An Alternative Proof of the Largest Number of Maximal Independent Sets in Connected Graphs Having at Most Two Cycles

G. C. Ying, Y. Y. Meng, B. E. Sagan, and V. R. Vatter [1] found the maximum number of maximal independent sets in connected graphs which contain at most two cycles. In this paper, we give an alternative proof to determine the largest number of maximal independent sets among all connected graphs of order n ≥ 12, which contain at most two cycles. We also characterize the extremal graph achieving this maximum value.


Introduction
Let ( ) be a simple undirected graph.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 set of all maximal independent sets of a graph G is denoted by ( ) MI G and its cardinality by The problem of determining the largest value of ( ) mi G in a general graph of order n and those graphs achieving the largest number was proposed by Erdös and Moser, and solved by Moon and Moser [2].It was then extensively studied for various classes of graphs in the literature, including trees, forests, (connected) graphs with at most one cycle, bipartite graphs, connected graphs, k-connected graphs, (connected) triangle-free graphs; for a survey see [3].Recently, Jin and Li [4] determined the second largest number of maximal independent sets among all graphs of order n.
There are results on independent sets in graphs from a different point of view.The Fibonacci number of a graph is the number of independent vertex subsets.The concept of the Fibonacci number of a graph was introduced in [5] and discussed in several papers [6] [7].In addition, Jou and Chang [8] showed a linear-time algorithm for counting the number of maximal independent sets in a tree.Jou and Chang [9] determined the largest number of maximal independent sets among all graphs and connected graphs of order n, which contain at most one cycle.Later B. E. Sagan and V. R. Vatter [10] found the largest number of maximal independent sets among all graphs of order n, which contain at most r cycles.In 2012, Jou [11] settled the second largest number of maximal independent sets in graphs with at most one cycle.G. C. Ying, Y. Y. Meng, B. E. Sagan, and V. R. Vatter [1] found the maximum number of maximal independent sets in connected graphs which contain at most two cycles.In this paper, we give an alternative proof to determine the largest number of maximal independent sets among all connected graphs of order 12 n ≥ , which contain at most two cycles.We also characterize the extremal graph achieving this maximum value.
For a graph ( ) , the cardinality of ( ) V G is called the order, and it is denoted by G .The neighborhood ( ) : and . The star-product of

Preliminary
The following results are needed.

1)
2) If x is a leaf adjacent to y, then Then ( ) Proof.The derivative of ( ) and ( ) , where ( ) , where ( ) where ( ) , B i j is the set of batons, which are the graphs obtained from a basic path P of 1 i ≥ vertices by attaching 0 j ≥ paths of length two to the endpoints of P in all possible ways (see Figure 1).Theorem 2. ( [9]) If F is a forest with , where ( ) , where ( ) , where ( ) , where , where ( ) , where , where ( ) (see Figure 2), where ( )

The Alternative Proof
In this section, we give an alternative proof to determine the largest number of maximal independent sets among all connected graphs of order 12 n ≥ , which contain at most two cycles.We also characterize the extremal graph achieving this maximum value.
, where ( ) A unicyclic graph is a connected graph having one cycle.The order of a unicyclic graph is at least three.The following lemmas will be needed in the proof of main theorem.Lemma 4. Suppose that G T H = ∪ is the union of a tree T and a unicyclic graph H, where . The equality holds if and only if Note that H has one cycle, then 3 H n k = − ≥ .We consider two cases.
Case 1. 6 n ≥ is even.By Lemma 2, Theorem 1 and Theorem 5, we have If the equality holds, then ( ) ( ) ( ) Hence the equality holds if and only if By Lemma 3 and since By Theorem 1 and Theorem 5, we have If the equality holds, then . Suppose that F is a forest of order 4 n ≥ having at most two components.Then ( ) ( ) and the equality holds if and only if ( ) Proof.Let F be a forest of order 4 n ≥ having at most two components such that ( ) mi F as large as possible.Then The equalities hold and ( ) ( ) The equality holds if and only if ( ) Then F has exactly one even component, we assume that 2 s ≥ is even.By Theorem 1, then . T h e equalities hold and . The equality holds if and only if ( ) The following is the proof of Theorem 6.
Proof.Let H be a connected graph of order 12 n ≥ with at most two cycles such that ( ) mi H as large as possible.Then , by Theorem 5, H have at least two cycles.That means that H have exactly two cycles and ( ) . Since H is connected, this means that v connects to every component of ( ) , 2 1,1 2 Note that H has two cycles, hence ( ) ( ) having at most one cycle.By Theorem ) . By Lemma 1 and Theorem 4, then . ( ) Note that H has two cycles and v is lying on some cycle.Thus v has two edges incident to some component of H v − .Since    , where T is a tree and H ′ is a unicyclic graph.By Lemma 4 and Theorem 3, then   ) ( ) for odd 13 n ≥ .□ mi G mi G mi G = .Lemma 3. Let , , a b n be integers such that 0 2 n a b < < < and let ( )

Figure 1 .G is a graph of order 2 n
Figure 1.The baton ( ) , B i j with

≠
. Let v be a vertex lying on some cycle such that ( ) deg v is as large as possible.Since graph H v − is a graph of order 1 n − with at most one cycle.We consider two cases.Case 1.

( ) 4
deg v = , the number of the components of H v − is at most three.Thus F is either a tree or the union of two trees.By Lemma 5, ( ) ( )

3
deg v = .Note that H has two cycles and v is lying on some cycle.Thus v has two edges incident to some component of H v − .Since ( ) 3 deg v = , the number of the components of H v − is at most two.Thus H v T H′ − = ∪ > .This contradicts to the claim, so n is odd.Thus