Title: Bivariegated graphs and their isomorphisms
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Title: Bivariegated graphs and their isomorphisms
Physical Description: v, 93 leaves : ill. ; 28 cm.
Language: English
Creator: Riddle, Fay Aycock, 1949-
Copyright Date: 1978
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Mathematics thesis Ph. D
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Statement of Responsibility: by Fay Aycock Riddle.
Thesis: Thesis--University of Florida.
Bibliography: Bibliography: leaves 91-92.
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BIVARIEGATED GRAPHS AND THEIR ISOMORPHISMS


By

FAY AYCOCK RIDDLE













A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY





UNIVERSITY OF FLORIDA

1978

































To Dennis

















ACKNOWLEDGEMENTS


The author would like to express her sincere

thanks for the leadership and guidance of Professor

A. R. Bednarek, who has been most patient and under-

standing in the long-range completion of this effort.

His insight into the open problems in graph theory has

provided the basic topics of this work.


The author also wishes to acknowledge the

remainder of her supervisory committee: Drs. J. E.

Keesling, M. P. Hale, Jr., T. T. Bowman, and W. D. Hedges

for their personal contributions to her academic training.


The author would like to thank her family for

allowing her the opportunity to attend the superior

elementary, secondary, and undergraduate institutions that

laid the foundation for higher academic pursuits.


Finally, the author would like to thank her

husband, Dennis, for his encouragement and support during

her graduate program.


















TABLE OF CONTENTS


Page


ACKNOWLEDGEMENTS ...............................

ABSTRACT ........................................

INTRODUCTION ...................................

CHAPTERS

I BASIC DEFINITIONS AND THEORY ...........

II MATRIX REPRESENTATION AND BIVARIEGATED
HUSIMI TREES ...........................

III GRAPH ISOMORPHISMS .....................

IV AUTOMORPHISM GROUPS ....................

Automorphism Groups of a Graph and
Its Factors ........................
Permutation Graphs ...................

V GRAPHS OF SEMIGROUPS AND SEMIGROUPS
OF GRAPHS ..............................

Graphs of Semigroups .................
Semigroups of Graphs .................

VI DISCUSSION .............................

APPENDIX I: BIVARIEGATED GRAPHS ...............

APPENDIX II: BIVARIEGATED HUSIMI TREES ........

REFERENCES .....................................

BIOGRAPHICAL SKETCH .............. ...............












Abstract of Dissertation Presented to the Graduate Council
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy




BIVARIEGATED GRAPHS AND THEIR ISOMORPHISMS


By

Fay Aycock Riddle

June, 1978



Chairman: A. R. Bednarek
Major Department: Mathematics



Characterizations of bivariegated graphs, in

particular, bivariegated Husimi trees, are presented.

The relationship between graph isomorphisms and length

of irreducible cycles in a bivariegated graph is

investigated. Also investigated are automorphism groups

of bivariegated graphs and the relationship between

bivariegated graphs and semigroups.

The largest class of graphs for which graph

isomorphisms can be characterized by all irreducible

cycles being of length four is the class of Husimi

trees.


















INTRODUCTION


A graph G is a finite nonempty set V of

vertices together with a set E of unordered pairs of

distinct vertices of V, called edges. A graph is

bivariegated if the vertex set can be partitioned into

two sets of equal size such that each vertex is adjacent

to one and only one vertex in the set not containing

it. This defines a bijection f between the two vertex

sets. Consider the following question: What is the

relationship between the structure of a bivariegated

graph G and the bijection f? In particular, when is f

an isomorphism between the two graphs into which G

is partitioned? Bednarek and Sanders [2] prove that

if these two graphs are trees, then f is an isomorphism

if and only if every irreducible cycle in G has length

four. This paper extends this theorem by proving it

for a class of graphs larger than the class of trees.

Furthermore, in studying the above question, other

information about bivariegated graphs is obtained.












Chapter I presents some basic definitions and

elementary results of graph theory that are useful

in later chapters. The definitions primarily follow

the notation used by Harary [6 ].

Chapter II gives a representation of the

adjacency matrix of a bivariegated graph and presents

some characterizations of bivariegated Husimi trees.

Chapter III answers the question posed above

by extending to a larger class of graphs a theorem

relating graph isomorphisms and bivariegated graphs.

Chapter IV examines the relationship between

the automorphism groups of a bivariegated graph and its

factors and presents a result obtained by considering the

case in which the factors are equal.

Chapter V examines the relationship between

bivariegated graphs and graphs of semigroups. Then,

a theorem relating semigroups to endomorphisms of a graph

is examined for the case in which the graph is bivariegated.

The final chapter discusses the results and

questions arising from them.

In the appendices are listings of bivariegated

graphs with 2, 4, 6, and 8 vertices and listings of bivariegated

Husimi trees with 2, 4, 6, 8, 10, and 12 vertices.

















CHAPTER I

BASIC DEFINITIONS AND THEORY



Definition 1.1: A graph G is a nonempty finite set of

points (or vertices), V, along with a prescribed set E

of unordered pairs of distinct points of V, known as

edges. We write G = (V,E).


If two distinct points, x and y, of a graph

are joined by an edge, they are said to be adjacent.

We write e = {x,y} to denote the edge between x and y,

and we say that x and e are incident with each other,

as are y and e.


A subgraph of G is a graph having all of its

points and lines in G. A spanning subgraph contains

all the points of G.


Two graphs G and H are isomorphic, denoted

G C H, if there exists a one-to-one correspondence,

called an isomorphism, between their point sets which

preserves adjacency.













Definition 1.2: A walk of a graph G is a finite sequence

of points such that each point of the walk is adjacent

to the point of the walk immediately preceding it and

to the point immediately following it. If the first

and last points of a walk are the same point, we say

the walk is closed, or is a cycle (provided there are

three or more distinct points and all points are distinct

except the initial and final points). A graph is acyclic

if it contains no cycles. A walk is a path if all the

points are distinct.


A complete cycle has every pair of points adjacent.

The complete graph K has every pair of its p points
------ p
adjacent.


A graph is connected if every pair of points

is joined by a path. A maximal connected subgraph of

G is called a connected component or simply component

of G.


Definition 1.3: The degree of a point v of G, denoted

deg v, is the number of lines incident with v.


The point v is isolated if deg v = 0; it is

an end point if deg v = 1.


A graph is regular of degree n if every vertex

has degree n.














Theorem 1.4: The sum of the degrees of the points of

a graph G is twice the number of lines.


The proof of Theorem 1.4 can be found in [ 6 ].


Definition 1.5: A cut point of a graph is one whose

removal increases the number of connected components.

A nonseparable graph is connected, nontrivial, and has

no cut points. A block of a graph is a maximal non-

separable subgraph.


The notation V {v} means V minus the vertex v

and all its incident edges.


Theorem 1.6: Let v be a point of a connected graph G.

The following statements are equivalent:

(1) The point v is a cut point of G.

(2) There exist points u and w distinct from

v such that v is on every u-w path.

(3) There exists a partition of the set of

points of V {v} into subsets U and W

such that for any points u e U and w E W,

the point v is on every u-w path.


Theorem 1.6 is proven in [6 ].


Definition 1.7: A tree is a connected acyclic graph.















A Husimi tree is a connected graph in which

every block of G is a complete graph.


The length of a walk (v ,v ,... v ) is n, the

number of occurrences of lines in it. The distance

d(u,v) between two points u and v in G is the length

of a shortest path connecting them, if any.


A metric space M is called a graph metric space

if there exists a graph G whose vertex set can be put

in one-to-one correspondence with the points in M in

such a way that the distance between every two points

of M is equal to the distance between the corresponding

vertices of G. A graph is Ptolemaic if its associated

metric space M is such that for any x,y,z,w e M, the

three numbers d(x,y)*d(z,w),d(x,z)*d(y,w), and d(x,w)-

d(y,z) satisfy the triangle inequality.


A graph G is said to be weakly geodetic if for

every pair u,v of vertices of G such that d(u,v) < 2,

there is exactly one shortest u-v path.


Kay and Chartrand [ 9] prove the following:


Theorem 1.8: A graph G is a Husimi tree if and only

if G is weakly geodetic and Ptolemaic.














Definition 1.9: A graph G = (V,E) is said to be

bivariegated if G = G1 f G2 = (V 1 u V2, E u E2 u

Ef), where G1 = (V1,E1), G2 = (V2,E2), VI n V2 = 0,

f: V, V2 is a bijection, and Ef = {{x,f(x)}jx VI},

where we let {x,f(x)} denote the edge incident with x

and f(x). Therefore, a graph G is bivariegated if and

only if its vertex set can be partitioned into two

disjoint equal sets such that each vertex is adjacent to

one and only one vertex in the set not containing it.

If G, = G2' we say that G is the permutation graph of G1

(or, equivalently, of G2).

Definition 1.10: An n-factor of a graph G is a spanning

subgraph of G which is not totally disconnected and is

regular of degree n. In particular, G has a 1-factor

if it has a spanning subgraph consisting of disjoint

edges.

Sanders [13] characterizes trees with 1-factors:

Theorem 1.11: A tree is bivariegated if and only if it

has a 1-factor.

Theorem 1.12: A bivariegated graph with p vertices has

at least p/2 and at most p2 /4 lines.















Proof: Let G = G1 f G2. The minimum number of lines

G can have is when both G1 and G2 are totally disconnected.

In that case there exist p/2 pairs of points in G edged

by lines in Ef. The maximum number of lines G can have

is when both G1 and G2 are complete. In that case, each

of G and G2 has n(n 1)/2 edges, where n = p/2. In

addition, there are n edges in E f. Thus, G has a total

of n(n 1) + n = n2 = p2/4 lines.


Definition 1.13: A coloring of a graph is an assignment

of colors to its points so that no two adjacent points

have the same color. An n-coloring of a graph G uses

n colors. The chromatic number X(G) is defined as the

minimum n for which G has an n-coloring.


