set intersection representations for almost all graphs

Set Intersect ion Representations for Almost All Graphs

Nancy Eaton DEPARTMENT OF MATHEMATICS

THE UNIVERSITY OF RHODE ISLAND KINGSTON, RI 02881 USA e-mail: ea ton @ma th. un. edi

David A. Grable

HUMBOLDT-UNIVERSITAT zu BERLIN INSTITUT FUR INFORMATIK

0-10099 BERLIN GERMANY e-mail: grableQinformatik. hu-berlin. de

ABSTRACT

Two variations of set intersection representation are investigated and upper and lower bounds on the minimum number of labels with which a graph may be represented are found that hold for almost all graphs. Specifically, if &(G) is defined to be the minimum number of labels with which G may be represented using the rule that two vertices are adjacent if and only if they share a t least k labels, there exist positive constants Ck and ck such that almost every graph G on n vertices satisfies

ckn2 c i n2 log2 n log2 n

I &(G) I -.

Changing the representation only slightly by defining Oodd(G) to be the minimum number of labels with which G can be represented using the rule that two vertices are adjacent if and only if they share an odd number of labels results in quite different behavior. Namely, almost every graph G satisfies

Journal of Graph Theory Vol. 23, No. 3, 309-320 (1996) 0 1996 John Wiley &Sons, Inc. CCC 0364-9024/96/030309-12

310 JOURNAL OF GRAPH THEORY

Furthermore, the upper bound on Oocicl(G) holds for every graph. 0 1996 John Wiley & Sons, Inc.

1. INTRODUCTION

An interesting problem in computational graph theory is to find a method of representing graphs that is efficient in terms of both space and time. One way to represent a graph is to assign a set of labels from some universal set to each vertex of the graph in such a way that two vertices are joined by an edge if and only if their label sets have nonempty intersection. E. Szpilrajn-Marczewski [ 121 first proved in 1945 that such an intersection representation exists for every graph.

Plainly, the efficiency of this representation depends mostly on the size of the universal set of labels. For certain classes of graphs, the least number of labels needed in a intersection representation is relatively small (graphs that are the edge-disjoint union of I cliques can be represented using only 1 labels and must use at least 1 labels and so trees on n vertices require only n - 1 labels), but most graphs require a relatively large number of labels. We use the expression “almost all graphs have property P” to mean that the proportion of graphs that do not have property P goes to 0 as the number of vertices goes to infinity. A result of Poljak, Rodl, and Turzik [ 111 shows that almost all graphs on n vertices require at least en2/ log2 n labels and a recent result of Frieze and Reed [8] shows that almost all graphs require no more than c’n2/ log2 n labels, where c and c’ are positive constants. The constant c’ is no smaller than 5 but the constant c is not given explicitly. Earlier, Erdos, Goodman, and Posa had shown that every graph can be represented with at most Ln2/41 labels [7].

In [9], Kou, Stockmeyer, and Wong showed that determining the number of labels required to represent a given graph is NP-complete. Combining their proof with the nonapproximability results of Arora, Lund, Motwani, Sudan, and Szegedy [2] demonstrates that, in fact, there exists a positive F so that no polynomial time algorithm can approximate the number of labels required to represent a given graph on n vertices to within a factor of nE unless P = NP.

In this paper, we address two variants of this representation method about which much less was known. These variants involve changing the intersection rule that determines the presence of an edge. In the first, two vertices are joined by an edge if and only if their sets intersect in at least k labels and, in the second, if and only if their sets intersect in an odd number of labels.

Being a bit more formal, for a positive integer k , a k-representation of a simple graph G = (V, E ) is a set U of labels and an assignment T : V + 2u of sets of labels to the vertices in such a way that two vertices u and 21 are adjacent if and only if Ir(u)nr(v)l 2 lc. The term odd-representation is defined by changing the intersection rule to “lr(u) n r(u)l is odd.”

As mentioned above, the key question is “how large must the universe of labels be in order to represent a given graph G?’ For k-representations, this number is called &(GI and, for odd-representations, it is called O,dd(G). Our primary purpose here is to give good upper and lower bounds on these parameters for almost all graphs. For an analysis of &(G) for some other classes of graphs the reader is referred to [51 and [61.

