Algebras of covers

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Thesis (Ph. D.)--University of Florida, 1980.
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Includes bibliographical references (leaves 65-66).
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ALGEBRAS OF COVERS


BY

ROBERT ERNEST OSTEEN
















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


1980












ACKNOWLEDGEMENTS


The computational experiments reported in Chapter 2

were made possible by a grant from the National Science

Foundation, NSF-MCS75-21130.

The many hours invested by the author in the company

of members of his supervisory committee -- in courses, in

seminars, and in informal meetings --have been helpful, en-

joyable, and sometimes even inspirational. I am especially

grateful to Professor A. R. Bednarek for his interest in this

work, for his guidance, and for his encouragement.

With only a heartbreakingly small number of days in

which to do a very demanding chore, Peggy Lopez was able to

type this volume -- and to do it well. She showed great char-

acter and determination in doing so, and I am proud and hap-

py to have her for a sister.













TABLE OF CONTENTS


PAGE

ACKNOWLEDGEMENTS ii

ABSTRACT iv

CHAPTER

1. INTRODUCTION 1

2. MAXIMAL RECTANGLES IN FINITE BINARY RELATIONS 5

2.1 Diclique Detection 5
2.2 Dicliques as Cliques 8
2.3 Experimental Timing Results 12

3. GRAPHICAL COVERS 19

3.1 Known Characterizations 19
3.2 A New Characterization 21

4. EFFICIENT COVERS 26

4.1 The Distributive Lattice of Efficient 27
Covers
4.2 Predecessors and Successors in E 30
4.3 The Cliques of the Composition of Two 45
Graphs
4.4 Related Algebras of Covers 49
4.5 A Metric for Z 56

APPENDIX A 62

REFERENCES 65

BIOGRAPHICAL SKETCH 67


iii








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


ALGEBRAS OF COVERS

By

Robert Ernest Osteen

August 1980

Chairman: A. R. Bednarek
Major Department: Mathematics


This study is concerned with graphs, directed graphs,

and covers.

A mapping from the class of directed graphs into the

class of (undirected) graphs is described, which mapping

has the following property: the family of maximal rectangles

(dicliques) of a given directed graph coincides precisely

with the family of maximal complete subgraphs (cliques) of

the associated graph.

A cover of a set is said to be efficient in case no two

of its members are related by inclusion. Any cover of a set

induces a graph on the set: a pair of distinct elements of

the set are adjacent in case there exists a member of the

cover to which both belong. A cover is graphical in case

the family of cliques of the graph induced by the cover co-

incides with the cover. It is shown that a cover is graphical

if and only if (1) it is efficient, (2) the median of any

triple of its members is a subset of one of its members, and

(3) given a chain refining the cover, there exists a member









of the cover which contains every member of the chain.

It is established that the family, E, of efficient cov-

ers of a finite set X -- ordered by the refinement rela-

tion -- is a distributive lattice.

The subfamily 0 of E, consisting of the graphical cov-

ers of X, is also a lattice under the refinement relation --

but not a sublattice, since the join of graphical covers

(as efficient covers) need not be graphical. The same is

true of the family of partitions of X.

The family G of graphs on X, ordered by the subgraph

relation, is also a subset of E -- under the identification

of G = (X, E) with the efficient cover of X consisting of

the maximal members of E u {{x): x E X} -- and G is a

sublattice of E.

The order structure of E is used to provide an algorithm

which makes essential use of the given families of cliques

of two graphs on X in order to compute the family of cliques

of that graph on X which is their composition as symmetric

reflexive relations on X.

Finally, a metric for the family of efficient covers

of a finite set is given: the distance between two covers

is the difference between the heights of their join and

their meet.











CHAPTER 1

INTRODUCTION


This work is concerned with binary relations, graphs,

directed graphs, and covers.

A binary relation from a set X to a set Y is a subset

R of X x Y. A relation on X is a relation from X to X.

A graph, G = (V, E), consists of a nonvoid set V and a

family E of subsets of V of cardinality two; members of V

are termed points or vertices, members of E are called lines

or edges, and to say that x and y are adjacent means that

x and y are distinct points and {x, y} E E. A directed

graph or digraph, D = (P, L), consists of a nonempty set, P,

of points or vertices and a set, L, of ordered pairs of

points, which are called directed lines or directed edges.

The difference between a digraph and a relation is only

terminological: a digraph, D = (P, L), is a relation, L,

on P. Also, a graph G = (V, E),may be viewed as a symmetric

irreflexive relation on V.

A cover of a set X is a collection of sets -- often

subsets of X -- whose union contains X. For example, the

family, Q(G), of all the cliques of a graph, G = (V, E), is

a cover of V consisting of subsets of V. A complete

subgraph of G is a subset W of V such that if x and y are

any two elements of W then {x, y}E E; a clique of G is an

inclusion-maximal complete subgraph of G. If C is a chain







(linearly ordered set) of complete subgraphs of G, then uC

is easily seen to be a complete subgraph of G; consequently,

by the Hausdorff Maximality Principle, every complete sub-

graph of G is contained in a clique of G -- that is, Q(G)

is, in fact, a cover of V, and every complete subgraph of G

is a subset of a member of Q(G).

The family, R(D), of dicliques of a digraph, D = (P, L),

is a cover of L by subsets of L. A rectangle in D is a sub-

set of the form A x B; a diclique of D -- or a maximal

rectangle in the relation L on P -- is an inclusion-maximal

rectangle in D.

In Chapter 2, a means is given whereby the problem of

finding the set of dicliques of a given finite digraph can

be transformed into the problem of identifying the set of

cliques of a graph associated with the digraph, such that

the two families are identical. Other diclique detection

algorithms are described, and the results of comparative

execution time experiments are discussed.

Chapter 3 is concerned with graphical covers. A family,

S, of sets is a graphical cover of uS in case the family,

Q(G(S)), of cliques of the graph, G(S), induced by S

coincides with S, where G(S) = (uS, E(S)) and a pair of

distinct points of G(S) are adjacent in case there exists

a member of S to which both belong. For example, if

G = (V, E) is a given graph and Q(G) is the family of its

cliques, then G(Q(G)) = G; and if S has a member which

properly contains a member of Q(G) then G(S) 0 G.





3

Under the assumption that no element of uS belongs to

infinitely many members of f, a set of (two) necessary and

sufficient conditions on S in order that S be graphical

was given by Bednarek and Stolarski [3]. It is shown in

Chapter 3 that the addition of a third condition on S -- a

chain closure condition -- makes the restrictive assumption

unnecessary.

It was observed in Bednarek et al. [2] that the notion

of graphical covers -- referred to there as "class cov-

ers" -- is a generalization of that of partitions, and

efficient covers generalize graphical covers. An efficient

cover is a cover in which no two members are inclusion-

comparable. In Chapter 4 the family of efficient covers of

a finite set is analyzed. It is shown to be a distributive

lattice under the refinement relation. (A family S of sets

refines a family T in case S E S implies the existence of

T E T such that S c T.) The relationships among the

families of graphs, graphical covers, efficient covers, and

partitions of a finite set X are explored. All but the

graphs are partially ordered by refinement; the graphs

are partially ordered by the subgraph relation: (X, E) <

(X, E') in case E c E'. The partitions are a subset of the

efficient covers and also form a lattice -- but not a sub-

lattice of the efficient covers, because the join (as

efficient covers) of a pair of partitions may not be a

partition. The same is true of the graphical covers.

The graphs on X may also be considered to reside among

the efficient covers, by means of an order isomorphism:





4

G = (X, E) is identified with the family of maximal members

of E u {{x}: x e X}. (Incid tally, since it is customary

to designate a singleton, {x}, by x, the set just mentioned

will henceforward be simply denoted by E u X.) Under this

identification, the graphs on X form a sublattice of the

efficient covers of X.

The structure of-the family of efficient covers of X

and its relationship with the family of graphs on X are used

to identify the family of cliques of the composition of two

graphs on X, given the families of cliques of the two

graphs. (By the composition of two graphs is meanV the

symmetric closure of their composition as symmetric re-

flexive relations, as in Bednarek et al., rather than in

the sense of Harary [8], in which the term is used with

a different meaning.)

Finally, the structure of the family of efficient

covers of a finite set provides a solution to a problem

raised by Osteen [14] in connection with the issue of

selecting one of several efficient covers representing

non-hierarchical classifications, namely, to define a

metric on the family of efficient covers of a finite set

by means of the order relation. The difference of heights

of the join and the meet of two elements of a finite

distributive lattice defines a metric.











CHAPTER 2

MAXIMAL RECTANGLES IN FINITE BINARY RELATIONS


A rectangle in a binary relation R c X x Y is a non-

empty subset of R of the form A x B; a maximal rectangle is

one which is not a subset of any other. Particularly when

X and Y are finite, such a relation may be regarded as a

digraph, D = (P, L), with P = X u Y and L = R; from this

point of view, maximal rectangles are termed, "dicliques."

This chapter is concerned specifically with algorithms

for the determination of the set of maximal rectangles of

a given finite binary relation -- or the dicliques of a

given (finite) digraph. Consequently, all relations are

assumed to be finite, even though many of the assertations

made below do not require that assumption.

Known diclique detection algorithms are first discussed,

after which a new algorithm is presented and validated.

Finally, the results of comparative timing experiments are

reported.


2.1 Diclique Detection


Algorithms for the identification of the maximal rec-

tangles in a given binary relation have been developed by

G. Fay [ 6 ], R. M. Haralick [ 7 ], E. M. Norris [ 12 ], and

a referee [131.







The basis for the Norris algorithm is the following

theorem.


Theorem 2.1.1. [Norris] Suppose R c X x Y, x E X, xR = 0,

and S c {x} x Y. Let T be the (disjoint) union of R and S,

M the family of maximal rectangles of R, and N the family

of maximal rectangles of T.

(1) {x} x xT E N if and only if for all C x D E M,

xT 4 D.

(2) Suppose C x D E M. (If D 4 xT then C x D E N.

If D c xT then C x D 4 N and C u {x} x D e N.