Clearly, the chromatic number of a tree is 2.

The chromatic number of the complete graph on p vertices

is p. Thus, a Husimi tree which is not also a tree has

as its chromatic number the length of its longest cycle.

















CHAPTER II

MATRIX REPRESENTATION AND
BIVARIEGATED HUSIMI TREES




It is the purpose of this paper to determine

some characterizations relating to bivariegated graphs.

Sanders [13 ] considers a subclass of these, the bivariegated

trees, and characterizes it. Thus, in this chapter we

attempt to further generalize the properties of bivariegated

graphs by considering a slightly larger subclass, the

bivariegated Husimi trees, and attempt to characterize

it. In the process of doing this, it became evident

that it would be extremely helpful in making our general-

izations if we were to have at our disposal a set of

examples of all bivariegated graphs with p or fewer

vertices, for some p. Thus, before we present our

observations concerning bivariegated Husimi trees,

we will first look at how the examples were generated.



Adjacency Matrices of Bivariegated Graphs

By definition, a bivariegated graph G = G1 f -

G2 with 2n vertices can be factored into graphs G1 and G2














with n vertices where each vertex in G1 = (V1,E1) is

edged with a unique vertex in G2 = (V2,E2). Thus, to

generate examples of all bivariegated graphs with 2n

vertices, one needs to consider all possible combinations

of pairs of graphs with n vertices, connected by all

possible bijections f: V1 V2. Let V1 = {ul,u2, ... ,un}

and V2 = {11v 2,...v }. For any bijection f, the vertices

of V2 can be renamed v1',V 2',...,v n' in such a way that

f(u.) = v.', i = 1,2,...,n. In light of this, one can

generate all bivariegated graphs having 2n vertices by

first considering all possible combinations of pairs

of graphs Gl and G2 with n vertices. Then, rather than

considering bijections between G1 and G2, we merely

consider all possible permutations of the vertices of

G2.

Because for rather small values of n this

becomes a tedious process, it was desirable to write a

computer program to perform the process above. The

program is based on the representation of a graph by

its adjacency matrix.


Definition 2.1: The adjacency matrix A = [a..] of a

labeled graph G with p points is the p x p matrix in

which a.. = 1 if v. is adjacent to v. and a., = 0 other-
1] 1 ] 13
wise.















Clearly there is a one-to-one correspondence

between labeled graphs with p points and p x p symmetric

binary matrices with zero diagonal. Now consider the

adjacency matrix A of a bivariegated graph. Let the

first n rows and columns of A correspond to ul,u2,...,un

and the second n rows and columns correspond to vl' ,v2' ...,

v as defined above. Partition A into 4 square sub-
n
matrices. Because f(u.) = v.', i = 1,2,...,n, it should

be clear that each of the upper right and lower left

submatrices is the identity matrix.


Now we need to generate the upper left and

lower right submatrices, corresponding to adjacency

matrices of G1 and G2, respectively. Consider the

adjacency matrix of G Because it is symmetric, we

need only to consider possible entries above its zero

diagonal. For an n x n matrix, there are 1 + 2 + ... +

(n-1) = n(n-l)/2 entries above the diagonal. Then all

possible graphs G1 can be found by generating all binary

sequences of length m = n(n-l)/2, of which there are

2m. Similarly, there are 2 possible adjacency matrices

for G2. The number of combinations of 2 different

elements from a set of p is p:/(2(p-2)!), and the number

of ways to choose two identical elements from a set of














size p is p. The sum of these can be easily shown to

be p(p+l)/2. Now, if p = 2 we have 2m- (2m+l)

adjacency matrices for G1 f G2. However, since

the representation of a graph by adjacency matrix is not

unique, this merely gives us an upper bound on the

number of bivariegated graphs with 2n vertices. Thus,

we have proven the following:


Theorem 2.2: Let G = G f G2 be a bivariegated graph

with 2n vertices. Then the vertices of G can be parti-

tioned in such a way that the adjacency matrix of G is

the 2n x 2n symmetric matrix,





I A2


with zeros on the diagonal, A1 and A2 the adjacency

matrices of GI and G2, and I representing the n x n

identity matrix.


Corollary 2.3: There are at most 2 m-(2m+l) bivariegated

graphs with 2n vertices, where m = n(n-l)/2.


Because the method outlined above generates

duplicates of the same graph, the computer program was















written to group the adjacency matrices that it

generates according to the number of non-zero entries

in each row (which corresponds to the degree sequence

of each graph). These groups were then checked by hand

for duplicates. From this we obtain the list in Appendix

I of all bivariegated graphs with at most 8 vertices.



Bivariegated Husimi Trees


Now we return to the purpose of this chapter,

that of characterizing bivariegated Husimi trees. Let

H be a Husimi tree. Obviously, if H is bivariegated,

it has an even number of vertices.


Sanders [13] proves that a tree is bivariegated

if and only if it has a 1-factor. Clearly it is true

that if H is bivariegated, then it has a 1-factor.

However, the converse is not true, as shown in Figure

II-1, below.





v 2

v3 4 5 6 7 8


FIGURE II-1















This Husimi tree has a 1-factor, consisting of the edges

{vl,v2}, {v3,v4, {v5 ,v6}, and {v7'v 8}, yet it is not

bivariegated. One can state, however, that if a Husimi

tree H has a 1-factor such that no edge in the 1-factor

is contained in a cycle, then H is bivariegated. The

proof is exactly as in the theorem by Sanders.


The-above exemplifies the following: all of

Sanders' characterizations of bivariegated trees cannot

be extended to bivariegated Husimi trees. Because of

this, an examination of the properties of the bivariegated

Husimi trees that were generated earlier was begun.

However, since there are so few bivariegated Husimi

trees with 8 or fewer vertices, little could be

generalized from these examples. Thus, it was necessary

to derive a scheme for generating examples of bivariegated

Husimi trees only. We will now present the results

obtained while studying this problem.


Proposition 2.4: Let G = G f G2 be a bivariegated

Husimi tree. If a is a cycle in G, then a must lie

completely within G1 = (Vl,E ) or G2 = (V2,E2).













Proof: Let a = (vl,v 2'...v" ,v 1. Suppose that v1 s V1

and that f(v1) = V2 E V2 is in the cycle a. Now consider

any other vertex, say v31 in a. Because G is a Husimi

tree, a is complete. Assume v3 E Vi. Because a is

complete, v3 and v2 are edged, contradicting G being

bivariegated, since v2 is also edged with v1 E V A

similar contradiction can be obtained if we assume

v3 e V2'


We also obtain the following result:


Proposition 2.5: Let G = G f G2 be a bivariegated

Husimi tree. Then both GI and G2 cannot be connected

graphs.


Proof: Suppose G = (VI,E ) is connected. Let ul,u2 E V1

and a = (u ,u ,ui ... ,ui+nU2) be a path in G1 between uI and

u2. Now f maps ul and u2 to f(u ) = v1 and f(u2 = v2

in G2 = (V2,E ). Suppose there exists a path B = (vl,vj,

vj+'....' j+m',v2) in G2 between v1 and v2. Then(ul,ui,...,

ui+nu2'v2'v j+m* ..vjvl'1l) is a cycle in G, with

vertices in both V1 and V2. This is complete because G

is a Husimi tree, contradicting G being bivariegated

(since, for example, u V1 is edged with both vI and

v2 V2)'













Corollary 2.6: Let G = G1 f G2 be a bivariegated

Husimi tree. If there is a path in G connecting the

vertices u1 and u2 in V1, then there is no path in G2

connecting the vertices f(u ) and f(u2) in V2.


Using the information gathered up to this

point to obtain examples of bivariegated Husimi trees,

it was noticed that the portions of these graphs that

are not cycles are bivariegated trees. This is for-

malized in the following theorem:


Theorem 2.7: Let G be a Husimi tree. Consider the graph

G' which consists of G minus each edge which is contained

in a cycle. Then G is bivariegated if and only if all

the components of G' are bivariegated trees.


Proof: Suppose one of the components of G' (call it T)

is not bivariegated. Clearly each component of G' is

a tree. Let W be the set of all vertices of T contained

in cycles of G. The set W is not empty, or else G has no cycles,

in which case G = G' is not bivariegated. If T is not

bivariegated, the connection of the vertices in W to cycles

in G can still not be bivariegated. This is because all

vertices in a cycle of a graph G = G1 f G2, if G

is to be bivariegated, must all be in either G or G2.

Thus, a vertex in W which is in, say G1, and which has no















"mate" in G2 contained in T does not get a mate by being

connected to a cycle. Thus, the addition of any number

of cycles to any number of vertices of components of

G' will not permit G itself to be bivariegated.


Now, suppose each component of G' is bivariegated.

Partition the vertices of a component T into two sets,

V1 and V2. corresponding to the bivariegation. Any

cycle connected to a vertex v of T will have all of its

vertices in the same set, say V1, as v. Further, each of

the other vertices of such a cycle is connected to a

bivariegated tree (otherwise, it has no mate in V2).

Because the bivariegation of a tree is unique, once one

of its vertices is assigned to V1 or V2, the remainder

of its vertices are uniquely bivariegated. Apply this

process to each cycle and bivariegated trees connected

to it, and a bivariegated Husimi tree will be obtained.


Sanders [13] proves that the bivariegation of

a tree is unique. We prove an analogous result below.


Theorem 2.8: The bivariegation of a Husimi tree is

unique.















Proof: If G is bivariegated, the components of G'

are bivariegated trees, where G' is defined as in

Theorem 2.7. Now consider any cycle of G. Suppose

the bivariegation places its vertices into, say VI.

