A finiteness theorem for maximal independent sets

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Graphs and

Combinatorics

A Finiteness Theorem for Maximal Independent Sets

Min-Jen Jou 1., Gerard J. Chang 1, Chiang Lin 2, and Tze-Heng Ma a

1 Department of Applied Mathematics, National Chiao Tung University, Hsinchu 30050, Taiwan. e-mail: gjchang@math.nctu.edu.tw

2 Department of Mathematics, National Central University, Chungli 32054, Taiwan 3 Institute of Information Science, Academia Sinica, Nankang, Taipei 11529, Taiwan

Abstract. Denote by mi(G) the number of maximal independent sets of G. This paper studies the set S(k) of all graphs G with mi(G) = k and without isolated vertices (except G ~ K1) or duplicated vertices. We determine S(1), S(2), and S(3) and prove that IV(G)[ < 2 ~-1 + k - 2 for any G in S(k) and k > 2; consequently, S(k) is finite for any k.

1. Introduction

All graphs in this paper are simple, i.e., finite, undirected, loopless, and without multiple edges. In graph G, an independent set is a subset of V(G) in which every two distinct vertices are nonadjacent. A maximal independent set is an independent set which is not a proper subset of any other independent set. A clique is a subset of vertices in which every two distinct vertices are adjacent. A maximal clique is a clique which is not a proper subset of any other clique. Let MI(G) denote the set of all maximal independent sets of G and mi(G) the size of MI(G).

Erd~Ss and Moser raised the problem of determining the maximum number f(p)

of maximal independent sets possible in a graph with p vertices and that of determining which graphs have this many maximal independent sets. Later, M o o n and Moser I7] gave a complete answer to this problem, which is that for p > 2,

f

Y, ifp = 3t for t > 1,

f ( p ) = 4 4 . y - 1 , i f p = 3 t + l f o r t > l , /

L 2 . 3 ' , i f p = 3 t + 2 f o r t > O ,

and mi(G) = f(p) if and only if

~

tK3, G - ~ ( t - 1)K 3 U K 4 o r ( t - 1)K 3U2K 2, / L t K 3 U

K2,

ifp = 3t, ifp = 3t + 1, i f p = 3 t + 2 .

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322 M.-J. Jou et al. ErdSs and Moser actually raised their problem in terms of maximal cliques, which are maximal independent sets in complement graphs. About two decades later, a number of authors studied the same problem for trees [4, 6, 8-I0], connected graphs [1, 2], triangle-free graphs [3], and bipartite graphs [5].

Instead of determining an upper bound for mi(G), this paper studies mi(G) from another point of view. For a fixed positive integer k, our problem is to determine all graphs G satisfying mi(G) = k. In a graph G, the neighborhood of a vertex x is

N~(x) = {y ~ V(G) : x is adjacent to y in G}.

A vertex x is isolated if NG(x) = ~. Two vertices x and y are duplicated if N~(z) =

Na(y). The following lemmas are trivial.

Lemma 1.1. I f x is an isolated vertex in G, then mi(G - x) = mi(G).

Lemma 1.2. I f x and y are duplicated vertices in G, then mi(G - x) = mi(G).

Proof. The lemma follows from the fact that for any A ~ MI(G), x ~ A if and only

if y ~ A. []

By Lemmas 1.1 and 1.2, deleting an isolated vertex or a duplicated vertex from graph G does not change mi(G), so we shall consider only those graphs without isolated or duplicated vertices. Denote by S(k) the set of all graphs G with mi(G) = k and without isolated vertices (except G-~ K1) or duplicated vertices. In this paper, we determine S(1), S(2), and S(3). We also prove that

I V(G)I _<

2 k-1 + k - 2 for any G in S(k) and k _ 2; consequently, S(k) is finite for any k.

2. Graphs G with mi(G) = k

In this section we first determine S(1), S(2), and S(3). The following idea is useful in this paper: For an independent set B of G there exists at least one A ~ MI(G) such that B _~ A.

