Parameters of 5-Chromatic Strongly Regular Graphs

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Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. SOME TOPICS IN COMBINATORIAL DESIGN THEORY AND ALGEBRAIC GRAPH THEORY

DISSERTATION

Presented in Partial Fulfillment of the Requirements for

the Degree Doctor of Philosophy in the Graduate

School of The Ohio State University

Nick C. Fiala, M . S. *****

The Ohio State University 2002

Dissertation Committee: Approved by Dr. Akos Seress, Advisor < 4 * £ * « , Dr. Dijen Ray-Chaudhuri Advisor Dr. Surinder Sehgal Department of Mathematics

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. UMI Number 3049023

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Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. ABSTRACT

This dissertation consists of three chapters, each one based upon papers by the author.

At the risk of being redundant, we have made each chapter self-contained for ease of

reading.

Chapter 1 is based upon two papers by the author and deals with a conjecture

in combinatorial design theory and extremal set theory known as the A-design con

jecture. A A-design on v points is a set of v subsets (blocks) of a u-element set

(points) such that any two distinct blocks meet in exactly A points and not all of the

blocks have the same size. Ryser’s and Woodall’s A-design conjecture states that all

A-designs can be obtained from symmetric designs by a certain complementation pro

cedure. In Chapter 1, we prove that the A-design conjecture is true when v = 6p + 1,

where p is any prime number, and when v = 8q -F 1, where q = 1 or 7 (mod 8) is a

prime.

Chapter 2 deals with the determination of all strongly regular graphs with chro

matic number equal to 5. A strongly regular graph is a finite simple regular graph,

not complete or edgeless, such that the number of common neighbors of any two

distinct vertices depends only upon whether or not they are adjacent. The chromatic

number of a graph is the fewest number of colors that are required to color its ver

tices such that adjacent vertices always receive different colors. Haemers determined

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. all strongly regular graphs with chromatic number 3 or 4. In Chapter 2, we use

eigenvalue techniques and computer enumerations to begin the determination of all

strongly regular graphs with chromatic number 5. We show that there are at most

43 possible sets of parameters for such a graph. We deal completely with 32 of these

sets and obtain a partial result for one additional set.

Chapter 3 deals with a generalization of strongly regular graphs that the author

calls strongly regular vertices and partially strongly regular graphs. A strongly regular

vertex in a finite simple graph is such that the number of common neighbors it has

with any other vertex depends only upon whether or not they are adjacent. A partially

strongly regular graph is a regular graph with at least one strongly regular vertex. In

Chapter 3, we prove some basic properties of strongly regular vertices and partially

strongly regular graphs, provide several constructions of partially strongly regular

graphs, determine all partially strongly regular graphs on at most 10 vertices, and

make several conjectures regarding partially strongly regular graphs.

iii

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. ACKNOWLEDGMENTS

I would like to thank my advisor, Dr. Akos Seress, for giving me the freedom to

work on whatever I liked. I would like to thank Dr. Ronald Solomon for serving

on my advisory committee, and Dr. Dijen Ray-Chaudhuri and Dr. Surinder Sehgal

for serving on both my advisory and dissertation committees. I would like to thank

Adam Wolfe for computer programming help, and Ted Spence for supplying me with

combinatorial data. I would also like to thank my family and God.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. VITA

October 11, 1973 ...... Born in Oakes, North Dakota

1996 ...... B. S. in Mathematics and Physics, minor in Economics, Rose-Hulman Institute of Technology, Indiana

2000 ...... M. S. in Mathematics, The Ohio State University

1996 - Present ...... Graduate Teaching Associate, The Ohio State University

FIELDS OF STUDY

Major field: Mathematics

Specialization: Combinatorics

Studies in Algebraic Graph Theory Dr. Akos Seress Design Theory and Coding Theory Dr. Dijen Ray-Chaudhuri Group Theory Dr. Surinder Sehgal Group Theory Dr. Ronald Solomon

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. TABLE OF CONTENTS

A b s trac t ...... ii

Acknowledgments...... iv

V i t a ...... v

List of Tables...... viii

CHAPTER PAGE

1 A-designs on 6p + 1 and 8p + 1 points ...... 1

Introduction ...... 1 Preliminary results ...... 4 The lonin-Shrikhande method...... 5 A-designs with g = 6 ...... 10 A-designs with g = 8 ...... 23 Conjectures ...... 36

2 5-chromatic strongly regular graphs ...... 38

Introduction ...... 38 Basic properties of strongly regular graphs ...... 40 Some families of strongly regular g ra p h s ...... 41 The Paley graphs ...... 41 The triangular graphs ...... 41 The lattice graphs ...... 42 Strongly regular graphs from quasi-symmetric designs ...... 42 Strongly regular graphs from latin squares...... 43 Strongly regular graphs from systems of linked symmetric designs . . 43 Strongly regular graphs from partial geometries...... 44 Strongly regular graphs from rank 3 permutation groups...... 44 Matrix-theoretic to o ls ...... 45

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Eigenvalues , cocliques, and chromatic number of strongly regular graphs 48 Parameters of 5-chromatic strongly regular graphs ...... 50 The parameter sets 1, 2, 4, 5, 6 , 7, 8, 9, 10, 11, 14, 16, 19, 20, 21, 25, 28, 29, and 4 3 ...... 54 The parameter set ( 7 7 ,1 6 ,0 ,4 ) ...... 55 The parameter set ( 3 5 ,1 6 ,6 ,8 ) ...... 56 The parameter set (2 6 6 ,4 5 ,0 ,9 ) ...... 59 The parameter set (1 6 5 ,3 6 ,3 ,9 ) ...... 61 Computer results...... 63 The parameter set ( 1 7 ,8 ,3 ,4 ) ...... 64 The parameter set (1 0 0 ,2 2 ,0 ,6 ) ...... 65 The parameter set (8 1 ,2 0 ,1 ,6 ) ...... 66 The parameter set (6 4 ,1 8 ,2 ,6 ) ...... 67 The parameter set (4 0 ,1 2 ,2 ,4 )...... 68 The parameter set (3 6 ,1 4 ,4 ,6 ) ...... 69 The parameter set (2 6 ,1 0 ,3 ,4 ) ...... 71 The parameter set (2 5 ,1 2 ,5 ,6 ) ...... 71 The parameter set (4 5 ,1 2 ,3 ,3 )...... 72 A partial result for the set (76,30,8,14) ...... 73 The remaining parameter sets...... 74 C o n c lu s io n ...... 76

3 Strongly regular vertices and partially strongly regular graphs ...... 78

Introduction ...... 78 Basic P ro p e rtie s ...... 80 Constructions ...... 85 Eigenvalues ...... 87 Another Result...... 91 Conjectures ...... 94 Small Graphs...... 96

Bibliography ...... 99

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. LIST OF TABLES

TABLE PAGE

3.1 Strictly partially strongly regular graphs on at most 10 vertices . . . 97

viii

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. CHAPTER 1

A-DESIGNS ON 6P + I AND 8P + 1 POINTS

This chapter in based on the papers [25] and [27] by the author. In this chapter, we

shall prove that all A-designs on 6p + 1 points, p a prime, are type-1, and that all

A-designs on 8q + 1 points, q = 1 or 7 (mod 8) prime, are type-1.

Introduction

Definition 1 .1 Given integers A and v satisfying 0 < A < u, a A-design D on u

points is a pair (X .S ), where X is a set of cardinality v whose elements are called

points and B is a set of v distinct subsets of X whose elements are called blocks, such

that

(i) For all blocks A, B € B, A ^ B, \A fl B\ = A, and

(ii) There exist blocks A,B G B with |yl| ^ |f?|.

Remark 1.2 If X is a u-set and 6 is a set of distinct subsets of X such that any

two of these subsets intersect in A > 1 elements of X , then the non-uniform Fisher

inequality [54], [50] says that |B| < v. Thus, A-designs are extremal set systems in

this sense.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. A-designs were first defined by Ryser [62], [63] and Woodall [85]. The only known

examples of A-designs are obtained from symmetric designs by the following comple

mentation procedure. Let (X , A) be a symmetric (v, k, ^)-design with n ± k/2 and

fix a block A 6 A. Put B — {A } U {A A B : B 6 A, B A }, where A denotes

the symmetric difference of sets (we refer to this procedure as complementing with

respect to the block A). Then an elementary counting argument shows that ( X,B )

is a A-design on v points with one block of size k and v — 1 blocks of size 2A, where

A — k — fx. Any A-design obtained in this manner is called a type-I A-design.

Remark 1.3 If p. = k/2, then all of the blocks of ( X , B) have the same size, violating

condition (ii) of Definition 1.1. In this case, (X , B) is a symmetric design. The reason

for condition (ii) is to exclude symmetric designs from our theory.

Example 1.4Complementing a projective plane of order three, i.e., a symmetric

(13,4, l)-design, with respect to a fixed line, we obtain a 3-design on 13 points with

one block of size 4 and 12 blocks of size 6 (by a 3-design, we mean a A-design with

A = 3, not a £-design with t = 3).

The X-design conjecture of Ryser [62], [63] and Woodall [85] states that all A-

designs are type- 1. The conjecture was proven for A = 1 by deBruijn and Erd"s [12],

for A = 2 by Ryser [62], for 3 < A < 9 by Bridges and Kramer [3], [51], [5], for A = 10

by Seress [68], for A = 14 by Tsaur [77], [ 6], and for all remaining A < 34 by Weisz

[81]. S. S. Shrikhande and Singhi [71] proved the conjecture for prime A and Seress

[69] proved it when A is twice a prime.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Investigating the conjecture as a function of v rather than A, Ionin and M. S.

Shrikhande [48], [49] proved the conjecture for v = p + 1 ,2p + 1 ,3p + 1, and 4p + 1,

where p is any prime, and Hein [42], [43] proved it for v — 5p + 1 , where p is a prime

not congruent to 2 or 8 modulo 15. The conjecture has also been verified by computer

for all v < 85 [86] (it is somewhat interesting that 86 = 5 • 1 7 + 1 and 17 = 2 (mod 15)

is prime, so this value of v is not covered by Hein’s result). Continuing along these

lines, in this chapter we will prove the following two results.

Theorem 1.5 A ll X-designs on u = 6p + 1 points, p a prime, are type- 1.

Theorem 1.6 A ll X-designs on v = 8p + 1 points, p = 1 or 7 (mod 8) a prime, are

type-1.

The method employed to prove Theorems 1.5 and 1.6 is a slight extension of

the method of Ionin and M . S. Shrikhande developed in [48] and [49] and used in

[42], [43]. In fact, Ionin and M. S. Shrikhande remarked in [48] that they thought

that the techniques developed in that paper might work in the v = 6p + 1 case as

well. However, whereas they were always able to reduce to the case of designs having

at most two distinct block sizes, in this paper we w ill have to deal with designs

potentially possessing three different block sizes. In fact, at one point we w ill have

to deal w ith a design potentially possessing five different block sizes. However, we

reduce this to three by using a minimal counterexample argument.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Preliminary results

D e fin itio n 1 .7 Given a A-design D = (X , B) and a point x 6 X, the replication

number of x is the number of blocks A € B which contain x.

Rem ark 1.8 If we complement a symmetric (v, k, /i)-design with respect to a block

.4, then the points lying in A will all have replication number v — k +1 in the resulting

type -1 A-design D, and the points lying outside A will all have replication number k

in D.

Ryser [62] and Woodall [85] independently proved the following theorem concern

ing these replication numbers.

Theorem 1.9 I f D = (X , B) is a A-design on v points, then there exist integers r > 1

and r ' > 1, r ^ r*. such that every point x 6 X has replication number r or r* and

r + r" = v + I. In addition, the integers r and r* satisfy the equation

(1.1)

We w ill also need the following three theorems concerning the integers r and r*.

The first was stated without proof in [86]. For a proof see [67].

Theorem 1.10 A A-design on v points with replication numbers r and r* is type- 1

if and only if r(r — l)/{v — 1) orr*(r* — l) /( v — I) is an integer.

T h e o re m 1.11 [48], [49] Let D be a A-design with replication numbers r and r* and

put g = gcd(r — l,r* — 1). I f g = 1, 2, 3, or 4, then D is type-1.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. T h e o re m 1.12 [72], [69] Let D be a \-design with re-plication numbers r and r*,

r > r*. Put g = gcd(r — l,r* — 1). I f gcd(A, (r — r*)/g ) = 1 or 2, then D is type-1.

Additionally, we will need the following three theorems concerning the validity of

the A-design conjecture for certain values of A.

T h eo re m 1.13 [12], [62], [3], [51], [5], [ 68] The X-design conjecture is true fo r X < 10.

T h eo re m 1.14 [81] The X-design conjecture is true fo r X = 12.

T h eo re m 1.15 [77], [6][81] The X-design conjecture is true fo r X = 14, 15, 16, and

18.

T h eo re m 1.16 [71] The X-design conjecture is true fo r prime A.

T h eo re m 1.17 [69] The X-design conjecture is true when X is twice a prime.

Rem ark 1.18 Although it is proved in [81] that the A-design conjecture is valid for

all A < 34, we will not need the full strength of this result here. The contents of

Theorems 1.13, 1.16, and 1.14 will suffice to prove Theorem 1.5, and these together

with Theorems 1.15 and 1.17 will suffice to prove Theorem 1.6.

The Ionin-Shrikhande method

Let D = (X,B) be a A-design on u points. Then Theorem 1.9 implies that every

point of D has replication number r or r* for some integers r ^ r*. Therefore, the

underlying set X of our A-design is partitioned into two subsets, E and E m, of points

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. having replication numbers r and r*, respectively. Let \E\ = e and |E*| = e*, so

e + e* = v. Also, for any block A e B, put r A = \A fl E\ and r A = |A fl E"|, so

r 4 + t\ = |.4|. We will frequently use the trivial inequalities 0 < r A < e for all A.

The following simple relation among these parameters is the starting point of the

Ionin-Shrikhande method developed in [48] and [49].

Lemma 1.19 Let D = (X , B) be a X-design on v points with replication numbers r

and r*. Then the following relation holds fo r all blocks A 6 B:

(r - 1)(|A| - 2ta ) = (v - 1)(|A| - A - r A). (1.2)

Proof. Fixing a block A € B, we will count in two different ways all of the pairs

(x, B), where x 6 X , B € B, B ± .4, and x € A fl B. This gives us the equation

r A(r — 1) + T^(r* - 1) = X(v — 1), which is easily transformed into equation (1.2). □

Now, let g = gcd(r — 1, r* — 1). Then, since (r — 1) + (r* — 1) = v — 1 by Theorem

1.9, we also have g = gcd(r — 1, v — 1) = gcd(r* — 1, v — 1). We put

v — 1 <7 = ------(1.3) 9 Then, since gcd((r — 1 )/g,q) = 1, equation (1.2) implies that q divides |.4| — 2 t .4 for

all blocks A £ B. Therefore, for each block A we define an integer aA by

qaA = \ A \ - 2 r A. (1.4)

Next, we define the quantity

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. « = 5 > . (1*5) A€B Also, equations (1.2) and (1.4) imply that

ta =z A - -

and

r ; = A + — (1.7) 9 for all .4. Adding equations (1.6) and (1.7) we obtain

|A| = 2A + (1.8) 9 for all .4.

Rem ark 1.20 Note that equation (1.8) implies that for any two blocks A,B € B,

I-4.J = |i?| if and only if a a — 0 b -

The next three equations are easily verified:

^ |A| = er+ e*r*, (1.9)

•4€S

^ rA - e r , (1.10) Aes and

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. £r; = eV. (1. 11) AeB

Equations (1.4), (1.9), and (1.10) then imply that sq = I^ €B(|A| - 2t a) = e*r* -

er = (v — e)(v — r + 1) — er, which can be transformed into

sq = gq(gq — e — r + 3) — (2e + r — 2). (1-12)

Equation (1.12) then implies that q divides 2 e + r — 2. Therefore, we define a positive

integer m by

qm = 2e + r — 2 . (1-13)

Similarly, equations (1.4), (1.9), and (1.11) imply that q divides 2e’ + r* — 2. Thus,

we define a positive integer m* by

qm' = 2e’ + r* - 2. (1.14)

Adding equations (1.13) and (1.14), we obtain

m + m* = 3 <7- (1-15)

Finally, equations (1.12), (1.13), and (1.15) imply that

s = g2q - g(e + r ) + 3 g - m . (1.16)

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Remark 1.21 Upon further manipulation of the above equations, we eventually ar

rive at

(r — r*)(m * — m) = g[v — (4A — 1)]. (1-17)

Note that equation (1.17) and the fact that r ^ r* imply that v = 4A — 1 if and only

if m = m *.

The next lemma establishes formulae for e and r in terms of the parameters A,g , q,

and m. They follow easily from equations (1.13) and (1.17).

Lemma 1.22 [48] I f D is a X-design on u ^ 4A — 1 points, then

_ gX - {g - m )-q + g - m 3 g — 2m

and

r _ (2g - m){gq -F 2) - 2gX 3 g — 2m

The next result gives a way of constructing new A-designs from old ones by com

plementing with respect to a fixed block. For a proof see [48].

Remark 1.23 In what follows, if we complement with respect to the block .4, the

parameters of the new design will be denoted by A(A), r(A), m(.4), etc.

Lemma 1.24 Let D = (X , B) be a X-design on v points with replication numbers r

and r* and let A 6 B. Put

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. B(A) = {A} U {A A B :BeB,B^A}.

Denote by D (A ) the complemented set system (X,f?(A)). Then we have

(i) If A — E or Em, then D(A) is a symmetric (u, |A|, |A| — A )-design,

(ii) If A =£ E and A ^ E *, then D (A ) is a A(A)-design on v points with r(A) = r,

r*(A) = r*, and m(A) = m + 2aA, where A (A) = |A| — A,

(Hi) I f A ^ E and A ^ E* and D is type-1, then D (A ) is also type-1, and

(iv) (D(A))(A) = D.

A-designs with g = 6

VVe are now in a position to prove Theorem 1.5. It w ill follow easily from the fol

lowing Theorem. In what follows, the computer program Mathematica [84] was used

extensively to carry out computations.