SET INTERSECTIONS REPRESENTATIONS 31 1

2. k-REPRESENTATIONS

The following relationship between &(G) and &(G), for k 2 2 holds for all graphs G.

k(OI(G))l/k 5 &(G) I &(G) + k - 1. e

The upper bound is obtained by starting with a 1-representation and adding the same k - 1 new labels to each vertex. The lower bound is obtained for an arbitrary graph G by considering T, a k-representation of G. For each of (“(f))k-sets of labels, L, create a new label and assign it to each vertex zi such that L C ~(w). The assignment of new labels is a 1-intersection representation.

One might expect that as one moves to larger values of k, fewer labels are required to represent a given graph, so that &(G) would be closer to the lower bound given above. This expectation is borne out by the following example: consider a graph G composed of 1 disjoint, nontrivial cliques. It is not hard to see that for such a graph, &(G) is the smallest positive integer such that (“(f)) 2 1.

It is perhaps mildly surprising that for most graphs, in order to find a k-representation, one cannot do significantly better than the upper bound for &(G) given above. This is the content of our first theorem.

Theorem 1. Let k 2 2 be a fixed positive integer. There exist positive constants Ck and c i depending only on k such that for almost every graph G with vertex set [n],

The upper bound follows immediately from the result of Frieze and Reed [8] and the technique of adding k - 1 new labels. It is the lower bound that is interesting.

The density of induced subgraphs of a graph will play an important role here. Let G be a graph and let X be a subset of its vertex set. The density of the subgraph G x of G induced by X is defined as density (Gx) = ( E ( G x ) ( / ( l f l ) . The following lemma concerning the density of subgraphs of almost all graphs plays a key role.

Lemma 2. Let 0 < E 5 1/24. Almost every graph with vertex set [n] has the property that, for every set of vertices X with 1x1 2 (4 /~ ’ ) logn,

1 1 - - E < density(Gx) < - + E . 2 2

Proof. Let G be a random graph with vertex set [n]. For now, let X be a fixed subset of [n] and J: = 1x1. The number of edges in G x is a binomially distributed random variable with parameters 1/2 and (5). Using a well-known result [3, Theorem 1.71 we see that as long as ($) 2 2 4 / ~ ,

In other words, the density of G x is outside of the interval (f - E , f + E ) only with this small probability.


Now let s be an arbitrary integer. The probability that there exists a set of vertices X with at least s elements such that the density of G x is outside the interval is at most

2 (;i> ( 2 2 -;&2 (;)} < C e x p { zlogn - 2 & 2 3 ( ; )}

z = s

TL

z = s

< exp{-(:&2( i ) - ( s + l ) l o g n ) } .

We are interested in determining when this quantity goes to zero as n goes to infinity. It turns out that any s 2 ( 4 / ~ ~ ) logn suffices. Namely, such a choice implies that s - ( 3 / c 2 ) logn 4 00 and therefore that : E ~ ( $ ) - (s + 1) logn -+ co, forcing (1) to go to zero. Consequently, if s 2 ( 4 , ’ ~ ~ ) logn, almost no graph has an induced subgraph with at least s vertices whose density is outside the interval (: - E , f + E ) . This is equivalent to the statement of the lemma. I

Consider a k-representation of a graph G that uses only O k ( G ) labels. For each label i, let C, denote the set of vertices having i as a label. The sets C, have the property that a pair of vertices is an edge in G if and only if they are contained together in at least k of the sets. For each i, let e, be the number of edges in the subgraph induced by C,, let n, be the number of nonedge pairs in the subgraph induced by C,, and let c, = e, + 77, = ( .I’ ) . Since each edge of G is contained in at least k of the C,

I(- I

i

and since each nonedge pair is contained in at most ( k - 1) of the C,,

It is well known that, for every fixed 6 > 0, almost every graph G has

Thus, almost every graph has

and

Therefore, almost every graph has

(c c, - c %) ( k - 1) (; + 6) cnz = ( k - 1)(1 + 26) k ( ; - 6 ) k ( 1 - 26) Eel 2

SET INTERSECTIONS REPRESENTATIONS 313

Simplifying and setting 6 = 1/(8k - 6) gives

k( 1 - 25) 2k c e - “ 2 k - 1 - 2 5 > ccz = mxcz (3)

for almost every graph. Here, we used that k 2 2.