Lemma 2.1. I f G is an induced subgraph of H, then mi(G) <_ mi(H).

Proof. For any B ~ MI(G), B is an independent set of H and so there exists at

least one A ~ MI(H) such that B ~ A. Therefore, there exists a function f from

MI(G) to MI(H) such that f(B) ~ MI(H) and B ~_ f(B) for any B ~ MI(G). Since B

is a maximal independent set of G and B c_ f(B),

B = f(B) f7 V(G). (2.1)

Consequently, f is a one-to-one function and so mi(G) < mi(H). []

Lemma 2.2. For any two disjoint graphs G and H, mi(G U H) = mi(G)mi(H). It is straightforward to check that mi(Kn) = n for any n >_ 1, mi(P2) = mi(P3) = 2, mi(P4) = 3, mi(P s) = 4, mi(Ca) = 3, mi(C4) = 2, and mi(Cs) = 5. For the values of

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Lemma 2.3. Suppose G is a graph without duplicated vertices. I f G has a cycle of

length >4, then mi(G) >_ 4. I f G has a cycle of length ___3, then mi(G) -> 3.

Proof. We first consider the case where G = (V, E) has a cycle of length _>4. Choose

such a cycle of minimum length n. For the case of n >_ 5, by the minimality of n, the cycle has no chord, i.e., C, is an induced subgraph of G. If n = 5, then mi(G) ->

mi(C~) = 5 > 4. If n -> 6, then mi(G) -> mi(C,) -> mi(Ps) = 4. Thus we may assume

that G has a 4-cycle C: v 1, v 2, v a, v 4, v 1 . Now consider the following three cases.

Case 1. C has two chords vl v3 and v2v4. In this case, {vl, v2, va, v4} is a clique and

so mi(G) -> rni(K4) >_ 4.

Case 2o C has exactly one chord, say vlv 3 ~ E and v2v4eE. Since G has no

duplicated vertices, there exists a vertex y not in C that is adjacent to exactly one vertex of (vl, v3}, say vly ~ E and ray ¢ E. Choose four maximal independent sets Az, A2, A3, and A4 of G that include {02}, {v4}, {v3,vl}, and {va,y}, respectively. Since {v 2, v3, v4} is a clique and vl is adjacent to y, these four maximal independent sets are distinct. Thus mi(G) -> 4.

Case 3o'C has no chord, i.e., vlv 3 ~ E and v2v ~ ~ E. Since G has no duplicated

vertices, there exist vertices y and z not in C that are adjacent to exactly one vertex of {vl,v3} and {v2,v4}, respectively, say, vly ~ E, v3y q~ E, v2z ~ E, and v,,z ~ E.

Choose four maximal independent sets Al, A2, A3, and A4 of G that include {v3,vl}, (v3,y}, {v4,v2}, and {v,,z}, respectively. Since v3 is adjacent to v4, vl is adjacent to y, and v2 is adjacent to z, these four maximal independent sets are distinct. Thus mi(G) > 4.

Finally, for the case where G has a cycle of length 3, mi(G) > mi(C3) = 3. []

Since any graph G with at least one edge has mi(G) > 2, S(1) = {K1}. Theorem 2.4. S(2) = {P2).

Proof. It is clear that mi(Pz) = 2 and P2 has no isolated or duplicated vertices. On

the other hand, suppose G is in S(2). By Lemma 2.2 and the assumption that G has no isolated vertices, G is connected. If G has a cycle, then mi(G) >_ 3 by Lemma 2.3, which is impossible. Since mi(P4) -- 3, the maximum distance between two vertices of G is at most two. Therefore G is a star and so in fact is P2, as G has no duplicated

vertices. []

Besides P4 and K 3, the two graphs G 1 and G 2 in Fig. 2.1 are such that

rni(G) = 3.

G1 G2

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324 M.-J. Jou et al. Theorem 2.5. S(3) = {P4, K3, G1, G2 }.