Theorem 1.25 Let D = (X ,B ) be a X-design on v points with replication numbers

r and r*. I f g = gcd(r — l,r* — 1) = 6 , then D is type-1.

Proof. If A < 13, then Theorems 1.13, 1.16, and 1.14 imply that D is type-1.

Therefore, we may assume that A > 14. By equation (1.3), we may write v = 6q + 1.

For each integer i, let an denote the number of blocks A € B w ith aA — i. VVe will

frequently use the trivial fact that a* > 0 for all i. Since the number of blocks is equal

to the number of points, we clearly have

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Y ^ a i = 6q + l. (1.18) ie z Also, equations (1.5), (1.16), and (1.17) and the formulae of Lemma 1.22 imply

that

W _ (3? + l)(m 2 — 18m + 72) + 18A /■, in \ / „ mi ~ Q (1-19) 9 — m iez if m jz 9.

Next, equation (1.4) implies that for any block A € B, we have |A|= 2 t a + qaA.

Using this and the formulae of Lemma 1.22, equation (1.1) is transformedinto

(m — 9)ai ^ A(m - 9) + i(2A - Zq - 1)

4(m — 9)2q2 1 ( 1.20) [q(m — 6) — 2A + lj[g(m — 12) -f- 2A — 1] A

if m ^ 9.

Now, since 6 divides r — 1, r is odd and equation (1.13) implies that m is odd as

well. Also, equation (1.15) implies that m + m* = 18. Without loss of generality, we

may assume that m < m*. Therefore, m = 1,3,5,7, or 9.

Case 1: m = 1.

In this case, the formulae of Lemma 1.22 imply that e = (6A — 25q + 5)/16,

r = (33

r A = A — [(5q + 2A — 1)/ 16]crA for any block .4 6 B. Then the inequalities 0 < t a < e

imply that 5 < crA < 16A/ (5q + 2A — 1) for all A. Now, 6 divides r — 1 and r > 1,

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. so r > 7. This gives us the inequality q > (2A + 15)/11. Combining the last two

inequalities, we obtain that crA = 5 for all A. Therefore, by Remark 1.20, all blocks

have the same cardinality, contradicting the definition of a A-design.

Case 2: m = 3.

In this case, Lemma 1.22 implies that e = (2A — 3 g + l)/4 , r = (9q—2A+3)/2, and

r* = (3 <7+ 2 A + l ) /2 . Also, equation (1.6) implies that ta — A—[(3qr-h2A—l)/12]cr>i for

any block A . Then the inequalities 0 < rA < e imply that 3 < aA < 12A/(3?-l-2A — 1)

for all .4. Next, r > 7 gives us the inequality q > (2A + ll)/9 . Combining the last

two inequalities, we obtain that aA = 3 or 4 for all A. Hence, a* = 0 for all i except

possibly 3 and 4. Then equations (1.18), (1.19), and (1.20) become

03+04 = 6 q + 1, (1.21)

3(9<7 + 2A + 3) 3a3 + 4a4 = (1.22) 2

and

Solving equations ( 1.21 ) and (1.22) yields

21 q - 6A - 1

and

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Inserting the above expressions for a 3 and a4 into equation (1.23) and manipulating

the result, we arrive at

(2A - 3 q - 1)2 (2A - 3 q + l)(21Xq - 36? - 6A2 + 5A - 4) = 0.

Now, e > 0 so 2A — 3q + 1 ^ 0. Also, 2A — Zq — 1 ± 0 since v ± 4A — 1 because m ± 9.

Therefore, we obtain that

21A<7 - 36

which can be transformed into

(21A - 36) (7? - 2A) = 37A + 28.

Now, 37A + 28 > 0 and 21A - 36 > 0, so 7q - 2A > 1. If 7q - 2A = 1, then A = - 4 ,

a contradiction. If 7q — 2 \ = 2, then r = (2q + 5 )/2 is not an integer, a contradiction.

Therefore, we must have 7q — 2A > 3. Consequently, 3(21A — 36) < 37A + 28, which

implies that A < 5, a contradiction.

Case 3: m = 5.

In this case, Lemma 1.22 implies that e = ( 6A —q r + l)/8, r = (21g—6A+7)/4, and

r* = (3g+6A+1)/4. Also, equation (1.6) implies that r A = A — [(g+2A — l ) / 8]o\i for

any block A. Then the inequalities 0 < ta < e imply that 1 < a A < 8A / (q + 2X— 1)

for all A. Also, r > 7 gives us the inequality q > (2A + 7)/7. Combining the last two

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. inequalities, we obtain that a a = 1,2, or 3 for all A. Hence, a* = 0 for all i except

possibly 1, 2, and 3. Then equations (1.18), (1.19), and (1.20) become

a i 4- a2 4- a3 = + 1, (1-24)

a, + 2a, + 3a, = 21l, + ^8A + 7, (1.25)

and

^ 1_ t=i 4A - *(2A - Zq - 1) (1 - q - 2A)(2A - 7q - 1) A' K '

Solving equations (1.24), (1.25), and (1.26) yields

O13?3 4" Oio?“ 4* On? 4- Quo a t = 8A(2A — 7 q— 1)(2A + q — l ) ' where a l3 = -459A + 126, o l2 = 162A2 - 447A - 66, o u = 204A3 - 12A2 - 225A - 54,

and q 10 = —72A4 + 20A 3 + 10A2 - 21A - 6,

O23?3 4- Ct22(r 4" O21 ? 4* O2o do = 2A(2A — 7? — 1)(2A + q — l ) ' where a03 = 51A - 63, Q 22 = -324A 2 + 359A + 33, o2i = 108A3 - 180A2 + 153A + 27,

and Q20 = 36A3 — 24A2 + 13A + 3, and

033?3 4- Q32?2 4- a 3i? 4- o30 a3 = 8A(2A — 7 q — 1)(2A -F <7 — 1) ’

where Q33 = -81A+126, o 32 = 558A2 - 7 5 7A -6 6, a 3i = -444A 3 + 44 4 A2 - 291A - 54,

and 030 = 72A4 - 132A3 + 54A 2 - 23A - 6 .

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Replacing q by a real variable x in the above expressions for a2, and a3, we

obtain three functions, ai(x), a2 (x), and a3 (x). Now, we already know q > (2A+7)/7,

and the inequality e > 1 implies that q < 6A — 7. This implies 2A — 7q — 1 <

0. Therefore, ai(x), a2 (x ), and a3 (x) are continuous functions of x on the interval

[(2A + 7 )/7 , 6A — 7].

Now, the function a 2 (x) has zeros only at —1/3, 6A + 1, and

6A2 - 5A + 3 221 ” 17A - 21 '

Clearly, —1/3, 6A + 1 £ [(2A + 7)/7,6A — 7], so a 2 (x) has at most one zero on this

interval. However,

,2A + 7, —5(3A + 14)(A2 - 14A + 21) n a2\ Z ) = ------T— < U v 7 ' 98A

and

. 6(2A — 3)(9A — 10) “ 2(6A ” 7 1 ------a(B X—"6)— ? °-

Therefore, a2 (x) has exactly one zero on the interval [(2A + 7 )/7 , 6A — 7] at z21. The

above inequalities then imply that a 2 (x) is negative on the interval [(2A + 7)/7, z21).

Hence, we must have q € [z2\ , 6A — 7].

Next, the function a 3 (x) has zeros only at (2A — 3)/9, 6A + 1, and

6A2 — 3A 2 231 ~ 9A - 14 '

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Clearly, (2A—3 )/9 , 6A+ 1 ^ [201 , 6A—7], so a 3 (x) has at most one zero on this interval.

However,

, . 3(12A2 - 13A - 3) n a3(zoi) = —------> 0 -U 17A - 21 and

. ,, —3(A — 2)(13A — 15) “3(6A_7) = ------

Therefore, a3 (x ) has exactly one zero on the interval [ 221 , 6A — 7j at 231 . The above

inequalities then imply that a3 (x) is negative on the interval ( 231 , 6A — 7]. Hence, we

must have q € [221 , 231 ].

Now, the function at(x) has zeros only at —(2A + l)/3 and

78A2 - 63A - 18 T 4v/36A4 - 252A3 - 87A2 + 108A + 36 ~n’~l2_ 153A-42

Clearly, —(2A + l) /3 £ [221 , 23 !], so a t(x) has at most two zeros on this interval.

However,

, . 2(13A + 3) n a‘('2l) = 17A - 21 > 0l

_ ,A , -(7A + 2)(5A 2 — 36A — 12 ) 2 = ------16A(5A — 2) ------< ° ’ and

. . 5(9A + 2) „ “ l(Z3l) = I a ^ T T > ° ’

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. where z2i < A /2 < 231 . Therefore, ai(x) has exactly two zeros on the interval [ 201 , 231 ]

at 2 u and 212 . The above inequalities then imply that ai(x) is negative on the interval

( - a . 212 ). Hence, we must have q G [221 , 2 u] U [212 , 231 ]-

Next, we easily obtain the inequalities 221 > (6A + 2)/17 and 231 < (6A + 8 )/9 .

Also, we have 36A4 — 252A3 — 87A2 + 108A+ 3 6 > ( 6A2 — 22A—46)2, which implies that

2 U < (54A2 + 25A + 166)/(153A - 42) and zl2 > (102A2 - 151A - 202)/(153A - 42).

This in turn implies that 2 n < (6A + 6 )/1 7 and 212 > (6A —9 )/9 . Therefore, we have

q G ((6A 2 )/1 7 , ( 6A + 6)/17) U (( 6A — 9)/9, (6A + 8 )/9 ). Thus, since q is an integer,

we must have q G { (6A + j ) / 17 : 3 < j < 5} U { (6A + Ar)/9 : — 8 < k < 7}.

I f q G {(6A + j ) / 17 : 3 < j < 5}, then r = (4q + j + 7 )/4 for j = 3,4, or 5.

But, clearly r is an integer only for j = 5. Therefore, we must have q = (6A + 5)/17.

However,

6A + 5 — (13A + 8)(52A2 - 1143A - 234) 17 34A(2A + 13)(10A — 3) <

for A > 23. Thus, we must have A < 22. But, for A < 22, q is an integer only for 2

and 19. So, we must have A = 19. Since 19 is prime, we conclude that D is type-1

by Theorem 1.16. Therefore, we may assume that q G {(6A + fc)/9 : —8 < fc < 7}.

Then r = (12q + k + 7 )/4 for some integer — 8 < k < 7. But, clearly r is an

integer only for k = —7, —3,1, and 5. VVe easily eliminate (6A — 7)/9, (6A + 1 )/9 , and

(6 A + 5 )/9 since these are not integers. This leaves us with q — (6A—3)/9 = (2A—1)/3.

However, this implies that v = 4A — 1, a contradiction by equation (1.17) since m # 9.

Case 4: m = 7.

In this case, Lemma 1.22 implies that e = (6A — q — l) /4 , r = (15? — 6A + 5 )/2 ,

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. and r* = (6A — 3 q — l)/2. Now, if A = E or E* and A € B, then Lemma 1.24 (i)

implies that D{A) is a symmetric design. But, then (D(A))(A) — D by Lemma 1.24

(iv) and D is type-1 by definition of a type-1 design. Therefore, we may henceforth

assume that E, E ' £13. If there exists a block A with a a < —1, then Lemma 1.24

(ii) implies that g(A) = 6 and m(A) < 5, so D(A) is type-1 by cases 1, 2, and 3. But,

then D is also type-1 by Lemma 1.24 (iii) and (iv). If there exists a block A with

a a > 3, then m (A) > 13, so m *(A) < 5 and once again we conclude that D is type-1.

Therefore, we may assume that a a = 0,1, or 2 for all blocks .4. Hence, a* = 0 for all

i except possibly 0, 1, and 2. Then equations (1.18), (1.19), and (1.20) become

ao + fli + + 1, (1-27)

_ 18A — 15q — 5 (1 . . . a i + 2a2 ------, (1.28)

and

y =±______= ______^21 I (i99) ^ 2A - i(2A - 3g - 1) (g-2A + l)(2A -5?-l) A' v }

Solving equations (1.27), (1.28), and (1.29) yields

oco2q~ + aoiq + Qoo a o = 2(2A — 5 q — 1)(2A — g — 1) ’

where Q02 = — 117A — 10, aoi = 132A2 — 78A — 12, and aroo = —36A3 + 36A2 — 9A — 2,

<*L3

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. where a l3 = 195, aio = —324A + 299, a n = 108A2 — 216A + 117, and Qio = 36A2 —

36A + 1 3 , and

_ <*23

where a22 = —135, a 22 = 297A - 207, a3i = — 192A2 + 222A — 81, and a 2o =

36A3 - 64A2 + 37A - 9.

Replacing q by a real variable x in the above expressions for ao, a i, and a2, we

obtain three functions, ao(x), ai(x), and ao(x). Now, the inequalities r > 7 and r* > 7

imply that (2A + 3)/5 < q < 2A — 5. This implies that (2A — 5q — 1)(2A — q — 1) <

0. Therefore, ao(x), ai(x), and a2(x) are continuous functions of x on the interval

[(2A + 3)/5,2A — 5].

Now, the function a i(x ) has zeros only at —1/3 and

54A - 39 2\/144A2 - 468A + 169 -11,212- g5

Clearly, —1/3 ^ [(2A + 3 )/5 ,2 A — 5], so a t(x ) has at most two zeros on this interval.

However,

OA + 3. —(3A + 7)(A 2 — 12A + 26) n = SfXTT) < °-

_ A (3 A + 2)(7A — 26) ‘"(2) = 4(3A — 2) > °'

and

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. where (2A -+- 3 )/5 < A /2 < 2A — 5. Therefore, a t(x) has exactly two zeros on the

interval [(2A + 3)/5,2A — 5] at and Then the above inequalities imply that

ai(x) is negative on [(2A + 3 )/ 5,z u )U (z i2 , 2 A — 5]. Hence, we must have <7 6 [211 , 212 ].

Next, the function

14A-9t2>/4A2-28A + 9 . 21, . 22 = .

Clearly, (A — l)/3 £ [211 , 212 ]. Therefore, a2 (x) has at most two zeros on this interval.

However,

, . 36A + 26 + 3 \ / 144 A 2 - 468A + 169 0 a2 (~ ll) = ------> 0,

/4Aa "(7A + 5)(12A 2 - 76A - 45) ^ n 5 10(2A + 1)(6A — 5) < ’

and

, . 36A + 26- 3\/144A2 - 468A + 169 „ 02 (^12 ) = ------^ ------> 0,

where 2 n < 4A/5 < z\o. Therefore, ao(x) has exactly two zeros on the interval

[211 , 212 ] at 221 and 222 - Then the above inequalities imply that a 2 (x ) is negative on

the interval ( 221 , - 2 2)■ Hence, we must have q 6 [211 , 221 ] U [222 , 212 ].

Suppose first that q € [222 , 212 ]. Now, we have the inequalities 4A 2 — 28A + 9 >

(2A — 8)2 and 144A2 — 468A +169 < (12A — 19)2. These imply that 222 > (6A — 9 )/5

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. and z12 < (6A — 5)/5. Therefore, q € ((6A — 9)/5, (6A - 5 )/5 ). Thus, since q is an

integer, we must have q = (6A—8 )/5 , (6A—7 )/5 , or ( 6A—6)/5. If q — (6A —8)/5, then

r = (10

is not an integer, a contradiction. If q = (6A — 6)/5, then r = (10g — I)/2 is not an

integer, a contradiction. Hence, we must have q € [zn, -21 ]-

Now, the function a o (x ) has zeros only at

66A2 - 39A - 6 + 2v/36A4 - 324A3 + 9A 2 + 36A + 4 -01, ^02 = 117A +10

Also, we have

Qo(-n ) =

3[2028Aa - 4596A 4- 1638 - (169A - 124)y/144A2 - 468A + 169) > 0, 10[97A - 52 - 6 n/144A2 - 468A + 169]

and

3[2028A2 - 4596A + 1638 + (I69A - 124)y/144A2 - 468A + 169] > 0, 10[97A - 5 2 + 6V144A2 - 468A + 169]

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. where zn < A /2 < 221 - Therefore, a0(x) has exactly two zeros on the interval [zu, 3 >i]

at zoi and z02. Then the above inequalities imply that a 0(x) is negative on the interval

(zoi, ^02 )- Hence, we must have q e [zu , ~oi] U [202 . Z21 ]-

Now, we easily obtain the inequalities zu > (6A — 1)/13 and zoi < (2A + 2 )/3 .

Next, the inequality 36A4 — 324A3 + 9A2 + 36A + 4 > ( 6A2 — 28A — 80)2 implies that

zoi < (54A2 + 17A + 154)/(117A + 10) and z 02 > (78A2 - 95A - 166)/(117A + 10).

This in turn implies that z0l < (6A + 3)/13 and zo2 > (2A —3 )/3 . Therefore, we have

q 6 ((6A — 1)/13, (6A + 3)/13) U ((2A — 3 )/3 , (2A + 2 )/3 ). Thus, since q is an integer,

we must have q € { (6A + j) /1 3 : 0 < j < 2} U {(2A + k ) j3 : — 2 < k < 1}.

If q € { (6A + j)/13 : 0 < j < 2}, then r = (2 q + j -1- 5)/2 for j = 0, 1, or 2 .

However, clearly r is an integer only for j = 1. Therefore, q = (6A + 1)/13, which

implies e = (6q —1 )/2 , which is not an integer, a contradiction. Hence, we must have

q € {(2A + k )/3 : —2 < k < 1}.

Then r = (6q + 3k + 5 )/2 for k = —2, —1,0, or 1. However, clearly r is an integer

only for k = — 1 and 1. If q — (2A —1)/3, then v = 4A — 1, a contradiction by equation

(1.17) since m /9 . If q = (2A + l)/3, then a2 (q) = 9/(A — 1). Since this must be an

integer, this implies that A — 1 divides 9, or A = 2,4, or 10, a contradiction.

Case 5: m = 9.