Specifically, define Next, set E = 1/(8(4k - 1)) 5 1/24. Let S be the set of labels i with C, “small.”

S = {ZJJCzJ < (4/€’)10gn}.

Let L contain the remaining labels. By lemma 2, almost every graph G has density ( G x ) < + E for every ‘‘large’’ set X. In particular, for every i E L,ei < (i + E ) c ~ . Therefore,

’ i t L \ - ztS i E L 1 t s

Combining (3) and (4) gives

This says that

c c2 S(2k - 1)

x c z 5 z t L ? E S

for almost every graph. Turning again to (2) we see that

1ES z t L

By the definition of S, for i E S,

1 8 2 &4

c, < -lCJ’ < -log2n.

Therefore, 1 6 k - 5 8 I S ( & log’n


which says that

Sk(2k - 2 ) n(n - 1) n2 > ck- Is’ > 216(4k - 3)(16k - 5)(4k - log2n log2n

for some constant C k . Since 1st is the number of labels with “small” Ci there are at least this many labels and so Bk(G) is bounded below by this expression for almost every graph.

I We mention also that one can use martingale tail inequalities such as Azuma’s Inequal-

ity (see, e.g., [l, Chapter 71) to show that &(G) is even more tightly concentrated than Theorem 1 indicates. Specifically, noting that as each new vertex is exposed at most kn new labels are required, Azuma’s Inequality implies that, for any positive constant A,

This completes the proof of Theorem 1.

Pr[JBk(G) - Ex[&(G)]I > Xkn3/2] < 2e-X2/2.

3. ODD-REPRESENTATIONS

In stark contrast to the k-representations, which require a tremendous number of labels, the odd-representations require relatively few. Our first result is that every graph on n vertices can be odd-represented with at most n - 1 labels. We present an explicit algorithm for finding the representation. Afterward, we show that almost all graphs require at least n - v‘% - [log n1 labels, demonstrating that our algorithm is essentially best possible.

Note that we do not have any complexity results concerning odd-representation. It is an open and perhaps interesting problem to determine whether finding &~, I (G) is NP- complete.

The next result originally appeared in the first author’s Ph.D. dissertation [4]. It states that given a graph G with vertex set [n], the following algorithm generates an odd-representation using the universal set U = { 1 1 , l 2 , . . . , 1 7 c - l } .

Algorithm OddRep

Inputs: A positive integer n and a graph G with vertex set [n].

Outputs: Subsets R1, . . . , R,, of U = { 1 1 , 1 2 , . . . , l n - ] } . begin for i +- 1 to n do

begin R, t 8. for J t 1 to z - 1 do

if ( 1 R, Ti RJ 1 is odd) xor (21 is an edge of G) then Add 1, to R,.

i f z < n then Add I , to R,. output R,. end

end

It is not too difficult to see that this algorithm always produces an odd-representation. Notice that the R, are built in-order and never altered and that, except for R,,, I , is always

SET INTERSECTIONS REPRESENTATIONS 31 5

the maximal element of R3. This ensures that once we decide whether or not to include I , in R,, no subsequent decision can affect the presence or absence of the edge ij.

Thus, every graph on n vertices can be odd-represented with at most n - 1 labels. But how many labels are necessary? In the case of &representations we were satisfied to find an answer with the same order of magnitude as the upper bound. That much is easy in the case of odd-representations. A simple counting argument shows that there are only 2tn odd-representations of graphs on n labeled vertices using t labels. Since there are 2”((”-1)/2 labeled graphs with n vertices, if we hope to represent almost all of them, we require at least n/2 - o(n) labels.

We found this answer unsatisfactory and decided to determine the constant coefficient of n. It is 1. By the way, we knew that such a constant coefficient must exist by Azuma’s Inequality. To apply Azuma’s Inequality, we notice that at most two new labels are required in an odd-representation as each new vertex is exposed in a graph. Add the two new labels to each previously exposed vertex that is adjacent to the new vertex, thus not disturbing the parity of the intersections of sets assigned to pairs of previously exposed vertices, and to the new vertex, only assign one of the two new labels.

