Maximal independent sets in graphs with at most one cycle

(1)

DISCRETE

APPLIED

Discrete Applied Mathematics 79 (1997) 67-73

MATHEMATICS

Maximal independent sets in graphs with

at most one cycle *

Min-Jen

Jou and Gerard

J. Chang *

Department of Applied Mathematics, National Chiao Tuny University Hsinchu 30050, Taiwan ROC

Received 22 June 1995; received in revised form 15 August 1996; accepted 5 March 1997

Abstract

In this paper, we determine the largest number of maximal independent sets among all connected graphs of order n, which contain at most one cycle. We also characterize those extremal graphs achieving this maximum value. As a consequence, the corresponding results for graphs with at most one cycle but not necessarily connected are also given.

Keywords: (Maximal) independent set; Cycle; Connected graph; Isolated vertex; Leaf; Baton

1. Introduction

In a graph G = (V,E), an independent set is a subset S of V such that no two vertices in S are adjacent. A maximal independent set is an independent set that is not a

proper subset

of any other independent set. The number of maximal independent sets of G is denoted by mi(G).

The problem of determining the maximum value of mi(G) in a general graph of order n and those graphs achieving the maximum value was proposed by Erdiis and Moser, and solved by Moon and Moser [7]. The problem was independently solved by Fiiredi [l] and Griggs et al. [3] for connected graphs; for triangle-free graphs by Hujter and Tuza [4]; for bipartite graphs by Liu [6]; for trees independently by Wilf [9], Sagan [8], Griggs and Grinstead [2], and Jou [5]. Sagan’s solution for trees uses an induction from a vertex whose neighbors are all leaves except possibly one. Jou’s method is to get the solution for forests and then use this to prove the results for trees. The main purpose of this paper is to study the problem for connected graphs with at most one cycle. We first give alternative proofs for the solutions to the problem in trees and forests by a method combining the ideas in Sagan’s and Jou’s papers. The idea then is used to solve the problem for connected graphs with at most one

* Supported in part by the National Science Council under Grant NSC83-020%M009-050. * Corresponding author. E-mail: gichang@math.nctu.edu.tw.

(2)

68 M.-J. Jou, G.J. Changl Discrete Applied Mathematics 79 (1997) 67-73 i

.

” 4 , jl

j2

W,

2)

B(4,3)

B(473)

j = j, +

j2 B(i, d Fig. 1. Batons.

cycle. The corresponding results for graphs with at most one cycle, but not necessarily connected, are also obtained.

2.

Trees and forests

This section gives an alternative proof for the solution to the problem of determining the maximum value t(n) (respectively, f(n)) of mi(G) for a tree (respectively, forest) of order n and those trees (respectively, forests) achieving this maximum value.

The neighborhood N(x) of a vertex x is the set of vertices adjacent to x and the

closed neighborhood N[x]

is {x} UN(x). A vertex x is an

isolated vertex

if N(x) = 4 and a

leaf

if IN(x)] = 1. For a graph G = (V,E) and S C V, the

deletion

of S from G is the graph G -S obtained from G by removing all vertices in S and all edges incident to these vertices. We use C,, to denote the cycle with

n

vertices.

Lemma 1

(Hujter and Tuza [4] and Jou [5]). Zf G

is a graph in which x is adjacent

to exactly one vertex y, then

mi(G) = mi(G - N[x]) + mi(G - N[y]).

Lemma 2

(Ftiredi [l]). rfn 2 6,

then

mi(C,)=mi(C,_~)+mi(C,_~).

Lemma 3

(Hujter and Tuza [4] and Jou [5]). Zf G

is the disjoint union of two graphs

G1 and Gz, then

mi( G) = mi(G1 )mi(Gz).

For simplicity, let Y = 4. For

i, j > 0,

define a

baton B(i, j)

as follows. Start with a

basic path P

with

i

vertices and attach j paths of length two to endpoints of

P; see

Fig. 1. Note that B(i, j) is a tree with

i + 2j

vertices.

Lemma 4. For any j > 0,

mi(B( 1, j)) = 2j, mi(B(2, j)) = 2j + 1

and

mi(B(4,j)) =

2j+r + 1.

Proof.

The lemma follows from repeatedly applying Lemma 1 to the leaves of the

(3)

M.-J. Jou. G.J. Changi Discrete Applied Mathematics 79 (1997) 67-73 69

We now give an alternative proof for the solution to the problem in trees.

Theorem 5

(Wilf [9], Griggs and Grinstead [2], Sagan [8] and Jou [5]). rf T is a tree

with n 2 1 vertices, then mi(T) 6 t(n), where

r ‘-’ + 1

t(n) = if n is even,

r n-l if n is odd.

Furthermore, mi(T) = t(n) if and only if T 2 T(n), where

T(n) =

B(2,y) or B(4, q) ifn is even,

B(l, 9) if n is odd.

Proof.