Each vertex of that cycle must be connected to a

bivariegated tree or both a bivariegated tree and a cycle,

or else it cannot be adjacent to a vertex in V2. Now,

once the bivariegation of a single vertex of a tree is

determined, the bivariegation of the remainder of its

vertices is uniquely determined. This tree may or may

not be connected to other cycles, but if it is, the

bivariegation of the cycles attached to it is uniquely

determined by the bivariegation of the vertex of the

tree to which it is attached. Therefore, because (i)

a bivariegated Husimi tree consists of bivariegated

trees and cycles, (ii) a tree is uniquely bivariegated,

and (iii) each cycle is uniquely bivariegated according

to the bivariegation of the vertex of the tree to which

it is attached, the entire Husimi tree is uniquely

bivariegated.


We now see that if a bivariegated Husimi tree

contains a cycle, then each vertex of that cycle is

attached to a bivariegated tree. Further, if it contains














more than one cycle, the block between any two cycles

is also a bivariegated tree. This motivates the use of

the following for constructing all bivariegated Husimi

trees: Sanders [13] constructs all bivariegated trees

by taking the smallest bivariegated tree (two vertices

connected by an edge) and successively adding remote

end vertices. A remote end vertex is defined as

follows:


Definition 2.9: Let e be an end point of graph G, and

let x be the unique point adjacent to e. If deg x = 2,

then e is called a remote end vertex.


We can construct all bivariegated Husimi trees

by taking the smallest bivariegated Husimi tree (which

is also the smallest bivariegated tree) and attaching,

one at a time, to a vertex a graph of one of the forms

in Figure 11-2, below.


FIGURE 11-2













That is, attach to the vertex either a remote end vertex

or a complete n-cycle with n-1 end points, where the

attachment is made at the circled vertex. Thus, we

obtain the following theorem.


Theorem 2.10: Let G be a bivariegated Husimi tree.

Construct G' as follows: let v be any vertex of G.

To v, attach either a remote end vertex or a complete

cycle such that the vertices other than v are edged

with an end vertex. Then G' is a bivariegated Husimi

tree.


Proof: Suppose G = G1 f G2 where GI = (V1,E2).

Suppose v s V1 and a remote end vertex is attached at

v. Then the remote end vertex can be added to V2 to

form the set V2', and the vertex x (as in Definition

2.9) can be added to V1 to form the set Vl'. Clearly

G' = G f G2', where GI' = (V1',E,') and G2' =

(V2',E2') is a bivariegated Husimi tree.


Now consider the case where a cycle of the

type described above is added to v c VI. Add every other

vertex in that cycle to V1 to form the set VI', and add

the end vertices attached to this cycle to V2 to form

V2'. Then G' = Gi' f G2' is a bivariegated Husimi tree.















Appendix II gives the results of generating

all bivariegated Husimi trees with 2, 4, 6, 8, 10, and

12 vertices by using the method in Theorem 2.10.


As Sanders [13) points out, the technique of

adding remote end vertices may produce duplicates of

a particular graph. The same is true for the method

outlined above, as shown in Figure 11-3.


FIGURE 11-3

















CHAPTER III

GRAPH ISOMORPHISMS




The contents of this chapter are the results

of attempts to extend to a larger class of graphs a

theorem relating graph isomorphisms and bivariegated

graphs.


Definition 3.1: Let a graph G have vertex set V and

edge set E. A finite sequence, a = (v ,vlv 2..."' n,

of distinct vertices of G is called a path provided

{v ,v. } I E for i = 0,1,2,... ,n-l. A cycle is a

sequence a = (v0,v, .... v n-_v n) such that v0 = vn

and (v0,v1 ....,v nl ) and (vl,v2 ....,v n) are paths.

A cycle is irreducible if it contains no diagonals;

that is, for v.,v. vertices of a, {v.,v.}E E implies

li-jl = 1 or n-l. If a = (v0,vI .... Vn), then the

length of a is n.

Bednarek [ 1 ] characterized tree isomorphisms

in terms of bivariegated trees:

Theorem 3.2: If T1 = (V1,E ) and T2 = (V2,E2) are

trees and if f: V, V2 is a bijection, then f is









23

an isomorphism if and only if all irreducible cycles

in T1 f T2 have length four.

One may wonder whether or not we may replace

"trees" in the preceding theorem by some larger class

of graphs. Bednarek [ 1] gives an example to show

that "trees" may not be replaced by "connected graphs."

This is demonstrated by Figure III-1, in which G1

and G2 are squares and f the isomorphism given by

yi = f(x ), i = 0,1,2,3, and in which


a = (xO,xl,x2',2,y3,y ,x0)


is an irreducible cycle of length six.


xo x1






Y y yl


y3 Y2


FIGURE III-1


It is one of the aims of this chapter to find

the largest class of graphs for which "trees" in

Theorem 3.2 may be replaced.















It can be proven that "trees" may be replaced

by "Husimi trees":


Theorem 3.3: If T1 = (V1,E1) and T2 = (V2,E2) are

Husimi trees and if f: V1 V2 is a bijection, then

f is an isomorphism if and only if all irreducible

cycles with edges in both T1 and T2 have length four.


Proof: Suppose f is not an isomorphism. Then either

there is an edge {x,y} E E for which {f(x),f(y)} / E2

or there exists an edge {s,t} c E2 such that {f (s),

f- (t)} / E1. Considering the first case we see that

because {f(x),f(y)} / E2, then f(x) and f(y) are not

in the same block of V2 since every block in a Husimi

tree is complete. So V2 is separable, and we can

decompose it into blocks B1,B2,.. .,B k. Let B1 be the

block such that f(x) E B1 and Bk the block such that

f(y) E Bk. Then there is a path a = (f(x),tl,...,t ,

f(y)) from B1 to Bk between f(x) and f(y) where each

ti, i = 1,2,...,n, is a cutpoint of a block through

which a passes. Because each block is complete, there

is only one edge between each cutpoint of a particular














block. Now consider the path


a' = (x,f(x),tl, .... tn,f(y),y,x).


This is a cycle of length greater than or equal to

5 and irreducible because of the way a was chosen.

To prove the necessity of the condition,

suppose f is an isomorphism. Let a = (v0,v 2,v"..,v k)

be the largest irreducible cycle in T1 f T2 with

edges in both T1 and T2. Let v0 = vk V1 and vk-1 E V2.

Thus f-1 (vkl) = vk = V0 e V, so f(vk) = f(v0) = Vkl.


We claim that Vk-2 e V2. For if not, then

Vk-2 C VVk-l C V2, and f(vk-2 = Vk-1 However,

this is impossible since f(v0) vk-l' contradicting

the one-to-oneness of f.

We next claim that v1 e V For if not, then

v1 E V2,v0 E V1, and f(v0) = vI. However, this

contradicts the functionality of f, as f(v0) = vk-

also.

We also claim that Vk-3 C V2. For if not,

then Vk-3 E Vlv k-2 C V2, and f(vk-3) -= Vk. However,













since {Vk-r' Vk-2 is an edge, {f- 1 k- f-(vk-2) =

{v0,v k-3 must also be an edge.

Finally, we claim that v2 E V For if not, then

v2 E V2 and f(v ) = v2. However, since {v0,v 1 is an

edge, {f(v0) f(v )1 = {v v is also an edge,

contradicting the irreducibility of a.

Now, consider f-1 (vk-2 ),f 1 (vk-3) V1 and

f(vl) ,f(v2) c V2. Construct the cycle

a' = (v0,vk ,f(vl),f(v2),v2,v3,... ,v k-3,

f-1(vk-3 ),f-1vk2),v0).

The length of a' is equal to the length of a plus two.


We claim that a' is irreducible. To prove this,

we will check each vertex in a' and show that it is adjacent

to no vertices in a' other than those immediately preceding

and succeeding it in the definition of a'.


Clearly v0 is not adjacent to either f(v1),

f(v2) ,v2 .. ,vk-3. Also {v0,f (vk-3)} is not an edge,

for if it were, {vk-lv k-3} would also be an edge,

contradicting the irreducibility of a.














Clearly, Vk-1 is not adjacent to either f(v2)

(since that would imply {v0,v2} is an edge), v2,...,

or vk-3. Referring to Figure III-2, it is clear that

vk-1 is not adjacent to either f- k-3) or f- (vk-2 '

as these vertices are in V.l


Also, f(vI) is not adjacent to either v2,v 3...,

or vk-3 (otherwise, there would be a cycle in V2, forcing

{vkl' ,v k-3} to be an edge, contradicting the irreducibility

of a). Since f-1 (vk-3) and f- (vk-2) are in V1 f(v1

is not adjacent to either of these.


As for f(v ) above, f(v2) is not adjacent to

either v3 .. k-3' 1 (vk-3), or f- (vk-2).


By definition of a, v2 is not adjacent to

either v4,., or vk-3. Also, {v2,f (vk-3) is not

an edge, for if it were, then


6 = (v0lv2' f- (vk3) ,f-(vk-2) -1 ) = vk = v0)


is a cycle in V. Since V1 is the set of vertices of

a Husimi tree, S is a complete cycle, forcing {v0,f-1

(v k-3)} to be an edge. However, we have shown earlier that

this cannot be an edge.












































Vk-2


FIGURE 111-2


















Similarly, {v2',f (vk-2)} is not an edge, for

if it were, then


V- = (v0vl,v2f (-1 k-2),f -1 Vk) = 0)


would be a cycle in VI. Then, completeness of cycles

in T1 = (V1,E-) forces v0,v 2} c E contradicting

the irreducibility of a.


By definition of a, it is clear that none of

v3" ... vk-3 is adjacent to another. Also, none of

v3 .... Vk-3 is adjacent to f- (vk-3) since f-1 (vk-3)

is only adjacent to vk-3 and f -l(vk-2). By the same

reasoning, none of v3,..., k-3 is adjacent to f-1 (v-2).


Therefore, a' is irreducible, contradicting our

definition of a as being the largest irreducible cycle.

Then it must be true that all irreducible cycles in

T1 f T2 have length four.