Thus, for a random graph G and any positive constant A,

In other words, we knew that almost all graphs have their values of Oodd in an interval only a little wider than fi. But of course this did not tell us where that interval is located.

Now, although we are really only interested in representing graphs without loops, it will be convenient to extend the concept to the class of looped graphs-that is, the class of graphs where a single loop is allowed (but not required) on each vertex. The extension of the representation is obvious: a vertex has a loop if and only if it has an odd number of labels (its set has odd intersection with itself). Use OLdd(G’) to denote the fewest number of labels required to represent a looped graph G’.

Since our usual odd-representation does not care about the cardinalities of the representing sets, we see that d,dd(G) = min6Add(G’), where the minimum is over all looped variants G’ of G. We will prove that, for any a > 0, the number of looped graphs G’ with OLdCl(G’) 5 n - a is at most 2(nZ+n-a2-a+2ar10gnl)/2, which gives us that the number of ordinary graphs G with O,dd(G) 5 n - a is also at most 2 ( n z + n - a 2 - - a f 2 a ~ 1 0 ~ n ~ ) ~ 2 (in the worst case that every bad G’ gives a different bad G).

If we set a 2 6 + [logn], this last result tells us that the number of graphs G with Oodd(G) 5 n - a is asymptotically smaller than 271(n-1 ) /2 , the total number of graphs. In other words, almost every graph has Oodd(G) > n - 6 - [logn].

We do not count looped graphs with low odd-representation numbers directly, though. First, we will show that for any looped graph G’,Ohdd(G’) 2 rank(Adj(G’)). Then we will show that at most 2 ( n 2 f n - a 2 - - a f 2 a r 1 0 g n l ) / 2 of the n x n symmetric matrices over GF[2] (adjacency matrices of looped graphs with n vertices) have rank at most n - a. Combining these facts, which we will prove as lemmas below, we see that there are at most 2(nZ+n-a2-a+2a~10gn1) /2 looped graphs G’ with OAdd(G’) 5 n - a.

Actually our first lemma proves more than we need. In addition to proving that OAdd(G’) 2 rank(Adj(G’)), we also prove that Ofdd(G’) 5 rank(Adj(G’))+l. As a result, the lemma is best possible. To see this, note that a single loopless edge S has @Add ( S ) = 3 = rank(Adj(S)) + 1 and a loopless triangle T has Okdd(T) = 3 = rank(Adj(T)).


Lemma 3. If G’ is a looped graph, then I ~ ~ , ~ ~ ~ , ( G ’ ) is equal to rank(Adj(G’)) or rank(Adj(G’)) + 1.

The idea of the proof is to demonstrate a vector space injection from the row space of the adjacency matrix into GF[2]e!~~1~l(G‘), which is at most one dimension short of being a bijection.

Let T be an odd-representation of an n vertex looped graph G’ using t = 19Ldd(G’) labels and consider it as a function r : V(G’) + GF[2It. Let R C GF[2It be the span of the image of V(G’) under T . Let a(.) E GF[2]“ be the row of the adjacency matrix corresponding to vertex u and let A C GF[2]” be the row space of the adjacency matrix. In other words, let a : V(G’) + GF[2]’l be defined by the adjacency matrix and let A be the span of its image.

Define a vector space injection f : A -+ R C GF[2]‘ by selecting a collection of vertices u l r . . . , u, such that u ( u l ) , . . . , a(u,.) form a basis for A and mapping the point

Proof.

p = c,a(v1) + . . . + c,a(v,)

to

f ( p ) = CIT(V1) + . . . + C,T(V,).

To see that f is an injection, it suffices to show that T ( V ~ ) , . . . ,r(u,) form an independent set in GF[2It. Suppose not. Suppose there were el , . . . , c,, not all zero, so that

f i r ( u 1 ) + . . . + c , T ( ~ , ) = 0’ E GF[2It. ( 5 )

Notice that for any two vertices u and u, r(u) . r ( v ) = 1 (mod 2) if u and u are adjacent and T ( U ) . ~ ( u ) = 0 (mod 2) otherwise. Thus in GF[2] , T ( U ) . ~ ( u ) = a(u),, = a(u),, where a(u), , is the vector coordinate of a(u) corresponding to vertex u. For each vertex u, dot both sides of (5) with ~ ( u ) to get

cla(u*)v + . . . + c,a(v,), = 0.