If there exists a block A with a & < —1, then m (A) < 7 and D is type-1 by cases

1, 2, 3, and 4. If there exists a block A with a a > 1, then m (A ) > 11, so m’ (A) < 7

and once again D is type -1 by previous cases. Therefore, we may assume that aA = 0

for all blocks A, a contradiction. This concludes the proof of Theorem 1.25. □

Corollary 1.26 A ll X-designs on v = 6p + 1 points, p a prime, are type-1.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Proof. If p does not divide g, then g = 1,2,3, or 6 and D is type-1 by Theorems

1.11 and 1.25. If p does divide g, then p divides r — 1 and r * — 1. Without loss of

generality, we may assume r > r * . Therefore, either r = 5p + 1 and r * = p + 1 or

r = 4p + 1 and r * = 2p -F 1.

First, suppose that r = 5p -F 1 and r* = p -F 1. If p = 2, then r = 11 and r* = 3,

so g = gcd(10,2) = 2 and D is type-1 by Theorem 1.11. If p = 3, then r = 16 and

r* = 4, so g = gcd(15,3) = 3 and D is type -1 by Theorem 1.11. If p = 1 (mod 6),

then 6 divides 5p+l, so r ( r - l) /( t /- l) = 5(5p+l)/6 is an integer and D is type-1 by

Theorem 1.10. If p = 5 (mod 6), then 6 divides p-Fl, so r*(r* — l)/(u — 1) = (p -F 1 )/6

is an integer and D is type-1 by Theorem 1.10.

Next, suppose that r = 4p + 1 and r* = 2p-F1. If p = 3, then r = 13 and r* = 7, so

g = gcd(12, 6) = 6 and D is type -1 by Theorem 1.25. If p = 1 (mod 3), then 3 divides

2p + 1, so r*(r* — l)/(u — 1) = (2p + l)/3 is an integer and D is type-1 by Theorem

1.10. If p = 2 (mod 3), then 3 divides 4p +1, so r(r — l)/(u — 1) = 2(4p + l)/3 is an

integer and D is type -1 by Theorem 1.10. □

A-designs with g = 8

In this section, we w ill prove that all A-designs w ith g = 8 are type-1. However,

because of an additional complication, we will slightly alter our proof strategy and

use a minimal counterexample argument. To this end, we will need the following

simple lemma.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Lemma 1.27 Let g be a fixed positive integer and suppose that there exists a non-

type-1 A-design with replication numbers r and r* and gcd(r — l,r* — 1) = g. Let D

be such a design with minim al A. Then all o f the blocks o f D have size at least 2A.

Proof. Suppose that D has a block A of size less than 2A. Complementing with

respect to A, we obtain the new design D(A). Now, D(A) is non-type -1 and has

g(A) = g by Lemma 1.24 (ii), (iii), and (iv). Also, D(A) has new A-value A(.4) =

|A| — A. However, \A\ — A < A since |A| < 2A, contradicting the m inim ality of D ’s

A-value. □

VVe are now in a position to prove Theorem 1.6. It will follow easily from the

following Theorem. In what follows, the computer program Mathematica [84] was

used extensively to carry out computations.

Theorem 1.28 Let D = (X , B) be a \-design on v points with replication numbers

r and r *. I f g = gcd(r — 1, r* — 1) = 8, then D is type-1.

Proof. Suppose that Theorem 1.28 is false. Then there exists a non-type-1 A-design

with g = 8. Let D = (X , B) be such a design w ith minimal A. Then by Lemma 1.27,

we know that |.4| > 2A for all blocks A € S. We also know that A > 20 by Theorems

1.13, 1.16, 1.14, and 1.15.

By equation (1.3), we may write v = Sq + 1. For each integer i, let a, denote the

number of blocks A € B with aA = i. We will frequently use the trivial fact that

a, > 0 for all t. Since the number of blocks is equal to the number of points, we

clearly have

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. J 2 a i= 8 q + l . (1.30) «e z Also, equations (1.5), (1.16), and (1.17) and the formulae of Lemma 1.22 imply

that

^ . (4q + l)(m 2 - 24m + 128) + 32A /t 2 ^ = - (L31) •ez

if m 7^ 12 .

Next, equation (1.4) implies that for any block A 6 5 , we have |A| = 2t a + q a A .

Using this and the formulae of Lemma 1.22, equation (1.1) is transformed into

(m — 12 )a< " A(m — 12) + i(2A — 4g — 1)

4(m — 12 ) 2q2 1 (1.32) [q(m — 8) — 2A + l][g(m — 16) + 2A — 1] A

if m 7^ 12 .

Now, since 8 divides r — 1, r is odd and equation (1.13) implies that m is odd as

well. Also, equation (1.15) implies that m + m* = 24. Without loss of generality, we

may assume that m < m *. Therefore, m = 1,3,5,7,9, or 11.

Case 1: m = 1.

In this case, the formulae of Lemma 1.22 imply that e = ( 8A — 49q + 7)/22,

r = (60q — 8A+15)/11, and r* = (28q+8A + 7 )/ll. Also, equation (1.6) implies that

ta = A — [(7? + 2A — l)/22]cr^ for any block A e B. Then the inequalities 0 < r A < e

imply that 7 < oA < 22A/(7 q + 2A — 1) for all A. Now, 8 divides r — 1 and r > 1,

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. so r > 9. This gives us the inequality q > (2 A + 21)/15. Combining the last two

inequalities, we obtain that aA = 7 for all A. Therefore, by Remark 1.20, all blocks

have the same cardinality, contradicting the definition of a A-design.

Case 2: m = 3.

In this case, Lemma 1.22 implies that e = (8A—25q+5)/18, r = (52qr—8A + 13 )/9 ,

and r * = (20q + 8A + 5)/9. Also, equation (1.6) implies that t a = A — [(5q + 2A —

1)/18)

18A/(5q + 2A — 1) for all A. Next, r > 9 gives us the inequality q > (2A + 17)/13.

Combining the last two inequalities, we obtain that aA — 5 or 6 for all .4. Hence,

ai = 0 for all i except possibly 5 and 6. Solving equations (1.30) and (1.31), we obtain

a5 = (172g - 32A - l l ) / 9 and a6 = 4(8A - 25q + 5 )/9 .

Now, oA > 0 and |A| > 2A for all .4, so we must have r > r* by equation (1.8).

This implies that q > A /2. However, this implies that a 6 < —2(9A — 10)/9 < 0, a

contradiction.

Case 3: m = 5.

In this case, Lemma 1.22 implies that e = ( 8A — 9qr+3)/14, r = (44g —8A + l l ) / 7 ,

and r* = (12q + 8A + 3)/7. Also, equation (1.6) implies that t a = A — [(3q + 2A —

1)/14]

14A/(3q + 2A — 1) for all A. Also, r > 9 gives us the inequality q > (2A + 13)/11.

Combining the last two inequalities, we obtain that aA = 3,4, or 5 for all A. Hence,

Oj = 0 for all i except possibly 3, 4, and 5. Solving equations (1.30), (1.31), and

(1.32) we obtain

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. _ a 3393 + a 32(I~ + Q!31<7 4" a 30 ° 3 “ 14A(2A - 11<7 - 1)(3 9 + 2A - 1) ’

where a& = - 10212 A + 7920, a 32 = 4909A2 - 6981A + 60, a 31 = -288A 3 + 1496A2 -

1588A - 720, and a 30 = -6 4A 4 - 152A3 + 107A2 - 59A - 60,

_ &43q2 + &42Q ' + a 4lQ + <*40 ° 4 “ 7A(2A - 11 q - l)(3 g + 2A - 1) ’

where q43 = 5328A - 7920, q 42 = -6221A 2 + 8099A - 60, q.u = 1392A3 - 2376A2 +

1752A + 720, and a 40 = -64A4 + 288A3 - 147A2 + 61A + 60, and

_ &53Q3 + a 52(r + ^51? + a 50 ° 5 “ 14A(2A - 11? - 1)(3<7 + 2A - 1) ’

where o;53 = —4140A + 7920, Q 50 = 5741A” — 8783A + 60, 0:51 = —2048A3 + 2584A" —

1692A - 720, and q 50 = 192A4 - 368A3 + 131 A2 - 49A - 60.

Replacing q by a real variable x in the above expression for as, we obtain the

function a 5( i ) . Now, aA > 0 for all A, so r > r*. This implies that q > A/2. Also,

the inequality e > 1 implies that q < (8A — ll)/9 . This implies 2A — llq — 1 < 0.

Therefore, as(x) is a continuous functions of x on the interval [A/2, ( 8A — ll)/9 j.

Now, the function as(x) has zeros only at (3A — 5)/20, ( 8A + 3)/9, and 251 =

(8A2 - 5A + 4)/(23A - 44). Clearly, (3A - 5)/20, ( 8A + 3 )/9 , 251 i [A/2, (8A - ll)/9[,

so a5(x) has no zeros on this interval. However,

.3A —(5A + 12)(12A + 5)(37A 2 — 112A — 16) ° s (T * “ ------14A(17A — 4)(25A + 4) ------< ° '

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. where A/2 < 3A/4 < (8A — ll)/9 . Therefore, a5(x) is negative on the interval

[A/2, (8A — ll)/9 ], a contradiction.

Case 4: m = 7.

In this case, Lemma 1.22 implies that e = ( 8A—g+l)/10, r = (36

r* = (4<7+ 8A + l ) / 5. Also, equation (1.6) implies that ta = A— [(?+2A— 1)/10]

any block A. Then the inequalities 0 < r A < e imply that 1 < oA < 10A/(g + 2A — 1)

for all A. Also, r > 9 gives us the inequality q > (2A + 9)/9. Combining the last

two inequalities, we obtain that crA = 1 ,2 ,3, or 4 for all A. Therefore, since oA > 0

for all A, we must have r > r*. FVom this we obtain the inequality A < 2 q. Now,

suppose there exists a block A with crA = 4. Then the inequality ta > 0 implies that

A > 2q — 2. Therefore, A = 2q — 2 ,2q — 1, or 2q. If A = 2q — 2, then (r — r * )/8 = 1, a

contradiction by Theorem 1.12. If A = 2q — 1, then r = (20q+17)/5 is not an integer,

a contradiction. If A = 2 q, then r = (20q + 9)/5 is not an integer, a contradiction.

Therefore, we must have a4 = 0. Hence, a* = 0 for all i except possibly 1, 2, and 3.

Solving equations (1.30), (1.31), and (1.32) we obtain

_ <*13

where q 13 = -924A+216, a l2 = 331A2-997A-138, a u = 512A3-280A2 - 444A-72,

and aw = -192A4 + 56A3 - 11 A2 - 35A - 6,

_ a 23?3 4" Q:2293 4" 4* Ocoq ° 2 = 5A(2A - 9 q - l)(g + 2A - 1) ’

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. where aoa = 168A-216, a 22 = -1387A2+1615A+138, a 21 = 336A3-40 8A 2+576A +72,

and q2o = 64A4 + 96A3 — 45A 2 + 41A + 6, and

_ Q3393 + ct32 g2 + 0:3 iq ■+■ Q30 ° 3 “ 10A(2A - 9 q - l ) { q + 2A - 1) ’

where Q33 = -132A + 216, a 32 = 1163A2 - 1683A - 138, q 3 1 = -864A3 + 616A2 -

548A - 72, and a 30 = 64A4 - 208A3 + 61A2 - 37A - 6.

Replacing q by a real variable x in the above expressions for a i and a3, we obtain

two functions, at(x) and a3 (x). Now, we already know that q > A/2. Also, the

inequality e > 1 implies that q < 8A — 9. This implies that 2A — 9q — 1 < 0.

Therefore, ai(x) and a3 (x) are continuous functions of x on the interval [A/2, 8A — 9].

Now, the function a-^x) has zeros only at (A — 3 )/12, 8A + 1, and

8A2 - 3A + 2 ~31 ~ 11A-18 ‘

Clearly, (A — 3)/12,8A + 1 £ [A/2,8A — 9], so a3 (x) has at most one zero on this

interval. However,

_ ,A, (A+ 2)(5A + 3)(15A + 2) = ------10A(5A — 2) ? °

and

., 0 , —4(A — 2)(19A — 21) ° 3(8A“ 9) = ------A (7A —"8)------< 0 '

Therefore, a3 (x) has exactly one zero on the interval [A/2,8A — 9] at 2 31. Then the

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. above inequalities imply that a 3 (x) is negative on the interval (z 3 i, 8A — 9]. Hence,

we must have q G [A /2 ,23 i].

Next, the function a i(x ) has zeros only at —(3A + l) / 4 and

128A2 — 116A - 24 + 5V64A4 - 680A3 + 49A 2 + 204A + 36 231A - 54

Clearly, -(3A + l)/4, zn [A/2, x31], so a^x) has at most one zero on this interval.

However,

,A, —(5A + 1)(5A2 — 52A — 12 ) _ “ l ( 2 ] = ------10A(5A - 2) < °

and

. . 7(13A + 2) n “,(-'3l) = 11A - 18 > °-

Therefore, ai(x) has exactly one zero on the interval [A/2, 23 i] at 212 . Then the above

inequalities imply that ai(x) is negative on the interval [A/ 2 , 212 ). Hence, we must

have q € [zl2, 2 3I].

Now, we easily obtain the inequality 2 3i < (8A + 12)/11. Next, the inequality

64A4 - 680A3 + 49A 2 + 204A + 36 > ( 8A2 - 43A - 144)2 impUes that zl2 > (168A2 -

331A — 744)/(231A — 54). This in turn implies that z\2 > (8A — 16)/11. Therefore,

we have q € ((8A — 16)/11, ( 8A + 12)/11). Thus, since q is an integer, we must have

q € { (8A + fc)/ll : -15 < k < 11}.

Then r = (200A + 36A: + 99)/55 and e = (80A + 11 — k ) / 110 for some integer

—15 < k < 1 1 . So, 5 must divide 36Ar + 99 and 10 must divide 11 — k . Therefore, we

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. must have k = —9,1 , or 11. I f q = (8A — 9 )/ll, then (r — l )/8 = (5A — 7 )/ll, so 11

divides 5A — 7. But, then r ( r — l) / ( v —1) = 5(5A — 7 )/l 1 is an integer, a contradiction

by Theorem 1.10. If q = (8A + 1)/11, then (r* - l )/8 = (3A — 1)/11, so 11 divides

3A - 1. But, then r * (r * — l) /( v — 1) = 3(3A - 1)/11 is an integer, a contradiction by

Theorem 1.10. So, we must have q = (8A + 11)/H- However,

.8A + 11. —4(A — 22)(17A + 33) n ------33A(5A + 11)------*= 0

for A > 23. Therefore, we must have A = 20,21, or 22. But, the only such value of

A that makes q integral is A = 22. Since 22 is twice a prime number, we obtain a

contradiction by Theorem 1.17.

Case 5: m = 11.

If there exists a block A with a a < —2, then m(A) < 7 and D(A) is non-type-1

with g(A) = 8 by Lemma 1.24 (ii), (iii), and (iv). Thus, we obtain a contradiction

by cases 1,2,3, and 4. If there exists a block A with a a > 3, then m(A) > 17, so

mm{A) < 7 and once again we are done by previous cases. Therefore, we must have

a a = —1 ,0 ,1, or 2 for all blocks A. Hence, ai = 0 for all i except possibly —1 ,0 , 1,

or 2. Now, if r > r * , then by equation (1.8) we must have 0 for all .4 since

1-4-1 > 2A for all A. Similarly, if r < r * , then we must have a a < 0 for all A. Thus,

we have the following two subcases.

Subcase 5a: aA < 0 for all A.

In this subcase, erA = —1 or 0 for all .4. Solving equations (1.30) and (1.31) we

obtain a_L = 60g — 32A + 15 and ao = 32A — 52q — 14. Now, r < r \ which implies

that q < (A —1)/2. However, this implies that a_i < —(2A + 15) < 0, a contradiction.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Subcase 56: crA > 0 for all A.

In this subcase, oA = 0 ,1, or 2 for all A. Solving equations (1.30), (1.31), and

(1.32), we obtain

oiQ2

where a02 = -3 2 3 A - 30, o0i = 288A2 - 136A - 1 6 , and aoo = -6 4 A 3 + 6 4 A2 - 13A - 2,

a l3 Q3 4- a l2?2 4* &11Q 4- Q:io fll = (2A — 5q — 1)(2A — 3q — 1) ’

where a 13 = 1140, Q12 = —1373A + 893, an = 528A2 — 696A + 228, and aio =

—64A3 + 132A2 - 87A + 19, and

3 o 003q + a 22<7" + &21Q 4" ^20 “ 2 = 2(2A — 5q — 1)(2A — 3q — 1) ’

where ao3 = —2040, aoo — 2813A — 1598, a 2i = — 1280A2 + 1432A — 408, and a 2o =

192A3 - 320A2 + 179A - 34.

Replacing q by a real variable x in the above expressions for at and a2, we obtain

two functions, ai(x) and ao(x). Now, the inequalities r > r* > 9 imply that A/2 <

q < (2A — 3)/3. This implies (2A — 5q — 1)(2A - 3q — 1) < 0. Therefore, at(x) and

a2 (x) are continuous functions of x on the interval [A/2, (2A — 3)/3|.

Now, the function a t(x) has zeros only at (A — l) / 4 and

136A - 76 + v/256A2 - 1292A + 361

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Clearly, (A — l) /4 , Zu [A/2, (2A - 3)/3], so a 1(2 ) has at most one zero on this

interval. However,

. A (A + 1)(3A — 38) ^ n i(2 = —— > 0

and

.2A — 3. — (5A - 9)(7A 2 - 56A + 114) n ------6(1 ^ 3) ------< °-

Therefore, a\{x) has exactly one zero on the interval [A/2, (2A — 3 )/3 ) at 212 . The

above inequalities then imply that at(x) is negative on the interval ( 212 , (2A — 3 )/3 ].

Hence, we must have q € [A /2 ,212 ].

Next, the function 02 (2 ) has zeros only at (3A — 2)/8 and

128A — 68 T \/64A 2 - 1088A + 289 -21 ,-22 - 255

Clearly, (3A — 2 )/8 , 221 £ [A /2 ,212 ], so a2 (x) has at most one zero on this interval.