(3) If A / {x} and A x B E N \ M, then A x B=

C u {x} x D n xT for some C x D E M.

(4) Let H = {{x} u C x D n xT: C x D E M,

D n xT i 0},L = eH (the maximal members of H),

and N* = N \ {{x} x xT}. Then N* is the

(disjoint) union of L and M n N.

Let Q be a given relation from X to Y and

X = {xl, x2, ..., Xn }. For 1 < i s n let Qi

({xl} x xlQ) u ({ x2} x x2Q) u ... u ({xi} x iQ). The

strategy of the Norris algorithm is to apply Theorem 2.1.1 --

for each i from 1 to n 1 -- to obtain the maximal rectan-

gles of T = Qi+l from those of R = Qi and S = xi+1Q.

The algorithms of Fay [ 6 ] and Haralick [ 7 ] are

essentially identical, although they were developed

independently. The Fay-Haralick algorithm is based on

the following theorem.





7

Theorem 2.1.2 [Haralick] Let R c X x Y, A c X, and B c Y.

Then A x B is a diclique of R if and only if for some U c X,

B = n {xR: x e U) and A = n (Ry: y e B).

The Fay-Haralick algorithm first finds the second

projections, B, of all the maximal rectangles, A x B; then

the first projections, A, are obtained by means of the

computation, A = n {Ry: y E B}. The second projections are

obtained as the nonempty intersections of subsets of

U1 = {xR: x e X}, in the following manner. Let Uk+ be

the set {D n E: D E U EE Uk, k+1 = U- u ... u Uk+l'

and W = U \ V; compute W ,W W until
-k+1 -k+l -k 1 2'
W = 0, at which point the set of second projections of
-t
the maximal rectangles is Vt.

Theorem 2.1.3 is the foundation of the algorithm of

the referee [131.


Theorem 2.1.3. Let R c X x Y, which X n Y = 0. Define

R* = R u R- u (X x X) \ 1X u (Y x Y) \ ly, where 1

and 1 are the identity relations on X and Y, respectively.
y
The family of maximal rectangles of R is the collection of

all nonempty sets of the form, (Cn X) x (C n Y), where C

is a clique of R*.

Since R* is an irreflexive symmetric relation on

X u Y, R* is an undirected graph and the reference to

"cliques" is justified.

The strategy of the algorithm is clear from the statement

of the theorem. Its main constituent is a clique detection

algorithm (any clique detection algorithm).











2.2 Dicliques as Cliques


The problem of finding the dicliques of a digraph (or

the maximal rectangles of a binary relation) may be trans-

formed into that of finding the cliques of a graph. The

family of cliques of a suitably defined undirected graph

coincides precisely with the family of dicliques of a given

directed graph.

The undirected graph, G = (V, E), associated with a

given digraph, D = (P, L), is defined as follows: V = L;

if (a, b) E L and (c, d) E L then {(a, b), (c, d)} e E in

case (a, b) X (c, d), (a, d) E L, and (c, b) E L. Thus,

distinct vertices, (a, b) and (c, d), of G are adjacent in

case D has one of the three subgraphs of Figure 2.1.

The underlying idea of the definition of G and the

theorem can be stated informally as follows. To say that a

set of directed lines "form a rectangle" means that A x B

is a rectangle, where A is the set of initial points of the

lines and B is the set of their terminal points. A vertex

of G represents a minimal rectangle of D, a directed line.

A pair of adjacent vertices of G correspond to a pair of

directed lines of D which form a rectangle; thus, G is the

binary symmetric relation "co-rectangularity" on the set of

directed lines of D. A complete subgraph of G represents a

set of directed lines each pair of which are co-rectangular;

in this case, the entire set of directed lines forms a

rectangle. The following theorem is exemplified in Figure 2.2.





9







b



a= c


d






a

b=d



c


Figure 2.1






























(1,2)


(4,5) (4,4)


Figure 2.2


(6,7) '


(3,2)




(3,4)


(6,8)





11

Theorem 2.2.1. Let D be a digraph, G the associated graph,

R(D) the family of dicliques of D, and Q(G) the family of

cliques of G. Then R(D) = Q(G).


Proof: First, observe that if A x B is a rectangle in D

then A x B is a complete subgraph of G. If (a, b) E A x B

then, since A x B is a rectangle in D, (a, b) E L = V; thus,

A x B c V. Also, if (a, b) and (c, d) are distinct ele-

ments of A x B, then (a, d) E L and (c, b) e L, since A x B

is a rectangle in D; thus {(a, b), (c, d)} E E. As each

pair of distinct vertices of G in A x B are adjacent, A x B

is a complete subgraph of G.

Secondly, if M is a complete subgraph of the graph G,

then HI(M) x H2(M) ={a: (a, b) e M)x {b: (a, b) e M} is a

rectangle in D, and M c 1 (M) x H (M). Let a E (M),

b E I (M); there exist a', b' E P with (a, b'), (a', b) e M.
2
If a = a' then (a, b) E M c V = L. Otherwise, (a, b') and

(a', b) are distinct vertices of G belonging to the complete

subgraph M, and so, adjacent in G; hence by the definition

E, (a, b) E L. Thus, n (M) x H2(M) c L. Since also

M H 0, I (M) x H2(M) X 0; hence, 1 (M) x H2(M) is a rectangle

in D. That M c H1(M) x 12(M) is obvious.

Now let A x B E R(D). A x B is a complete subgraph of

G, so A x B c M for some M c Q(G). A x B c Mc HT(M)

x 12(M), which is a rectangle in D; by maximality of A x B,

Hn(M) x H2(M) = A x B. That is, A x B = M E Q(G), which

proves that R(D) c Q(G).

Conversely, let M e Q(G). Since nH(M) x n2(M) is a







rectangle in D, HI(M) x n2(M) c A x B for some A x B E R(D).

Now A x B is a complete subr-aph of G, and M c H (M) x

I (M)I c A x B; by the maximality of M, M = A x B. Thus

M = A x B E R(D), which establishes that Q(G) c R(D) and

completes the proof. D


2.3 Experimental Timing Results


The four diclique detection algorithms discussed in

the preceding sections were implemented by the author in

PL/I. In the interest of brevity (especially for the

labelling of the figures below) the algorithms will be re-

ferred to below by the names, Algorithm A, B, C and D,

where the order corresponds to the order of their presenta-

tions -- for example, "Algorithm A" refers to the Norris

algorithm.

The clique detection procedure used with Algorithms C

and D was an implementation by the author of the algorithm

of Bron and Kerbosch [ 5 ].

Random digraphs were generated by means of a random

number generator. Digraphs with point set sizes, p = 16,

32, .and 64, and with various line densities were obtained

(the line density is the ratio of the number of directed

lines to the maximum possible number of directed lines, p2).

Specifically, the range of line densities with p = 16 was

0.05, 0.10, ..., 0.45, and 0.50; with p = 32 it was 0.05,

0.10, ..., 0.35, and 0.40; and with p = 64 it was 0.05,

0.10, 0.15, 0.20, and 0.25.





13

Because the line densities were relatively small, the

number, r, of dicliques increased with increasing line

density, d. Moreover, it is not unreasonable to judge the

performance of the algorithms on the basis of execution

time versus the size of the output, namely, the number of

dicliques. Therefore, the timing results for the random

digraphs of 16, 32, and 64 points are presented graphically

in Figures 2.3, 2.4, and 2.5 (respectively) in the form of

execution times versus number of dicliques. Because of the

large ranges of numbers of dicliques and execution times,

both axes are scaled logarithmically. Thus, a straight

line with slope k would represent an execution time depend-

ency on the number of dicliques of the form, T = Ark, where

A is a constant; shifting a line upwards would correspond

to multiplying A by a number greater than unity.

The graphs of log(time) versus log(number of dicliques)

of Algorithms A, B, and D in Figures 2.3, 2.4, and 2.5

approximate parallel straight lines with slope two, so the

execution times for each is proportional to the square of

the number of dicliques. Furthermore, a little arithmetic

based on the vertical translations of the graphs for

Algorithms B and D above that for Algorithm A provides the

estimates, TB = 2TA and TD = 6TA. In short, Algorithm A

clearly outperforms Algorithms B and D.

Turning now to a comparison of Algorithms A and C,

Figure 2.6 exhibits just the timing results for these two

algorithms -- but for all the digraphs of the experiment.
































20


0
10



5
*H


Number of Dicliques


20 50 100 200 500 1000


Figure 2.3


0.5


10.

10


























500 B




200 ; A

0 D
50





E


20


5.
20 C












2


Number of Dicliques



10 20 50 100 200 500 1000 2000


Figure 2.4






















500


20U0

c)
50


C


D


















Number of Dicliques


10 20 50


0 Z O0 500 1 000 2000 500U


Figure 2.5






























m
T3
0
CQ
O

U,



.H







A (64 points


A (32 points)




C (64 points)






/C (32 points)


A (16 points)



C (16 points)









Number of Dicliques


I -


10 20 50 100 200


Figure 2.6


100



50




20


5




2



1.0



0.5




0.2



0,1


5 0 1
500 1000


5000


-c


, !1





18

The slopes of the graphs for Algorithm C are only about

0.6, whereas those of Algorithm A are about 2.0. On the

other hand, Algorithm A is faster on the digraphs of low

density, so one is tempted to conclude simply that

Algorithm A is superior to Algorithm C for sparse digraphs.

However, the line densities at which Algorithm C overtakes

(with respect to speed) Algorithm A are about 0.40, 0.25,

and 0.15 for p = 16, 32, and 64 (respectively). Consequent-

ly, it appears that Algorithm C might be faster than

Algorithm A for large digraphs, unless the line density is

exceedingly small. For example, one might guess -- on the

basis of just those three pairs of numbers -- that the

critical line density is approximately given by (0.55)e-0.02p

by this estimate, Algorithm C should outperform Algorithm A

on all digraphs of density at least 1/1000 -- provided that

the number of points is greater than about three hundred.