In trying to extend the theorem to a larger

class of graphs, it was noted that much of the proof

above depends on the existence of diagonals, that is,














edges {vi,v } in a cycle (v0,vI ... ,v = 0) such that

either li-jl i 1 or n-l. There is a characterization

of Ptolemaic graphs in terms of number of diagonals,

proven by Howorka [ 8 ] :


Theorem 3.4: A graph G is Ptolemaic if and only if

every n-cycle of G where n = 2k + r for r = 0,1, has

at least 3k +- 2r 5 diagonals.


It was conjectured that "trees" in Theorem

3.2 could be replaced by "Ptolemaic." However, Figure III-3

shows a counter example. Both G1 with V1 = {xl,x2,x31x4}

and G2 with V2 = {y1,y2,y3,y4} are Ptolemaic since their

four cycles have 1 diagonal. However, a = (xl,x4,x3,y3,

y2',y,x1) is an irreducible cycle of length six, and the
map f(x ) = y., i = 1,2,3,4, is an isomorphism.


The following theorem gives the additional

requirement in order for Theorem 3.2 to hold for Ptolemaic

graphs:


Theorem 3.5: If G1 = (V1,E1) and G2 = (V2,E2) are

Ptolemaic graphs such that every cycle of length four

induces a complete subgraph, and if f: G, G2 is a























































FIGURE III-3


xl


X3


Y3


Yf















bijection, then f is an isomorphism if and only if all

irreducible cycles in G1 f G2 with edges in both

G1 and G2 have length four.


Proof: The sufficiency of the condition is proved as

in the proof of Theorem 3.2.


To prove the necessity of the condition, we

construct a cycle a' exactly as in the proof of Theorem

3.2. By appeals to (i) the irreducibility of a, and

(ii) the one-to-oneness of f, we can prove that neither

v0,vk-l f(v ), nor f(v2) is edged with another vertex

in a' as to contradict the claim that a' is irreducible.

Similarly, v2 is not edged with either v3,v4,..., or

Vk-3'

If {v2,f (v k-3)} were an edge, then (v0,vl,

v2f1 (vk-3 ),f1 (vk-2 ),v0) is a five-cycle in G Since

G1 is Ptolemaic, there must be at least three diagonals.

Possible diagonals are (i) {vf- (vk3) i, (ii) {vi

f-1 k-3), (iii) {vlf1 (vk2), or (iv) {v2, f1 (vk-2

However, case (i) cannot be a diagonal, or else it would

force {vk-l ,k-3} to be an edge. Therefore, the other














three cases must be the diagonals. This gives us a

cycle (v,vv 2',f-1 (vk-2 ),v0) of length four, which by

our hypothesis, must be complete. This is a contradiction

because it would force {v',v 2 to be an edge.


Also, {v2,f (Vk-2)} is not an edge, for if

it were, we would have a cycle (v0,v,v 2,f (v k-2),

f- (k- )) of length four, again with completeness

forcing {v0,v21 to be an edge.


Similarly, none of v3, .. vk-3 is adjacent to

any vertex of a' such that there would be a contradiction

to the irreducibility of a'. Thus, a' is irreducible,

contradicting a being the largest such cycle. In view

of this, the theorem is proved.


It turns out that Theorem 3.3 and 3.5 are

equivalent because of the following theorem:


Theorem 3.6: G is a Husimi tree if and only if

G is Ptolemaic and every cycle of length four induces a

complete subgraph.


Proof: If G is a Husimi tree, then every cycle is

complete and satisfies the diagonal requirement for

being Ptolemaic.















Conversely, suppose G is Ptolemaic and every

cycle of length four induces a complete subgraph.

Let P (n) be the proposition that every cycle of length

n induces a complete subgraph. Clearly, P(4) is true.

Now suppose that P(m) is true for m = 5,...,n-l.

Consider a cycle a = (vl,v 2,...,v ,v1) of length n.

There is at least one diagonal, and we may assume

without loss of generality that this is the edge

{vlVk }. Divide the vertex set of a into two subsets:


A = {v ,v2,...,vk} and B = {vk ... n'v 1}


By the inductive hypothesis, each of these sets is

complete. We claim that there is an edge between any

v. E A and v. E B. For consider the cycle (v ,vl,v,vkv i)

of length four. By completeness, we see that {v.,v.}

is an edge. Thus P(n) is true, and therefore, P(m)

is true for all m greater than or equal to 5.


The following theorem proves that the largest

class of graphs for which Theorem 3.2 holds is the

class of Husimi trees:













Theorem 3.7: Suppose G1 = (VI,E ) and G2 =(V2,E2)

are connected graphs and f: V,1 V2 is a bijection.

Further, suppose that f is an isomorphism if and only

if all irreducible cycles in G1 f G2 with edges in

both G1 and G2 have length four. Then G1 and G2 must

be Husimi trees.


Proof: We will show that if G1 and G2 are not Husimi

trees, then f being an isomorphism is not equivalent

to the requirement on the length of irreducible cycles.


Suppose G1 is not a Husimi tree and that

(a v ',v,. ,v n) is a cycle in G1 with at least one

missing diagonal. Without loss of generality, assume

that v0 is one of the vertices of a missing diagonal in

a. Let j be the smallest index such that {v0,vj} is not

an edge. A shortest path from v0 to v is (v0,vj-_,Vj),

since clearly {v0,vj. _l} El or else a contradiction of

the choice of j would exist. Now, consider a shortest

path from f(v ) to f(vn) of increasing index. The

length of 8 is at least two since {f(v ),f(vn)} Z E2.

No diagonals exist, or else the shortness of the path

would be contradicted. Consider the cycle


', = (v0,v j ,vj,f(vj) ,..,f(vn),vn














where the portion of a' between f(v ) and f(v n) is

exactly as in S. Because f is an isomorphism, v. 1

is edged with no vertex in 8, v0 is edged only with

f(v) = f(0) in 8, and v. is edged only with f(v.)

in B. Thus a' is an irreducible cycle, and its length

is 6.


So far, we have only proven variations of

Theorem 3.2 for finite graphs. The theorem may be

expanded to the case of infinite trees:


Theorem 3.8: If TI = (V-,EI) and T2 = (V2,E2) are

infinite trees and if f: V1 + V2 is a bijection, then

f is an isomorphism if and only if all irreducible

cycles in T1 f T2 have length four.


Proof: The sufficiency of the condition is proven as

in the proof of Theorem 3.2. The necessity of the

condition is proven differently than in that proof because

its use of a longest irreducible cycle is inapplicable in

the case of infinite trees.


To prove necessity, suppose f is an isomorphism.

Let a = (v,v2,... ,v n) be an irreducible cycle in T1 f T2

of length greater than 4. Some of the vertices are in

VI, and some are in V2. Let v1 = vn a V1 and f(vI) = vn-l.














Write a as

(V ...,vnl,vnl+ n2' .'. v 1,v n.+l -'-,vnk ,v )
v1 ~-1 "2 k-l nk 1

6 V C V2 6 V2

where n1 / 2 and f(vI) = v Consider the closed walk
k
a' in V2, where


a' = (f(v ) ,...,. ,f(vn ),v nl+2 .... n2 f(vn2+2) ,...,vnk)

1 2- 2
where f(v ) = v n.+l for i odd, v = f (v n.+l) for
1 1 1 1
i even, and f(v1) = v
nk

Claim: The length of a' is greater than two. Note that v3

is not in V2 because if it were, then f(v2) = v3. However,

since f is an isomorphism and {vlv 2} E1, it would imply

that {f(v1),f(v2)} = {vn-' ,v3} E2, contradicting the

irreducibility of a. Thus vl,v21V3 are all in VI, and


a' = (f(vl),f(v2),f(v3) ... f(vl))


has length greater than 2 because f(v3) and f(v1) do not

form an edge (by the one-to-oneness of f).


Claim: The walk a' is a trail. For if not, there is

an edge e = {x,y} that is repeated in a'. If e is an














edge in a and there do not exist v. and v. in a such

that f(v ) = x and f(v ) = y, then e is repeated in a,

contradicting the fact that a is a cycle.


So there exist v. and v. such that f(v.) = x

and f(v.) = y. But this would imply that {v.,v.} is

an edge in a, contradicting the irreducibility of a
-1
since x and v. are in a and {x,f (x) = v.} is an edge

in T1 f T2. Thus, e = {x,y} is an edge in a', not a.

So there exist v. and v. in V such that f(v.) = x

and f(v.) = y. If e is repeated, then either (i)

{v.,v.} is repeated in a (but this would contradict a

being a cycle), or (ii) two edges in T1 being mapped

to e (contradicting f being one-to-one).


Therefore, no edge in a' is repeated, and

thus a' is a trail. Moreover, a' is a circuit. This

means that a' contains a cycle, contradicting the fact

that T2 is a tree. This completes the proof.


One can define an equivalence relation R on

vertices of bivariegated trees:















Definition 3.9: Let x and y be vertices in T1 f T2'

and define x = y modulo R if and only if x = f(y), y = f(x),

or x = y.


This equivalence relation identifies each

vertex in T1 with its image in T2 such that one can

consider T1 collapsing onto T2 to form a new graph

T1 f T2/R.


If we let E be the set of edges in T1 f T2/R

we can define adjacency as follows:


Definition 3.10: Let [x] and [y] be the equivalence

classes of the vertices x and y, respectively, of

T1 f T2 under R. Then {[x],[y]} E E if and only

if there exists a p e [x] and a q [y] such that {p,ql

E E16E2 where E1 and E2are the edge sets of T1 and T2,

respectively.


Using this equivalence relation, we can

prove another version of Theorem 3.2:


Theorem 3.11: If T1 = (VI,E ) and T2 = (V2,E2) are

trees and if f: V,1 V2 is a bijection, then f is an

isomorphism if and only if T1 f T2/R is a tree.