However,

A —A2 + 16A + 68 „ 03 2 " 2(A — 2) < °

and

, , 64A + 19 - 4\/256A2 - 1292A + 361 „ 02(212) = ------gg------> ° -

Therefore, ao(x) has exactly one zero on the interval [A / 2 , 212 ] at z-w. The above

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. inequalities then imply that a2 (i) is negative on the interval [A/ 2 , 222 )- Hence, we

must have q € [222 . - 12 ]-

Now, the inequalities 64A2 - 1088A + 289 > ( 8A - 96)2 and 256A2 - 1292A + 361 <

(16A — 40)2 imply that 222 > (8A —10)/15 and 212 < (8A—6)/15. Therefore, since q is

an integer, we must have q = (8A-9)/15, (8A-8)/15 or (8A-7)/15. If q = (8A—9)/15,

then

., 8a - 9, —(2A — 51)(19A — 42) n 15 = 15(A — 3)(A + 2) < °

for A > 26. Therefore, we must have 20 < A < 25. However, no such value of A

makes q integral, a contradiction. If q = (8A — 8)/15, then e = (16A + 9)/10 is not

an integer, a contradiction. If q = (8A — 7)/15, then

„ . . , 8A - 7 , (7A - 38)(17A - 13) . „ 5 <«,(— ) = 15(A_ 2)(A + i,

Therefore, we must have ^ ((S A - 7)/1 5 ) = 6 or 7. If aL(( 8A - 7)/15) = 6, then A

is not an integer, a contradiction. If ai(( 8A — 7)/15) = 7, then A = 44. But, then

(r — r *)/8 = 5 and we apply Theorem 1.12 to obtain a contradiction.

Case 6: m = 9.

If there exists a block .4 with

previous cases. If there exists a block A with 4, then mm(A) < 7 and we are

done. If there exists A with <7.4 = 1, then m(.4) = 11 and we are finished by case 5. If

there exists A with crA = 2, then mm(A) = 11 and we are finished. Therefore,

or 3 for all A. Hence, a* = 0 for all i except possibly 0 and 3. Solving equations

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. (1.30) and (1.31), we obtain a

Inserting these expressions for a0 and a3 into equation (1.32) and manipulating the

result, we arrive at

(2A - 4 q - 1)2[175

Now, 2A — Aq — 1 ^ 0 since v ^ 4A — 1 because m ^ 12. Therefore, we obtain that

175q2 - (256A - 200)? + 64A2 - 64A + 25 = 0.

Now, the left-hand-side of the above equation is a quadratic polynomial in q, and

therefore its discriminant must be a perfect square. This yields the equation

576A2 - 1600A + 625 = N 2

for some integer N. This equation can then be transformed into

(31104A — 6N — 43200)(31104A + 6N - 43200) = 17500.

However, by considering all possible ways of factoring 17500 into the product of two

integers, the above equation can be shown to have no integral solutions, a contradic

tion. This concludes the proof of Theorem 1.28. □

Corollary 1.29 All \-designs on v = 8p + 1 points, p = 1 or 7 (mod 8) a prime,

are type- 1.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Proof. If p does not divide g, then g = 1,2,4, or 8 and D is type-1 by Theorems

1.11 and 1.28. If p does divide g, then p also divides r — 1 and r * — 1. Without loss

of generality, we may assume r > r * . Therefore, either r = 7p + 1 and r ' = p + 1,

r = 6p + 1 and r* = 2p + 1, or r = 5p -I-1 and r * = 3p + 1.

I f r = 6p + 1 and r* = 2p + 1, then (r — r*)/g = 2 and D is type -1 by Theorem

1.12. If r = 5p + 1 and r* = 3p + 1 , then (r — r*)/g = 2 and D is type-1 by Theorem

1.12. Thus, we may assume that r = 7p -I-1 and r* = p -I- 1.

If p = 1 (mod 8), then 8 divides 7p + 1, so r(r — l)/(w — 1) = 7(7p + l )/8 is an

integer and D is type-1 by Theorem 1.10. If p = 7 (mod 8), then 8 divides p + 1, so

r *(r* — l) /( v — 1) = (p + l )/8 is an integer and D is type -1 by Theorem 1. 10. □

R e m a rk 1.30 The author has also proven that all A-designs with g = 7 are type-1

[26j. However, a proof attempt analogous to that of Corollary 1.26 or 1.29 breaks

down in the v = 7p + 1. Consequently, the author was unable to establish whether

or not all A-designs on 7p + 1 points are type-1.

Conjectures

It is the author’s great hope to one day see the A-design conjecture completely re

solved. However, this appears to be an exceedingly difficult problem. Therefore, we

make the following three weaker conjectures, which are hopefully more vulnerable to

attack.

Conjecture 1.31 Every \-design has at least one block o f size 2A.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Conjecture 1.32 A ll X-designs have exactly two block sizes.

Conjecture 1.33 A ll X-designs with only two block sizes, one o f which is 2A, are

type-1.

Obviously, the combination of Conjectures 1.31, 1.32, and 1.33 would prove the A-

design conjecture. However, each conjecture is still very interesting on its own. It has

been verified by computer that all A-designs on v < 713 points with only two block

sizes are type -1 [86]. One could prove Conjecture 1.33 by showing that the other

block size occurs only once [85].

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. CHAPTER 2 5-CHROMATIC STRONGLY REGULAR GRAPHS

This chapter is based on the paper [29] by the author. In this chapter, we will begin

the determination of all primitive strongly regular graphs with chromatic number

equal to 5.

Introduction

Before we begin, we must establish some terminology and notation. All graphs con

sidered are finite and simple (no loops or parallel edges). The vertex set of a graph

G will be denoted by V(G). The complement of G will be denoted by G. Given

x 6 V(G), a vertex adjacent to x is called a neighbor of x. The set of all neighbors

of x is called the neighborhood of x and will be denoted by N (x). If |iV(x)| = k for

all x € V{G), then G will be called k-regular. Given distinct x ,y € V(G), the set

iV(x) fl N (y) is called the set of common neighbors of x and y.

A strongly regular graph with parameters v, k (1 < k < v — 2), A, and p., or a

srg(v, k, A, p.), is a A:-regular graph on v vertices such that every pair of adjacent ver

tices has exactly A common neighbors and every pair of distinct nonadjacent vertices

has exactly p. common neighbors. For a good introduction to the theory of these

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. graphs, see [18], [33], or [80]. For a survey of strongly regular graphs, see [ 11], [16],

or [66].

A set of mutually nonadjacent vertices in a graph G is called a coclique of G. A

maximum coclique in G is a coclique of maximum size in G. The size of a maximum

coclique in G is called the coclique number of G and will be denoted by a(G).

A proper vertex coloring, or just a coloring, C of a graph G is a function C :

V(G) —► { 1,...,/} such that C{x) ^ C(y) whenever x and y are adjacent in G. Such

a coloring will be called an /-coloring of G. Thus, an /-coloring of G is just a partition

of V(G) into / cocUques, which we w ill call color classes.. If there exists an /-coloring

of G, then G is called l-colorable. The smallest / such that G is /-colorable is called

the chromatic number of G and will be denoted by ,\(G)- If x(G) = /, then G w ill be

called /-chromatic.

A strongly regular graph that is connected with connected complement is called

primitive. A strongly regular graph that is not primitive is called imprimitive. The im-

primitive strongly regular graphs are precisely the disjoint unions of complete graphs

of the same size and the complete multipartite graphs with all color classes of equal

size. VVe regard the imprimitive strongly regular graphs as trivial and we will not

mention them further.

In his thesis [36], [37], Haemers proved the following result.

Theorem 2.1 Given a positive integer the number of primitive strongly regular

graphs with chromatic number equal to x is finite.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Haemers then proceeded to determine all primitive strongly regular graphs with chro

matic number equal to 3 or 4 [36], [37]. The purpose of this chapter is to continue with

this work and begin the determination of all 5-chromatic primitive strongly regular

graphs.

Basic properties of strongly regular graphs

In this section, we present some basic properties of strongly regular graphs that we

w ill need in the forthcoming sections. This material is all standard and can be found

in [18], [33], or [80]. The first two results are proved by simple counting.

Proposition 2.2 A graph is a srg(v, k, A, p.) i f and only i f its complement is a

srg(v,v — k — 1, v — 2k + p. — 2,v — 2k + A).

Proposition 2.3 I f a srg(v, k , A, p) exists, then k(k — A — 1) = (v — k — 1 )p.

The last result of this section is a well-known characterization of the primitive

strongly regular graphs as the regular graphs with exactly three distinct eigenvalues.

Theorem 2.4 .4 regular graph G is a primitive srg(v,k,X,p) if and only if it has

exactly three distinct eigenvalues. Moreover, these eigenvalues are k and

A - p ± y/(X - p)2 + 4(fc - p) r,s = 2

with multiplicities 1 and

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. respectively. Additionally, i f G has a nonintegral eigenvalue, then k = (v — l) /2 ,

r,s = ( —1 ± \/v )/2 , v = 1 (mod 4), and v is not an integral square.

Some families of strongly regular graphs

In this section, we w ill describe several families of strongly regular graphs that will

arise in the forthcoming sections.

The Paley graphs

Given a prime power q with q = 1 (mod 4), the Paley graph P(q) is the graph with

V(P(q)) = GF(q), with two vertices being adjacent if their difference is a nonzero

square in G F(q). It can be shown that P(q) is a srg(q, (q — l ) / 2 , (q — 5 )/4 , {q — l) /4 )

[80].

The triangular graphs

Given an integer n > 4, the triangular graph T (n) is the line graph of the complete

graph K n. It is easy to show that T(n) is a srg(n(n — l)/2,2(n — 2),n — 2,4). The

following result says that the triangular graphs are characterized by their parameters

except when n = 8.

Theorem 2.5 [19], [20], [21], [46], [80] Let G be a srg(n(n — l)/2 ,2(n — 2), n — 2,4)

fo r some n ^ 8. Then G = T (n).

When n = 8, there are precisely three additional strongly regular graphs with the

same parameters as T(8). These are known as the Chang graphs.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. The lattice graphs

Given an integer n > 2, the lattice graph Loin) is the line graph of the complete

bipartite graph K n,n- It is easy to show that Lo(n) is a srg(n2,2 (n — l),n — 2 , 2 ).

The following result says that the lattice graphs are characterized by their parameters

except when n = 4.

Theorem 2.6 [70], [80] Let G be a srg(n 2,2 (n — 1), n — 2,2) fo r some n ^ 4. Then

G = Lo{n).

When n = 4, there is precisely one additional strongly regular graph with the same

parameters as Lo(4). This graph is known as the Shrikhande graph.

Strongly regular graphs from quasi-symmetric designs

A quasi-symmetric design [64] is a t-{v, k, A) design, t > 2, such that every pair of

distinct blocks intersects in either x or y points for some x and y (the intersection

numbers), where x < y. Given a quasi-symmetric design D, the block graph of D is

the graph with the blocks of D as its vertices, with two vertices being adjacent when

they intersect in y points. Goethals and Seidel [34] proved that the block graph of a

quasi-symmetric design is strongly regular.

Theorem 2.7 The block graph of a quasi-symmetric t-(v, k, A) design D urith inter

section numbers x and y is a s rg {i/,k f, A',/jf). where vf = b, kf = [k{r — 1) — x{b —

1)1/(3/ —x )> X = Ar/ + r/ -i-s/ -i-r/s/, andp! = kf+r's', where r* = {r — \ — k + x ) /( y — x)

and s' = — (k —x ) /{ y —x) are its nonprincipal eigenvalues and b and r are the number

of blocks o f D and the replication number o f D , respectively.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Strongly regular graphs from latin squares

Given a latin square L of order n, the latin square graph G (L) of L is the graph with

the n2 cells of L as its vertices, with two distinct vertices being adjacent if they are

in the same row, in the same column, or contain the same symbol. It is easy to show

that G(L) is a srg(n2,3(n — l) , n ,6) [80].

Given a latin square LL orthogonal to L, the n symbols of LL give rise to a

partition of V(G{L)) into n cocliques of size n. Also, since it is clear that a coclique

of G(L) can contain at most one cell from each row of L, we must have a(G(L)) < n.

Thus, a pair of orthogonal latin squares of order n gives rise to a pair of n-chromatic

strongly regular graphs.

Strongly regular graphs from systems of linked symmetric designs

A system o fn linked symmetric (v, k, A) designs [15] is a set {Xo,..., X „} of disjoint

sets together with incidence relations i , j = 0, ...,n, i ^ j , between each pair

of distinct sets such that (i) for all i , j = 0,...,n, i ^ j , the incidence structure

(Xi,Xj,Xij) is a symmetric (v, k, A) design, and (ii) for any three distinct sets X t,

X j, and X[, and any two points p 6 X j and q 6 X i, the number of points r 6 X t such

that r is incident with both p and q is x if p and q are incident and is y otherwise. It

follows that we must have (x — y)2 = k — X and y(k + x — y) = kX.

The incidence graph of such a system has the elements of U"_0X i as its vertices,

with two vertices being adjacent when the corresponding points are incident. This

incidence graph is strongly regular if and only if nA = y(n — 1), in which case it is a

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. s rg (i/,k ', X',p!), where 1/ = (n-Fl)u, kf = nk, A' = kf+r'+s'+r's', and p! = kZ+r's',

where r 1 = k /n and s' = —k are its nonprincipal eigenvalues.

Strongly regular graphs from partial geometries

A partial geometry [2], [11], [80] with parameters (s ,t,a ), s , t , a > 1, or a p g(s,t,a),

is a partial linear space (P, L) such that each point of P is incident with exactly t -f 1

lines of L, each line is incident with exactly s + 1 points, and for each nonincident pair

(p, I) € P x L there are exactly a lines through p that intersect I. If a = 1, then the

partial geometry is called a generalized quadrangle [60] and we w rite GQ(s, t) instead.

The point graph of such a partial geometry has vertex set P, with two distinct

vertices being adjacent if they are collinear. It was shown by Bose [2] that the

point graph of a pg(s, t,a ) is a srg((s + l)(sf + a )/a ,s(t + l),s — 1 + (a — l) t , (t +

l)a). A strongly regular graph that is the point graph of a partial geometry is

called a geometric graph, and a strongly regular graph with the same parameters

as a geometric graph is called pseudo-geometric. A strongly regular graph this is

pseudo-geometric for a p g (s,t,a) is geometric if and only if every edge is contained

in a complete subgraph of size s + 1. It can be a very difficult problem to determine

when a pseudo-geometric graph is actually geometric.

Strongly regular graphs from rank 3 permutation groups

General references for the material of this subsection are [44] and [17].

Let T be a transitive permutation group on a set X. I f x € X, then the orbits of

the point stabilizer Tx are called suborbits and their number is called the rank of T.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. The orbits of T on X x X in its diagonal action are called orbitals. There is a natural

bijection between the orbits of Tx and the set of orbitals of T, and so the rank of T

is also equal to the number of its orbitals.

The orbital graph associated with a nontrivial orbital A is the directed graph with

vertex set X and w ith an edge directed from x to y if (x, y) £ A. This graph is

connected if T is primitive. The orbital A is called self-paired if (x, y) £ A implies

(y, x) £ A. If A is self-paired, then its orbital graph may be regarded as an undirected

graph.

Suppose T is primitive of rank 3. Let A be one of the two nontrivial orbitals and

suppose A is self-paired (this is the case if [T| is even). Then the orbital graph of A

is a primitive strongly regular graph.

Matrix-theoretic tools

In this section, we w ill present some matrix-theoretic results that we w ill need in the

forthcoming sections and that axe of general use in algebraic graph theory.

Given a real symmetric u x v m atrix A, we w ill denote the eigenvalues of A (which

must be real) by At (A ) > ... > A„(A). If A is the (0, l)-adjacency matrix of a graph

G, then we sometimes write Ai(G ) instead of Aj(A), i = 1,... ,v. If B is a v\ x v\

matrix with ui < v, then we say that the eigenvalues of B interlace the eigenvalues

of A if B has only real eigenvalues and if A4(A ) > K {B ) > A„_„l+t for i = 1,..., ui.

We say that the interlacing is tight if there exists an integer I, 0 < I < Vi, such that

Ai(A ) = Ai(B ) for i = 1,..., I and A„_0l+ i(A ) = A*(£?) for i = I + 1,..., vt .

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. The first two matrix results we will need give examples of eigenvalue interlacing.

The first of these is sometimes referred to as the Cauchy inequalities [56].

Theorem 2.8 Let A be a real symmetric v x v matrix partitioned as follows:

where A i,i is square. Then the eigenvalues o f A\,i interlace the eigenvalues o f A.

The next result is a very useful generalization of a result that has been called the

Higman-Sims technique by Payne and Haemers [35], [37], [39].

Theorem 2.9 Let A be a real symmetric v x v matrix partitioned as follows:

f \

.4 =

where .4,,, is square fo r i = l,...,t/i. Let b ij be the average row sum of .4tJ fo r

i, j = 1,... ,u i. Let B = (b ij). Then the eigenvalues o f B interlace the eigenvalues

o f A, and i f the interlacing is tight, then A ij has constant row and column sums fo r

i , j = l,...,n .

The next result, due to Weyl [81], gives us inequalities for the eigenvalues of the

sum of two symmetric matrices.

Theorem 2.10 Let A and B be real symmetric v x v matrices. Then

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Xi+jiA) + K-AB) < AAA + B) < K-AA) + Ai+1(B)

fo r i = 1,..., v and j = 0,..., m in {i — 1, v — i}.

Actually, we will only need a special case of Theorem 2.10. Henceforth, we will denote

by J a square matrix of the appropriate size consisting of all 1’s.

Corollary 2.11 Let A be a real symmetric v x v m atrix and let c be a positive real

number. Then

Ai(A - cJ) < AAA)

fo r i = 1,..., v.