Finally, on the basis of the assumption that the execu-

tion times of the two algorithms may be estimated by the

products of powers of the number, p, of points and the

number, r, of dicliques of the digraph, the data suggest

that TA = (k) (p0.38) (r2) and TC = (2.2k) (p2.5) (r.6),

where k is a constant depending on the computing system

(seventeen microseconds for the data reported above).











CHAPTER 3

GRAPHICAL COVERS


Given a family S of sets, there is a natural way to

use it as the basis for a definition of a graph, G(S). The

set of points of G(S) is V(S) = uS; the set E(S) of lines

of G(S) consists of all pairs, {x, y}, of distinct points

such that there exists a member, A, of S to which both

points belong. It is an immediate consequence of the defi-

nition of G(S) that each member of S is a complete subgraph,

that is, S c K, where K denotes the family of all complete

subgraphs of G(S). Since also the family Q of cliques of

the graph is contained in K and is refined by K -- K Q --

S c K Q c K and S < Q.

The family S is said to be graphical -- or a graphical

cover of uS -- in case Q = S, that is, the cliques of the

graph induced by S are precisely the members of S.

It is with characterizations of graphical covers that

this chapter is concerned.


3.1 Known Characterizations


A characterization of graphical covers was given by

Zelinka [16].





20

Theorem 3.1.1. [Zelinka] A family S of sets is graphical

if and only if

(i) if M0 e S, T c S, and M0 cu T, then nT c MO; and

(ii) if A c uS and {A} 4 S, then there exists a two-

element subset B of A such that {B} 4 S.


Now the contrapositive of (ii) is as follows: if each

pair of elements of A is contained in a member of S then A

is contained in a member of S, that is, if A is a complete

subgraph of G(S) then A is contained in a member of S, or

K S. Moreover, K S if and only if Q < S, since K s Q

and Q c K, so (ii) is equivalent to Q < S.

Condition (i) implies that S is efficient. Of course Q

is efficient, and S Q by the way G(S) is defined. The

further assumption in condition (ii) that Q < S gives the

consequence that S is graphical with virtually no argument:

S Q S with both S and Q efficient implies S = Q.

Because one of the hypotheses (Q < S) is so close to

the consequent (Q = S), this characterization is not fully

satisfying.

A. R. Bednarek and S. J. Stolarski [3 ] gave a char-

acterization of graphical covers of finite sets. The state-

ment of the theorem requires some further definitions.

If (A, B, C) E SxSxS, the median of (A, B, C) is de-

fined to be m(A, B, C) = (A n B) u (A n C) u (B n C). Also,

the set of all medians of triples from S is denoted by m(S),

and eS = {A E S: if A c B e S then A = B}.





21

Theorem 3.1.2. [Bednarek-Stolarski] Assume that S is a

finite set. Then S is graphical if and only if

(i) S = eS, that is, S is efficient; and

(ii) mS 5 S.


What was in fact, proved in [ 3 ] was the following

more precise statement.


Theorem 3.1.3 If S is graphical then S is efficient and

mS 5 S. If S is efficient and mS < S, then if Q is a finite

clique of G(S) then Q E S.


In the next section a condition is given, which to-

gether with conditions (i) and (ii) of Theorem 3.1.2 are

necessary and sufficient conditions for S to be graphical --

irrespective of the cardinality of uS.


3.2 A New Characterization


In the statements of all of the following propositions,

S is a family of sets, G(S) is the graph induced by S, K is

the family of all complete subgraphs of G(S), and Q is the

family of all cliques of G(S).


Lemma 3.2.1 Suppose mS S. If A E K and A is finite then

{A} s S.


Proof: If A E K and A has no more than two members, then

it is clear that A is contained in a member of S.

Suppose by way of contradiction that there are finite

members of K which are contained in no members of S. Let A








be one such having least cardinality. Then 3 5 IAI, so let

al, a2, and a3 be three distinct elements of A. For i = 1,

2, and 3, IA {ai}I = |Al 1 < |AI, so there exists Mi in

S such that A {a} c Mi. Now A = m(A-al, A-a2, A-a3) c

m(Ml, M2, M3). But since mS 5 S, there exists MO E S such

that m(M1, M2, M3) c M0, so that A c M0 E S -- contradict-

ing the supposition that {A} t S. D


Corollary 3.2.2. Suppose mS 5 S. If A e Q and A is finite,

then A E S.


Proof: Let A be a finite clique. Since A is a finite mem-

ber of K, {A) S S by Lemma 3.2.1. Thus, there is a member B

of S such that A c B. Since S c K, A c B E K. However, A is

a maximal member of K, so A = B. Hence, A E S. 0


Corollary 3.2.3. Suppose uSI is finite. Then S is graphic-

al if and only if

(i) S = eS; and

(ii) mS < S.


Proof: Suppose first that S is graphical, that is, Q = S.

Since no two cliques are inclusion-comparable, S is ef-

ficient. Let A, B, and C be members of S, and suppose that

x and y are two elements of m(A, B, C). Each of x and y

belongs to at least two of A, B, and C, so there is some

one of A, B, and C to which both belong. Thus, x, y are

adjacent in G(S); hence, m(A, B, C) e K. K:Q= S, so mS S.









Now suppose S = eS and mS s S. By the preceding corol-

lary, each finite clique of I(S) is a member of S. Since G(S)

is finite, this means Q < S. But S < Q and Q = eQ in any case,

and S is efficient by hypothesis; hence, S = Q, that is, S

is graphical. D


A family of sets S is said to satisfy the refining chain

condition in case given any chain C which refines S, there

is a member of S which contains every member of the chain,

that is, {uC} S.

In the terminology of [15], a family T of sets is closed

in case given any chain D in T, uD e T. Clearly, S satis-

fies the refining chain condition if and only if kS is

closed, where kS denotes the family of all subsets of all

members of S.

A family H of sets is a hereditary family provided that

A c B E H implies A E H.


Proposition 3.2.4. If H is a hereditary family of sets,

G is a closed subfamily of H, and G contains all the finite

members of H, then G = H.


Proof: Suppose by way of contradiction that H \ G is not

empty. Let A be a member of H \ G having least cardinality.

Since G contains all finite members of H, A is an infinite

set. Therefore, A is the union of a chain C of subsets of
A, each member of which has strictly smaller cardinality





24

than that of A. If B E C, B c A e H, so since H is heredi-

tary, B E H. Since IBI < IAI, B j H \ G, so B E G. But then

A = uC e G -- a contradiction. O


Lemma 3.2.5. If mS S and S satisfies the refining chain

condition then K < S.


Proof: The family kS of all subsets of members of S is a

subset of K, and K is a hereditary family.

Since mS < S, if A E K and A is finite then by the

Lemma 3.3.1,{A}< S, which is to say that A E kS.

Finally, because S satisfies the refining chain con-

dition, kS is closed. Hence, by Proposition 3.2.4, kS = K,

that is, each member of K is contained in a member of S. O


Theorem 3.2.6. S is graphical if and only if

(i) S = eS;

(ii) mS S; and

(iii) S satisfies the refining chain condition.


Proof: The necessity of (i) and (ii) was shown in the

proof of Corollary 3.2.3 (without the use of the finite-

ness. assumption). If C is a chain in kS c K then the

union of the members of the chain is clearly a member of

K and therefore contained in a member of Q = S.

Sufficiency follows from Lemma 3.2.5 and the hypothesis





25

that S is efficient: Q c K < S, so Q 5 S; S s Q by the def-

inition of G(S). Since S ai;: Q are efficient and Q S S Q,

it follows that S = Q, that is, S is graphical. 0


Assume now that X = u S is finite. Then any member A of a

cover C of X is contained in a maximal member of C, so eC

covers X; in particular, if S is a cover of X, then S < emS

and emS covers X. If S is graphical, then emS < S, so

emS = S. Conversely, if emS = S then S is certainly effi-

cient; and since finiteness assures that mS emS, S is

graphical. This establishes the following proposition.


Theorem 3.2.7. Suppose u S is finite. Then S is graphical

if and only if emS = S.


This suggests that the three conditions of Theorem

3.2.6 -- S is efficient, mS 5 S, and the refining chain

condition -- might be reduced to two conditions by combining

the first two into one, as in Theorem 3.2.7.


Conjecture 3.2.8. A family S of sets is graphical if and

only if

*(i) emS = S, and

(ii) if C is a chain refining S then {u C} S.


In spite of the fact that the conjecture seems so close

to Theorem 3.2.6 -- especially in view of Theorem 3.2.7 --

the author has so far only established the truth of the

conjecture for the countably infinite case, and the general

case remains open.











CHAPTER 4

EFFICIENT COVERS


Let X be a nonempty finite set, X = {1, 2 ..., n}.

By a cover of X is meant a family C of subsets of X such

that u C = X. An efficient cover A of X is a cover no two

members of which are inclusion-comparable. Let E denote

the family of covers of X anj E the family of efficient

covers of X.

Since any cover C is a finite partially ordered set

(poset) with respect to the inclusion relation, each mem-

ber of C is contained in a maximal member of C. Hence,

e(C) = {A e C: if A c A' E C then A = A'} is a cover -- as

well as being an efficient family of sets -- so that eC is

an efficient cover, and the efficiency operator, e, defined

above is a function from E into E. Since E c E and eA =

A for A e E, e: -E is surjective.

The binary relation < of refinement on E is easily seen

to be reflexive and transitive, and its anti-symmetry fol-

lows from the efficiency of the members of E. Suppose

A B and B A. If A E A then since A < B there is a mem-

ber of B of B such that A c B; since B E B : A, there is a

member A' of A such that B c A',thus, A c B c A', with A

and A' in A, an efficient cover, so A = A' and A = B;






27

thus A E B, and A c B. That B c A follows from the same

argument, hence, if A < B v i B B A then A = B.

Not only is (EZ,) a poset, but it will be shown to be a

lattice, indeed, a distributive lattice.


4.1 The Distributive Lattice of Efficient Covers

A poset (P,5), is a lattice in case each pair of its ele-

ments has a least upper bound and a greatest lower bound.

The least upper bound of elements x and y is the (neces-

sarily unique) element z such that x 5 z, y < z, and if

x < w and y < w then z w. The greatest lower bound is de-

fined dually.


