, determine the mostfrequently occurring element in a list of lengthm with symbols chosen from a setof sizen (for the block labels) and from a list of lengthm with symbols chosenfrom a set of sizem (for the shifts). By lemma 8.3.1 above, the first of these willtake O(n+mlogn) time units, the second O(m+mlogm). As this has to be donein n blocks, this gives a total of O(n2 +mnlogn+mn+mnlogm).

The third stage is, within each block, to performm operations to reconstructthe symbols. This is integer arithmetic, which we assumed takes constant timeper symbol, so this requires a total of O(mn) time. So the total time required isO(mn) + O(n2 + mnlogn+ mn+ mnlogm) + O(mn). If m≥ n, this reduces toO(mn+mnlogm) = O(mnlogmn), while if m≤ n it reduces to O(n2 +mnlogn),as required.

We should also consider the space complexity of the alternative algorithm.This time, we require a look-up table for our integer arithmetic, and there are alsothings that have to be stored whilst performing the algorithm.

Proposition 8.3.3. The amount of storage space required by algorithm 6.6.1 isO(mn2), and the space required to perform the algorithm isO(mn).

Proof. We need to store a look-up table, where for each of themnsymbols, forn possible divisors we record a quotient/remainder pair. This requires a total of2mn2 = O(mn2) storage units. To perform the algorithm, we need to storemnquotient/remainder pairs, then the two auxilliary lists (one of lengthm and one oflengthn) to find their most frequently-occurring element, and needmn units tostore the reconstructed group element. This gives a total of 2mn+m+n+mn=O(mn) units.

8.4 Comparison of the two algorithms

We conclude this chapter by comparing the two algorithms’ time and space com-plexities. We begin by evaluating the time and space complexity of algorithm


1.4.5 for the special case where the group isCm oSn.

Proposition 8.4.1.The time required by algorithm 1.4.5, where the group involvedis CmoSn, is O(n2m2). The amount of storage space necessary isO(m3n3) and thespace required to perform the algorithm isO(mn).

Proof. From section 2.2, we have that the base size ofCm oSn is n, the degreeis mn and the size of the required uncovering-by-bases isbm−1

2 + 1c (which isO(m)). Substituting these values into theorem 8.1.4, we obtain O(m)O(mn2) =O(m2n2) for the time complexity. Substituting these values into theorem 8.1.5gives O(m)O(n(mn)2) = O(m3n3) for the storage space, and O(mn) space unitsrequired to perform the algorithm.

From this, we can deduce the following.

Corollary 8.4.2. For decoding the group CmoSn, the alternative algorithm (6.6.1)requires less time and less storage space than the uncovering-by-bases algorithm(1.4.5).

Proof. Theorem 8.3.2 states that the time complexity of the alternative decodingalgorithm (6.6.1) is O(mn+ mnlogm) (if m≥ n) or O(n2 + mnlogn) (if m≤ n).From proposition 8.4.1 above, the time complexity of algorithm 1.4.5 in this caseis O(m2n2). Now, we have

O(mn+mnlogm) < O(mn+m2n) = O(m2n) < O(m2n2)

andO(n2 +mnlogn) < O(n2 +mn2) = O(mn2) < O(m2n2).

In both cases, algorithm 6.6.1 requires fewer time units than algorithm 1.4.5.

For the storage space, the comparison is much easier: we simply have O(mn2)for the alternative algorithm, and O(m3n3) for the original algorithm.

We comment that actuallyperformingthe two algorithms requires the sameamount of space, O(mn), in each case. However, as far as both time and storagespace are concerned, the alternative algorithm (6.6.1) is definitely an improve-ment. As we are only considering a special case (i.e. a particular family of groups),it is perhaps not surprising that a more specialised algorithm performs better inthat situation. However, the uncovering-by-bases algorithm is still useful, as itworks in a much broader range of circumstances.

Appendix A

SomeGAP programs

The following programs have all been referred to in the text. The language usedis GAP [21]; the version used being 4.4.

A.1 The decoding algorithm for sharply k-transitive groups

The following GAP function, Decode, will carry out the algorithm described insection 1.3.G is a sharplyk-transitive group of degreen and correction capabilityr, U the required uncovering, andw the received word.

Decode:=function(G,U,n,k,r,w)local z,x,y,i,h;for x in U do

y:=[];for i in [1..k] doy[i]:=w[x[i]];

od;if Size(Set(y))=k thenh:=RepresentativeAction(G,x,y,OnTuples);z:=OnTuples([1..n],h);if DistanceVecFFE(z,w)<=r thenreturn z;

fi;fi;

od;end;

The commandRepresentativeAction finds the element of the groupG map-ping thek-tuplex to thek-tupley, and theOnTuples command following it con-

129

SOME GAP PROGRAMS 130

verts this permutation into list form. TheDistanceVecFFE command calculatesthe Hamming distance of two lists.

