A Simple Proof of an Inequality Connecting the Alternating Number of Independent Sets and the Decycling Number

If alpha=alpha(G) is the maximum size of an independent set and s_{k} equals the number of stable sets of cardinality k in graph G, then I(G;x)=s_{0}+s_{1}x+…+s_{alpha}x^{alpha} is the independence polynomial of G. In this paper we provide an elementary proof of the inequality claiming that the absolute value of I(G;-1) is not greater than 2^phi(G), for every graph G, where phi(G) is its decycling number.

💡 Research Summary

The paper investigates a fundamental relationship between two graph invariants: the alternating value of the independence polynomial, denoted I(G;‑1), and the decycling number φ(G), which is the smallest number of vertices whose removal makes the graph acyclic. The independence polynomial of a graph G is defined as I(G;x)=∑{k=0}^{α(G)} s_k x^k, where s_k counts the independent (stable) sets of size k and α(G) is the independence number. Substituting x=‑1 yields the alternating sum I(G;‑1)=∑{k=0}^{α(G)} (‑1)^k s_k, often called the alternating number of independent sets.

Historically, the inequality |I(G;‑1)| ≤ 2^{φ(G)} was proved using algebraic topology, specifically through homological arguments that relate the independence complex of G to its Betti numbers. While powerful, those proofs require sophisticated machinery that is not readily accessible to most combinatorialists. The present work offers a completely elementary proof based solely on elementary graph theory and induction, thereby broadening the audience and providing new insight into why the bound holds.

The authors begin by recalling the standard deletion–contraction recurrence for the independence polynomial:
I(G;x) = I(G−v;x) + x·I(G−N