Inverse problems for the number of maximal independent sets

In the notation of the current paper, Linek proved [4] the strong ι-completeness of B. It justifies the consideration of the optimization problem of finding L B ι, ν (n). The trivial lower bound is L B ι, ν (n) log 2 n (which follows from the inequation ι(G) ≤ 2 ν(G) ). A graph constructed in [4] to realize a given natural number has the maximal possible sizes of parts: ⌊log 2 n⌋ and ⌊log 2 (n -2 ⌊log 2 n⌋ + 1)⌋. For n = 2 k -1 such graph would have 2k -2 vertices, which is double the expected optimal size. Some n of the above form can be realized more economically, as the following statement shows.

Theorem 1. For k = 2 t we have L F ι, ν (2 k -1) k. Proof. Just note that

At the same time ν( t-1 j=0 K 2 j ,1 ) = 2 t + t -1 k. The existential inverse problem for (B, ι m ) is trivial, as any n ≥ 4 can be realized as the number of m. i. s. in corona-graph K ′ n-2,n-2 . If we consider ψ to be the number of vertices of a graph, we come up with an optimizational inverse problem: “for natural n find minimal L(n) such that there exists a graph on L(n) vertices having n maximal independent sets”. The remaining part of the paper is dedicated to estimating L(n).

ι m ,ν (n)

Lemma 1. Let G be bipartite with parts L G , R G and without isolated vertices. Let G be a bipartite graph, vertex-disjoint with G, and let U 1 and U 2 be some subsets of first and second parts of G respectively. Let G ′ be a graph obtained by connecting all vertices in U 1 (resp. U 2 ) to all vertices in L G (resp. R G ). Then we have

where ι m ( G + U 1 + U 2 ) stands for the number of m. i. s. of G having non-empty intersections with both U 1 and U 2 .

Proof. The statement of the lemma can be checked by direct counting. If an m. i. s. of G ′ contains no vetices of G, then it must contain at least one vertex from both U 1 and U 2 , the number of such sets being ι m ( G + U 1 + U 2 ). If a m. i. s. of G ′ contains vertices from both parts of G, then it is disjoint with U 1 ∪ U 2 , and its subsets in G and G must themselves be maximal independent sets in G and G respectively. Thus the number of such m. i. s. equals

Let G be bipartite with U 1 and U 2 being some subsets of its parts. Put

Lemma 2. Let Γ be a finite set of bipartite graphs with selected subsets in their parts, such that {h

Proof. The lemma is proved by induction on n with a help of lemma 1. Let Γ 0 be an arbitrary finite set of bipartite graphs having {ι m (G) | G ∈ Γ 0 } ⊇ [1, n 0 ]. For example, as Γ 0 we can take the set

Let ν 0 be the maximal number of vertices of graphs in Γ 0 . It suffices to prove that for any n the following inequality holds:

which would imply the statement of the lemma. The inequality (1) trivially holds for n ≤ n 0 . Consider an arbitrary n ′ , n ′ > n 0 , and assume that (1) holds for all n less than n ′ . By the conditions of the lemma, there exists some G ∈ Γ and some natural k, such that

By ( 2) and k

Proof. The lower bound of (3) follows from the observation that a number of m. i. s. in a bipartite graphs cannot exceed the number of subsets of any of this graph’s parts.

To obtain the upper bound we apply lemma 2 with Γ being equal to the following set of graphs (subsets U 1 , U 2 are marked as bold vertices; pairs of numbers (h ′ G , h ′′ G ) are scribed under the graphs):

• ~~~~~~~~@ @ @ @ @ @ @ @

• ~~~~~~~~@ @ @ @ @ @ @ @

• @ @ @ @ @ @ @ @ • @ @ @ @ @ @ @ @

(18, 17) It can be checked, that such Γ meets the conditions of lemma 2 and for this set the parameter γ would equal 12(log 2 18) -1 < 2.88. It implies the lower bound in (3).

Remark. The inequality (3) remains valid without O(1) summand, which can be proven in the same way as in theorem 2. Upper bound in (3) may be directly improved by finding a better set Γ. To find such Γ one can apply an exhaustive computer search (which in fact was used to find Γ that we provide above).

We feel certain that the following is true:

Thought we were unable to prove the above conjecture, theorem 3 approves it for some special class of naturals. Next we need to prove some auxillary statements. Proof. Apply lemma 1, with whole parts of G selected as U 1 and U 2 . Thus we obtain G ′ on (ν(G) + ν( G) + 4) vertices with (ι m (G) + ι m ( G) -2) maximal independent sets. It suffices to apply lemma 4 to G ′ . Lemma 6. Let G and G be bipartite without isolated vertices, and let s, t ∈ N. Then there exists bipartite graph without isolated vertices having

maximal independent sets and no more than ν(G) + ν( G) + 2s(t + 1) + 3 vertices.

Proof. For s = 1 the statement follows from lemma 5 (before applying the lemma add matching on 2t vertices to G). So for the rest of the proof we assume that s ≥ 2. We also assume that

Parts of G and G will be denoted as L G , R G and L G , R G respectively. We shall consider a graph G ′ which is const