Lemma 4.1.1 If A E E and B E Z then e(A u B) is the least

upper bound of A and B.


Proof: First, e(A u B) E E because A u B is a cover of the

finite set X.

Suppose A E A. Then A E A u B and A c M for some maxi-

mal member M of A u B. Thus, A c M E e(A u B); hence,

A s e(A u B). By the same argument, B 5 e(A u B). There-

fore, e(A u B) is an upper bound of A and B.

Now suppose that C E E is an upper bound of A and B.

Since A < C and B C, A u B refines C. Since also

e(A u B) c A u B, e(A u B) 5 C. Therefore e(A u B) is the

least upper bound of A and B. D


If A and B are families of sets, let i(A, B) be defined

as follows: i(A, B) = {A n B: (A, B) e A x B}.






28

Lemma 4.1.2 If A E E and B E E then ei(A, B) is the

greatest lower bound of A and B.


Proof: To see that i(A, B) is a cover of X, let x e X.

There exist Ax E A and Bx E B such that x e Ax and x e B x

Thus, x e Ax n Bx E i(A, B), i(A, B) is a cover of the

finite set X, and ei(A, B) E Z.

For any A n B E i(A, B), A n B c A E A and A n B c

B e B so i(A, B) refines both A and B; since ei(A, B) c

i(A, B), ei(A, B) is a lower bound of A and B in E.

Now suppose that C is a lower bound of A and B.

If C e C then there exist A E A and B E B such that

C c A and C c B, so C c A n B E i(A, B). Let M be a max-

imal member of i(A, B) such that A n B c M. Then C c A n

B c M E ei(A, B), so that C < ei(A, B). Hence, ei(A, B)

is the greatest lower bound of A and B. D


Theorem 4.1.3 (E, <) is a lattice with join operator

A v B = e(A u B) and meet operator A A B = ei(A, B).


Proof: Lemmas 4.1.1 and 4.1.2. D


In the sequel use will be made of the following facts

concerning all lattices. The proofs of these familiar prop-

erties -- which can be found in Birkhoff [4] -- will not

be repeated here. The meet and join operators are idem-

potent, commutative, and associative, and they satisfy the

absorptive laws, a A (a v b) = a and a v (a A b) = a, for

all elements a and b. Also, a b if and only if





29

a v b = b. The distributive inequality, (a A b) v (a A c) <

a A (b v c), holds for all elements a, b, and c.

A distributive lattice is a lattice in which

a A (b v c) = (a A b) v (a A c) for all elements a, b,

and c. O


Theorem 4.1.4 (E,<) is a distributive lattice.


Proof: By the distributive inequality and the anti-

symmetry of <, all that must be shown is that A A (B v C) 5

(A A B) v (A A C), for all A, B, and C in E.

Let D E A A (B v C). Then D = E n F for someE E A

and F E B u C. Suppose F E B; then D = E n F E i(A, B) <

ei(A, B) = A A B, so {D} refines A A B. Similarly, if

F e C then {D} refines A A C. Since every singleton sub-

set of A A (B v C) refines A A B u A A C, A A (B v C)

refines A A B u A A C. But A A B u A A C < (A A B) v

(A A C) = e(A A B u A X C), so A A (B V C) < (A A B) v

(A A C). Hence, (E,<) is distributive. D


If a lattice has a greatest element -- that is, an

element, 1, such that x 5 1 for every element, x -- that

element (necessarily unique by anti-symmetry) is called the

unit of the lattice. Dually, the zero of a lattice -- if

it has one -- is its least element. The unit of (E,5) is

U = {{1, 2, 3, ..., n}}, and the zero is Z = {{l},{2},{3, ...,

{n}}: if A E then Z < A < U.

A complemented lattice is a lattice with a unit and a

zero in which for each element, x, there is an element, x'





30

(the complement of x), such that x v x' = 1 and x A x' = 0.

It will now be shown that E Is not a complemented lattice.

Let H = {x \ {i}: i e x}. It is easy to see that if

A E \ {U} then A < H, so A v B = U implies B = U; but

then A A B = A U = A, so unless A = Z, A has no

complement.

A sublattice of a lattice is a subset which is closed

under joins and meets. An interval, Ex, y] = {s: x 5 s 5 y}

of a lattice is a sublattice. The interval [Z, H] of E is

not complemented, either. Consider A = {X \ {il:

i = 1, 2, ..., n 1} e [Z, H]. For A' E [Z, H] and A v A' =

H requires {1, 2, ..., n 1} e A'; but then

{1, 2, ..., n 2} e i(A, A') < A A A'. If 4 n then this

implies that A A A' contains a member of cardinality at

least two, whence A A A' $ Z. Furthermore, neither Z \ {U,H}

nor E\{Z} even has both a unit and a zero, so neither is

complemented.

Thus, Z is a distributive lattice, but not a comple-

mented lattice -- nor can it be made complemented by a

suitable deletion of one or two elements from its extremes.


4.2. Predecessors and Successors in E


Suppose (P,) is a poset, x < y and x z y implies

z = x or z = y. This condition is often rendered "y covers

x." However, since the elements of E are "covers" -- in a

different sense -- this condition will here be rendered,

"x precedes y" and "y succeeds x."






31

The problems addressed in this section are the

characterizations of the predecessors and successors of a

given member of E.


Lemma 4.2.1 If A precedes B then IB \ Al = 1.


Proof: If IB \ Al < 1 then B \ A = B c A; but then

A < B 5 A and A = B -- a contradiction. Hence, 1 < IB \ Al.

Now suppose 1 < IB \ Al and let B and B'.be distinct

members of B \ A. Clearly, C = e(A u {B}) E E and A 5 C.

Also, since A < B and B E B, C < B. Thus, A 5 C 5 B.

Assume C = A. Since B 4 C, B is properly contained in

some member A of A. In view of A 5 B, however, this con-

tradicts the efficiency of B. Therefore, C $ A.

Suppose C = B. Then B' E C = e(A u {B}), so since

B' i B, B' E A -- contradicting B' r B \ A. Hence, C / B.

But A 5 C B B, A $ C, and C $ B contradicts the hypo-

thesis that A precedes' B. Therefore, the supposition,

1 < IB \ Al, is false. Since 1 : IB \ Al has already been

established, IB \ A = 1, as asserted. O

For A E A E E with 1 < IAI, let P(A, A) be defined by

P(A, A) = e((A \ {A}) u {A \ {x}: x A}) -- the efficient

cover of X consisting of the maximal members of the cover

obtained as follows: delete A from A, then adjoin the

maximal proper subsets of A.

Clearly,-P(A, A) A. It will be shown that P(A, A)

precedes A, and that every predecessor is such a cover.






32

Lemma 4.2.2 If 1 < IAI and A E A E E, then A \ {A} c

P(A, A).


Proof: Let A' E A \ {A}. Since A \ {A} is an efficient

family, A' is properly contained in no member of A \ {A}.

Also, A' 4 A, since A is efficient, A E A, and A' E A.

If x E A than A' P A \ {x} c A. Therefore, A' E (A \

{A}) u {A \ {x}: x e A}, and A' is contained no other

member of that family, so A' e P(A, A). D


Lemma 4.2.3 If A e A E E and 1 < IAI then P(A, A)

precedes A.


Proof: Since P(A, A) E E and P(A, A) < A, what remains to

be shown is that there is no member of E strictly between

P(A, A) and A.

Assume that B E and P(A, A) < B < A.

First it will be shown that A \ {A} c B. Let

A' E A \ {A}; then A' E P(A, A) by Lemma 4.2.2. Since

P(A, A) < B < A, there exist B' e B and A" E A such that

A'c B'c A". But A is efficient, so A' = A", whence

A' = B'. Hence, A \ {A} c B.

Assume first that the inclusion is not proper, that is,

that B = A \ {A}. Since B = A \ {A} c P(A, A), B < P(A, A).

Because P(A, A) < B, B = P(A, A). Thus, B = A \ {A}

implies B = P(A, A).

Assuming now that A \ {A} is properly contained in B, let

B e B \ (A \ {A}), A' E A, and B c A'. If A' $ A then

A' E A \ {A}c B; B c A' with B E B and A' E B implies A' = B;





33

but B A \ {A), contradicting A' E A \ {A}. Thus,

B E B \ (A \ {A}) implies B r A.

If A 4 B then every B E B \ (A \ {A}) is properly con-

tained in A, so that {B} refines P(A, A). Since also

A \ {A} c P(A, A), B < P(A, A). But P(A, A) B, so

B = P(A, A).

If A E B, then since A \ {A} c B, A c B. But

A, B E E and A c B A implies A = B.

Hence, P(A, A) 5 B A implies B = P(A, A) or B = A,

which is to say that P(A, A) precedes A, as claimed. O


Theorem 4.2.4. B precedes A in E if and only if B'= P(A, A)

for some non-singleton member A of A.


Proof: That each such P(A, A) precedes A was established by

Lemma 4.2.3.

Assume that B is a predecessor of A. By Lemma 4.2.1,

IA \ BI= 1.

Suppose first that B c A. Then A = B u {A} with A 4 B.

Since A and B are efficient covers, 1 < |Al. Now

B = A \ {A} and A \ {A} c P(A, A) by Lemma 4.2.2, so

B < P(A, A) < A. Since B precedes A, and P(A, A) A,

B = P(A, A)

Now suppose that B 4 A. Let B E B \ A, A' E A,

and B c A'. Since B 4 A, B # A': since B is efficient,

A' a B. Thus A' e A \ B. However, A \ B = {A} for some

A E A, so that B \ A refines {A}. In particular, each

B' E B \ A is properly contained in A, since A k B \ A,

so B \ A refines P(A, A). Since by assumption there exists





34

at least one member B of B \ A, 1 |BI < I|A, so A is not

a singleton. A = (A n B) u 'A \ B) = (A n B) u {A}, and

A 4 B, so A n B = A \ {A}; as A \ {A} c P(A, A) by

Lemma 4.2.2 and B \ A refines P(A, A), B = (B n A) u

(B \ A) < P(A, A). But as before, B < P(A, A) A implies

B = P(A, A), and the characterization of the predecessors

of a member of E is proved. O


Corollary 4.2.5. The number of predecessors of A is the

number of its non-singleton members.