Proof: Suppose f is an isomorphism. We want to show

that T1 is isomorphic to T1 f T2/R. For every

v.i VI, define g(v ) = [v ]. This is clearly a one-

to-one and onto mapping. If {x,y} c El, then

{g(x),g(y)} = {[x] [y] } E. Also, if {g(x),g(y) }

E, there is a p e [x] and a q c [y] such that {p,q}

c E 1E2. If {p,q} E then p = x and q = y, so

{x,y} c E1. If {p,q} e E2, then since f is an isomorphism,

{f-1 (p)},f (q)} = {x,yl} E1. Therefore, g is an

isomorphism, and T1 f T2/R is a tree since it is

isomorphic to one.


Now suppose f is not an isomorphism. Assume

that {x,y} E E, but {f(x),f(y)} / E2. In T1 f T2/R

this corresponds to {[x],[y] } E and the existence of

a path a from [f(x)] = [x) to [f(y)] = [y] of length

greater than one, where


a = ([f(x)],tl ... tn, [f(y)] [f(x)]).


No t. = [x] or [y], so a is a cycle, and, therefore,

T1 f T2/R is not a tree.

















CHAPTER IV

AUTOMORPHISM GROUPS




In this chapter we examine the relationship

between the automorphism groups of G = G1 f G2

and its factors and, in conclusion, quote a result

obtained by considering the case in which G1 = G2'


Automorphism Groups of a Graph and Its Factors

Definition 4.1: An automorphism of a graph G = (V,E)

is a permutation h: V V such that if {u,v} E E, then

{h(u),h(v)} c E. The set of all automorphisms of G

forms a group, which will be denoted Aut(G).


It is an interesting problem to look at the

relationship between the automorphism groups of the

bivariegated graph G = G f G2 and its factors,

G1 = (V1,E1) and G2 = (V2,E2). For, suppose h is an

automorphism of one of the factors of G, say G1. Then,

unless h is the identity map, h maps at least one vertex,

say u, to another. Because G is bivariegated, u is

edged with some vertex v in G2. Now, in order














for h to correspond to an automorphism on the entire

graph G, h must similarly move v. This motivates the

following theorem:


Theorem 4.2: Let g be any automorphism of one of

the factors of G = G1 f G2 (without loss of

generality, the factor GI). If g is an automorphism

on G that agrees with g on G1 (that is, g restricted

to G1 is identically g), then it must be true that

for any vertex v c V 2'


g(v) = f(g(f-l(v))).

Proof: If g is an automorphism of G, then it preserves

edges in G. Then, because G = (V,E) is bivariegated,

for any v c V2, there exists a u c V1 such that f(u) = v

and {u,v} c E. So it must also be true that {g(u),

g(v)} E However, since g restricted to G1 is iden-

tically g, g(u) = g(u), and g(v) e V2 (since all vertices

in V1 are already images of some vertex under the

map g). But g(u) e V1 is edged with only one element

of V2, namely f(g(u)), because of the bivariegation.

Thus,


g(v) f(g(u)) = f(g(u)) = f(g(f-1(v))).














This theorem suggests that if we are trying

to relate automorphism groups of the factors of G =

G1 f G2 to Aut(G), then we can extend the auto-

morphisms of the factors of G to ones on G itself and

examine whether or not these extensions are automorphisms

of G. Furthermore, as suggested by Theorem 4.2, the

extension of an automorphism g on G1 (or G2) to a

possible automorphism g of G is uniquely determined

by the bivariegation.


To obtain an understanding of this problem,

it is helpful to look at a few examples. In all the

examples, let V = {1, 2' ,v3} and V2 = {v4,v ,v 6.

For a mapping g, the notation (iI i2 3 ... i n) means

that g(v. ) = v. g(v )= v. ,..., and g(v. ) = v. .
1 2 2 3 n 1
If a particular subscript, say i., is omitted, then

assume g(v. ) = v. Further, the notation id will
1. 1.
refer to the identity map.


Example 4.3: Consider the bivariegated graph in

Figure IV-1.













V2


vl 2 v 3



v4 v6


v 5


FIGURE IV-1


Aut(G1) = {id,(l 3)}. Clearly, the identity on G1

induces the identity map on G. Now let g be the

automorphism (1 3). Theorem 4.2 forces g to be the

map (1 3) (4 6). It should be clear that Aut(G) consists

only of g as defined above and the identity. Thus,

Aut(G ) induces all the automorphisms in Aut(G).

Similarly, Aut(G2) induces Aut(G).


Example 4.4:


In the graph below,

v
2


FIGURE IV-2















Aut(G1) = {id, (1 3)}. If g is the automorphism (1 3)

on Gi, it forces the automorphism g on G to be (1 3)

(4 6). This, together with the identity, is Aut(G).

Thus, Aut(G) is completely determined by Aut(G ).

However, Aut(G2) = {id, (5 6), (4 5), (4 5 6), (4 6 5),

(4 6)} = S3, the symmetric group on three elements.

Only one of these, (4 6), induces an automorphism on

G.


Example 4.5: Consider the graph in Figure IV-3.

v
V
2






v4 v6


v5

FIGURE IV-3


Aut(CG) = {id,(l 3)}. As in the previous examples,

these induce the identity and the automorphism (1 3)

(4 6) on G. However, Aut(G) = {id, (l 3)(4 6), (1 4)

(2 5) (3 6), (1 6) (2 5) (3 4)}. Thus, although every

automorphism on GI, and similarly G2, induces one on














G, Aut(G ) does not completely determine Aut(G).


Example 4.6:


Consider the graph in Figure IV-4:












v4 v6
vI v3






5

FIGURE IV-4


Aut(G2) = {id, (4 5)}. These induce the identity and

the automorphism (4 5) (1 2) on G. However, Aut(G) =

{id,(4 5) (1 2),(3 6), ( 2)(4 5)(3 6)}, so Aut(G2)

does not completely determine Aut(G).


Now consider Aut(GI) = S3. Only two of these

six automorphisms, the identity along with (1 2), induce


maps on G which are automorphisms.












Example 4.7: Consider the graph in Figure IV-5:

V2










v
5


FIGURE IV-5


Aut(Gl) = {id,(2 3)}. This forces the identity and

the map (2 3)(5 6) on G. This is not an automorphism,

however, since {v4,v5} 5 EE but {g(v4),g(v5)} =

{v4,v6} / E. Thus, only the identity on G induces

an automorphism on G.


The examples above point out the following:

in some cases, the automorphism aroup of a factor of G

completely determines Aut(G); in some cases, the

automorohism qroup of a factor determines some (but

not all) of the automorphisms of Aut(G). In other

instances, only the identity induces an automorphism

on G.














The following theorem generalizes when one

of the above cases occurs.


Theorem 4.8: Let G = G1 f G2 be a bivariegated

graph with 2n vertices. If Aut(G2) (or equivalently

Aut(G )) is Sn, then every automorphism of G induces

a unique automorphism of G.


Proof: It is clear that the automorphisms must be

unique, for any map so induced according to Theorem

4.2 must be uniquely defined by the structure imposed

on G by the bivariegation. So, let g c Aut(Gl). Extend

g to g on G by defining for v e V2, g(v) = f(g(f (v))).


Claim: The map g preserves all edges of G. Clearly it

preserves edges in G1 since g restricted to G1 is defined

to be g. Now, consider any u E V1 and v e V2 such that

f(u) = v. Then since {u,v} s E, {g(u),g(v)} must be

an edge. But g(v) = f(g(f-l(v))) = f(g(u)), and by

definition of bivariegation {g(u),g(v)} = {g(u),f(g(u))} =

{g(u),f(g(u))} E E. Thus, g preserves edges joining

G1 and G2. Finally, g preserves edges in G2. For,

consider {s,t} c V2. Since Aut(G2) = Sn', any permutation

of elements in G2 preserves edges in G2. In particular,

g does. Thus, {g(s),g(t)} E V2.













It should be clear at this point that if Aut(G) =

Sn and Aut(G ) 7 Sn, then every automorphism of G2 does

not induce an automorphism of G. Example 4.4 exhibits

this. It should also be clear that Aut(G ) (or equivalently

Aut(G2)) is S if and only if GI is either complete or

totally disconnected.


If one is to attempt to answer the question of

when an automorphism of G1 extended to G is not an

automorphism of G, one can notice the following: when-

ever there is an automorphism g of Gl such that for

s,t e VI, {f(s),f(t)} s E but {f(g(s)),f(g(t)) } E,

then g is not an automorphism of G. For example, in

the graph in Example 4.4, let g be the automorphism of

G2 corresponding to (4 5 6). Then for v5 and v6'

{f-1 (v5 )f -(v6)} {v2,v3} E but {f-1 (g(v5)) f1

(g(v6)) = {f- 1 v 1f-l(14)} v {V3,V} 4 E.


Figure IV-6 provides a table of all bivariegated

graphs with six or less points and their automorphism

groups. The automorphism group of G1 or G2 is of type I

if only the identity induces an automorphism on G, type

II if every automorphism induces an automorphism on G,

and type III if the automorphisms induce all of the









50




automorphisms on G. Further, the automorphism group

of G1 or G2 is of type IV if it contains at least one

non-identity element that induces an automorphism on

G and at least one element which does not induce an

automorphism on G. For each graph, the top half

represents G1 and the bottom half G2.