Taking these equations together, we get the vector equation

cla(u1) + . . . + c,a(v,) = 0’ E GF[2]”.

This contradicts the fact that a ( v l ) , . . . , u(u,) are independent (as they must be since they form a basis for A).

Let R’ & R C GF[2It be the image of f . Since f is an injection, rank(A) = rank(R’) 5 rank(R) 5 t . This is half of the statement of the lemma.

To prove that rank(A) 2 t - 1, define another representation r’ : V(G’) -+ R’ GF[2]’ by ~’(v) = f ( a ( v ) ) . It is straightforward to see that T’ also represents G’ as an odd representation. Let u and u be vertices. We will show that a(u),, = ~’(u) . ~ ’ ( u ) . Write a(.) and a(.) in terms of the basis for A:

a(.) = CIa(u1) + . . . + c,a(v,)

u (v ) = d l U ( U l ) + ’ . . + dza(vz ) .

T ’ ( U ) = f(a(.)) = clr(v1) + ‘ . . + C,T(U,)

First note that


and that

T’(.u) = f ( a ( v ) ) = d l T ( W 1 ) +.. . + d,r(v,) = d l T ’ ( W l ) + . ‘ . + dJ’(7JJ.

Thus, for any basis element v,

’T(Vz) = a(u),” = (c14v1) + . . . + %4u.c))?,t = (CIT(U1) + . ‘ ‘ + C,T(ZI,)) . T(ZIZ)

= (ClT(V1) + ’ . . + C,T(V,) ) . T’(ZI,)

= T’(U) . ?-’(?I,).

4 v ) u = dla(vl)?L + . . . + d.L4%), = d l ( ? - ( U l ) . ?- (U) ) + ’ ’ . + &(?-(v,) ‘ ?-(U))

= dl(T’(W1) . T ’ ( U ) ) + ’ . ’ + d T ( T ’ ( U E ) ’ T’ (U) )

= ?-’(?I) ’ T’ (U) ,

Finally, this implies that

as desired. Next we show that rank(R’) 2 t - 1. Let M be the matrix whose rows are ~ ‘ ( w ) for

v E V(G’). R’ is the row space of M and we show that rank(R’) = rank(M) 2 t - 1 by showing that no even-sized collection of columns of M have sum 0’ E GF[2In. Suppose not. Suppose C1 , . . . , C2k were columns of M with sum 0’. Let C be the submatrix of M formed by these columns and replace any row of C whose last element is 1 with its complement. Call the new matrix formed C’, and let the matrix M’ be the same as M with the columns in the submatrix C replaced by C’. It is easy to see that the interrow dot products of C’, and hence of M’, are preserved by this operation since the complement of any set with respect to an even set has the same parity as the original and every row of C has even parity. But the resulting matrix M‘ then has a column of Os, which may be removed, giving an odd-representation for G’ using t - 1 labels, contradicting the definition of t = OA,,(G’). Since there are no even-sized zero-sum sets of columns, the columns of M must contain at most one minimal dependent set (necessarily of odd cardinality) and so rank(M) 2 t - 1.

We note that a single odd-sized zero-sum set of columns of M cannot be ruled out because the complement of an even set with respect to an odd set is odd. We can, how- ever, rule out the presence of two different odd-sized zero-sum sets since their symmetric difference is an even-sized zero-sum set.

To summarize, we proved that

t - 1 5 rank(M) = rank(R’) = rank(A) 5 rank(R) 5 t.

This proves that t = OE,,(G’) is either equal to rank(A) = rank(Adj(G’)) or rank(Adj(G’))+

Lemma 4. The number of n x n symmetric matrices over GF[2] with rank less than or 1, as desired. I

equal to n - a is at most 2(n2+n-a2-a+zar10gn1)/2.