Proof. First of all, mi(P¢) = mi(K3) = mi(G1) = mi(G2) = 3 and P4, Ka, G1, and G 2

have no isolated or duplicated vertices. On the other hand, suppose G is in S(3). By Lemma 2.2 and the assumption that G has no isolated vertices, G is connected.

First, G has at most one block B which is not K2 and all other blocks intersect B; otherwise, G contains 2K 2 as an induced subgraph, which implies mi(G)>

mi(2K2) = 4 > 3, a contradiction. Second, B has 2 or 3 vertices, otherwise G has a

cycle of length > 4, which implies mi(G) > 4 > 3 by Lemma 2.3, again a contradiction. For the case where B is K2, there are exactly two other blocks which are K 2 and intersect B at different vertices. This gives/'4. For the case where B is K 3, there are exactly 0, 1, or 2 blocks which are K2 and intersect B at different vertices. This

gives K3, GI, and G 2. []

To generalize the graphs in Theorems 2.4 and 2.5, we consider split graphs. A graph G is split if its vertex set can be partitioned into a clique C = {el, v2,..., vk} and an independent set I =-{ul,u 2 .... ,urn}. For the case of [.) N ( u i ) ~ C,

l < i ~ m

mi(G) = k and the maximal independent sets are {vi} U (I - N ( v i ) ) , 1 < i <_ k. For

the case of U N(ut) = C, mi(G) = k + 1 and besides the above k maximal inde-

l ~ i ~ m

pendent sets, I is the (k + 1)th maximal independent set. Note that the graphs P2, P4, K3, G1, and G2 are all of this form.

For k > 4, it becomes hard to determine S(k). However, we can prove that I V(G)I < 2 k-~ + k - 2 for any G in S(k); consequently, S(k) is finite for any k. Theorem 2.6. I f k > 2 and G ~ S(k), then ! V(G)I < 2 k-t + k - 2.

Proof. Without loss of generality, we may assume that G is in S(k) and has as many

vertices as possible. Let MI(G) = {A1, A2 . . . Ak} and

B(v) = {i: v ~ Ai ~ MI(G)}

for all v ~ V(G). It is clear that each B(v) # Z. Also, B(v) v~ i l , 2 . . . k} since G has no isolated vertices. For any u # v, N(u) ~ N(v) since G has no duplicated vertices. Assume that there exists some vertex x ~ N(u) - N(v). Then { x, v} c_ Aj for some

A~ ~ MI(G). Thus j ~ B ( v ) - B(u). This proves that B(u) # B(v) whenever u ¢ v.

Denote by ~ = {B(v): v ~ V(G)}. Then ] V(G)[ = [~[.

For any i ~ {1,2,...,k}, we claim that {i} ~ ~. Otherwise, suppose {i} ~ ~ and consider the graph G* obtained from G by adding a new vertex v*, which is adjacent to all vertices in

V(G)-

Ai. Note that MI(G*) is the same as MI(G)

except that A~ is replaced by A~ U iv*}. Also, G* is without isolated or duplicated vertices, a contradiction to the choice of G. Thus {i} ~ ~ for all 1 < i < k. We may assume that

V(G) = {v~ ... Ok,..., Vm} and B(v,) = {i} for 1 < i < k.

If viv i ~ E(G), then v~ and vj are not both in the same independent set; i.e.,

B(vi) fl B(vj) = E~. On the other hand, suppose rio j ~ E(G). Then ivy, vj} is an inde-

pendent set and so is a subset of some A~ ~ MI(G); i.e., r ~ B(v~) N B(vj). In conclu- sion, viv j ~ E(G) if and only if B(vi) fl B(vj) = ;g.