Proof: If A and A' are distinct non-singleton members of A,

then A E P(A, A') and A 4 P(A, A), so P(A, A) X P(A, A').

The number of predecessors is therefore not less than the

number of non-singleton members. But by Theorem 4.2.4, the

number of predecessors is not greater than the number of

non-singleton members. O


For A E E let gA denote the graph induced by A, as in

Chapter 3: gA = (X, E(A)), with {x, y} e E(A) if and only

if x and y are distinct elements of X and there exists

A e A such that {x, y} c A. Since X u E(A) covers X,

e(X u E(A)) E E. The family, Q(gA), of cliques of gA is

also a member of E.Moreover, it is clear that e(X u E(A)) <

A < Q(gA), and B belongs to the interval [e(X u E(A)),

Q(gA)] of E if and only if gB = gA.

Consider a predecessor, P(A, A) = e((A \ {A}) u

{A \ {x}: x e A}), where 1 IAI and A E A.





35

Suppose A = {x, y}, a set of two elements. Since A is

efficient, {A} 4 A \ {A), that is, if A' E A \ {A} then

{x, y} A'. Therefore, it is clear that {{x, y}} 4

P(A, A), so that {x, y} e E(A) \ E(P(A, A)). Of course

E(P(A, A) c E(A), as P(A, A) A. Hence, if IAI = 2 then

gP(A, A) is a proper subgraph of gA. If {u, v} is any

other line of gA then {u, v} 4 A, so {u, v} c A' for some

other A' E A. A glance at the definition of P(A, A) shows

that {u, v} e E(P(A, A)).

Thus, E(A) has just one line, {x, y},which does not

belong to E(P(A, A).

Now assume that 3 IAI. Let {x, y} E E(A); there

exists A' E A with {x, y} c A'. If A' = A, let z E A \

{x, y); then {{x, y}}< {A \ {z}} P(A, A). If A' # A

then {A'} s P(A, A), so again, {{x, y}} < P(A. A). Thus,

{x, y} e E(P(A, A)), so that E(A) c E(P(A, A); as

P(A, A) A, E(P(A, A)) c E(A); hence gP(A, A) = gA.

The preceding conclusions are recorded in the following

statement.


Proposition 4.2.6. Let P(A, A) be a precedessor of A.

If 3': IAl then gP(A, A) = gA, and if 2 = IAI then

E(P(A, A)) = E(A) \ {A}.

The problem of characterizing the successors of a

member of will now be taken up.

Suppose B succeeds A. Then gA is a subgraph of gB

since A B. Either gA = gB, or gA is a proper subgraph

of gB.





36

Lemma 4.2.7. Suppose gA gB. Then B succeeds A if and

only if B = e(A u {C}) for some C c X such that ICI = 2

and {C} A.


Proof: Assume first that B succeeds A.

Let C = {x, y} be a line of gB which is not in E(gA).

Then {C} J A, and A C = e(A u (C}); also, A i C,

since C E C \ A.

Now if C' E C then either C' E A or C' = C. If C' = C

then since C E E(gB), {C') < B. If C' e A then since

A B, {C'} B. Therefore, C < B.

Since A C C < B, B succeeds A, and C A, it follows

that B =C=e(A u {C}), where ICI = 2 and {C} t A, as

required.

Now suppose B = e(A u {C}) for some two-element set,

C = Ix, y}, such that {C} t A. Then A s B and C E B \ A.

Let D = P(B, C). If DE D then either D E A or D is a

one-element subset of C; in either case, {D} < A, so that

D S A. Thus D s A s B, with D a predecessor of B and A X B;

hence, A = D, that is, B is a successor of A. This com-

pletes the proof. D


A characterization of successors which induce the same

graph will be given, followed by a complete characterization

of all successors of a given member of E. The development

is facilitated by the following lemma.


Lemma 4.2.8. If B succeeds A then B = e(A u {B)) for

some B E B such that {B} A.





37

Proof: Since B succeeds A, A < B and B $ A. Let B E B

with B t A, and C = e(A u {B}). Then C E A C B,

and since B E C \ A, C / A. Therefore, B = C. D


Lemma 4.2.9. Suppose gA = gB. Then B succeeds A if and

only if B = e(A u {E}), where {E} is a minimal subset of

some D E mA with respect to the property, {E} $ A.


Proof: Assume first that B succeeds A. By Lemma 4.2.8,

B = e(A u {E}) for some E E B with {E} t A.

Let F be a proper subset of E, and let C = e(A u {F}).

Then C E Z and A C < B. E 4 F and {E} $ A, so {E} $ C;

thus B S C, so B X C. Since B succeeds A, A = C, whence

{F} A.

Since gA = gB, 3 : (El, so let x, y and z be three

members of E, and let A, B, and C be members of A containing

E \ {x}, E \ {y}, and E \ {z}, respectively. E = m[E \ {x},

E \ {y},E \ {z}] c D ='m[A, B, C] E mA.

Thus B = e(A u {E}) E c D e mA, and every proper sub-

set of E is contained in a member of A, as asserted.

Now assume B = e(A u {E}), with E c D e mA, {E} t A,

and every proper subset of E is contained in a member of A.

Then A s B and B t A.

Let C = P(B, E) = e((B\ {E}) u {E \ {x}: x e E}).

For x e E, {E \ {x}} < A by the minimality of E with respect

to the property of being contained in no member of A.

If F E B \ {E} = e(A u {E}) \ {(F then F E A; therefore,

C A. Since C precedes B, C < A < B, and A X B it

follows that A = C. Hence, B succeeds A. D





38

Theorem 4.2.10. B succeeds A if and only if B = e(A u {B})

for some B E B such that {B)} A while every proper subset

of B is contained in some member of A. In this case, more-

over, IBI = 2 if and only if gA is a proper subgraph of gB;

otherwise, 3 < IBI and there exists D E mA such that B c D.


Proof: Suppose B succeeds A. By Lemma 4.2.8, B = e(A u {B})

for some B E B such that {B} A. Let C be a proper subset

of B and C = e(A u {C}); then A 5 C B. Since {B} $ A,

B < C and B E B, it follows that B C, so B $ C. Because

B succeeds A, this necessitates A = C = e(A u {C}), so C

is contained in a member of A. This proves the necessity of

the conditions given in the first statement.

Now suppose that B = e(A u {B}), B E B, {B} 4 A, and

every proper subset of B is contained in some member of A.

A B and B E B \ A, so A $ B. Since {B} 4 A, 2 < IBI,

so let C = P(B, B). If C E C then either C E B \ {B}

e(A u {B}) \ {B}c A, or C = B \ {x} for some x e B, in

which case {C} 5 A by the minimality of B with respect to

being contained in no member of A.

Thus, C 5 A 5 B, C precedes B, and A i B. Therefore,

A = C = P(B, B), and the proof of the first statement is

complete.

Turning to the second statement, since A 4 B and A

is a cover of X, 2 < IBI.

Suppose IBI = 2. Since {B} t A, B E E(B) \ E(A), so

gA 6 gB. Converserly, if gA f gB, then IBI = 2 by

Lemma 4.2.7.






39

Now suppose 3 : IB|. Then every two-element subset of

B is contained in some memb ,, of A, so that gA = gB. Con-

versely, if gA = gB, then the necessity of 3 s IBI was

shown in the proof of Lemma 4.2.9, which also established

the existence of D E mA such that B c D.

Denoting e(A u {C}) by S(A, C), each successor of A

is S(A, C) for some C'-- just as each predecessor A

is P(A, D) for some D. A consequence of the following

proposition is that A = P(S(A, C), C) and B =S(P(B, D), D) --

provided, of course, that C and D satisfy the conditions

which make S(A, C) a successor of A and P(B, D) a prede-

cessor of B. U


Proposition 4.2.11. A precedes B if and only if there

exists C c X such that A = P(B, C) and B = S(A, C).


Proof: The existence of C such that A = P(B, C) implies

that A precedes B by Theorem 4.2.4.

Conversely, assume that A precedes B. Again by

Theorem 4.2.4, A = P(B, C) for some non-singleton C E B.

It remains to be shown that B = S(A, C).

,Let C = S(A, C) = e(A u {C}). Then A < C. Also, C s B,

since A < B and C E B. Since {C} s C but {C} A, it fol-

lows that A C C. Therefore, since A precedes B and

A s C s B, B = C = S(A, C). O


It is clear from Theorem 4.2.4 and Theorem 4.2.10

that finding successors is considerably more difficult than

finding predecessors. That is, it is harder to ascend in E

than to descend.





40

Furthermore, while the number of predecessors and their

identities are readily specified in Colollary 4.2.5, no such

explicit specification for the successors is known.

Even if one needs only successors inducing the same

graph, the situation does not improve noticeably.

Figure 4.1 illustrates the fact that given D E mA such

that {D} t A, D might have more than one subset E which is

minimal with respect to the property, {E} $ A. Figure 4.2

shows that there can be distinct C and D in mA such that E

is a minimal subset of each not contained in any member

of A.

It would be surprising indeed, however, if ascending

in E were not computationally costly. Given a graph,

G = (X, E), e(X u E) E Z and Q(G) -- the family of cliques

of G -- is in E. Ascending through [e(X u E), Q(G)] from

bottom to top amounts to finding the cliques of a given

graph -- an NP-complete problem [ 9 1.


Lemma 4.2.12. Suppose A E Z, and let K(s) =

{K e K: IKI s}, where K is the family of complete subgraph

of gA and s is a non-negative integer. Then K(s + 2)

(em)S(A).


Proof: The argument proceeds by induction.

Because of the gA is defined, K(2) A = (em) (A).

As inductive hypothesis, assume K(s + 2) <

(em)S(A) for s = 0, 1, ..., t 1, where t is a positive

integer.