Si, II
Aut(G) = S2
S1, I


S 2' III
Aut(G) = S2[E2]
S2, III


I ] 3' II
Aut (G)
S3' II


= group of
order 48


S + El, I
Aut(G) = S2[E4]
S2 + El, I
S2 + El, II
Aut(G) = group of
S2 + E, II order 16


S2 + El, I
Aut(G) = S2 [E4]
S2 + El, I


S2 + El, II
Aut(G) = K4
S2 + E II


S2 + El, III
Aut(G) = S2 + E4
S3' IV
S3, II
Aut(G) = D6
S3, II


0



0



0


S3, IV
Aut(G) = K4
S2+ E1, II
S3, IV
Aut(G) = S2[E4]
S2+E1 III
S2 + El, III
Aut(G) = S2[E4]
S2 + El, III
S III
Aut(G) = group of
S3, III order 6
S2 + El, I
Aut(G) = K4
S2 + El, I

S2 + E III
Aut(G) = S2 [E4]
S3, IV


FIGURE V-6


S2' II
Aut(G) = D4
S2, II
S2 II
Aut(G) = D4
S2, II


w


0


Li-


I'
















Permutation Graphs


It appears that in order to obtain a feel

for the automorphism groups of a bivariegated graph

and its factors, one needs to examine this problem

by looking at a smaller class of bivariegated graphs.

We will conclude this chapter by looking at some

results obtained by doing this.


Rather than examine the automorphism groups

of the general class G1 f G2 of bivariegated graphs,

M. Borowiecki [ 3 1 considers the case where G1 = G =

G. First, he makes the following definitions:


Definition 4.9: Let G be a graph whose vertices are

labeled vl,...,vn and let a be a permutation on the

set {l,...,n1. Then by the a-permutation graph P, (G)

of G is meant the graph consisting of two disjoint,

identically labeled copies of G, say G and G', together

with n additional edges e., 1 < i < n, where e. ioins

the vertex labeled v. in G with v () in G'. For

clarity, we label a vertex G' as v.'. Denote G = (V,E)

and G' = (V',E'). Let a* be the permutation of V such

a*(v.) = v. if and only if a(i) = j. We define the














mapping a: V V' by a(v.) = v.' if and only if

a(i) = j.


Thus, P (G) corresponds to G f G, with a

corresponding to f. Now, Borowiecki states the following

problem of Frechen [5 ]: Consider the class H(P,G)

of all permutations a* which preserve a property P

of a graph G under transformation from G to Pa(G).

Determine the properties P and graphs G for which

(a) H(P,G) is a subgroup of S n, n = IVI

(b) H(P,G) = Aut(G).

Borowiecki considers the above problem for the

chromatic number X (G) of a graph G and proves the

following:


Theorem 4.10: If X(G) > 2, then H (X,G) is a group.


Corollary 4.11: If G is not totally disconnected,

then Aut(G)c H(x,G)C S n
n

















CHAPTER V

GRAPHS OF SEMIGROUPS AND SEMIGROUPS OF GRAPHS




A metric characterization of the graph of

a semigroup prompted an examination of the relationship

between bivariegated graphs and graphs of semigroups.

Then, a theorem relating semigroups to endomorphisms

of graphs motivated a search for a connection between

this theorem and bivariegation.



Graphs of Semigroups


In the following, let S represent a semigroup,

and S*the set of proper subsemigroups of S.


Definition 5.1: The graph E(S) of a semigroup S is

the graph whose vertices are the proper subsemigroups

of S with two of these, say, A and B, edged if and

only if AnB / 0.


Example 5.2: Suppose S is the semigroup given by

the multiplication table














a b

a a a

b b b


Then S* = {{a},{b}}, and E(S) is a graph consisting of

two disconnected vertices.


Example 5.3: Suppose S is the semigroup given by


a b c

a a a a

b b b b

C c c c

Then S* = {{a},{b},{c},{a,b},{a,c},{b,c}}, and E (S)

is the graph in Figure V-1.



{b}




{a,b} {b,cl




{a} ac. f{c}


FIGURE V-I















Bosak [4 ] proved some results concerning

graphs of semigroups, and posed two problems: (1)

Does there exist a semigroup with more than two

elements whose graph is disconnected? (2) Find a

necessary and sufficient condition for a graph to be

the graph of a semigroup.


The first problem was answered in the negative

in a theorem proven by Lin [10]:


Theorem 5.4: Every semigroup with more than two

elements has a connected graph.


Pondelicek [12] also solved Bosak's first

problem, and he further obtained a metric characterization

of the graph of a semigroup:


Theorem 5.5: The diameter of the graph of a semigroup

does not exceed three.


Because of this metric characterization,

and the work in this paper with graphs that have distinct

metric characterizations, it was thought that it might

be interesting to look at the relationships (if any)

between the graphs of semigroups and bivariegated graphs.














In particular, it was hoped to partially solve the

second Bosak problem by answering some of the following

questions: Which finite semigroups have bivariegated

graphs? Which connected bivariegated graphs are the

graphs of semigroups?


First, the graphs of all semigroups of order

four or less were examined, using the scheme given

by Petrich [11]. For any class C of semigroups, a

possibly larger class may be constructed by performing

one or both of the following types of operations on

the members of C: (1) adjunction of an identity or

zero, (2) inflation, and (3) forming direct products.

Inflation may be defined as follows:


Definition 5.6: To every element s of a semigroup S,

associate a set Z such that: the sets of Z are pair-

wise disjoint and Z nS = {s}. The set V = u Z ,

together with the multiplication x y = ab if x Za

and y E Zb' is an inflation of S.


The remaining semigroups that cannot be

obtained by one of the above operations are listed by

Petrich according to either their multiplication tables













or the configuration of their greatest semilattice
decompositions and number of elements in each class.

Definition 5.7: A commutative semigroup in which all
elements are idempotent is called a semilattice.

Clearly, the trivial semigroup of order one
has no graph since it has no proper subsemigroup.
The semigroups of order two have as their graphs
either one or two disconnected points. From the graphs
of semigroups of order three, only two are bivariegated.
First, consider G2, the cyclic group of order two, with
identity a and multiplication table defined by
a a = a = b b and a b = b a = b. Inflate
about the non-identity element b, obtaining the addi-
tional element c, with multiplication table

a b c

a a b b
b b a a
c b a a

Clearly, {a} and {a,b} are proper subsemigroups,
producing a graph consisting of two edged vertices
which is a bivariegated graph.






___ ii















In addition, the semigroup defined by the

multiplication table


a b c

a a a a

b a a a

c a a b

has {a} and {a,b} as its proper subsemigroups, producing

the same graph as above.


For the semigroups of order four, the same

graph as above is obtained by inflating G3 about each

of its non-identity elements. In addition, the same

graph is obtained by forming the graph of G4.


From these results, conjectures were made.

It is clear that the graph of a cyclic group is complete

because each subgroup contains the identity. The only

bivariegated complete graph is the one consisting

of two edged vertices. Thus, for the graph of a

cyclic group to be bivariegated, it must have exactly

two proper subsemigroups. Recalling that the number

of subgroups of G is equal to the number of distinct

divisors of n, it can be seen that the graph of G ,















where n has three distinct divisors, is bivariegated.

Three distinct divisors are necessary since one

divisor of n will be the number one, corresponding to

the trivial subgroup, and one divisor of n will be

n itself, corresponding to the improper subgroup Gn-

That leaves one nontrivial proper subgroup to correspond

to the second vertex of the bivariegated graph. Now,

the integer n has three distinct divisors only when

it is the square of a prime number. Thus, the following

theorem has been proven:


Theorem 5.8: If n is the square of a prime number,

then the graph of Gn is bivariegated.


It was stated above that the graphs of infla-

tions of certain cyclic groups are bivariegated. By

definition of inflation, multiplication by the element

obtained by inflation is the same as for the element

from which the inflation is generated. Thus, in the

multiplication table, the rows and columns corresponding

to the new element and the inflating element are

identical. If n is prime, then Gn has no nontrivial

proper subgroup. If the inflating element is not

the identity of Gn, then no additional subgroups are















obtained by adding the new element since its mul-

tiplication table is identical to that of the inflating

element. (This is not true if the inflating element

is the identity of G since the new subgroup consisting

of the identity and the new element is formed.) Clearly,

Gn itself is a proper subsemigroup of the semigroup

obtained by inflation about a non-identity of G n

If n is prime, then this semigroup contains only one

other proper subsemigroup, the trivial one. Thus, the

following theorem has been proven:


Theorem 5.9: If n is prime, then the graph of the

semigroup obtained by inflating G about a non-identity
n
element is bivariegated.


After examining a number of graphs of semi-

groups, the following was conjectured and proven:


Theorem 5.10: Suppose G is the graph of some semigroup

S (i.e., I(S) 2 G). If G is bivariegated, then G

has exactly two vertices.


Corollary 5.11: If E(S) is bivariegated, then S has

exactly two proper subsemigroups.














Proof: Let S1 be a vertex of G = E(S). Since G = G f -

G2 is bivariegated, there is another vertex S2 G2

with which S1 E G1 is edged. By definition of E(S),

S1 and S2 are edged if and only if they have at least

one element in common; call it xl. In order for S2

and S1 to be distinct, one of these must contain

another element, say x2. Without loss of generality,

let x2 e S2 and x2 / SI. If G has exactly two vertices,

there is nothing to show. So, assume that G has more

than two vertices. Let S3 be a third vertex which

is connected to either S1 or S2. Without loss of

generality, assume S2 and S3 are edged. It is clear

that S3 C G2. Assume x2 E S2 n S3. Note that xI / S3.

Now, if G is bivariegated, there exists another vertex

S4 such that f(S4) = S3. If S4 and S3 are edged, they

have an element in common. This element cannot be

x1, or else S4 e G1 would be edged with both S2 and

S3. Similarly, the element in common cannot be x2.

So, both S4 and S3 contain some other element, say x3.

Notice that S2 n S3 is a subsemigroup (as it is the

intersection of two subsemigroups) containing x2 and

possibly other elements. Thus we have a fifth sub-

semigroup, and thus a fifth vertex. Call S5 = S2 n S3.