Proof. Apply the right-hand side of Theorem 2.10 with i = 1,.... v and j = 0 and

use the fact that the eigenvalues of —cJ are 0 and —cv with multiplicities v — 1 and

1, respectively. □

The last m atrix result we shall need concerns symmetric matrices with just two

distinct eigenvalues.

T h e o re m 2.12 [37] Let A be a real symmetric v x v matrix partitioned as follows:

where A i,i is Vi xvi. Suppose that A has eigenvalues r and s, r > s, of multiplicities

f and v — f , respectively. Then

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. r if I < i < f — Vi,

Aj(A2 ,2 ) = i s i f f + 1 < i < v — Vi,

r + s — otherwise.

Eigenvalues , cocliques, and chromatic number of strongly reg ular graphs

In [36], [37], Haemers determined all primitive strongly regular graphs with chromatic

number equal to 3 or 4. He found there to be three such graphs w ith chromatic number

3, and 18 such graphs with chromatic number 4. His result is as follows.

Theorem 2.13 If G is a Z-chromatic primitive strongly regular graph, then G is one

o f the following graphs: the pentagon P ( 5) (the unique srg(5,2 ,0 ,1)), the Petersen

graph T ( 5) (the unique srg( 10,3,0,1)), or Z,2 (3) (the unique sr^(9,4,1,2)). I f G

is a 4-chromatic primitive strongly regular graph, then G is one o f the following

graphs: T ( 6) (the unique srg( 15,6,1,3)), Lo(4) (one o f the two srg( 16,6,2,2)), £o(4)

(one o f the two srg( 16,9,4,6)), the Shrikhande graph (the exceptional s rg (l 6 , 6,2 ,2 )),

the Clebsch graph (the unique s rg (l 6 , 5,0,2) [65]), the Hoffman-Singleton graph (the

unique srg(50,7,0,1) [47]), the Geurirtz graph (the unique srg( 56,10,0,2) [31], [10]),

or one of the 11 incidence graphs o f a system of 3 linked symmetric (1 6 ,6 ,2 ) designs

[57],

The following result will be used to determine all possible parameter sets of a

5-chromatic primitive strongly regular graph.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. T h e o re m 2 .1 4 [37] I f G is a primitive srg(v, k, A, p) with eigenvalues k, r, and s

(r > s) integral, then

- r ( X(G) - 1) < s < - 2 , (2.2)

and

3 < k < -s(X(G) ~ 1). (2.3)

The next result will be very useful for obtaining contradictions to the existence

of 5-chromatic strongly regular graphs with various parameter sets.

T h eo re m 2.15 [37] I f G is a graph on v vertices, then

x(0 - 2 - £ K -i(G ) > A ,(G ). t=0 The next two results give upper bounds on the size of a coclique in a primitive

strongly regular graph. The first is an unpublished result of Hoffman.

T h e o re m 2.16 [37], [39] Let G be a primitive srg(v, k, A, p) with smallest eigenvalue

s. Let C be a coclique in G. Then |C| < vs/(s — k) with equality if and only if each

vertex outside C is adjacent to exactly —s vertices in C.

T h e o re m 2 .1 7 [22] Let G be a primitive strongly regular graph with smallest eigen

value o f m ultiplicity g. Then a(G) < g.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. The next result concerns primitive strongly regular graphs possessing a coclique that

achieves both of the bounds given in Theorems 2.16 and 2.17

Theorem 2.18 [37] Let G be a primitive s rg (v ,k ,\,p .) with eigenvalues k, r, and

s (r > s) o f multiplicities 1, / , and g, respectively. Let C be a coclique in G with

|C| = vs/(s — k) = g. Then the subgraph o f G induced on V(G) \ C is a srg(v —

g, k + s, k -F (r + 2)(r + s), k + s + r ( r + s)).

The last result of this section concerns primitive strongly regular graphs that pos

sess a coloring for which all color classes are cocliques satisfying Hoffman’s bound

given in Theorem 2.16. Such a coloring has been called a Hoffman coloring by

Haemers and Tonchev [41].

T h eo re m 2.19 [41], [78] Let G be a primitive s rg (v ,k ,\,p ) with eigenvalues k, r,

and s (r > s) of multiplicities 1, f , and g, respectively. Suppose that G has a Hoffman

coloring. Then

with equality i f and only i f G is the incidence graph of a system o f —k/s linked

symmetric (v s /(s — k), —s, s(s + l)(s — k )/[k -i- s(v — 1)]) designs.

Parameters of 5-chromatic strongly regular graphs

In this section, we find all possible parameter sets for strongly regular graphs with

chromatic number equal to 5. The computer program Mathematica [84] was used in

the proof of the following result.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Theorem 2.20 A b-chromatic primitive strongly regular graph G must have one of

the following 43 parameter sets:

1. (5 ,2 ,0 ,1 )

2. (13,6,2,3)

3. (17,8,3,4)

4. (21,10,3,6)

5. (25,16,9,12)

6 . (16,5,0,2)

7. (15,6,1,3)

8. (16,9,4,6)

9. (10,3,0,1)

10. (9 ,4 ,1 ,2 )

11. (10,6,3,4)

12 . ( 100, 22 , 0, 6)

13. (7 6 ,3 0,8 ,1 4 )

14. (75,32,10,16)

15. (81,20,1,6)

16. (76,21,2,7)

17. (77,16,0,4)

18. (64,18,2,6)

19. (57,14,1,4)

20 . (49,16,3,6)

21 . (56,10,0,2)

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 22 . (40,12,2,4)

23. (36,14,4,6)

24. (35,16,6,8)

25. (5 0 ,7 ,0 ,1 )

26. (26,10,3,4)

27. (25,12,5,6)

28. (16,6,2,2)

29. (15,8,4,4)

30. (266,45,0,9)

31. (225,48,3,12)

32. (196,39,2,9)

33. (210,33,0,6)

34. (165,36,3,9)

35. (176,25,0,4)

36. (125,28,3,7)

37. (162,21,0,3)

38. (115,18,1,3)

39. (96,19,2,4)

40. (85,20,3,5)

41. (99,14,1,2)

42. (45,12,3,3)

43. (2 8 ,8 ,3 ,2 )

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Proof. Let G be a 5-chromatic primitive srg(v, k, A, fi) w ith eigenvalues k, r, and s

(r > s), with multiplicities 1, /, and g, respectively. If G has a nonintegral eigenvalue,

then by Theorem 2.4, k = (v - l) /2 , s = ( —1 — y/v)J2, v = 1 (mod 4), and v is

not an integral square. Now, by Theorem 2.16, we have a(£) < y/v. Therefore,

v/y/v = y/v < v/a(G) < \{G) = 5, and so v < 25. Thus, we must have v = 5, 13, or

17. The only strongly regular graphs on 5, 13, or 17 vertices have parameter sets 1,

2, and 3 in the list above.

Now, suppose r and s are both integers. Then the inequalities (2.1), (2.2), and

(2.3) become

1 < r < 3, (2.4)

—4r < s < —2, (2.5)

and

3 < k < -4 s . (2.6)

Also, by Proposition 2.3 and Theorem 2.4, we may write

(k — r)(k — s) v = ------(2.7) k + rs

A = fc + r + s + rs, (2.8)

and

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. H = k + rs. (2.9)

Now, using inequalities (2.4), (2.5), and (2.6), we obtain 442 possible (r, s,k) triples.

For each such triple, we check that A > 0, n > 1, and that / and g are both integers.

Then using equations (2.7), (2.8), and (2.9), we obtain the parameter sets 4 through

43 in the list above. □

The parameter sets 1, 2, 4, 5, 6, 7, 8, 9, 10, 11, 14, 16, 19, 20, 21, 25, 28, 29, and 43

In this section, we will deal with some of the easier parameter sets.

By Theorem 2.13, all of the graphs with parameter sets 1, 6, 7, 9. 10, 21, 25,

and 28 are 4-coIorable and so are not 5-chromatic. Also, there are two graphs with

parameter set 8, namely L 2 (4) and the complement of the Shrikhande graph. £o(4)

is 4-chromatic by Theorem 2.13, but the complement of the Shrikhande graph has

coclique number 3, and hence it has chromatic number at least f 16/3] = 6.

There is a single graph with parameter set 2, namely P(13). Clearly, {0,2,7} U

{1 ,6 ,8 } U {3 ,5 ,1 1 } U {4 ,9 } U {10,12} is a partition of V(P(13)) into five cocliques,

and so P(13) is 5-chromatic.

By Proposition 2.2 and Theorem 2.5, there is only one graph for each of the

parameter sets 4, 11, and 29, namely T{7), T(5), and T( 6), respectively. T{5) and

T(6) are both 5-chromatic since they both have edge chromatic number 5. T (7 ) is

also 5-chromatic since T ( 6) is 4-chromatic by Theorem 2.13.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Similarly, by Proposition 2.2 and Theorem 2.6, there is also a unique graph for

each of the parameter sets 5 and 43, namely Lo(5) and Lo(5), respectively. 1-2 (5) is

5-chromatic since has edge chromatic number 5. Lo(5) is also 5-chromatic since

if we choose a color class of Ks$ and for each vertex in this color class we color its

five incident edges w ith the same color, but using different colors for each of the five

different vertices in this class, then this becomes a 5-coloring of Z,o(5).

A srg(75,32,10,16) has eigenvalues k — 32, r = 2, and s — —8, with multiplici

ties 1, / = 56, and g = 18, respectively. Therefore, by Theorem 2.16, such a graph

has cocliques of size no greater than 75(— 8 ) /( —8 — 32) = 15. So, a 5-coloring of

such a graph would be a Hoffman coloring. Thus, by Theorem 2.19, a 5-colorable

srg(75,32,10,16) would have to be the incidence graph of a system of 4 linked sym

metric (1 5 ,8 ,4 ) designs, since 56 = —(3 2 /(—8))(3 2 /(—8) + 18). However, this is not

possible since 4 — 1 = 3 does not divide 4 • 4 = 16. This fact was observed by Haemers

in [41].

There does not exist a strongly regular graph with parameter set 16, 19, or 20

[38], [83], [13],

The parameter set (77,16,0,4)

In this section, we prove the following Theorem.

Theorem 2.21 A srg(77,16,0,4) is 5-chromatic.

Proof. Let G be a srg(77,16,0,4). Brouwer [7] proved that such a graph is unique,

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. namely G must be the complement of the block graph of the quasi-symmetric 3-

(2 2 ,6, 1) design D w ith intersection numbers 0 and 2. By Theorem 2.4, the eigenvalues

of G are k = 16, r = 2 , and s = —6, with multiplicities 1, / = 55, and g = 21 ,

respectively. By Theorems 2.16 and 2.17, a(G) < 7 7 (-6 )/(-6 — 16) = 21 = g. Now,

the number of blocks containing a given point in D is 3 (^ITi1) / (3 Z1) = 21, and so G has

a coclique C of size 21 consisting of all the blocks through a fixed point. Therefore, by

Theorem 2.18, the subgraph of G induced on V(G)\C is a srg(56,10,0,2). However,

the srg(56,10,0,2) is 4-chromatic by Theorem 2.13, and so G is 5-chromatic. □

The parameter set (35,16,6,8)

There are exactly 3854 srg(3 5 ,1 6,6 ,8 ) [58], [76]. However, we w ill show in this section

that none of them are 5-chromatic.

Theorem 2.22 There does not exist a 5-chromatic srg(35, 1 6 ,6 ,8 ).

Proof. Let G be a srg(36,16,6,8). Then, by Theorem 2.4, G has eigenvalues k = 16,

r = 2, and s = —4, with multiplicities 1, / = 21, and g = 14, respectively. By

Theorem 2.16, a (G ) < 36(—4 ) /( —4 — 16) = 7. Suppose that G is 5-chromatic. Then

a 5-coloring of G would have to consist of five cocliques of size 7. Let G be 5-colored

and let Ci and Co be two of the color classes. Partition V(G) into C\, Co, and

V(G) \ (C i U Co). Let A be an adjacency matrix for G. Then we can assume that A

is in the following form:

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 0 A j>2 A i,3

A = A f 2 0 A2,3

\, ^ 1A ,3 T n * At*2,33 ^ A 3 ,33.3 y,

where the two 0’s on the diagonal are each 7 x 7 and A ii2, Aj[2, A£3, and Ao 3 all

have constant row sums of —s = 4 by Theorem 2.16. Let

Let G\ be the induced subgraph of G with adjacency matrix At. Then G\ is a 4-

regular bipartite graph on 14 vertices, and so At(G i) = 4 and A 14(G i) = —4, since the

spectrum of a bipartite graph must be symmetric about the origin [ 1]. By Theorem

2.8, we also have A2 (A) = 2 > Ao(Ai), and so G\ has eigenvalue 4 with multiplicity 1

and so is connected. Now, the matrix A — [(k — r)fv\J = A — [(16 — 2)/35 ]J = A —

(2/5) J has just two distinct eigenvalues, namely r = 2 and s = —4, with multiplicities

/ + 1 = 22 and g = 14, respectively. Therefore, by Theorem 2.12 with i = 20,21, we

have

2 2 2 A2 o(A3,3 — —J) = r + s — Xz(Ai — — J) = —2 — As(Ai — -J) 0 0 0

and

A2i(.4 3,3 — - J ) —r + s — Ao(Ai - -J) = - 2 - A 2(A t - -J). 5 5 5

Now, by Corollary 2.11 with i = 20 , 21 , we have

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. -^20 (^3,3 — 7 J) ^ A2o(^3,3) 3

and

2 A21 (Aj.3 — 7 ^ ) < A21 (^ 3 ,3 )- 3

Let G3 be the induced subgraph of G with adjacency matrix A 3 ,3 . Then G3 is 3-

chromatic and the average row sum of A 3,3 is 8, and so by Theorems 2.15 and 2.9 we

must have

Ao(-4i — ^ J ) + A3( Ai — 7 J ) + 4 > —A2 i(G 3) — A2 o(G3) > A i(G 3) > 8 5 5

which imphes that

A2 (A i — 7 J) + A3 (A i — 7 J) > 4. 5 3

Now, A ! and A i — (2 /5 ) J have the same eigenvalues except for the eigenvalue 4

of A j which becomes 4 — 14(2/5) = —8/5 for A i — (2 /5 ) J. Therefore, Ao(A[ —

(2/5) J), A3 (A i-(2 /5 )J ) < 2, which implies that A2 (A i—(2/5)7) = A 3 (Ai~(2/5)7) =

2. Thus, G i has eigenvalue 2 with m ultiplicity at least 3. Since G i is bipartite, it also

has eigenvalue —2 with m ultiplicity at least 3. Now, the sum of the squares of the

eigenvalues of G i is equal to 14 • 4 = 56. Also, 1 -4 2 + 3 • 22 + 3 • ( —2)2 + 1 • ( —4)2 = 56.

Therefore, the 6 remaining eigenvalues of G\ must all be 0. Now, G\ is a connected

and balanced (color classes of equal size) bipartite graph. Such a graph has a perfect

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. matching if and only if its spectrum does not contain 0 [1], [53]. and so G\ does not

have a perfect matching. However, this is a contradiction since a regular bipartite

graph always has a perfect matching [23]. □

The parameter set (266,45,0,9)

According to [8], it is not known whether or not a sr^(266,45,0,9) exists. However,

in this section we w ill show that no such graph can be 5-chromatic.

Theorem 2.23 There does not exist a 5-chromatic srg( 266,45,0,9).

Proof. Let G be a sr^(266,45,0,9). Then, by Theorem 2.4, G has eigenvalues

k = 45, r = 3, and s = -12, with multiplicities 1, / = 209, and g — 56, respectively.

Suppose that G is 5-chromatic. Then the size of the largest color class in a 5-

coloring of G must be at least [266/5] = 54. Also, by Theorems 2.16 and 2.17,

a(G) < 266(—12)/(—12 — 45) = 56 = g. Therefore, the size of the largest color class

in a 5-coloring of G is 54, 55, or 56. Let G be 5-colored and let C be the largest color

class in this coloring. Partition V (G) into the sets C and V(G) \ C. Let .4 be an

adjacency matrix for G. We can assume that A looks as follows:

where the 0 in the upper left-hand comer of A is |C| x |C|. Let Go be the induced

subgraph of G with adjacency matrix A2,2- Clearly, Go must be 4-chromatic.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Suppose \C\ = 56. Then, by Theorem 2.18, the graph G 2 is a 4-chromatic

srg(210,3 3 ,0 ,6 ). However, by Theorem 2.13, this is a contradiction since such a

graph is not 4-colorable.

Next, suppose \C\ = 55. Now, the matrix A — [(45 — 3)/266] J = A — (3/19) J has

just two distinct eigenvalues, r = 3 and s = —12, with multiplicities / 4-1 = 210 and

g = 56, respectively. Therefore, by Theorem 2.12 with i — 209,210, 211, we have

^209(^ 2,2 — — */) = r 4- s — Ao(—— J ) = 3 — 12 — 0 = —9,

A210 (^ 2,2 — = r 4* s — At(—— J) = 3 — 12 — 0 = —9,

and

3 A211 (*4-2.2 — Yg<^) = s = - 1 2 .

So, by Corollary 2.11 with i = 209, 210, 211, we have Aoo<)(.4o 2 ), Aoio(-4-),2) > - 9

and A211 (^ 2 ,2 ) > —12. Also, since the average row sum of .42,2 is (266 • 45 — 2 • 55 •

45)/(266-55) = 7020/211, we have by Theorem 2.9 that A ^^o) > 7020/211. Then,

by Theorem 2.15 applied to the 4-chromatic graph Go, we have

7000 12 + 9 4- 9 = 30 > — Aou(Go) — Aoio(G2) — A209(G2) > Ai(G2) > ,

a contradiction.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Finally, suppose \C\ = 54. In this case, we have by Theorems 2.12 and 2.9

and Corollary 2.11 that A2 io(C2 ) ^ —9, A2 n (G 2 ), ^212 ( ^ 2 ) ^ —12, and A i(G2 ) ^

(266 • 45 — 2 • 54 • 45)/(266 — 54) = 3555/106. Therefore, by Theorem 2.15, we have

3555 12 + 12 + 9 = 33 > —A2i2(G2) — A2n(C'2) — A2io(£*2) ^ Aj.(C?2) ^ 106 ’

a contradiction. □

The parameter set (165,36,3,9)

Theorem 2.24 There does not exist a 5-chromatic sr^(165,36,3,9).