Proof. In a broad sense, the proof is based on the Incompressibility Method of Kol- mogorov Complexity (see e.g. [lo]), but really it just amounts to compression and simple


counting. Let A be an n x n symmetric matrix over GF[2] and, to simplify notation, assume for now that the first n-a rows form a basis for the row space. In addition, for each i such that n - a < i 5 n, let c % , J , . . . , c , , ~ - ( ~ E GF[2] be such that A, = c , , ~ A ~ + . . . + c ~ , ~ - ~ ~ A , - , . As is usual, we denote the ith row by Ai and the i, jth entry by Ai , j .

The following simple algorithm outputs a compressed version of A:

Algorithm Matrixcompress

Inputs: Integers n and a with 0 2 a 5 n, a symmetric n x n matrix over GF[2] whose first n - a rows form a basis for its row space, and for each n - a < i 5 n, coefficients c ~ , ~ , . . . , c ~ , ~ , - ~ ~ E GF[2] such that A,i = ci,lAl + . .. + ci , lL-aAlr-a. Output: A bit string of length (n - a)(. - a + 1) /2 + a ( n - a) .

begin for i + 1 to n - a do

for j + i to n - a do

f o r i + n - a + l tondo for j c 1 to n - a do

output Az, j .

output C, ,? .

end

In words, the algorithm puts out the upper right triangle of the upper left (n - a) x ( n - a) submatrix of A and all of the coefficients ci,j.

A can be easily reconstructed from the resulting bit string in the following manner. first fill in the upper left (n - a ) x (n - a ) square submatrix by placing the given entries of A symmetrically. Then use the c ~ , ~ information to reconstruct the first n - a columns of rows n - a + 1 through n. Reflect these entries to complete the first n - a rows and then reconstruct the lower right submatrix. Explicitly, the decompression algorithm is as follows.

Algorithm MatrixDecompress

Inputs: Integers n and a with 0 5 a _< n and a bit string of length ( n - a ) ( n - a + Output: An n x n matrix A over GF[2]. 1)/2 + a(n - a).

begin for i t 1 to n - a do

for j + i to n - a do begin input Ai ,J . i f i # j then A,,, t end

begin for k + 1 to n - a do Ai,k c 0. for j c 1 to n - a do

fori+-n-a+l t o n d o

begin input qj. for k + 1 to n - a do A,,k +- Ai,k + c i , jAj ,k .


end end

for k t 1 to n - a do Ak,i t A i , k .

begin for k t n - a + 1 to n do Ai,k c 0. for j t 1 to n - a do

for i + n - a + 1 to n do

for i t n - a + 1 to n do

begin for k + n - a + 1 to n do A i , k t Ai , k + ci, jAj,k.

end end

end

Observe that this compression method is valid for any symmetric matrix whose rank is at most n - a simply by including additional nonbasis vectors among the first n - a vectors and never referring to them later.

In the general situation, the basis vectors need not be (among) the first n - a rows, so we need to include a few more bits to indicate which rows do form the basis. The rows and columns of the matrix can then be permuted to bring the basis rows to the top and the given algorithms used or, alternatively, slight variations of the algorithms can be developed.

To choose the method for indicating which rows are part of a basis, we take advantage of the fact that we know that the lemma will be applied for small values of a. We choose to give the binary representations of the a rows that are not part of the basis. This requires a [log n1 bits.

All together, we require only (n - a ) ( n - a + 1)/2 + a ( n - a ) + a[lognl = (n2 + n - a2 - a + 2a[lognl)/2 bits to store a symmetric n x n matrix with rank at most n - a. This means that there may be no more than 2 ( 7 L a + ~ ~ ( 1 2 - a + 2 a r ' o g n 1 ) / 2 such matrices. This

I To summarize, these two lemmas and the discussion preceding them prove the following

finishes the proof of the lemma.

theorem.

Theorem 5. Almost every graph G with n vertices satisfies

TI - 6 - [lognl < O,dd(G) 5 TI - 1.

Furthermore, the upper bound holds for every graph G.

4. ACKNOWLEDGMENT

The first author acknowledges support by NSF Grant No. DMS-9310064.

ical Sciences and The Alexander von Humboldt Foundation for their support. The second author would like to thank Clemson University's Department of Mathemat-

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