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Now, choose a m a x i m a l chain C:

B(Vr,)D B(v,:)= " " = B(vr)

in the poset defined on ~ under set inclusion. N o t e that

B(v~,)

= {r~} and s < k - i. Partition 2i~.2 ... k} _ {B(v,,), ~} into C~, C2 . . . Cs, where C~ is the set of all subsets S such that S - B(v,,) # ~ b u t S _ B(v,,_,), where

B(Vro)

= {1,2 . . . . ,k}. F o r each 1 < i < s, partition C~ into pairs {Sj, Tj} such that Sj - B(v,,) = Tj -

B(v,,) # ;~

and

B(v,~)

is the disjoint union of BIN B(v,,) and TiN B(v,,). Consider the following cases:

Case I. S i fq B(vr, ) # ;g

and T i n B(v~,) # Z~.

Suppose Si ~ ~ and Tj ~ ~, say, S i =

B(x)

and Tj =

B(y).

N o t e that

B(x), B(y),

B(v,,) are pairwise non-disjoint. T h e n {x,y, Vr,} -- Aq for some 1 < q < k. By the definition,

q ~ B(x) N B(y) f3

B(v,,) = ~, a contradiction. Thus either Sj or T~ is not in ~.

Case 2. S~nB(v,,)

= ~ or T~NB(vr,)= ~ , say,

S~NB(v,,)

= B(v,,) and

TjNB(vr) =

In this case, S~ _ B(v,,_,). If S i is a p r o p e r subset of

B(v,,_l ),

by the choice of the chain C, Sj is n o t in ~. IfSj = B(v,,_,) and i = 1, then S i = {1,2 . . . k} is n o t in ~. If

S i = B(v,,_,)

and 2 < i < s, then Sj and T~ m a y be b o t h in N.

F r o m the discussions in Cases 1 and 2, 2 ~ 2 ... k} _ {B(v,,),N} can be partitioned into 2 k-~ - 1 pairs {Sj, T~} such that at least one in {S~, Tj}, except possibly s - 1 pairs, is not in ¢~. T h u s

I V ( G ) [ = t ~ [ < I + ( 2 k - l - 1 ) + s - 1 < 2 k - l + k - 2 . [] T h e upper b o u n d in T h e o r e m 2.6 is sharp as the following example shows. Consider the split graph G* whose vertex set

V(G)*

is partitioned into a clique

C = {vl, v2,..., Vk}

and an independent set I =

{Us: fg # S ~

{1, 2 , . . . , k - 1}} such that

ViUs ~ E(G)*

if and only if i e S. It is clear that

G ~ S(k)*

and [V(G*)t = 2 k-1 + k - 2. N o t e that T h e o r e m 2.4 is also a consequence of T h e o r e m 2.6.

References

1. Fiiredi, Z.: The number of maximal independent sets in connected graphs. J. Graph Theory 11, 463-470 (1987)

2. Griggs, J.R., Grinstead, C.M., Guichard, D.R.: The number of maximal independent sets in a connected graph. Discrete Math. 68, 211-220 (1988)

3. Hujter, M., Tuza, Z.: The number of maximal independent sets in triangle-free graphs. SIAM J. Disc. Math. 6, 284-288 (1993)

4. Jou, M.-J.: The Number of Maximal Independent Sets in Graphs, Master's thesis, Dept. of Math., National Central Univ., Taiwan (1991)

5. Liu, J.: Maximal independent sets in bipartite graphs. J. Graph Theory 17, 495-507 (1993)

6. Meir, A., Moon, J.W.: On maximal independent sets of nodes in trees. J. Graph Theory 12, 265-283 (1988)

7. Moon, J.W., Moser, L.: On cliques in graphs. Israel J. Math. 3, 23-28 (1965)

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326 M.-J. Jou et al. 9. Wilf, H.S.: The number of maximal independent sets in a tree. SIAM J. Alg. Disc. Meth.

7, 125-130 (1986)

I0. Zito, J.: The structure and maximum number of maximum independent sets in trees. J. Graph Theory 15, 207-211 (1991)

Received: October 5, 1994 Revised: August 21, 1995

A finiteness theorem for maximal independent sets

Graphs and

Combinatorics