Proof. Let G be a sr

k = 36, r = 3, and s = —9, with multiplicities 1, / = 120, anf g = 44, respectively.

By Theorem 2.16, we have a(G) < 165(—9 ) /( —9 — 36) = 33. Suppose that G is

5-chromatic. Then a 5-coloring of G must consist of five cocliques of size 33. Let G

be 5-colored and let C i and C i be two of the color classes. Partition V(G) into C i,

Ci, and V(G ) \ (C i U C2 ). Let 4 be an adjacency m atrix for G. Then we can assume

that .4 is in the following form:

( n A \ 0 4i,2 -41,3

A = .4^0 0 42,3

^ A I, 4 o 3 43,3 ^

where the two 0’s on the diagonal are each 33 x 33 and 4 1i2, 4^2, 4£3, and .4o3 all

have constant row sums of — s = 9 by Theorem 2.16.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Let

Let G\ be the induced subgraph of G with adjacency matrix A\. Then G\ is a Ir

regular bipartite graph on 66 vertices, and so Ai(Gi) = 9 and A66(Gi) = —9. By

Theorem 2.8, we also have Ao(A) = 3 > A2 (Ai), and so G\ has eigenvalue 9 with

multiplicity 1 and so is connected. Now, A — [(36 — 3)/165] J = A — (1 /5 ) J has just

two distinct eigenvalues, namely r = 3 and s = —9, with multiplicities / + 1 = 121

and g = 44, respectively. Therefore, by Theorem 2.12 with i = 98,99, we have

A98(-^3,3 — ~J) = r + S — Aj4 (>li — —J) = —6 — A04 (-4.1 — — J) 0 5 5

and

Agg(> 1.3,3 ~ t J) = t + s — A;a(i4i —- J) ~ —6 — X^ziAi — - J)- 5 5 5

Now, by Corollary 2.11 with i = 98,99, we have

Ag8(^3.3 ~ t J) < Ag8(i43,3) 5

and

Agg(i43i3 — — J) < Agg(433). 5

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Let Gz be the induced subgraph of G with adjacency matrix ^ 3,3 • Then G3 is 3-

chromatic and the average row sum of ^ 3,3 is 18, and so by Theorems 2.15 and 2.9

we must have

•^23(^1 — 3«/) + ^04(^1 — — J) -f- 12 > — Agg(G3) — ^9&(Gz) > Ai(G3) > 18 5 0

which implies that

^23(^1 — g J ) + Ao-jC-Ai — - J) > 6.

Now, A i and .4i — (1/5)J have the same eigenvalues except for the eigenvalue 9

of A i which becomes 9 — 66(1/5) = —21/5 for .4i — (1 /5 )J. Therefore, Ao3 (A ! —

(1/5)J), Ao4(A i — (1 /5 ) J) < 3, which implies that Ao3(-4i - (1 /5 ) J) = A24(-4i -

(1 /5 ) J) = 3. Thus, G\ has eigenvalue 3 with m ultiplicity at least 24. Since G i is

bipartite, it also has eigenvalue —3 with m ultiplicity at least 24. Now, the sum of

the squares of the eigenvalues of G i is equal to 66 • 9 = 594. Also, 1 • 92 + 24 • 32 +

24 • ( —3)2 + 1 • ( —9)2 = 594. Therefore, the 16 remaining eigenvalues of G i must all

be 0. Since G i is a connected and balanced bipartite graph, we arrive at the same

contradiction as in the proof of Theorem 2.22. □

Computer results

In this section, we use the computer algebra system GAP [30] and the share package

GRAPE [73] for computing with graphs to determine the 5-chromatic strongly regular

graphs w ith parameter sets 3, 12, 15, 18, 22, 23, 26, 27, and 42.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. The philosophy behind GRAPE is that a graph G comes equipped with a user-

specified group of automorphisms T < Aut(G) which is used to make GRAPE’s graph-

theoretic algorithms run more efficiently. GRAPE possesses a function that will return

a set of representatives of the set of complete subgraphs of G of size n, where n is

specified by the user. This function will return at least one complete subgraph from

each orbit of the set of K n subgraphs under the action of T. Thus, if V is the trivial

group, this function w ill return all complete subgraphs of a given size of G. However,

we do lose the increased efficiency that a large group of automorphisms would provide.

GRAPE also possesses a function to compute the complement of a given graph. Since

a complete subgraph of G is a coclique of G, we can use these two functions to find

all cocliques of a given size in G. This allows us to determine with certainty if a given

graph has a coloring with a particular combination of color class sizes, and hence if

it has a given chromatic number. This can be time-consuming, but the graphs that

we consider in this section are all small enough to make this approach feasible.

The parameter set (17,8,3,4)

There is a unique sr^( 17,8,3,4), namely P( 17). Using the definition of the Paley

graphs, P(17) was constructed within the GRAPE system. This is accomplished

by constructing an adjacency matrix A for P(17) and then executing the GRAPE

command

P17:=Graph(Gamma,[1..17],OnPoints,function(x,y) return A[x][y]=1;end,true);

where T is the trivial group (Gamma: =Group( () ) ; ) . This ensures that the command

CompleteSubgraphs0fGivenSize(ComplementGraph(P17),n);

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. will return every complete subgraph of P(17) of size Ti

l l P(17) was 5-colorable, then it would have to contain a coclique of size at least

[17/5] = 4. However, it can be shown with GRAPE that a(P(17)) = 3. Hence, P(17)

is not 5-colorable.

Theorem 2.25 There does not exist a 5-colorable s rg (l7 , 8 ,3 ,4 ).

Rem ark 2.26 Using GRAPE, it was shown that P(17) does possess six pairwise

disjoint cocliques of sizes 3, 3, 3, 3, 3, and 2. Therefore, by Theorem 2.25, P(17) is

6-chromatic.

The parameter set (100,22,0,6)

There is a unique strongly regular graph with parameter set 12 [32]. This graph

is known as the Higman-Sims graph and can be constructed from the Higman-Sims

group, a primitive rank 3 simple group [45].

By Theorem 2.4, the multiplicity of the smallest eigenvalue of this graph is g —

22. Therefore, by Theorem 2.17, this graph has cocliques of size no greater than

22. Using the library of primitive groups that is available in GAP, the Higman-

Sims graph was constructed within GRAPE as an orbital graph of the Higman-Sims

group (H S := P rim itiveG ro u p (1 0 0,3 );). It was found that this graph has exactly 100

cocliques of size 22. It was also found that this graph does not have a set of three

pairwise disjoint cocliques of size 22, and that any two coliques of size 22 intersect in

exactly 0 or 6 vertices. Therefore, by deleting three vertices in all possible ways from

the 100 cocliques of size 22, we see that this graph has at least 22 • 21 • 20 • 100/6 =

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 154000 cocliques of size 19. However, using GRAPE it was shown that this graph has

exactly 154000 cocliques of size 19. Therefore, any coclique of size 19 is contained in

a coclique of size 22. This implies that the Higman-Sims graph is not 5-colorable.

Theorem 2.27 There does not exist a 5-colorable sr^(100,22,0,6).

Rem ark 2.28 The Higman-Sims graph can also be constructed as follows [45], [11].

Let (X , B) be the quasi-symmetric 3-(22,6,1) design with intersection numbers 0 and

2 and take as vertices the set {oo}uX u B . Let oo be adjacent to all of the points of X

and to none of the blocks of B, let each point be adjacent to the blocks that contain it

and to none of the other points, and let two blocks be adjacent when they are disjoint.

Easy counting arguments show that the graph so constructed is a sr <7(100,22,0,6).

Since the unique srg(77,16,0,4) is 5-chromatic by Theorem 2.21, we clearly see that

this graph is 6-colorable. Therefore, by Theorem 2.27, the Higman-Sims graph is

6-chromatic.

The parameter set (81,20,1,6)

It was shown by Brouwer and Haemers [9] that there exists a unique srg(81,20,1,6).

In [9], seven different constructions of this graph are given. We w ill use the description

of this graph that was given in [79]: it is the graph with the elements of G F (81) as its

vertices, with two vertices being adjacent if their difference is a nonzero fourth power

in GF(81). Using this description, this graph was constructed in GRAPE. Now, if this

graph was 5-colorable, then it must possess a coclique of size at least [81/5] = 17.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. However, using GRAPE, it can be shown that this graph has coclique number 15.

Hence, this graph is not 5-colorable.

Theorem 2.29 There does not exist a 5-colorable srg(81, 2 0 ,1 ,6 ).

The parameter set (64,18,2,6)

Haemers and Spence [40] have done a complete enumeration of strongly regular graphs

with parameters (64,18,2,6). They found there to be exactly 167 such graphs. Adja

cency matrices of these graphs are available at [76], and in what follows we will denote

these graphs by G ^,,, i = 1 , . . . , 167, where the ordering is the same as it is on [76].

Using the adjacency matrices on [76], these 167 graphs were constructed in GRAPE.

Now, a 5-colorable srg(64,18,2,6) must have a coclique of size at least f64/5] = 13.

It was found that the 16 graphs Gw.i, i = 5, 6, 7, 10, 16, 32, 36, 47, 48, 50, 59, 79,

83, 85, 86, 128, do not have a coclique of size 13 and are therefore not 5-colorable. It

was also found that only the 11 graphs Gm,*, i = 1, 2, 3, 4, 8, 9, 13, 14, 18, 19, 20,

have a coclique of size 16. Since there are exactly 11 srg(64,18,2,6) with chromatic

number 4 by Theorem 2.13, these graphs must be 4-chromatic. Additionally, the two

graphs G64.27 and G&1.54 both have 5-colorings with four color classes of size 13 and

one color class of size 12, the 45 graphs G 64,i, i = 15, 17, 21, 22, 23, 28, 29, 30, 31, 34,

35, 39, 40, 41, 42, 43, 44, 45, 46, 60, 62, 64, 65, 66, 68, 70, 71, 74, 75, 76, 78, 90, 94.

95, 96, 97, 100, 109, 117, 118, 119, 132, 133, 139, 158, all have 5-colorings w ith four

color classes of size 14 and one color class of size 8, and the 13 graphs G& 4,,, i = 25,

38, 56, 57, 61, 63, 69, 84, 91, 108, 110, 130, 137, all have 5-colorings with four color

classes of size 15 and one color class of size 4. By trying all possible combinations of

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. color class sizes, it can be shown that the remaining 80 graphs are not 5-colorable.

Thus, there are exactly 60 srg(64,18,2,6) with chromatic number 5. We also obtain

the following characterization of the 4-chromatic srg(64,18,2,6).

Theorem 2.30 There are exactly 60 srg(64,18,2,6) with chromatic number 5. A

srg(64, 18,2,6) has chromatic number 4 if and only i f it has a coclique of size 16.

Remark 2.31 In [55], the author claims to prove that a srg(64,18,2,6) must be

geometric for a GQ(3,5), which is unique [60]. In view of the computer enumeration

done in [40], this result is obviously in error.

The parameter set (40,12,2,4)

Spence [74] has done a complete enumeration of strongly regular graphs with param

eters (40,12,2,4). He found there to be exactly 28 such graphs. Adjacency matrices

of these graphs are available at [76], and in what follows we w ill denote these graphs

by G\jo,i, i = 1 ,..., 28. Now, a 5-colorable sr

at least 40/5 = 8. Using GRAPE, it was shown that the only graphs possessing a co

clique of size 8 are the 11 graphs G.io,i, i = 1 , . . . . 11. We then used GRAPE and found

that each of these 11 graphs possesses a set of five pairwise disjoint cocliques of sizes

9, 9, 8, 8, and 6. Hence, there are exactly 11 srg(40,12,2,4) with chromatic number

5. We also obtain the following characterization of the 5-chromatic srg(40,12,2,4).

Theorem 2.32 There are exactly 11 srg(40,12,2,4) with chromatic number 5. A

srg(4Q, 12,2,4) has chromatic number 5 i f and only i f it has a coclique o f size 8.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. The parameter set (36,14,4,6)

We begin this section with a lemma.

Lemma 2.33 A srg(36, 14,4,6) with three pairwise disjoint cocliques of size 8 is

5-chromatic.

Proof. Let G be a srg(3 6 ,1 4 ,4 ,6 ) w ith three pairwise disjoint cocliques of size 8,

say C i, C

A be an adjacency matrix of G. We can assume that A looks as follows:

^ 0 Ai,o A ,3 A it4

A i 2 0 Ao,3 .42,4 A = -4[ 3 4^3 0 A3,4

^ *4^4 -4^4 A 3 4 A t,4 J

where the 0’s on the diagonal are all 8 x 8. Now, by Theorem 2.4, G has eigenvalues

k = 14, r = 2, and s = -4 , with multiplicities 1, / = 21, and g = 14, respectively.

So, by Theorem 2.16, a(G) < 36(—4 )/(—4 — 14) = 8. Therefore, we know that the

block matrices A li2, .41>3, A j<2, -42i3, Aj[3, A£3i A^4, A 2a, and A jA all have constant

row sums of —s = 4. Let G4 be the induced subgraph of G with adjacency matrix

At,4. Then G 4 must be 2-regular, and hence is a disjoint union of cycles. Let c be

the length of one of these cycles. If c = 12, then G 4 is bipartite and so 2-chromatic,

and so G is 5-chromatic. Suppose c < 12. Partition V (G i) into the vertices of this

cycle and the remaining ones. This induces a 5 x 5 partitioning of A, the average row

sums of which are given by the m atrix

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. C 0 4 4 4 6 - f 4 0 4 c 4 6 — f 4 4 0 c 4 e - f 4 4 4 2 0

4 4 4 0 2

Now, the eigenvalues of B are Xi(B ) = 14, \ 2(B) = 2, A3 (2 ?) = A4( 5 ) = A5(B ) = —4.

We see that the eigenvalues of B interlace the eigenvalues of A tightly. Thus, by

Theorem 2.9, the block matrices in this partitioning of A all have constant row and

column sums. Therefore, c/4 is an integer and so c is even. This implies that G\ is

bipartite and so G is 5-chromatic. □

McKay and Spence [58] have done a complete enumeration of strongly regular

graphs with parameters (3 6 ,1 4 ,4 ,6 ). They found there to be exactly 180 such graphs.

Adjacency matrices of these graphs are available at [76], and in what follows we will

denote these graphs by G 36 ,., i = 1, • - •, 180. Now, if such a graph is 5-colorable, it

must possess a coclique of size [36/5] = 8. However, using GRAPE, it was shown

that only the graphs G 36.48 and £36,77 have a coclique of size 8. W hat is more, these

two graphs each have a set of three pairwise disjoint cocliques of size 8, and so by

Lemma 2.33, they are both 5-chromatic. Thus, there are exactly two 5-chromatic

srg(36, 14,4,6). We also obtain the following characterization of the 5-chromatic

srg( 36,14,4,6).

Theorem 2.34 There are exactly two srg(36, 14,4,6) with chromatic number 5. A

srg(36, 14,4,6) has chromatic number 5 i f and only i f it has a coclique o f size 8.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. The parameter set (26,10,3,4)

There are exactly 10 srg(26, 10,3,4). These were found by Paulus [59] and shown to

be all such graphs by Rozenfel'd [61]. Adjacency matrices of these graphs are available

at [76], and in what follows we will denote these graphs by G26,i, i = 1 , . . . , 10. Now,

a srg(26, 10,3,4) has smallest eigenvalue s = —3, and so it has cocliques of size no

greater than 26(—3 )/(—3 — 10) = 6 by Theorem 2.16. Using GRAPE, it was shown

that the only graph with a pair of disjoint cocliques of size 6 is G26 ,8- Thus, a 5-

coloring of any of these graphs except G26,8 must consist of one color class of size 6

and four color classes of size 5. Using GRAPE, it was shown that of these graphs,

only C?26,i and Go6,2 possess such a coloring. It was also shown that Gog,a has such a

coloring as well. Thus, there are exactly three 5-chromatic srg(26,10,3,4).

Theorem 2.35 There are exactly three 5-chromatic srg( 26,10,3,4).

Remark 2.36 The remaining 7 srg(26,10,3,4) that are not 5-colorable all have six

pairwise disjoint cocliques of sizes 6, 4, 4, 4, 4, and 4. Therefore, by Theorem 2.35,

these 7 graphs are all 6-chromatic.

The parameter set (25,12,5,6)

There are exactly 15 srg(2 5 ,1 2 ,5 ,6 ). These were found by Paulus [59] and shown to

be all such graphs by Rozenfel'd [61]. Adjacency matrices of these graphs are available

at [76], and in what follows we w ill denote these graphs by i = 1 , . . . , 15. Now,

a srg(2 5 ,1 2 ,5 ,6 ) has smallest eigenvalue s = —3, and so it has cocliques of size no

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. greater than 25(—3 )/(—3 — 12) = 5 by Theorem 2.16. So, if such a graph were 5-

colorable, it would have to possess five pairwise disjoint cocliques of size 5. Using

GRAPE, it was shown that the only graphs with such a coloring are Gos.n and Gos.is-

Hence, there are exactly two 5-chromatic srg(25,12,5,6). Now, there is exactly one

pair of orthogonal latin squares of order 5, and so the two 5-chromatic srg(25,12,5,6)

are precisely the graphs arising from these two latin squares.

Theorem 2.37 There are exactly two 5-chromatic srg(25, 1 2 ,5 ,6), those being the

two that arise from the two latin squares of order 5.

Remark 2.38 The remaining 13 srg(25,12,5,6) that are not 5-colorable all have six

pairwise disjoint cocliques of sizes 5, 4, 4, 4, 4, and 4. Therefore, by Theorem 2.37,

these 13 graphs are all 6-chromatic.