A.2 The “uncovering-by-bases” decoding algorithm

As mentioned in section 1.4, this works in the same way as that described above,except that it includes a step to check that an element agreeing with the receivedword w in the positions labelled by a basex actually exists. It also allows for thepossibility that the bases in the uncoveringU have different sizes.

UBBDecode:=function(G,U,n,r,w)local z,x,y,i,h;for x in U do

y:=[];for i in [1..Size(x)] doy[i]:=w[x[i]];

od;if Size(Set(y))=Size(x) thenh:=RepresentativeAction(G,x,y,OnTuples);if h<>fail thenz:=OnTuples([1..n],h);if DistanceVecFFE(z,w)<=r thenreturn z;

fi;fi;

fi;od;

end;

A.3 Programs for investigatingM12

These programs are described in section 6.4, and were used to investigate 4-errorpatterns in the Mathieu groupM12.

# Preliminaries


M:=Group((1,2)(3,4)(5,6)(7,8)(9,10)(11,12),(1,3,2)(4,7,5)(8,9,11));g:=[1,2,3,4,5,6,7,8,9,10,11,12];P:=Combinations([5..12],4);

# Program to list all codewords agreeing with identity permutation# in positions 1,2,3,4Locate:=function(M,g)local h,T;T:=[];for h in M do

if 1ˆh=1 and 2ˆh=2 and 3ˆh=3 and 4ˆh=4then Add(T,OnTuples([1..12],h));fi;

od;return T;

end;

# Program to construct all possible configurations of type (c)

KConstruct:=function(M,P,g)local p,q,i,y,h,j,k,w,K,L,W,S,T;T:=Locate(M,g);L:=Difference(T,[g]);K:=[];for h in L do

for p in P doq:=Difference([5..12],p);w:=[1,2,3,4];for i in p dow[i]:=g[i];

od;for y in q dow[y]:=h[y];

od;S:=[w[p[1]],w[p[2]],w[p[3]],w[p[4]],w[q[1]]];if Size((Set(S)))=5 thenj:=RepresentativeAction(M,[p[1],p[2],p[3],p[4],q[1]],S,OnTuples);k:=OnTuples([1..12],j);if k[q[2]]=w[q[2]] and k[q[3]]=w[q[3]] and k[q[4]]=w[q[4]]then Add(K,[g,h,k]);

fi;


fi;od;

od;return K;

end;

This gives the following output:

[ [ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ],[ 1, 2, 3, 4, 6, 7, 8, 5, 10, 12, 9, 11 ],[ 2, 3, 4, 1, 5, 6, 7, 8, 10, 12, 9, 11 ] ],

[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ],[ 1, 2, 3, 4, 6, 7, 8, 5, 10, 12, 9, 11 ],[ 4, 1, 2, 3, 6, 7, 8, 5, 9, 10, 11, 12 ] ],

[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ],[ 1, 2, 3, 4, 7, 8, 5, 6, 12, 11, 10, 9 ],[ 3, 4, 1, 2, 5, 6, 7, 8, 12, 11, 10, 9 ] ],

[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ],[ 1, 2, 3, 4, 7, 8, 5, 6, 12, 11, 10, 9 ],[ 2, 1, 4, 3, 5, 8, 7, 6, 9, 11, 10, 12 ] ],

[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ],[ 1, 2, 3, 4, 7, 8, 5, 6, 12, 11, 10, 9 ],[ 4, 3, 2, 1, 5, 8, 7, 6, 12, 10, 11, 9 ] ],

[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ],[ 1, 2, 3, 4, 7, 8, 5, 6, 12, 11, 10, 9 ],[ 4, 3, 2, 1, 7, 6, 5, 8, 9, 11, 10, 12 ] ],

[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ],[ 1, 2, 3, 4, 7, 8, 5, 6, 12, 11, 10, 9 ],[ 2, 1, 4, 3, 7, 6, 5, 8, 12, 10, 11, 9 ] ],

[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ],[ 1, 2, 3, 4, 7, 8, 5, 6, 12, 11, 10, 9 ],[ 3, 4, 1, 2, 7, 8, 5, 6, 9, 10, 11, 12 ] ],

[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ],[ 1, 2, 3, 4, 8, 5, 6, 7, 11, 9, 12, 10 ],[ 4, 1, 2, 3, 5, 6, 7, 8, 11, 9, 12, 10 ] ],

[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ],[ 1, 2, 3, 4, 8, 5, 6, 7, 11, 9, 12, 10 ],[ 2, 3, 4, 1, 8, 5, 6, 7, 9, 10, 11, 12 ] ],

[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ],[ 1, 2, 3, 4, 9, 11, 12, 10, 7, 6, 8, 5 ],


[ 3, 4, 2, 1, 5, 11, 7, 10, 9, 6, 8, 12 ] ],[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ],

[ 1, 2, 3, 4, 9, 11, 12, 10, 7, 6, 8, 5 ],[ 4, 3, 1, 2, 9, 6, 12, 8, 7, 10, 11, 5 ] ],

