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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, NSFMCS7521130. 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 inclusionmaximal 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 inclusionmaximal 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 ofthe 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 nonhierarchical 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 FayHaralick 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 FayHaralick 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 "corectangularity" on the set of directed lines of D. A complete subgraph of G represents a set of directed lines each pair of which are corectangular; 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 subraph 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)e0.02p 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. [BednarekStolarski] 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(Aal, Aa2, Aa3) 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 inclusioncomparable, 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 inclusioncomparable. 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 antisymmetry 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 antisymmetry) 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 nonsingleton 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 < IA, 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 nonsingleton members. Proof: If A and A' are distinct nonsingleton 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 nonsingleton members. But by Theorem 4.2.4, the number of predecessors is not greater than the number of nonsingleton 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 twoelement 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 oneelement 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 twoelement 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 nonsingleton 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 NPcomplete 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 nonnegative 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)tl(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)t2(A), where t is the size of a largest clique of gA. Proof: By Lemma 4.2.12, K(t) (em)t2(A). But K = K(t), since every complete subgraph of gA has t or fewer points. Also, (em)t2(A) < Q(gA), so that Q(gA) c K < (em)t2 < t2 Q(gA), which implies that Q(gA) = (em) t2(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 kelement 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 S1 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 twoelement 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)P2. 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 antisymmetry 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 twoelement 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 welldefined: [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 welldefined. 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 k2 (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 orderpreserving!) 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 NPcomplete [ 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 nonsingleton 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 nonsingleton 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 InclusionExclusion 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 joinirreducible elements. An element, x, of a lattice is joinirreducible 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 joinirreducible elements. A set is joinirredundant 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 joinirreducible, 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, joinirreducible elements are those with exactly one predecessor. If E, therefore, the joinirreducible elements are those covers having just one nonsingleton. These are easy to count: there are C(n, k) whose only nonsingleton 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 nonsingleton 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 nonsingletons 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 integervalued 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 JordonDedekind 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 realvalued 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 nonnegative. 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. 08A051, 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, AddisonWesley, 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., "Multilevel 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 a UNIVERSITY OF FLORIDA hIllllllffII111111 3 1262 07332 042 5 A a 