Let K e K (t + 2).











































A =

B =

C =

m(A, B, C) =

{{1, 2, 3}}11

{{1, 2, 4}} s


11, 2, 3, 6}

{1, 2, 4, 5)

{3, 4, 71

{1, 2, 3, 4}

{A, B, C}

{A, B, C}


Figure 4.1




















A

7.




F

6 D
\1 \ 9
B




10

5 E

c
C


m(A, B, C) = {2, 3, 4}

m(D, E, F) = {1, 2, 3, 4}


Figure 4.2





43

If IKI | t + 1, then K e K(t + 1), which refines

(em)t-l(A) by the induction hypothesis. Since always

B < (em)(B), this means that K is contained in a member

of (em)t(A).
Otherwise, IKI = t + 2, 3 IKI, so let x, y and z be

three distinct elements of K, and let A = K \ {x},

B = K \ {y}, and C = K \ {x}. Then {A, B, C} c K(t + 1),

so there are members A', B', and C' of (em) t-(A) such that

A c A', B c B', and C c C'. Thus K = m(A, B, C) c

m(A', B', C') E m(em)t-(A), so that {K} < (em)t(A).

Hence, K (t + 2) refined (em)t(A), which completes

the induction. O


Theorem 4.2.13. Q(gA) = (em)t-2(A), where t is the size

of a largest clique of gA.


Proof: By Lemma 4.2.12, K(t) (em)t-2(A). But K = K(t),

since every complete subgraph of gA has t or fewer points.

Also, (em)t-2(A) < Q(gA), so that Q(gA) c K < (em)t-2 <
t-2
Q(gA), which implies that Q(gA) = (em) t-2(A), as was to

be shown. D

It is evident that one could use the theorem as the

basis for a clique detection algorithm. Given a graph,

G = (X, E), apply the operator, (em), repeatedly -- begin-

ning with A = e(X u E) -- until its application produces

no change.







The number of iterations would be t -1, where t is the

size of a largest clique, because the value of t would not

be known in advance.

Although the number of iterations would be small, the

quantity of computation per iteration could be rather large.

Since B c mB and m(A, B, B) = B, mB is the union of B and

the set of medians of distinct triples of members of B. Let

the number of k-element subsets of a set of p elements be

denoted by C(p, k). Then the number of median computations

required to compute mB would be C(IBI, 3), = IBI (IBI 1)

(IBI 2)/6. The number of comparisons needed to compute

e(mB) from mB would be twice the number, C(ImBI, 2), of

pairs of distinct members of mB. Since ImBI 5 IBI +

C ( IBI, 3), the number of comparisons is bounded by

(2)(C(IBl + C(IBI, 3), 2)), and the sum of the number of

median computations and the number of comparisons is a

polynomial in IBi of degree six, roughly approximated by

BI6 / 36.

Furthermore, IBI can be quite large. On the first

iteration, JBI is the number of lines of the graph --

assuming for the purpose of estimation that the graph has

no isolated points. On the final iteration, IBI is the

number of cliques of G, which Moon and Moser [10] have

shown can be as large as 3n/3, where n is the number of

points. Thus, IB6 can be as large as about 9n, which is

indeed quite large.





45

4.3 The Cliques of the Composition of Two Graphs


If X and Y are sets then a (binary) relation from X

to Y is a subset R, of X x Y. If Y = X then R is simply

said to be a relation on X. The condition, (x, y) e R is

often written, xRy; also, xR = {y E Y: xRy}and Ry =

{x E X: xRy}.

A relation R on a set X is symmetric in case xRy

implies yRx; R is reflexive in case xRx for all x, and

irreflexive in case xRx for no x.

A graph, G = (X, E), may be regarded as a symmetric

irreflexive relation, E, on X. It is sometimes advanta-

geous to regard G as a symmetric reflexive relation on X,

namely, E u 1, where 1 is the identity relation on X.

The composition, R o S, of a relation R from X to Y

and relation S from Y to Z is the relation from X to Z,

R o S = {(x, z): from some y c Y, xRy and ySz}.

The square, G2, of a graph, G = (X, E), is defined in

Mukhopadhyay [il as the graph with point set X in which a

pair of (distinct) points, x and y, are adjacent in case

x and y are adjacent in G or both are adjacent in G to

some third point Regarding G as the reflexive symmetric

relation, E' = E u 1, as above, the square of G is the

composition of E' with itself -- but not the composition of

E with itself, as illustrated in Figure 4.3

Now let G = (X, E) and H = (X, E') be graphs on the

same set, X, of points. Let R = E u 1 and S = E' u 1,

where 1 is the identity relation on X. Define the













SI


R <



S






RS


S'R'


RS u SR < >


R'S' u S'R' Q:(C


Figure 4.3


SR k


R'S' q






47

composition of G and H to be the graph, G o H = (X, F),

where F = (R o S u S o R) \ 1. The minimal reflexive

symmetric relation containing R o S is R o S u (R o S)-1

R o S u S-1 o R1 = R o S u S o R. The fact that R o S

may not be symmetric although R and S are symmetric and

reflexive is illustrated in Figure 4.3, which also il-

lustrates the fact that this definition of G o H general-

izes the notion of the square of a graph. The reflexivity

of R and of S assures that G and H are subgraphs of G o H.

Suppose now that R and S are symmetric reflexive re-

lations on a set X and that the families Q(R) and,Q(S) of

cliques of the "graphs", R and S, are known. A. R. Bednarek

[1] has posed the following question: how may Q(R) and

Q(S) be used to find the family Q(T) of maximal complete

subgraphs of T = R o S u S o R? The following theorem pro-

vides an answer.


Theorem 4.3.1. Let R and S be reflexive symmetric relations

on a finite set X of n elements, and let T = R o S u S o R.

Then Q(T) = (em)t(eC) for some t < p 2, where p is the

size of a largest clique of T and C={A u B: (A, B) e

Q(R) x Q(S) and A n B B }.


Proof: Let R = e( {{x, y}: (x, y) e R} u X), and

S = e( {{x, y}: (x, y) e S} u X), and T = e( {{x, y}:

(x, y) e T} u X). Then R, S, T, and D = eC are efficient

covers of X -- as is Q(T) = Q(gT).








Any singleton in T is contained in some member of D.

Let {x, y) be a two-element member of T. Then (x, z) E R

and (z, y) E S for some z e X, so there exist A E Q(R)

and B E Q(S) such that {x, z} c A and {z, y} c B, so

that {x, y} c A u B E C D. Therefore, T < D.

It will now be shown that D < Q(T). Let A u B e D,

with A E Q(R), B E Q(S'), and x e A n B. Suppose a and b

are two members of A u B; if both belong to A or to B, then

a and b are adjacent in T, a ,supergraph of R and of S.

Otherwise, a E A and b E B; since x e A n B, {a, x} c A

and {x, b} c B, so (a, b) e T. Thus, each pair o elements

of A u B are adjacent in T, so A u B is contained in a

clique of T.

Therefore, T : D D Q(T).

Let p be the size of a largest clique of T = gD. By

Theorem 4.2.13, Q(T) = (em)P-2.


Incidentally, it is clear from the proof that Q(R) and

Q(S) could be replaced (respectively) in the statement of

the theorem by any efficient covers A and B of X such that

gA = R and gB = S -- eC is still between T and Q(T).

In the setting of Theorem 4.3.1, the number of itera-

tions of the operator, (em), required to reach Q(T) can be

substantially smaller than the number required in the con-

text of Theorem 4.2.13 -- the trip from eC up to Q(T) is

shorter than the trip from e(T u X) up to Q(T). Because

an algorithmic implementation of Theorem 4.3.1 makes use





4 9

of the known Q(R) and Q(S) to obtain Q(T), it might be fast-

er than a general clique det. tion algorithm applied to

T -- even if the algorithm of Theorem 4.2.13 is slower than

that other clique detection algorithm.


4.4 Related Algebras of Covers


Let G denote the .family of all graphs, G = (X, E),

where X is a fixed finite set. G' = (X, E') is a subgraph

of G = (X, E) if E' c E. Because of the reflexivity,

transitivity, and anti-symmetry of the inclusion relation,

it is immediate that (G, <) is a poset, where denotes the

subgraph relation on G. It is equally clear that G v G' =

(X, E u E') is the least upper bound of G and G' and that

G A G' = (X, E n E') is their greatest lower bound. More-

over, because of the distributivity of the set operators, u

and n, G A (G' v G") = (G A G') v (G A G"), so (G,) is

a distributive lattice'.

Now G has a unit, G, = (X, U) -- the complete graph

on X, where U is the set of all two-element subsets of E;

and G has a zero, GO = (X, 0), the totally disconnected

graph on X. The complement of G = (X, E) is defined to be

G = (X, E), where a two element subset, {x, y},belongs to

E if and only if it does not belong to E. Since

G v G = (X, E u E) = (X, E1) = G1 and G G =

(X, E n E) = (X 0) = GO, (G,) is a complemented distribu-

tive lattice -- (G, v, A, -) is a Boolean algebra.





50

The association of the graph gA = (X, E(A)) with a

given efficient cover A of X, which was mentioned in

Chapter 2 in connection with graphical covers, defines

a function, g: E + G.


Proposition 4.4.1. The function g: E G, is an order-

preserving mapping from (Z,5) onto (G,5).


Proof: Suppose A B in E. If {, y} E E (A) then there

exists A E A that {x, y} c A. Since A < B, there exists

B e B such that A c B: {x, y} c B e B implies {x, y} E(B).

Therefore, gA < gB.

The fact that g is surjective is seen as follows.

If G = (X, E) E G, e(X u E) E E and g(e(X u E)) = G. 0


The relation of graphical equivalence on E is defined

as follows: A 'x P in case gA = gB. It is evident that n is

an equivalence relation and that the family of %-equiva-

lence classes is E/% = {g- G: G E G}. For A E E the

equivalence class, tB: gB = gA}, is denoted [A]. Thus

A e [A] = g1 gA).


Proposition 4.4.2. If A E then [A] is the interval

sublattice, [e(X u E(A)), Q(gA)], of E.


Proof: If B E [A] then E(X u E(B)) = E(X u E(A)) and

Q(gB) = Q(gA). Since e(X u E(B)) s B Q(gB), B is a mem-

ber of the interval, [e(X u E(A)),Q(gA)].





