### 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 con-
nected 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

*is a subset S of*

**independent set***such that no two vertices in S are adjacent. A*

**V***is an independent set that is not a*

**maximal independent set**### 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.

0166-218xi97/$17.00 0 1997 Elsevier Science B.V. All rights reserved
**PZZ SO 166-2 18X(97)00033-4 **

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 **Lemma 4. For any j > 0,**

*and *

mi(B(4,j)) =
2j+r + 1.

**Proof. **

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

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

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

**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). }

_{= 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 **

**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. *

*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. }

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

*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 *

**.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.