Since x2 E S5, it must be in G2' or else S2 would

be edged with two vertices in GI. Then there must

be an S6 in G1 such that f(S6) = S5. S6 cannot

contain xl, or else S2 e G2 would be edged with two

vertices in GI. Similarly, x2 / S-, or else S6 would

be edged with both S2 and S5 in G2. Also, x3 i S6,

or else both S4 and S6 in G1 would be edged with S3

in G2. So, S6 must contain an additional element,

say x4, that is also an element of S5. But S5 =

S2 n S3. Thus, x4 must also be in both S2 and S3'

which forces S6 e G1 to be connected to S2' S3, and

S5, all of which are in G2. This contradicts the

fact that G is bivariegated. Thus, G has no more than

two vertices.


This theorem partially answers Bosak's second

problem, for it says that there are quite large classes

of graphs that are not graphs of semigroups. That is,

out of the entire class of bivariegated graphs, there

is only one, the two-point bivariegated graph, that

is the graph of a semigroup.
















Semigroups of Graphs


Now we will examine another problem relating

to graphs and semigroups. In an article written by

Z. Hedrlin and A. Pultr [7 ] there is a discussion of

relations with given finitely generated semigroups.


Definition 5.12: Let X be a set and R a binary rela-

tion of X, R C X x X. We write xRy if (x,y) s R,

x,y e X. Further, R = (R,X) is used to explicitly

show the set X. In graph theoretic terms, every

relation (R,X) may be associated with a graph G,

where X is the set of vertices of G and R is the set

of all edges of G.


A transformation f of X is called compatible

with a relation (R,X) if xRy implies f(x)Rf(y) for

all x,y e X. The set of all compatible transformations

with a relation (R,X) is denoted by C(R,X). In graph

theoretic terms, a transformation of X is called an

endomorphism of G if it sends each edge into an edge.

Evidently, f is an endomorphism of G if and only if

f is compatible with (R,X). Further, the set C(R,X),

the set of all endomorphisms of the graph associated















with R, is a semigroup under composition of

transformations.


Hedrlin and Pultr prove the following

theorems:


Theorem 5.13: Let S be a finitely generated semi-

group with unity element. Then there exists a

relation RT on a set XT such that C(RT,XT) under

composition of transformations is isomorphic with S.

Moreover, there exist infinitely many non-isomorphic

relations RT with this property.


The above theorem can be translated into

the language of graph theory as follows:


Theorem 5.14: For each finitely generated semigroup

S with unity element, there exist infinitely many

non-isomorphic graphs G such that S is isomorphic

with the semigroup of all endomorphisms of G.


Earlier in this chapter the relationship

between semigroups and bivariegated graphs was

examined. Thus, it became an interesting problem

to look at the following question: Of the infinitely














many graphs that arise in connection with Theorem 5.14,

how many of these are bivariegated? Further, what are

the necessary and sufficient conditions on S to insure

that G is bivariegated? In order to attempt to answer

these questions, the construction of (RTXT ) in the

proof of Theorem 5.13 must be examined.


Hedrlin and Pultr [7] define {(R.,X), i = 1,2,...,n}

to be a system of relations on X, n > 2, and T to be a

sequence from the following lemma.


Lemma 5.15: Let n be a natural number. Then there

exist infinitely many sequences T = {tl,t, ..., t n} of

natural numbers such that

t. / t. for i / j

t. + t. > tk for all i,j,k c {l,2,...,n}.


The proof, stated in [7], is based on the

definition t. = i + h(n + 1), where h is an arbitrary

natural number. It is clear from the definition of t.
1
that if n > 2, then t. > 2 for i = 1,2,...,n.

Continuing the description of the construction in

[7], denote by Y the set of all quadruples (x,y,i,j) such

that x R. y, 1 < j < t., where t. is the i-th member of
1 1 1















the sequence T. Let U = {uo,u }, V = {voVlv2}

be sets. Define XT = XUYuUuV, the union of mutually

disjoint sets. Let two elements in X be in the

relation RT in the following cases:


x RT(x,y,i,l),

(x,y,i,j) R T(x,y,i,j + 1) for j = 1,2,... ,ti 1,

(x,y,i,t ) R y,

UoRTX,voRTX for all x e X,

uoRTUlulRTuo'

voRTV ,vlRTV2 ,V2R Tv


The remainder of this chapter is a record

of what was attempted in order to find some relation-

ship between Theorem 5.14 and bivariegated graphs.

First, we wanted to examine the following question:

Given a finitely generated semigroup S with identity,

what additional properties of S are necessary in order

that there exist a bivariegated graph G such that S

is isomorphic to the semigroup of endormorphisms of

G?


The first attempt at obtaining a hint to the

nature of these properties was to choose examples of















bivariegated graphs, form their semigroups of

endomorphisms, and check the additional properties

of these semigroups. This attempt proved futile

because graphs of very small order normally have a

very large number of endomorphisms, causing the

corresponding semigroup to be too large to deal

with efficiently.


Next, the characteristics of the graph

corresponding to (RTX T) were examined to determine

at least necessary conditions for it to be bivariegated.

In particular, since the minimum requirement for a

graph to be bivariegated is that it have even number

of vertices, the conditions that cause the set XT

to have an even number of elements were sought. The

following results were obtained:


Lemma 5.16: Let m be the number of elements in S

and let {R., i = 1,2,...,n) be a system of relations

on S. Then XT contains m + mn(n + 1)(2h + 1)/2 + 5

elements.


Proof: Recall XT = XUYUUUV, the union of mutually

disjoint sets. The set X corresponds to S, having m elements.














To determine the number of elements in Y = {(x,y,i,j):

xR.y, 1 < j < ti.}, note that for each R. there are

mt. members of Y because j takes on all values between
1
1 and t., of which there are t. values. Thus, the order
1 1
of Y is

n
mt1 + mt2 + ... + mt = mt .
1 n i=1


But by Lemma 5.15, t. = i + h(n + 1), so the above
1
equals
n n n
mZ(i + h(n + 1)) = m(Zi + Zh(n + 1))
i=l i=l i=l

= m(n(n + 1)/2 + hn(n + 1))

= mn(n + 1) ( + h)

= mn(n + 1) (2h + 1)/2.


Now, clearly U = {u ou } has 2 elements and

V = {v ,v,v 2 1 has 3. Therefore XT has m + mn(n + 1)

(2h + 1)/2 + 2 + 3 elements.


As a result of the above lemma, we obtain

a necessary condition in order for the construction

of (R T,X T) to produce a bivariegated graph:


Theorem 5.17: A necessary condition for the graph

corresponding to (RT ,XT) to be bivariegated is that S















have odd order and the number n of relations on S

be such that either n or n + 1 is divisible by 4.


Proof: If m + mn(n + 1)(2h + 1)/2 + 5 must be even,

then m + mn(n + 1)(2h + 1)/2 must be odd. Consider the

following cases:


Case 1: The number m is even.

Then mn(n + 1)(2h + 1)/2 must be odd. Note that m/2

is a whole number. No matter what n is, either n

or n + 1 is even. If one factor in a product is even,

the product is even, so (m/2) (n) (n + 1) (2h + 1) is

even. Thus if m is even, (RT,X ) does not correspond

to a bivariegated graph.


Case 2: The number m is odd.

Then mn(n + 1)(2h + 1)/2 must be even. Suppose n is

even. Then n + 1 is odd. Let p = n/2. Clearly, p

is a whole number. Now, it is necessary for mp(n + 1)

(2h + 1) to be even. For any h, 2h + 1 is odd. So

each of m, n + 1, and 2h + 1 are odd. Thus we must

have p = n/2 even, implying that n is divisible by 4.


Now suppose that n is odd. Then n + 1 is

even, and 2h + 1 is odd. Let p = (n + 1)/2, a whole number.














It is necessary for mnp(2n + 1) to be even. Since

each of m, n, and 2h + 1 is odd, p = (n + 1)/2 must

be even. This forces n + 1 to be divisible by 4.


Thus, for the graph corresponding to (RTXT)

to be bivariegated, m must be odd and either n or n + 1

must be divisible by 4.


Although the restriction that a graph have

an even number of vertices is far from being a sufficient

condition for bivariegation, it nevertheless imposes

quite a restriction on the semigroup S. Thus, it

became interesting to examine the restriction on S

caused by the limitation on the number of lines of a

bivariegated graph.


Recall that in the construction of (RTXT)

above, the first relations defined were of the form


xRT(x,y,i,l).


This produces mn lines because there are m elements in

X and n values of i (each corresponding to {(R, i = 1,

2,...,n)). From the relations of the form


(x,y,i,j)R (x,y,i,j + 1) for j = 1,2,...,ti 1,
















we obtain m(t. 1) relations for each i. This produces

n n
Xm(t. 1) = mEt. mn
i=l 1 =1


which, from the proof of Lemma 5.16, is


= mn (n + 1)(2h + 1)/2 mn.


From the relations of the form


(x,y,i,ti)R y


there are clearly mn different relations. Further, from


uoRTx,v oRT for all x e X


there are 2m relations. Finally, from


uoR lulR uo'

VoRTVl TVlRTV2 V2RTVo


there are 4 relations. Thus, there is a total of

mn + mn(n + 1)(2h + 1)/2 mn + mn + 2m + 4

or


mn + mn(n + 1)(2h + 1)/2 + 2m + 4


relations.















Now the question is whether or not this lies

between the minimum number p/2 and the maximum number

p /4 of lines that a bivariegated graph may have. For

sake of simplicity, let


s = mn(n + 1) (2h + 1)/2.


Then, the first question is whether or not the number

s + mn + 2m + 4 of relations corresponding to the

construction of (RTp,XT) is greater than or equal to

half the number s + m + 5 of elements in X T; that is,

it should be true that


s + mn + 2m + 4 > (s + m + 5)/2.


This is clearly true. In addition, the second question

is whether or not the number of relations is less than

one-fourth of the square of the number s + m + 5 of

elements in XT; that is, it should be true that


s + mn + 2m + 4 < (s + m + 5) 2/4.