[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ],[ 1, 2, 3, 4, 10, 9, 11, 12, 8, 7, 5, 6 ],[ 2, 4, 1, 3, 5, 9, 7, 12, 8, 10, 11, 6 ] ],

[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ],[ 1, 2, 3, 4, 10, 9, 11, 12, 8, 7, 5, 6 ],[ 3, 1, 4, 2, 10, 6, 11, 8, 9, 7, 5, 12 ] ],

[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ],[ 1, 2, 3, 4, 11, 12, 10, 9, 6, 5, 7, 8 ],[ 3, 1, 4, 2, 5, 12, 7, 9, 6, 10, 11, 8 ] ],

[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ],[ 1, 2, 3, 4, 11, 12, 10, 9, 6, 5, 7, 8 ],[ 2, 4, 1, 3, 11, 6, 10, 8, 9, 5, 7, 12 ] ],

[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ],[ 1, 2, 3, 4, 12, 10, 9, 11, 5, 8, 6, 7 ],[ 4, 3, 1, 2, 5, 10, 7, 11, 9, 8, 6, 12 ] ],

[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ],[ 1, 2, 3, 4, 12, 10, 9, 11, 5, 8, 6, 7 ],[ 3, 4, 2, 1, 12, 6, 9, 8, 5, 10, 11, 7 ] ] ]

A.4 The “alternative” decoding algorithm

This is decribed in section 6.6.

# Program to "split up" an element of C_m Wr S_n (or a received# word) into n "blocks" of size m

SplitUp:=function(word,m,n)local blocks,i,j;blocks:=[];for i in [1..n] do

blocks[i]:=[];for j in [1..m] do

Add(blocks[i],word[m*(i-1)+j]);od;


od;return blocks;

end;

# Program to convert entries of a given block into (b,c) pairs

Pairs:=function(block)local pairs,m,i,j,q,s,b,c;m:=Size(block);pairs:=[];for j in [1..m] do

i:=block[j];q:=QuoInt(i,m);s:=RemInt(i,m);if s<>0 then b:=q;

else b:=q-1;fi;c:=(s-j) mod m;pairs[j]:=[b,c];

od;return pairs;

end;

# Program to compute the most frequently occurring element in a# list (assuming that a unique maximum exists)

Majority:=function(list)local L,freqs,i,maj;L:=Collected(list);freqs:=[];for i in [1..Size(L)] do

Add(freqs,L[i][2]);od;for i in [1..Size(L)] do

if L[i][2]=Maximum(freqs) thenmaj:=L[i][1];

fi;od;return maj;

end;


# Program to reconstruct a "corrected" permutation

Reconstruct:=function(blockperm,blockshifts,m,n)local newblocks,block,i,j;newblocks:=[];for i in [1..n] do

block:=[];for j in [1..m] do

block[j]:=blockperm[i]*m+j+blockshifts[i];if block[j]>(blockperm[i]+1)*m thenblock[j]:=block[j]-m;

fi;od;Append(newblocks,block);

od;return newblocks;

end;

# Program to perform the decoding algorithm

AltDecode:=function(word,m,n)local blocks,blockperm,blockshifts,pairs,transpairs,decoded,i;blocks:=SplitUp(word,m,n);blockperm:=[];blockshifts:=[];for i in [1..n] do

pairs:=Pairs(blocks[i]);transpairs:=TransposedMat(pairs);blockperm[i]:=Majority(transpairs[1]);blockshifts[i]:=Majority(transpairs[2]);

od;decoded:=Reconstruct(blockperm,blockshifts,m,n);return decoded;

end;

A.5 Programs for verifying the single orbit property

TheCodeParameters function return the degreen, minimum distanced and cor-rection capabilityr of a permutation groupG.


CodeParameters:=function(G)local n,d,r,ch,ch2;ch:=Set(NaturalCharacter(G));n:=Maximum(ch);ch2:=Difference(ch,[n]);d:=n-Maximum(ch2);r:=QuoInt(d-1,2);return([n,d,r]);

end;

TheIsUncovering function returnstrue if a given listU of k-subsets of[1..n]forms an(n,k,r)-uncovering.

IsUncovering:=function(U,n,r)local R,S,s,i;R:=Combinations([1..n],r);for i in [1..Size(U)] do

S:=[];for s in R doif Intersection(U[i],s)<>[]then Add(S,s);

fi;od;R:=ShallowCopy(S);

od;if R=[] then return true;

else return false;fi;

end;

So for a given groupG, we can verify the single orbit property as follows:

CP:=CodeParameters(G);b:=BaseOfGroup(G);U:=Orbit(G,b,OnSets);IsUncovering(U,CP[1],CP[3]);

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people.math.carleton.capeople.math.carleton.ca/~robertb/thesis.pdf · 5 3.3 A worked example in GAP . . . . . . . . . . . . . . . . . . . . . 42 4 Base-transitive groups of rank 3