51

Conversely, if e(X u E(A)) 5 B Q(gA), then

gA = g(e(X u E(A)) < gB 5 g(0(gA)) = gA follows from the

isotonicity of g. Hence, gB = gA, which proves the converse

inclusion. D


Proposition 4.4.3. Graphical equivalence is a congruence

relation on (E,V,A).


Proof: Suppose A, B and C are members of E and that A q B.

It is to be shown that A v C B v C and A A C B A C.

Suppose {x, y} e E(A v C); there exists D E A v C such

that {x, y} c D. Since A v C c A u C, D E A or D E C.

If D E C c B u C then D is contained in a maximal member D'

of B u C, so that {{x, y}} B v C, whence {x, y} E E(B v C).

If D E A than {x, y} e E(A) = E(B) since A % B, and E(B) c

E(B v C), so {x, y} E E (B v C). Hence E(A v C) c E(B v C).

Interchanging the roles of A and B in the preceding

argument, E(B v C) c E(A v C). Therefore, A v C, B v C.

Now assume {x,y} E E(A A C). By the definition of

A A C, there exist A E A and C E C such that {x, y} c A u C.

Since A n B, there exists B c B with {x, y} c B, so that

{x,.y} c B n C e i(B, C) and {{x, y}} < ei(B, C) = B A C.

Hence, {x, y} E E(B A C). Again by a parallel argument,

E(B A C) c E(A A C), as well. D


In view of the preceding proposition, the following

operations on E/% are well-defined: [A] v [B] = [A v B].

[A] A [B] = [A A B]. That (E/',V,A) is a distributive lattice

follows at once from these definitions and the fact that





52

(E,v,A) is a distributive lattice. Defining, as usual,

on E/% by [A] < [B] if and only if [A] v [B]= [B], is a

partial order with greatest lower bound g.l.b.([A], [B]) =

[A] A [B] and least upper bound l.u.b.([A], [B]) =

[A] v [B].


Theorem 4.4.4. The mapping, ^([A]) = g(A), is a lattice

isomorphism.


Proof: Because B E [A] and C E [A] implies gc = gB, g is

well-defined. That g is surjective follows from

g[Q(G)] = G; it is injective since [A] $ [B] implies gA $

gB.

Suppose [A] [B]. Then [A] = [A] v [B] = [A v B],

so that gCA] = gA < g(A v B) = A[A v B] = g(CA] v [B]) =

g [B ] = gB.

Suppose g[Al3 g[B]; gA gB. And e(A u gA) 5

e(X u gB) in E, so e(X u gA) v e(X u gB) = e(X u gB). Also,

e(X u gA) E [A] and e(X u gB) e [B], so [A] v [B]=

[e(X u gB)] = [B], whence, [A] < [B]. O


In view of the preceding theorem, G may be regarded as

a quotient of E. A very natural way to view G as a subset

of E is to identify G = (X, E) with e(X u E), from which it

is almost indistinguishable anyway. Moreover, e(X u E) is

the least member of g- 0 -- and the only member which could

reasonably be considered to be a graph, since all other

members have subsets of X of cardinality larger than two.





53

Finally, the order structure of G is of course preserved

under this identification: e(X u E) refines e(X u E') if

and only if (X, E) is a subgraph of (X, E').

The other natural transversal for E/" is to represent

each A by its greatest member, Q(gA). By Theorem 3.1.2,

this subset of Z is precisely 0, the family of all graphi-

cal covers of X. Plainly, if A and B are members of 0 then

A refines B -- as members of E -- if and only if gA is a

subgraph of gB, so that (0,5) is order isomorphic with

(G,<). In other words, 0 is not only a poset with respect

to the refinement relation inherited from Z, but it is a

distributive complemented lattice under that relation.

The greatest lower bound of two graphical covers,

A and B, was identified by Bednarek [ 1].


Proposition 4.4.5. [Bednarek] If A E B E Z, and both

are graphical,then A A B is graphical.


Proof: Let Al n Bl, A2 n B2, and A n B be members

of ei(A, B),with the Ai e A and the B. e B, and assume

x e m(A1 n BV, A2 n B2, A n B ) = (A1 n B1 n A2 n B2) u

(A1 n B n A n B ) u (A n B n A n B ). For some two

members i and j of {1, 2, 31, x e Ai n A, so x E

m(A1, A2, A3). Similarly, x E m(B1, B2, B3).

Since A and B are graphical, there exist A E A and

B e B with m(A1, A2, A3) c A and m(B1, B2, B3) c B. Hence,

m(A1 n B1, A2 n B2, A3 n B3) c A n B i(A, B), so

m(A AB) A A A B and A A B is graphical.





54

The join in Z of two graphical covers, however, need

not be graphical: A = {{1, ", 3), {3, 4}}, B = {{1, 2},

{1, 3}, {2, 4}, {3, 4}}, and A v B = {{1, 2, 3}, {2, 4},

{3, 4}} is not graphical since {2, 3, 4} E m(A u B).

Of course the join in E of graphical covers A and B

is graphically equivalent to their join in 0, so their

join in 0, Q(A v B), is known by Theorem 4.3.1 to be
k-2
(em) k(A v B), where k is the largest cardinality of a

clique of g(A v B).

There is another algebra of covers residing in E

which should be mentioned, namely, the family H of the

partitions of X. Like 0, n is not a sublattice of Z,

but it is a lattice under refinement. In this case,

the join and meet of members of H can both be described

in terms of the two members. Clearly, if A and B are

partitions, no two members of ei(A, B) meet, so the join

in n of a pair of partitions is their join in Z. The

least upper bound of A and B is {u K.: i = 1, 2, ..., r},

where {Ki: i = 1, 2, ..., r} is the family of connected

components of the intersection graph of the family A u B --

that is, distinct members C and D of A u B are adjacent in

case C n D i 0. Although this description of the join in I

may seem rather indirect, the computation of the components

of a graph is not very costly. Indeed, it is almost cer-

tain that the join in H is easier to compute than that

in e -- although certainly more difficult to evaluate than

the join in E.





55

Summarizing, n c 0 c Z, and G = 0 = E/n; the meet of a

pair of members, A and B, oT H, 0, or E is given by

ei(A, B), while the joins are computed differently in each.

Apparently, the join in o is more costly to compute than

in either n or E -- in spite of the fact that n c 0 c E.

(Evidently, "complexity" is not order-preserving!) Fur-

thermore, the join is.most easily computed in G, and G = 0.

This latter apparent paradox dissipates quickly, however,

when one considers that the "mapping" from G to 0 is the

computation of the cliques of a given graph -- a problem

known to be NP-complete [ 9 ].

Consider now the family, A, of hereditary covers of X,

partially ordered by inclusion, A = (A c 2X: X = u A, and

if A c B E A then A E A}. Since every singleton of X be-

longs to every member of A, the intersection of two mem-

bers, C and D, of A is again a cover of X. Moreover, if

A c B E C n D then A e C n D, so C n D E A. That C u D e A

is clear. Thus, (A,c) is a lattice.


Theorem 4.4.6. The mapping, e: A Z,is a lattice

isomorphism.


Proof: If AE A then eA E E, since X is finite. If A, Be A

and A c B, then eA eB is immediate.

Let k: EZ A be defined by kA = {B: B c A e A}.

If Ac E then ekA = A, and if BEA then keB = B. D


If A E A E E and 2 < IAI then k(P (A, A)) = kA \ {A}.

A predecessor of a member B of A is obtained by the





56

deletion from B of one of its maximal non-singleton members.

Some use of these observations about A will be made in the

application described in the following section.


4.5 A Metric for Z


Because E is a finite distributive lattice, it is, in

particular, a modular. lattice of finite length. Conse-

quently, by Theorem 4.5.1 below, d(A, B) = h(A v B) -

h(A A B) is a metric on E, where h(A) -- the height of A --

is the length of any maximal chain in E joining the zero

of E and A.

Although the proof of the theorem can be found in

Birkhoff [ 4 ], it is distributed over various sections of

the book, so an outline of the proof is given in Appendix A.


Theorem 4.5.1. If h is the height function on a modular

lattice L of finite length, then d:L x L R, defined by

d(x, y) = h(x v y) h(x A y), is a metric on L.


A straightforward implementation of the definition of

d requires the computation of the join and meet of the two

members of E, two height computations in E, and a subtrac-

tion. Joins and meets in Z are not difficult: A v B =

e(A u B), A A B = ei(A, B). The following proposition pro-

vides a computational specification for the height of a

member of E.


Theorem 4.5.2 Let 1XI = n, be the lattice of efficient
IBI+1 InBI
covers of X, and A E E. Then h(A) = Z (-1) 2
0B c A
(n + 1).





57

Proof: Because E and A are isomorphic, the height functions,

h and H, are the same -- th;tA is, if A E E then H(kA)= hA,

and if B E A then h(eB) = H(B).

The zero, 0, of A is {{1}, {2}, ..., {n},0}, and

h(0) = 0, so h(0) = 101 (n + 1). By Theorem 4.4.6 and

Proposition 4.2.11, if E and F are members of A then E pre-

cedes F if and only if E = F \ {M}, where M is a maximal

non-singleton member of F. In that case, H(F) = H(E) + 1.

By induction, if E c F then H(F) H(E) = IF \ El. In

particular, with E = 0, H(F) = IF \ 01 = FI (n + 1).

Thus, if A E then h(A) = IkAl (n + 1).

Now, IkAl is the number of subsets of members of A,

which by the Inclusion-Exclusion Principle, is given by

the summation in the statement. D


The number of terms of the sum is just one less than

the 2 However, an implementation need not compute the

terms for supersets C of subsets B of A such that nB = 0.

Computing terms by order of ascending cardinalities of B,

it would not be difficult to avoid the futile computation,

2k = (1 + l)k -- the latter being evaluated by means of

the Binomial Theorem.

The following corollary is an immediate consequence of

Theorem 4.5.2


Corollary 4.5.3. Let (E, 5) be the lattice of efficient

covers of a set X of cardinality n. Then the length of E

is given by 1(E) = 2n (n + 1).