First, note that mi(T(n)) = t(n) by Lemma 4. We shall prove the theorem by

induction on n. The theorem is obviously true for n < 3. Assume that it is true for all

n’ <n. Suppose T is a tree of order n B 4. Choose an end vertex x of a longest path

in T. Then x is a vertex adjacent to exactly one vertex y such that T - N[x] 2 T’ U iK1 for some i 3 0 and a tree T’ with n - 2 - i vertices; and T - N[y] 2 T” UjK, U kK2

for some j, k > 0 and a tree T” with n - 3 - i

- j

- 2k vertices. (We may assume that T” = 0 or T” contains at least 3 vertices.) Note that t(m) < r”-’ for m # 2. By Lemmas 1 and 3 and the induction hypothesis,

mi(T) = mi(T-N[x])+mi(T-N[y]) < t(n - 2 - i) + t(n - 3 - i -j - 2k)2” d i t(n - 2 - i) + r2k if T” = 8, i.e., 2k = n - 3 - i -j, t(n - 2 - i) + r n-3-I-j-2k-lr2k otherwise 6 i t(n-2)+r”-3 if n=2k+3, t(n - 2) + rnp4 otherwise d t(n).

Moreover, the equalities holding imply that either n = 2k + 3 or n is even with i = j = 0,

T-N[x]rT(n-2), and T-N[y]rB(l, v) U kK2 by the induction hypothesis.

For the case of n =2k + 3, T rB(l, K$) = T(n). For the later case, No(y) = {x,z} and z is an endpoint of the basic path of the baton T-N[x] g T(n - 2) =B(2, ?$)

or B(4, F), except possibly when T-N[x] ?B(2,1)=B(4,O)=P4. For the excep- tional case, we can view P4 as a suitable B(2,l) or B(4,O); and still assume that z is an endpoint of the basic path of the baton. Thus, T E+ B(2, F) or B(4,

y

), i.e.,

T= T(n). Cl

(4)

70 M-J. Jou, G.J. ChangIDiscrete Applied Mathematics 79 (1997) 67-73

n

is even

n

is odd

Fig. 2. H(n).

Furthermore, mi(F) = f(n) if and onZy if F = F(n), where

F(n) =

;K2

if n is even,

B(L +)uKf

s 2 orsomes

withO<s<q

ifnisodd.

Proof. First of all, mi(F(n)) = f(n) by Lemmas 3 and 4. Suppose F = sK2

U (Ur=, Ti),

where s > 0, m 2 0, and each I;: is a tree with ni # 2 vertices. Note that t(n) < r”-’

when n # 2. By Lemma 3 and Theorem 5,

mi(F) < 2’ fi t(q) < r2’ fir”‘-’

_{= rnPm 6 f(n).}

i=l i=l

Furthermore, if the equalities hold, then either m = 0 or m = 1 with nr odd and

t(nl )

=

r”l-‘. For the former case, F 2 ;K2 = F(n). For the later case, by Theorem 5, Tr Z

B(l, 9)

and so F ZF(n).

Cl

3. Graphs with at most one cycle

This section gives solutions to the problem of determining the maximum value h(n)

(respectively, g(n)) of mi(G) in a connected graph (respectively, general graph) of

order n that contains at most one cycle.

Define the graph H(n) of order n as follows, see Fig. 2. For even n, H(n) is the

graph obtained from B( 1, y)

by adding a KS and a new edge joining a vertex of

K3 and the only vertex in the basic path of B(

1, 9).

For odd n, H(n) is the graph

obtained from B(

1, 9)

by adding a KS with one vertex identified with the only vertex

in the basic path of B(

1, 9 ).

Theorem 7. If G is a connected graph with n 3 3 vertices such that G contains at

most one cycle, then mi(G) < h(n), where

3rnP4

h(n) =

if n is even,

r”-’ + 1 if n is odd.

(5)

M.-J. Jou, G.J. ChanglDiscrete Applied Mathematics 79 (1997) 67-73 71

Proof. By repeatedly applying Lemma 1, we have mi(H(n)) = h(n). We shall prove

the theorem by induction on n. The theorem is clearly true for 3 < n < 5. Assume that it is true for all n’ <n. Suppose G is a connected graph with H > 6 vertices such that G contains at most one cycle. If G is the cycle C,,, then, by Lemma 2 and the induction hypothesis,

mi(G) = mi(C,) = mi(C,_2) + mi(C,_s)

i 5 if n = 6, 6 h(n-2)+h(n-3) if n > 7 1 5 if n = 6, d 3T6 + (rnm4 + 1) if n 2 8 is even, (T3 + 1) + 3rnp7 if n 2 7 is odd = { 5 if n = 6, 5rnp6 + 1 if n > 8 is even, 7rnM7 + 1 if n 3 7 is odd < h(n).