The parameter set (45,12,3,3)

Spence [75] has done a complete enumeration of strongly regular graphs with param

eters (45,12,3,3). He found there to be exactly 78 such graphs. Adjacency matrices

of these graphs are available at [76], and in what follows we w ill denote these graphs

by (? 45,i, i = 1 ,..., 78. Now, a srg(45,12,3,3) has smallest eigenvalue s = —3, and

so it has cocliques of size no greater than 45(—3)/(—3 — 12) = 9 by Theorem 2.16.

So, if such a graph were 5-colorable, it would have to possess five pairwise disjoint

cocliques of size 9. Using GRAPE, we showed that all of these graphs except £ 45,24 ,

£45,48, and £45,71 have such a 5-coloring. Hence, there are exactly 75 srg(45,12,3,3)

with chromatic number 5.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Theorem 2.39 There are exactly 75 srg(45,12,3,3) with chromatic number 5.

Remark 2.40 The remaining three srg(45,12,3,3) that are not 5-colorable each

have six pairwise disjoint cocliques of sizes 9, 9, 8, 8, 8, and 3. Therefore, by Theorem

2.39, these three graphs are each 6-chromatic.

A partial result for the set (76,30,8,14)

Theorem 2.41 There does not exist a srg(76,30,8,14) with two disjoint cocliques

o f size 16.

Proof. Let G be a sn?(76,30,8,14). Then, by Theorem 2.4, G has eigenvalues

k = 30, r = 2, and s = —8, with multiplicities 1, / = 57, and g = 18, respectively.

Also, by Theorem 2.16, a(G) < 76(—8)/(—8 — 30) = 16. Suppose that G contains

two disjoint cocliques, say C\ and Co, of size 16. Partition V(G) into Ci, Co, and

V(G) \ (Ci U Co). Let A be an adjacency matrix of G. Then we can assume that A

has the following form:

Let

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. By Theorem 2.16, Ai has constant row sums of —s = 8. Thus, A i is the adjacency

matrix of an 8-regular bipartite graphG\ on 32 vertices. So, we must have A i(G i) = 8

and A3 o(Gi) = —8. Also, by Theorem 2.8, Ao(A) = 2 > Ao(Ai), and so all of the

other eigenvalues of G\ lie in the interval [—2,2]. Therefore, the sum of the squares

of the eigenvalues of G\ is at most 1 - 82 +15 • 22 +15 • (—2)2 + 1 • (—8)2 = 248. This is

a contradiction, since the sum of the squares of the eigenvalues of G\ must be equal

to 32 • 8 = 256. □

Corollary 2.42 A 5-coloring of a srg( 76,30,8,14) must consist of one color class

o f size 16 and fo u r color classes o f size 15.

The remaining parameter sets

The parameter sets that remain to be dealt with are numbers 13, 31, 32, 33, 35, 36.

37, 38, 39, 40, and 41. Below we make some observations about these 11 sets.

According to [8], it is not even known if a single strongly regular graph with param

eter set 13, 31, 32, 33, 35, 37, 38, or 41 exists. Parameter set 13 is the fourth smallest

set for which existence is in question, the three smallest sets being (65,32,15,16),

(69,20,7,5), and (75,32,10,16), none of which could be 5-colorable. We have Corol

lary 2.42 for parameter set 13, but this seems to be of little help. A srg(76,30,8,14)

could come from a two-graph on 76 vertices. The graph obtained by deleting three

cocliques of size 22 from a srg(99,14,1,2) would be a 2-regular graph on 33 vertices,

and so is not bipartite. Hence, a 5-coloring of a srg(99,14,1,2) can not have three

color classes of size 22. This simple observation seems to be of no help.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. By Theorems 2.16 and 2.19, a 5-chromatic strongly regular graph with parameter

set 31 must be the incidence graph of a system of 4 linked symmetric (45,12,3)

designs w ith x = 1 and y = 4. Unfortunately, these are large and have not been

enumerated like the systems of 3 linked symmetric (1 6 ,6 ,2 ) designs were in [57].

Parameter sets 33, 35, and 37 are the three smallest triangle-free sets for which

existence is in question, the fourth smallest set being (26 6,45,0,9), which we have

seen is not 5-colorable. There are only seven triangle-free primitive strongly regular

graphs known to exist, and there seems to be nothing known about parameter sets

33, 35, and 37. There also seems to be nothing known about sets 32, 38, and 41.

There is at least one strongly regular graph with parameter set 39, obtained using

a general construction of Haemers that is described in [11].

A 5-coloring of a strongly regular graph with parameter set 36 or 40 would have

to be a Hoffman coloring. Set 40 is on Haemers’s and Tonchev’s list [41] of small

strongly regular graphs that are feasible for a Hoffman coloring. It is currently the

second smallest set for which existence of such a coloring is unknown, the smallest set

being (63,30,13,15) (the set (36,15,6,6) was settled in the negative in [14]). Perhaps,

the techniques of [14] could be used to enumerate all strongly regular graphs with

parameter sets 36 and 40 that possess a Hoffman coloring. However, these graphs are

much larger and so this might not be feasible.

Parameter set 36 is also pseudo-geometric for a G Q (4 ,6), of which there is a

unique example known [60]. However, there may exist many such graphs besides the

geometric one(s). Similarly, set 40 is pseudo-geometric for a GQ(4 ,4 ), which is unique

[60]. However, such a graph need not be geometric. Indeed, there are at least two

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. such graphs known [8]. A Hoffman coloring of the geometric srg(&5,2 0 ,3 ,5 ) would

correspond to a partition of the points of GQ(4 ,4 ) into ovoids (a fan). However,

GQ(4A) does not have a fan [60].

Conclusion

In this final section, we will summarize our results thus far.

Theorem 2.43 Let G be a 5-chromatic primitive strongly regular graph. Then G is

isomorphic to one of the graphs on the following list:

(i) P(13) (the unique srg( 13,6,2,3)),

(ii) T(7) ( the unique srg(2l, 1 0 ,3 ,6 )),

(iii) Lo(5) ( the unique srg(25,16,9,12)),

(iv) T(5) ( the unique s rg (l0 ,6 ,3 ,4 )),

(v) the complement of the block graph o f the quasi-symmetric 3-(22,6,1) design (the

unique srg(77,16,0,4)),

(vi) one of the 60 srg(64,18,2,6) G ^ i (* = 15, 17, 21, 22, 23, 25, 27, 28, 29, 30,

31, 34, 35, 38, 39, 40, 41, 42, 43, 44, 45, 46, 54, 56, 57, 60, 61, 62, 63, 64, 65,

66, 68, 69, 70, 71, 74, 75, 76, 78, 84, 90, 91, 94, 95, 96, 97, 100, 108, 109, 110,

117, 118, 119, 130, 132, 133, 137, 139, 158) with coclique number less than 16,

(vii) one of the 11 srg(40,12,2,4) G4o,i (i = 1 ,..., 11) with a coclique o f size 8,

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. (viii) one o f the two srg(36, 14,4,6) G ^ i (i = 48,77) with coclique number 8,

(ix) one o f the three srg(26,10,3,4) Go6,i (i = 1 ,2 ,8 ),

(x) one o f the two srg(25,12,5,6) G ^ i (i = 11,15) arising from the two latin

squares o f order 5,

(xi) T(6) (the unique srp(15,8,4,4)),

(xii) one o f the 75 srg(4 5 ,12 ,3 ,3 ) CTis.i (i ^ 24,48,71) with a Hoffman coloring,

(xiii) Ii2(5) (the unique srg(25,8,3,2)),

or G is possibly the incidence graph of a system o f 4 linked symmetric (45,12,3)

designs (a srg(225,48,3,12)), a non-geometric srg( 85,20,3,5), or G has one of

the nine parameter sets (76,30,8,14), (196,39,2,9), (210,33,0,6), (176,25,0,4),

(125,28,3,7), (162,21,0,3), (115,18,1,3), (96,19,2,4), or (99,14,1,2).

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. CHAPTER 3 STRONGLY REGULAR VERTICES AND PARTIALLY STRONGLY REGULAR GRAPHS

This chapter is based upon the paper [28] by the author. In this chapter, we will

explore the concepts of strongly regular vertices and partially strongly regular graphs.

Introduction

A ll graphs considered will be finite and simple (no loops or parallel edges). The vertex

and edge sets of a graph G will be denoted by V(G) and E(G), respectively (edges

w ill be regarded as 2-subsets of V{G)). The complement of G will be denoted by G.

Given two vertices x and y in a graph, we use the notation x ~ y to denote that x

and y are adjacent, and x y to denote that x and y are not adjacent. If we wish

to emphasize that x and y are adjacent (non-adjacent) in a particular graph G , we

use the notation x ~ c y (x y)- If x ~ y, then the edge between x and y w ill be

denoted by xy or yx, and y is called a neighbor of x (x is also a neighbor of y). The

set of all neighbors of x is called the neighborhood of x, and is denoted by N (G ;x),

or just N (x) if there is no danger of confusion. The size of the neighborhood of x is

the degree of x, and will be denoted by dc(x) or just d (x). If x ~ y, then the number

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. of vertices adjacent to both x and y, or the number of common neighbors of x and

y, will be denoted by Ac(x, y) or just \(x ,y ). If i * y, i ^ y, then the number of

common neighbors of x and y will be denoted by pc(x, y) or p(x, y).

A strongly regular graph with parameters v, k (1 < k < v — 2), A, and p, or a

srg(v, k, A, p), is a fc-regular graph G on v vertices such that for all x, y € V (G ), x ^ y ,

A(x, y) = A if x ~ y, and p(x, y) = p ii x y. For a good introduction to the theory

of these graphs, see [18] or [33]. For a good survey of strongly regular graphs, see [11]

or [66].

In this chapter, we treat the concept of strong regularity as a property of vertices

rather than as a property of graphs as a whole. Accordingly, we make the following

definition.

Definition 3.1 Given a graph G on v vertices, a strongly regular vertex in G with

parameters A and p, or a srv(X,p) (s rv c (\,p ) if we wish to make G explicit) is a

vertex x G V(G) with 1 < d(x) < v — 2, and such that A (x, y) = A for all y ~ x,

and p(x, y) = p for all y * x, y ^ x. Two strongly regular vertices in G will be said

to have the same parameters if their parameter tuples are equal regarded as ordered

pairs in No x No, where No is the set of non-negative integers.

Clearly, a graph is strongly regular if and only if it is regular and all of its vertices

are strongly regular with the same parameters.

In the forthcoming sections, we will prove some of the basic properties of these

vertices and the graphs that contain them, as well as provide some constructions of

regular graphs that are not necessarily strongly regular, but do nonetheless possess, in

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. some cases many, strongly regular vertices. We also make some conjectures and find

all regular graphs on at most ten vertices with at least one strongly regular vertex.

Basic Properties

In this section, we will prove some simple properties of strongly regular vertices and

the graphs that contain them. We will see that several results concerning strongly

regular graphs have analogues in the context of a regular graph with a strongly regular

vertex.

A graph G is a srg(v, k, A, p) if and only if G is a srg(v, v — k — l, v — 2k + p —

2, v — 2k + A). The following proposition and its corollary show that a similar result

holds for strongly regular vertices in regular graphs.

Proposition 3.2 Let G be a graph on v vertices with a srvG(X,p), x, such that

every neighbor of x in G has degree ki in G, and every non-neighbor of x in G,

except possibly x itself, has degree in G. Then x is a srvg(v — dc{x) — k o + p —

2,v - dG{x) — kx + A).

Proof. If y x, then y * G x ,y ± x, and so A&{x,y) = |iV(G;x) D iV(G;y)| =

\N (G ;x) n i m y ) \ ~ 2 = |AT(G,x)UJV(G;y)| - 2 = v - \N(G;x) U N {G ;y )| - 2 =

w - (|AT(G;x)| + |iV(G;y)| - \N {G ;x) fl N (G ;y) |) -2 = v - dG(x) - k2 + p - 2.

Similarly, if y ^ x, y ^ x, then y ~ c x, and so /ig(x,y) = |iV(G;x) fl N (G :y)\ =

v - \N (G ; x)| - | N(G; y)| + |iV(G; x ) n N(G; y) \= v - dG(x) - kt + A. □

Corollary 3.3 A vertex x in a k-regular graph G with v vertices is a srvG(X, p) if

and only i f x is a srv^(v — 2k + p — 2 ,v — 2k + A).

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Corollary 3.4 Let G be a k-regular graph, on v vertices with a srv(X,p). Then

v — 2k + p — 2, v — 2k + A > 0 .

Proposition 3.5 Let G be a graph on v vertices with a srv(A, p), x. Then d(x)[d(x)—

A — 1] = [w — d(x) — l]p.

Proof. We will count pairs ( y,z) €. V(G) x V(G) such that x ** y *** z Xj z ^ x r

in two different ways. We can first choose y in d(x) ways, and then choose z in

d(x) - 1 — A ways, for a total of d(x)[d(x) - A — 1] such pairs. Or, we can first choose

z in v — 1 — d(x) ways, and then choose y in p ways, for a total of [u — d(x) — l]/x

pairs. □

Applying Proposition 3.5 to a regular graph with a strongly regular vertex, we obtain

the following familiar equation for strongly regular graphs.

Corollary 3.6 Let G be a k-regular graph on v vertices with a s rv (\,p ). Then

k(k — A — 1) = (v — k — l)p.

Proposition 3.7 Two strongly regular vertices in a graph G have the same degree if

and only i f they have the same parameters.

Proof. Let x be a srvc(A*, px), and let y ^ x be a srvc(Aj,, p f). Then, by Proposi

tion 3.5, we have the two equations

d(x)[d(x) — A* — 1] = [v — d{x) — 1 \px (3.1)

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. and

d{y)[d(y) - \y - 1] = [v - d{y) - 1 ]^ . (3.2)

Suppose that d(x) = d(y) = d. Then the two equations (3.1) and (3.2) become

d(d — Ax — 1) = (v — d — \)nx and d(d — Xy — 1) = (v — d — l)y y. (3.3)

If x ~ y, then clearly Ax = Xy = A, so the equations (3.3) become

d(d — A — 1) = {v — d — l) y x and d(d — A — 1) = (v — d — l)ny. (3.4)

Since v — d — 1 > 0, we can solve the equations (3.4) for fix and y.y to obtain

/xr = d(d —X — l)/(v — d — 1) = /Jy, and so x and y have the same parameters. If

x ** y, then clearly fj.x = jiy = /i, so the equations (3.3) become

d(d — Xx — I) = (v — d — l)/x and d(d — Xy — 1) = (v — d — 1 )/z. (3.5)

Since d > 0, we can solve the equations (3.5) to obtain Ax = d—(y —d—l) n /d — \ = Xy,

and again x and y have the same parameters.

Next, suppose that Ax = Aj, = A and y.x = Hy = y.. Then the two equations (3.1)

and (3.2) become

d(x)[d(x) — A — 1] = [u — d(x) — l]/z (3.6)

and

d{y)[d(y) - X - 1] = [v - d(y) - l]/i. (3.7)

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Now, if n = 0, then d(x) — A — 1 = d(y) — A —1 = 0, since d(x), d(y) > 0. This implies

that d[x) = d(y), so we can assume that /z > 0. Solving equations (3.6) and (3.7) for

d(x) and d(y) gives us the following four possible solution sets:

Since /z > 0, we have (A — p + 1 — y/(X — p + l)2 + 4/z(u — l))/2 < 0. This eliminates

all but one of the above solution sets. Thus,

, X - p + l + y j (A - p + l)2 + 4^(t; - 1) d{x) = ------=d{y),

and so x and tj have the same degree. □

Applying Proposition 3.7 to a regular graph gives us the following results.

Corollary 3.8A ll strongly regular vertices in a regular graph have the same param

eters.

Corollary 3.9A graph is strongly regular if and only if it is regular and all of its

vertices are strongly regular.

Corollary 3.10 A graph is strongly regular if and only if all of its vertices are

strongly regular with the same parameters.

The graph obtained by pasting together n > 2 copies of at a com m on vertex

x appears at first to be a counter-example to Corollary 3.10, since every pair of

distinct vertices has exactly one com m on neighbor, but the graph is not regular since

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. d{x) = 2n while the other vertices have degree two. However, x is not considered to

be a srv( 1,1) since it is adjacent to every other vertex.

In view of Corollary 3.8, the following definition makes sense.

Definition 3.11 A partially strongly regular graph with parameters v, k (1 < k <

v — 2), A, and p, or a psrg(v,k, X,p), is a fc-regular graph G on v vertices with at

least one srvc(X, p). We will sometimes call such a graph s-partially strongly regular

and use the notation s-psrg(v, k, X,p) if G has exactly s srv(A, p). If s < v, then we

say the graph is strictly partially strongly regular.

Thus, partially strongly regular graphs are a generalization of strongly regular graphs.

Clearly, a srg(v, k, A, p) is just a v-psrg(v,k,X,p). Other such generalizations of

strongly regular graphs have also been studied. For instance, in [24], the authors

study what they call Deza graphs. These are regular graphs in which the number

of common neighbors of two distinct vertices takes on one of only two values, not

necessarily depending on the adjacency of the two vertices.

The following well-known theorem says that the strongly regular graphs can be

completely characterized in terms of two matrix equations. In what follows, A denotes

a (0, l)-adjacency matrix of a graph, and I, J , and 0 denote an identity matrix, a

square matrix of all l ’s, and a square matrix of all 0’s, respectively.

Theorem 3.12 A graph, not complete or edgeless, is a srg(v,k,X , p) i f and only i f

AJ = kJ and A2 - XA - p{J - I - A) - k l = 0.

The following similar results provide a way to identify strongly regular vertices and

partially strongly regular graphs.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Proposition 3.13 A vertex x with 1 < d(x) < v — 2 in a graph on v vertices is a

srv(X, p) i f and only i f the row corresponding to x consists o f all 0 ’s in the matrix

A2 - \A - p{J - I - A) - d(x)I.

Corollary 3.14 I f G is a s-psrg(v,k, X, p), then s is equal to the number o f rows of

all 0 's in the m atrix A 2 — XA — p (J — I — A) — k l.

Constructions

In this section, we provide some constructions of partially strongly regular graphs.