58

The corollary could also have been established by

reference to a more general theorem [ 4 ], which asserts

that the length of any finite distributive lattice, L, is

the number of its join-irreducible elements. An element, x,

of a lattice is join-irreducible in case x / 0 and

x = y v z implies x = y or x = z. Moreover, every element,

x, of L has a unique representation as a join of a join-

irreduntant family of join-irreducible elements. A set

is join-irredundant in case the join of all its elements

is strictly greater than the join of the elements of any

proper subset of itself.

If x and y are distinct predecessors of z, then z is

not join-irreducible, since z = x v y; if z has only one

predecessor, w, then the join of any family of elements

strictly less than z is less than or equal to w, and so is

not z; thus, join-irreducible elements are those with

exactly one predecessor.

If E, therefore, the join-irreducible elements are

those covers having just one non-singleton. These are easy

to count: there are C(n, k) whose only non-singleton has k

elements, and the total number of them is therefore given
n n
by E C(n, k) = 2n (C(n, 0) + C(n, 1)) = 2n (1 + n).
k=2
Another expression for the distance between members

of E is given in the following proposition.


Proposition 4.5. The distance between A and B in Z is the

cardinality of the symmetric difference of kA and kB:

d(A, B) = IkA A kB|.





59

Proof: d(A, B) = h(A v B) h(A A B) = H(k(A v B)) -

H(k(A A B)) = H(kA v kB) H(kA A kB) = H(kA u kB) -

H(kA n kB) = (IkA u kBI (n + 1 )) (IkA n kBI (n + 1))

= IkA u kBI IkA n kBI = 1(kA ukB) \ (kA n kB)t =

IkA A kBI. D


Finally, d(A,B) could be computed by constructing a

maximal chain from A v B down to A A B, with the aid of

Theorem 4.2.4 and the following observation. Suppose U < V

and W is a non-singleton member of V; then U P(V, W) if

and only if {W} $ U. Therefore, it would not be necessary

to blindly compute predecessors, P(V, W), until the condi-

tion, U < P(V, W), were satisfied. Instead, the condition,

{W} U, could be checked (over non-singletons belonging to V)

until one was found satisfying the condition; then P(V, W)

would be computed in the certainty that U P(V, W) and

P(V, W) precedes V.

To illustrate many of the notions of this chapter, a

diagram of the lattice of efficient covers of {1, 2, 3, 4}

has been prepared. Figure 4.4 has those covers A such

that 5 5 height(A); Figure 4.5 has those members A such

that height(A) < 5.































































Figure 4.4































































Figure 4.5











APPENDIX A

OUTLINE OF A PROOF OF THEOREM 4.5.1


The required definitions are as in Birkhoff [4 ].

Let P be a poset with a least element, 0. The height

of an element x of P is given by h(x) = sup{l(C): C is a

chain in P, OE C, and x is the greatest member of C},

where 1(C) = ICI 1 is the length of a chain, C. P is

graded by an integer-valued function, r:P Z, in case

(i) x < y in P implies r(x) : r(y), and (ii) if ytis a

successor of x then r(y) = r(x) + 1.

The Jordon-Dedekind Chain Condition (JDCC) for a poset

P is the condition that all maximal chains joining each

pair of elements have the same finite length. P is upper

semimodular in case the existence of a common predecessor

of two elements implies the existence of a common successor;

lower semimodularity is defined dually.

A valuation on a lattice L is a real-valued function,

v: L R, such that v(x) + v(y) = v(x v y) + v(x A y) for

all x and y of L; v is isotone in case x y implies

v(x) v(y), and positive in case x < y implies v(x) < v(y).

The structure of the outline is as follows. First, a

sequence of lemmas is given which establish (Theorem A.6)

that the height function for a modular lattice of finite

length is a positive valuation.





63

This is followed by a sequence of lemmas which lead to

the following conclusion (Theorem A.13): if v is a positive

valuation on a lattice, L, then v(x v y) v(x A y) is a

metric on L.

Theorem 4.5.1 -- which states that if h is the height

function for a modular lattice of finite length then

h(x v y) h(x A y) is a metric -- follows at once from

Theorem A.6 and Theorem A.13.


Lemma A.1. A poset P with Osatisfies JDCC if and only

if P is graded by h.


Lemma A.2. A modular lattice is upper semimodular and lower

semimodular.


Lemma A.3. If a lattice, L, of finite length is upper

semimodular or lower semimodular then L satisfies JDCC.


Lemma A.4. If L is a lattice of finite length then if L is

upper semimodular or lower semimodular then L is graded by

h.


Proof: Lemmas A.1, A.2, and A.3. O


Lemma A.5. Let L be a lattice of finite length. Then L is

upper semimodular if and only if h(x) + h(y) 2 h(x v y) +

h(x A y), and lower semimodular if and only if h(x) + h(y) <

h(x v y) + h(x A y), for all x and y of L.


Theorem A.6. If L is a modular lattice of finite length,

then h is a positive valuation.







Proof: Lemmas A.4 and A.5. D


In the remaining propositions, L is a fixed lattice, v

is a valuation on L, and d(x, y) = v(xvy) v(x A y).


Lemma A.7. d(x, x) = 0, and d(x, y) = d(y, x). If v is

isotone then d is non-negative. If v is isotone, then v

is positive if and only if 0 < d(x, y) for distinct x

and y.


Lemma A.8. If al < a2- ...< an then d(al, an)

d(al, a2) + d(a2, a3) + ... + d(anI, an).


Lemma A.9. d(x, y) = d(x v y x A y) = d(x v y, y) +

d(y, x A y).


Lemma A.10. If v is isotone than a b < c d implies that

d(b, c) 5 d(a, b).


Proof: Lemmas A.7 and'A.8. D


Lemma A.11. If v is isotone than d(a v b, a v c) +

d(a A b, a A c) 5 d(b, c).


Proof: Lemmas A.7, A.8, and A.9. D


Lemma A.12. If v is isotone then d(x, z) 5 d(z, y) +

d(y, x).


Proof: Lemmas A.8, A.10, and A.11.


Theorem A.13. If v is positive then d is a metric.


Proof: Lemmas A.7 and A.12.










REFERENCES


[1] Bednarek, A. R., private communication, March 1978.

[2] Bednarek, A. R., K. D. Magill, Jr., and E. M. Norris,
"Binary relations," Mathematics Technical Report
no. 08A05-1, University of South Carolina, 1974.

[3] Bednarek, A. R.'and S. J. Stolarski, "On a certain
class of covers," Nieuw Archief voor Wiskunde,
16 (1968), 85 89.

[4] Birkhoff, Garrett, Lattice Theory, American Mathe-
matical Society Colloquium Publications, vol.
25, 1940.

[5] Bron, C. and J. Kerbosch, "Finding all cliques of an
undirected graph," Commun. ACM, 16 (1973),
575 577.

[6] Fay, G., "An algorithm for finite Galois connec-
tions," Technical Report 73, Institute for In-
dustrial Economy, Organization and Computation
Technique, 1973, Budapest.

[7] Haralick, R. M., "The diclique representation of
binary relations," J. ACM, 21 (1974), 356 366.

[8] Harary, F., Graph Theory, Addison-Wesley, Reading,
1969.

[9] Karp, R. M., "Reducibility among combinatorial prob-
lems," Technical Report 3, University of Cali-
fornia, Berkeley, April 1972.

[10]. Moon, J. and L. Moser, "On cliques in graphs," Israel
J. Math., 3 (1965), 23 28.

[11] Mukhopadhyay, A., "The square of a graph," J. Comb.
Theory, 2 (1967), 290 295.

[12] Norris, E. M., "An algorithm for computing the maxi-
mal rectangles in a binary relation," Rev. Roum.
Math. Pures et Appl., 23 (1978), 243 250.







[13] Norris, E. M., private communication of a diclique
algorithm described by an unidentified
referee of a ;bmitted paper of Norris, 1978.

[14] Osteen, R. E., "Multi-level automatic
classification for sequential reference
retrieval," doctoral dissertation, University
of Florida, 1972.

[15] Sierpinski, W., Cardinal and Ordinal Numbers,
Polska Akademia Nauk Monografie Matematczne,
Tom 34, 1958, Warszawa.

[16] Zelinka, B., "A remark on systems of maximal
cliques of a graph," Czechoslovak Math-
ematical Journal, 27 (1977), 617 618.











BIOGRAPHICAL SKETCH


Robert Ernest Osteen, the son of Robert Truby Osteen

and the former Jennie Louise Hasenbalg, was born in Saint

Augustine, Florida, June 13, 1936. His childhood was lived

in Saint Augustine, where he graduated from Ketterlinus

High School in 1954.

From September 1954 to September 1958, R. E. Osteen

served as an aviation electronics technician in the United

States Navy. In September 1958 he entered the University

of Florida, majoring in electrical engineering. In June

1962 he was awarded the degree B. E. E., with honors.

From July 1962 to March 1966, R. E. Osteen was a member

of the technical staff of the Electronic Switching Systems

Development Division of Bell Telephone Laboratories at

Holmdel, New Jersey. As a trainee in the Communications

Development Training program, he was enrolled as a part-

time graduate student in New York University, earning the

degree M. E. E. in June 1964.

From March 1966 to August 1967, Mr. Osteen was employed

as a senior engineer by the Defense, Space, and Special

Systems Group of the Burroughs Corporation at the Great

Valley Laboratory, Paoli, Pennsylvania. In September 1967

he returned to the University of Florida to continue his

studies in electrical engineering; he was awarded the

degree, Doctor of Philosophy, in August 1972.







He worked for a while as Assistant Professor of

Electrical Engineering at the Florida Institute of

Technology.

Then in 1974 Osteen returned, yet again, to the

University of Florida to resume his graduate studies --

this time in 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 ad luate, 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. K/Brooks
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.




N. L. White
Associate 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.




C. V. Shaffer
Professor of Electrical Engineering










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





S. Ulam
Graduate Research Professor
of Mathematics


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.


Dean, Graduate School


August, 1980


































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