Now, assume that G y C,. Either G contains a unique cycle C, or else G is a tree in which a leaf C is chosen. Choose a vertex x which is farthest to C. Then x is a leaf adjacent to y such that G - N[x] is the union of i 3 0 isolated vertices and a connected graph G’ with n - 2 - i vertices. Note that the connected subgraph G’ contains at most one cycle. By the induction hypothesis, mi(G’) < h(n - 2 - i). Thus, by Lemma 3,

mi(G - N[x]) = mi(iKi ) mi(G’) d h(n - 2 - i) 6 h(n - 2).

Also, mi(G - N[x]) = h(n - 2) implies that i = 0 and G - N[x] g H(n - 2) by the in- duction hypothesis. On the other hand, G - N[y] has at most n - 3 - i vertices, which is either a forest or the union of a forest F and a connected subgraph G” with t ver- tices, 3 < t 6 n - 3 - i, that contains exactly one cycle. So mi(F) 6 f(n - 3 - i - t) by Theorem 6 and mi(G”) d h(t) by the induction hypothesis. Thus,

mi(G - N[yl) < f(n - 3 -i) if G - N[y] is a forest, f(n - 3 -i - t)h(t) if G - N[y] is not a forest

max{rnP4, r”-4-‘(3r1-4), rn--3-t(r’-’ + 1)) if n is even, < max{rnP3, rn-3--t(3r’-4), r”-4--r(r’-’ + 1)) if n is odd 3rnw6 if n is even, d r n-3 _{if n is odd.}

(6)

72 M.-J. Jou, G.J. Chang IDiscrete Applied Mathematics 79 (1997) 67-73 Also, the equalities holding imply that ]G - N[y] I= n - 3 and

G-N[y]Z

KS

U

yK2

if

n

is even,

?Kz

if n is odd,

by the induction hypothesis and Theorem 6. So, by Lemma 1, mi(G) = mi(G-N[x])+mi(G-N[y]) if n is even, 6 { h(n - 2) + 3 . F6 h(n - 2) + Y”-3 if

n

is odd 3 . F6 + 3. rnp6 if

n

is even, = _i (rnP3 + 1) + P-’ if n is odd =

h(n).

And the equalities holding imply that G

EH(n),

since G - N[x]

gHH(n -

2) and G - N[y] is as above. 0

Theorem 8. Zf G is a graph with n >,

1

vertices such that G contains at most one

cycle, then

mi(G) <

g(n), where

if

n is even,

if n is odd.

Furthermore,

mi(G) =

g(n) if and only if G 2 G(n), where

G(n) =

;K2

if n is even,

K3 u 9K2

if n

is odd.

Proof.

By Lemma 3, it is clear that mi(G(n)) =

g(n).

For the case when G is a forest,

by Theorem 6, mi(G) <

f(n) < g(n);

and the equalities holding implies that

n

is even and GS

F(n) = G(n).

For the case when G is connected with exactly one cycle, by Theorem 7, mi(G) <

h(n) < g(n);

and the equalities holding implies that

n =

3 and G

NK3 = G(3). So we

may suppose that

G=F

UH, where

F

is a forest of order

nl

and

H

is a connected graph of order

n2

with exactly one cycle. By Lemma 3 and the above two cases,

mi(G) = mi(F)mi(H) <

f(nl)h(nz)

<

r”‘g(n2) =

s(n)

if

n1

is even,

r”‘-‘g(nz)<g(n)

if

nl

is odd. Furthermore, if the equalities hold, then

n1

is even with mi(F)

=f(nl)

=g(nl)

and mi(H)=

h(nz)=g(ng).

By the first two cases,

FE SK2

and

H NK3,

i.e.,

G”G(n).

0

(7)

.M.-J. Jou, G.J. Changl Discrete Applied Mathenzutics 79 (1997) 67-73 73

Acknowledgements

The authors thank two anonymous referees for their constructive suggestions that significantly shorten the proofs of Theorems 5 and 6.

References

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

[2] J.R. Griggs, C.M. Grinstead, 1986, unpublished result.

[3] J.R. Griggs, C.M. Grinstead, D.R. Guichard, The number of maximal independent sets in a connected graph, Discrete Math. 68 (1988) 21 l-220.

[4] M. Hujter, Z. Tuza, The number of maximal independent sets in triangle-free graphs, SIAM J. Discrete Math. 6 (1993) 284-288.

[5] M.J. Jou, The Number of Maximal Independent Sets in Graphs, Master Thesis, Department of Mathematics, National Central University, Taiwan, 199 I.

[6] J. Liu, Maximal independent sets in bipartite graphs, J. Graph Theory I7 (1993) 4955507. [7] J.W. Moon, L. Moser, On cliques in graphs, Israel J. Math. 3 (1965) 23328.

[8] B.E. Sagan, A note on independent sets in trees, SIAM J. Discrete Math. I (1988) 105-108.

[9] H.S. Wilf, The number of maximal independent sets in a tree, SIAM J. Algebraic Discrete Methods 7 (1986) 125-130.