C o n s tru c tio n 3.15 Let m > 1 and n > 3 be integers, and let be any

non-empty collection of non-complete (n — l)-regular graphs with v \ vertices,

respectively. Let G = m K n U (U(= li2i). Then G is clearly (n — l)-regular with

mn + Vi vertices, and the mn vertices that reside in the m copies of K n are all

srv c in — 2,0). The remaining vertices are not strongly regular in G. Thus, G is a

mn-psrg(mn + £ i = i vu n ~ 1, n — 2 ,0 ).

Construction 3.16 Let H be a srg(v, Ar, A,p) with p > 0, which is the case as long

as H is not a disjoint union of complete graphs, and with w ,x ,y ,z € V{H) such that

w x,yz 6 E(H), but wy,wz,xy,xz E(H). Let G be the graph with V(G) = V(H)

and E(G ) = (E(H ) \ { wx, yz} ) U {w y,xz}. Then G is fc-regular on v vertices, and

the vertices in \Jue{wj:^:}N(H;u)LS(N(H;w)r\N(H',z))\J(N(H;x)r\N(H;y)) will

be sruc(X,p). The remaining vertices will not be strongly regular in G.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. It is quite possible that Uu6{1UtX,y,z}iV(/f; u) = 0 in Construction 3.16. However, since

p > 0, (N (H ;w ) n N (H ;z )) U (N (H ;x ) D N (H ;y )) ^ 0. Thus, this construction

always provides at least one strongly regular vertex, and so G is a psrg(v, k, A, p).

Construction 3.17 Let k > 3 be an integer and choose k distinguished vertices

from the graph K = kKk, one from each copy of Kk. Call this set D. Create a

new vertex x and join it to every vertex in I? by an edge. Call the resulting graph

H. Now, partition the set V (K ) \ D into 2 -subsets such that each 2 -subset consists

of two vertices residing in different copies of Kk (this can always be done in many

ways). Call this partition P. Finally, define the graph G by V(G) = V(H) and

E(G) = E(H) U P. Clearly, G is fc-regular and has k2 + 1 vertices. In addition, the

vertex x is an sn>c(0, 1). The remaining vertices are not strongly regular in G. Thus,

G is a 1 -psrg(kr + 1, k, 0 , 1).

Construction 3.18 Let k > 4 be an even integer and let K = Kk+ijc- Let Ci and

Co be the color classes of K of sizes k + 1 and k, respectively. Let x be any vertex

of C i. Now, let Ci \ {x} = {!,..., Ar} and Co = (1 ',..., k1}, and let P be a partition

of Ci \ {x} into 2-subsets. Finally, let G be the graph defined by V(G) = V(K) and

E(G) = (E{H) \ { ii': 1 < i < k}) U P. Then G is fc-regular w ith 2k + 1 vertices, and

x is a srvc(0, k — 1). The remaining vertices are not strongly regular in G. Thus, G

is a l-psrg(2k + 1, Ar, 0, k — 1).

Construction 3.19 Let v, k, A > 0, and p > 0 be integers such that A: — A — 1 > 0,

k and A are not both odd, v — k — l> k — p> 0 , v — k — 1 and k — p are not both

odd, and k(k — A — 1) = (v — k — 1 )p. Let H be a A-regular graph on k vertices, and

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. let K be a (A: — ^-regular graph on v — k — 1 vertices. Create a new vertex x and

join it to every vertex of H by an edge. Now, draw a bipartite point-block incidence

graph of a l-(k , p, k — A — 1) design between H and K. Cali the resulting graph G.

Then G is a fc-regular graph with v vertices, and x is a srvc(A, p). The other vertices

may or may not be strongly regular in G. Therefore, G is a psrg(v , k, X,p).

Combining Construction 3.19 with Corollary 3.6, we obtain the following result.

Corollary 3.20 Let v, k, A > 0, and p > 0 be integers such that k — A — 1>0, k

and X are not both odd, v — k — 1 > k — p> 0 , and v — k — 1 and k — p are not both

odd. Then there exists a psrg(v, k, A, p) if and only if k(k — A — 1 ) = (v — k — 1 )p.

O f course, by Corollary 3.3, we can complement any of the above constructions to

obtain more examples of partially strongly regular graphs.

Eigenvalues

In this section, we briefly explore the eigenvalues of partially strongly regular graphs.

The eigenvalues of a graph contain much information about the graph. In par

ticular, graphs with a great deal of regularity tend to have few distinct eigenvalues.

Conversely, graphs with few distinct eigenvalues quite often exhibit much regularity

and symmetry.

The following well-known theorem says that we can recognize if a regular graph

is strongly regular simply by examining its number of distinct eigenvalues.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Theorem 3.21 A regular graph, not complete or edgeless, is strongly regular i f and

only i f it has at most three distinct eigenvalues.

The eigenvalues of a strongly regular graph are completely determined by its

parameters v, k, A, and p. However, the parameters v, k, A, and p, together with

the number s, do not determine the eigenvalues of a partially strongly regular graph.

For instance, there are two 2-psr

Although a strictly psrg(v, fc, A, p), G, must have more than three distinct eigen

values by Theorem 3.21, we w ill see that the eigenvalues of a srg(v, k, A, p) are also

eigenvalues of G. So, although the eigenvalues of a partially strongly regular graph

are not determined by its parameters, the parameters do determine three of the

eigenvalues. First, we need a lemma [37], [39].

Lemma 3.22 Let .4 be a real symmetric matrix partitioned as follows:

( •Ai.i A • • • .ni,m A ^

.4 =

\ Am,I • • • -4m

and let B — {b ij). I f A ^ has constant row sums fo r all i , j = 1,... ,m , then every

eigenvalue of B is also an eigenvalue o f A.

Lemma 3.23 Let G be apsrg(v,k,X,p). Then

A — p ± (A - p )2 + 4(fc — p) 2 are both eigenvalues o f G.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Proof. Let x be a srvc(X,p). Now, partition V(G) into three subsets, {x }, N (x),

and N {x) \ {x }. This induces a partition of the adjacency m atrix A of G, the average

row sums of which are given by

\ Ok 0

B = 1 A k-X-l

0 p k — p

The eigenvalues of B are k and (A — p ± y/(X — p)2 + 4{k — p))/2. Since x is strongly

regular, the block matrices in this partitioning of A have constant row sums. There

fore, by Lemma 3.22, these are also eigenvalues of A. □

Theorem 3.24 There does not exist a connected partially strongly regular graph with

exactly fo u r distinct eigenvalues.

Proof. Let G be a connected psrg(v, k, A, p) with exactly four distinct eigenvalues.

Since G is connected and regular, G has an eigenvalue k w ith m ultiplicity one. Also,

by Lemma 3.23, we know that

A - p. ± y/(X — p)2 + 4(fc - p) r , t =

are eigenvalues of G. Clearly, r ^ t, so let u be the one remaining eigenvalue of G,

and let /, g, and h be the multiplicities of r, t, and u, respectively.

Now we obtain the following three equations by counting eigenvalue multiplicities

and by taking the traces of A and A2.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 1 + / + g + h — v, (3.8)

k + f r + gt + hu = 0, (3.9)

and

k~ + f r 2 + gt2 + hu2 = vk. (3.10)

Since ur + (p — X)u+ p — k = 0 if and only if u = r or f, and r, t , andu are all distinct,

we can solve equations (3.8), (3.9), and (3.10) for / , g, and h to obtain

_ (v — k — 1 )n — k(k — A — 1) u2 + {fi — X)u + ii — k (we do not give the ugly expressions for / and g). However, (u—k—l)fi—k(k—X—l) —

0 by Corollary 3.6. Therefore, h = 0, a contradiction. □

Corollary 3.25 A connected graph is strongly regular i f and only i f it is partially

strongly regular and has at most fo u r distinct eigenvalues.

Corollary 3.26 A connected strictly partially strongly regular graph has at least five

disinct eigenvalues.

Applying Construction 3.15 with m = 1, n = 4, and R — { ^ 3 ,3 }, one obtains

a +psrg(lQ, 3 ,2 ,0 ) with exactly four distinct eigenvalues. Thus, the assumption

of connectedness is necessary in Theorem 3.24. Applying Construction 3.15 with

m = 2 , n = 3, and R = {C 4}, where C4 is the cycle on four vertices, and taking

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. the complement, we obtain a connected 6-psrg(10,7,4,7) with exactly five distinct

eigenvalues, all of which happen to be integers. It would be interesting to try to

obtain some sort of combinatorial characterization of connected partially strongly

regular graphs w ith five eigenvalues.

Another Result

In this section, we will prove a slight strengthening of Corollary 3.9.

Theorem 3.27 A graph on v vertices is strongly regular if and only i f it is s-partially

strongly regular with s > v — 3.

Proof. Let G be a s-psrg(v, k, A, p.) with s > v —3. Let S be a set of v —3 srvc(A, p),

and let V{G ) \ S = {x ,y , 2 } be the remaining three vertices. We must show that x,

y, and z are also s rv c {\,p )- By Proposition 3.3, it suffices to consider the following

two cases.

Case 1. Suppose that x ~ y ~ z ~ x. We must show that A(x,y) = A(x, z) =

A{y, z) = A.

We will count the number of pairs (u, u/) € V(G) x V(G) such that x ~ u ~ w

x, w # x, in two different ways. There are (k — 2)(k — 1 — A) such pairs with u ^ y , z .

There are also k — 1 — A(x, y) pairs with u = y, and k — 1 — A(x, z) pairs with u = z.

This gives us a total of (A: — 2 ){k — A — 1) + 2{k — 1) — A(x,y) — A(x, z) such pairs.

Or, we can just choose w in v — k — 1 ways, and then choose u in p ways, for a total

of (v — k — l)p pairs. This gives us the equation

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. (k — 2){k - A — 1) + 2(k — 1) — A(x, y) — A(x, z) = (v — k — 1)^. (3.11)

Similarly, counting pairs (u, w) such that y ~ u ~ w y ,w ^ y, in two ways gives

us the equation

(k - 2)(k - A - 1) + 2{k - 1) - A(x, y) - A(y, z) = (u - k - 1)^. (3.12)

Finally, counting pairs (u, w) such that z ~ u ~ w z,w ^ z, gives us the equation

(k — 2)(k - A — 1) + 2(k - 1) — A(x, z) — A(y, z) — { v - k — 1)^. (3.13)

Solving equations (3.11), (3.12), and (3.13) for A(x,y), A(x, z), emd A(y,z) gives us

A(x, y), A(x, 2 ), Ato, r) = ^ ~ k(k-I) + \(k-2)

Using Corollary 3.6, we can replace (v — k — l)/z with k(k — A — 1) in the above

expression. This gives us A(x, y), A(x, z), A(y, z) = A, and so G is strongly regular.

Case 2. Suppose that x ~ y z x. We must show that A(x,y) = A and

li{x ,z ) = fi(y, z) = fi. We will count in two different ways the number of pairs

{u,w ) 6 V(G) x V(G) such that x ~ u ~ w ** x,w x. The number of pairs with

u ^ y is (k — 1)(A: — 1 — A), and the number of pairs with u = y is k — 1 — A(x, y).

This gives us a total of (k — l)(fc — A — 1) + k — A(x,y) — 1 pairs. Also, the number

of pairs with w ^ z is (v — 2 — k)n, and the number of pairs with w = z is /x(x, -).

This gives us (v — k — 2)y.(x, z) such pairs. Therefore, we have the equation

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. (Ar — 1)(A: — A — 1) + k — A(x, y) — 1 = (v — k — 2)n(x, z). (3-14)

Similarly, counting pairs (u,w) such thaty ~ u ~ w *> y,w ^ y, gives us the equation

{k - l)(fc - A - 1) + k - A(x, y) -1 = (v - k - 2)n(y, z). (3.15)

Finally, we will count the number of pairs (u, w) such that z ~ u ~ w z,w ^ z, in

two ways. The number of pairs with w ± x, y is (v — 3 — k)fi, the number of pairs

with w = x is fj.(x, z), and the number of pairs with w = y is fj.(y, z). This gives us

a total of {v — k — 3)fj. + n(x, z) + n(y, z) such pairs. Or, we could just choose u in

k ways, and then choose w in Ar — 1 — A ways, for a total of A:(A: — A — 1) such pairs.

This gives us the equation

(v - k - Z)y, + n {x, z) + n(y, z) = k{k - A - 1). (3.16)

Solving equations (3.14), (3.15), and (3.16) for A (x,y),ii(x, z), and ft{y,z) gives us

and

/ ^ / \ A:(At — A — 1 ) — (v — Ar — 3 )^ tt(x,z),nfaz) = ------.

Using Corollary 3.6, we can substitute A;(A: — A —1 ) fo r (v — k — l)/i in the above

expression for A(x,y), and we can substitute(v — k — 1 )fi fo r k(k — A — 1) in the above

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. expression for n (x, z) and fi(y, z). This gives us A(x, y) = A and /z(x, z) = /z(y, z) = /z,

and again G is strongly regular. □

Applying Construction 3.15 with n = 3 and R = {C4}, we can obtain (t/ — 4)-

partially strongly regular graphs on v vertices for arbitrarily large v. Thus, Theorem

3.27, as weak as it is, is best possible. By Corollary 3.3, the assumption of connect

edness would not change the situation. One wonders if the assumption that both G

and G are connected would allow a stronger theorem to be proved.

Conjectures

In this section, we make several conjectures concerning partially strongly regular

graphs. The first three conjectures are very similar to Theorem 3.27.

First, we conjecture the existence of a subtractive term, larger than the three in

Theorem 3.27, for connected partially strongly regular graphs with connected com

plement.

Conjecture 3.28 There exists an integer c > 4, independent of v, such that a

connected s-partially strongly regular graph on u vertices with connected complement

is strongly regular if and only if s > u — c.

The next conjecture is a stronger multiplicative version of Conjecture 3.28.

Conjecture 3.29 There exists a constant c < 1, independent of v, such that a

connected s-partially strongly regular graph on v vertices with connected complement

is strongly regular if and only if s > cv.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Specializing Conjecture 3.29, we obtain the next conjecture, which the author’s intu

ition tells him must be true (so it probably is not).

Conjecture 3.30 A connected s-partially strongly regular graph on v vertices with

connected complement is strongly regular if and only if s > v/2.

By Theorem 3.27, a strictly s-partially strongly regular graph on v vertices must

have s < v — 4. Our next conjecture concerns the structure of such graphs that are

extremal with respect to this inequality.

Conjecture 3.31 Let G be a (v — Impartially strongly regular graph on v vertices.

Then G = mK$ U C4 or G = mKz U C4 for some integer m > 1.

Corollary 3.26 states that a connected strictly partially strongly regular graph

has at least five distinct eigenvalues. Our final conjecture is an upper bound on the

number of disinct eigenvalues of a partially strongly regular graph.

Conjecture 3.32 An s-partially strongly regular graph on v vertices has at most

2 fn /s] + 1 distinct eigenvalues.

O f course, Conjecture 3.32 says nothing when s = 1 or 2 . However, it is probably not

possible to say much about graphs with such a small s-value anyway. For instance,

there is a l-psrg(9,4 ,1 ,2 ) with all nine eigenvalues distinct.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Small Graphs

In this final section, we list all strictly partially strongly regular graphs on at most

ten vertices. The following table lists, for each of the 31 such graphs, the parameters

(v, fc, A,/z), the number s, and the number of distinct eigenvalues (L (G ) denotes the

line graph of a graph G, and C is the unique cubic graph with six vertices and girth

three).

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Table 3.1: Strictly partially strongly regular graphs on at most 10 vertices

no. V k (A,/*) s distinct eigenvalues comments

1. 7 2 ( 1 , 0 ) 3 4 integer Kz U C\

2 . 7 4 (1,4) 3 5 integer Kz U C4

3. 8 2 ( 1 , 0 ) 3 4 (2 integer) Kz U C5

4. 8 5 (2,5) 3 5 (3 integer) Kz U C5

2 5. 9 ( 1 , 0 ) 3 4 integer Kz U Cq

6 . 9 4 (0,3) 1 7 (1 integer) Construction 3.18, 3.19

7. 9 4 ( 1 , 2 ) 1 7 integer Construction 3.19

8 . 9 4 ( 1 , 2 ) 1 7 (5 integer) Construction 3.19

9. 9 4 ( 1 , 2 ) 1 7 (5 integer) complement of# 8

1 0 . 9 4 ( 1 , 2 ) 1 7 (5 integer) Construction 3.19

1 1 . 9 4 ( 1 , 2 ) 1 9 (5 integer) Construction 3.19

1 2 . 9 4 ( 1 , 2 ) 3 5 integer Const. 3.16(H = L(Kz,z))

13. 9 4 ( 1 , 2 ) 3 5 integer complement o# f1 2

14. 9 4 (2 , 1 ) 1 7 (1 integer) complement o# f6

15. 9 6 (3,6) 3 5 integer Kz U Cis

16. 1 0 2 ( 1 , 0 ) 3 5 (2 integer) KzUC7

17. 1 0 2 ( 1 , 0 ) 6 4 integer 2K3u Ca

18. 1 0 3 (0 , 1 ) 1 6 (3 integer) Construction 3.17, 3.19

19. 1 0 3 (0 , 1 ) 2 7 (5 integer) Construction 3.19

2 2 0 . 1 0 3 (0 , 1 ) 7 (5 integer) Construction 3.19

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. no. V k (A,/*) s distinct eigenvalues comments

21. 10 3 (0,1) 4 5 integer Const. 3.16H ( = L(KS))

22. 10 3 (2,0) 4 4 integer K\ U

23. 10 3 (2,0) 4 5 integer K au C

24. 10 6 (2,6) 4 5 integer K\ u Kz,z

25. 10 6 (2,6) 4 6 integer K iU C

26. 10 6 (3,4) 1 6 (3 integer) complement o f #18

27. 10 6 (3,4) 2 7 (5 integer) complement of #19

28. 10 6 (3,4) 2 7 (5 integer) complement of #20

29. 10 6 (3,4) 4 5 integer complement of #21

30. 10 7 (4,7) 3 6 (3 integer) Kz U Cj

31. 10 7 (4,7) 6 5 integer 2Kz U C\

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