However,

4s + 4mn + 8m + 16 < s2 + 2sm + m2 + 10m + 10s + 25

4mn < s2 + 2sm + m2 + 2m + 6s + 9.














At least one of the terms, 2sm, alone is larger than

4mn because


2sm = 2m 2n (n + 1)(2h + 1)/2

= m2n(n + 1)(2h + 1)

> mn(2) (3)

> 4mn.


Thus, the bounds on the number of lines in a bivariegated

graph places no additional restrictions on the semigroup

S.


Now, notice that in the definition of R ,

V0RTVI, VlRTV2, V2RTV0, and V0RTx for all x c X.

Suppose (R ,X ) is to be bivariegated by partitioning

some of the elements into G1 and some into G2 in such

a way that (RT,XT) can be transformed into G1 f G2'

Without loss of generality, let v0 e G Recall v0R v .

Then suppose v1 E G2. Because v1R Tv2 v2 must be in

G2 or else v would be related to two different elements

in G However, V2RT 0, which forces v2 to be related

to two different elements in GI. Thus, the assumption

that v1 e G2 was incorrect. Now since v1 e G if

v2 c G2, it would be related to two different elements














in Gi, so v2 e G However, since v0 is the only

element of V related to elements outside V and all

of the elements of V are in GI, it must be true that

neither v1 nor v2 is related to an element in G2.

Thus, the construction (RT,XT) does not produce a

bivariegated graph.


Therefore, it has been shown that in Theorem

5.14, none of the infinitely many graphs G such that

S is isomorphic to the semigroup of endomorphisms of

G is bivariegated. Thus, there must be many graphs

with the same semigroup of endomorphisms.


Finally, with regard to the relationship between

semigroups and bivariegated graphs, we have shown that

given a semigroup, its graph, if bivariegated, can only

be the connected tree on two vertices. Additionally,

we have shown that for a given semigroup S, no bivariegated

graph G can be constructed from it such that S is

isomorphic to the endomorphism semigroup on G.

















CHAPTER VI

DISCUSSION



Although some of the results of Sanders [13]

related to bivariegated trees were extended in this paper

to the case of bivariegated Husimi trees, it would be

desirable to obtain some further characterizations of

bivariegated Husimi trees. Sanders proves the following:


Theorem 6.1: A tree with 2n vertices has a 1-factor

if and only if the largest maximal independent set of

vertices of T contains n vertices.


Theorem 6.2: A tree is bivariegated if and only if

T has a 1-factor.


Thus, a bivariegated tree is characterized

in terms of independent sets of its vertices. There-

fore, the question arises: What is the relationship

between a bivariegated Husimi tree and independent sets

of vertices?


Another question arises due to the relationship

between trees and Husimi trees: Is there a way to















shrink the cycles of a bivariegated Husimi tree to

obtain a tree, and, if so, when is this tree bivariegated?

Or, a similar question is, can we expand the vertices

of a tree (possibly bivariegated) in some way to obtain

a bivariegated Husimi tree?


In one part of this paper we looked at the

relationship between Aut(G), Aut(G ), and Aut(G2),

where G = G1 f G2. It would be desirable to have

a better characterization of this relationship. One

way to proceed in solving this problem is in the manner

of Borowiecki [3 ], that is, by restricting G1 and G2

to be equal and the bijection f to be a part of a

smaller class of bijections. Then one can attempt to

answer some of the questions posed by Frechen [5 ]:

What are the properties P of G (=G2) which are preserved

in G1 f G2? What are the properties P and graphs

G (=G2) for which the class of all bijections preserving

P is a subgroup of the symmetric group of order n or

is isomorphic to Aut(G1l)?


Another topic presented in this paper is the

relationship between semigroups and bivariegated graphs.

It would be interesting to further study the restriction















on the diameter of the graph of a semigroup with respect

to some of the metric characterizations presented by

Howorka [ 8 ], and establish any connection with some

subclass of bivariegated graphs.


Finally, the semigroup of endomorphisms of a

bivariegated graph was studied. Hedrlin and Pultr [7 ]

present a construction which produces a graph from a

semigroup. A question which arises is: Can we derive

directly from the bivariegated graph (from which we

have obtained a semigroup) the graph produced by this

construction? Further, what is the relationship between

these graphs?







APPENDIX 1: BIVARIEG7ATED GRAPHS


p = 2



p 4



p = 6










p = 8
I J


(h/Il


I





p = 8 (continued)













p = 8 (continued)






p = 8 (continued)


FTI






p = 8 (continued)









rf h h l'
N J Mzz \J/ \..1/






p = 8 (continued)
/ K 1\ \--^ /\ A\






fc^ C0 00(
~~\ /_ \x^ BJ/\







APPENDIX II: BIVARIEGATED HUSIIMI TREES
p = 4


p = 2
I


p = 6


p= 8
I A t

II


IWr


I L


4,
L

-T






p = 10





- i
[Sd


IE












-M L



-V m




p = 12




r "'I p^ ^ -

rm1 En 1 :1



21117l 31711

TVV1 ET1Vf
I7\ TlT]L !
iiAn hi~i J\^T




p = 12 (continued)



iL






A 1


7L AL
mi_


1TE E










TL\




p = 12 (continued)
A-\ R-A L L
Err Kr- E -]


-Ah
AIL1



m

zzc'


Lv- Z7





p : 12 (continued)
FIKZ
IZE7

I^Z77


N^-n



















REFERENCES


1. A. R. Bednarek, A note on tree isomorphisms, J.
Comb. Theory, (B) 16 (1974), 194-196.

2. A. R. Bednarek and E. L. Sanders, A characterization
of bivariegated trees, J. Discrete Math., 5
(1973), 1-14.

3. M. Borowiecki, On the a-permutation graphs, Recent
Advances in Graph Theory, (Proc. Second
Czechoslovak Sympos., Prague, 1974), Academia,
Prague, 1975, 89-92.

4. J. Bosak, The graphs of semigroups, Theory of
Graphs and its Applications, (Proc. Sympos.
Smolenice, June 1963), Academic Press, New
York, 1965, 119-125.

5. J. B. Frechen, On the number of cycles in permutation
graphs, The Many Facets of Graph Theory, Springer-
Verlag, Berlin, 1970, 83-87.

6. F. Harary, Graph Theory, Addison-Wesley, Reading, 1969.

7. Z. Hedrlin and A. Pultr, Relations (graphs) with
finitely generated semigroups, Monatshefte
fur Math., 68 (1964), 213-217.

8. E. Howorka, On metric properties of certain clique
graphs, to appear.

9. D. C.Kay and G. Chartrand, A characterization of
certain ptolemaic graphs, Canad. J. Math.,
17 (1965), 342-346.

10. Y.-F. Lin, A problem of Bosak concerning the graphs
of semigroups, Proc. A.M.S., 21 (1969), 343-
346.









92









11. M. Petrich, Introduction to Semigroups, Charles E.
Merrill Pub. Co., Columbus, 1973.

12. B. Pondelicek, Diameter of a graph of a semigroup,
Casposis Pest. Mat., 92 (1967), 206-211.

13. E. L. Sanders, Independent sets and tree structure,
Doctoral Dissertation, University of Florida,
1975.

















BIOGRAPHICAL SKETCH


Leita Fay Aycock Riddle was born on December 22,

1949, to Thomas G. and Leita M. Aycock of Atlanta, Georgia.

In June, 1967, she was valedictorian of her graduation

class at The Arlington Schools in Atlanta. In the fall

of 1967, she entered H. Sophie Newcomb College of Tulane

University. On June 20, 1970, she married Dennis Lee

Riddle of Madison, Wisconsin. While in attendance at

Tulane, she was appointed to Tulane University Scholars

and Chi Beta honorary society. In May, 1971, she graduated

cum laude with the degree of Bachelor of Science in

mathematics. The following fall she joined the mathematics

faculty at Evans Junior High School in Spartanburg, South

Carolina. In September, 1972, she entered the Graduate

School of the University of Florida in mathematics. In

August, 1974, she received the degree of Master of Science.

From August, 1975, to date she has been a member of the

faculty at the University of South Carolina at Spartanburg,

where she teaches mathematics and computer science. At

the same time, she has continued her study in pursuit of

the degree of Doctor of Philosophy.

Leita Fay Aycock Riddle is a member of the American

Mathematical Society and the Honor Society of Phi Kappa Phi.












I certify that I have read this study and that
in my opinion it conforms to acceptable standards of
scholarly presentation and is fully adequate, in scope
and quality, as a dissertation for the degree of Doctor
of Philosophy.





A. R. Bednarek, Chairman
Professor of Mathematics




I certify that I have read this study and that
in my opinion it conforms to acceptable standards of
scholarly presentation and is fully adequate, in scope
and quality, as a dissertation for the degree of Doctor
of Philosophy.





J. E. Keesling
Professor of Math matics




I certify that I have read this study and that
in my opinion it conforms to acceptable standards of
scholarly presentation and is fully adequate, in scope
and quality, as a dissertation for the degree of Doctor
of Philosophy.





M. P. Hale, Jr.
Associate Professor of
Mathematics


II














I certify that I have read this study and that
in my opinion it conforms to acceptable standards of
scholarly presentation and is fully adequate, in scope
and quality, as a dissertation for the degree of Doctor
of Philosophy.





T. T. Bowman
Assistant Professor of
Mathematics




I certify that I have read this study and that
in my opinion it conforms to acceptable standards of
scholarly presentation and is fully adequate, in scope
and quality, as a dissertation for the degree of Doctor
of Philosophy.





W. D. Hedges
Professor of Education




This dissertation was submitted to the Graduate Faculty of
the Department of Mathematics in the College of Arts and
Sciences and to the Graduate Council, and was accepted
as partial fulfillment of the requirements for the degree
of Doctor of Philosophy.

June, 1978


Dean, Graduate School




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