[Paper Review] Cycles are determined by their domination polynomials

Saieed Akbari, Mohammad Reza Oboudi|ArXiv.org|Aug 23, 2009

Advanced Graph Theory Research3 references3 citations

TL;DR

This paper proves that cycles are uniquely determined by their domination polynomials, meaning no other non-isomorphic graph shares the same domination polynomial. Using recursive formulas, p-adic valuation analysis, and derivative evaluations at specific points, the authors show that if a graph has the same domination polynomial as a cycle $C_n$, it must itself be isomorphic to $C_n$, establishing $C_n$ as ${\cal D}$-unique for all $n \geq 1$. This resolves an open problem in domination polynomial theory.

ABSTRACT

Let $G$ be a simple graph of order $n$. A dominating set of $G$ is a set $S$ of vertices of $G$ so that every vertex of $G$ is either in $S$ or adjacent to a vertex in $S$. The domination polynomial of $G$ is the polynomial $D(G,x)=\sum_{i=1}^{n} d(G,i) x^{i}$, where $d(G,i)$ is the number of dominating sets of $G$ of size $i$. In this paper we show that cycles are determined by their domination polynomials.

Motivation & Objective

• To resolve the open question of whether cycles are uniquely determined by their domination polynomials.
• To establish that $C_n$ is ${\cal D}$-unique for all positive integers $n$, meaning no non-isomorphic graph shares its domination polynomial.
• To extend this result to show that the wheel graph $W_n = K_1 \vee C_{n-1}$ is also ${\cal D}$-unique.
• To use algebraic and recursive techniques to analyze the structure of graphs with identical domination polynomials.
• To prove that the only graph with the same domination polynomial as $C_n$ is $C_n$ itself, using invariants derived from polynomial evaluation and derivatives.

Proposed method

• Employed the recursive formula $D(C_n, x) = x(D(C_{n-1}, x) + D(C_{n-2}, x) + D(C_{n-3}, x))$ for domination polynomials of cycles.
• Used $p$-adic valuation ${\rm ord}_3(D(C_n, -3))$ to analyze the structure of graphs with the same domination polynomial.
• Evaluated the domination polynomial and its first and second derivatives at $x = -1$ to derive invariants $\alpha_n = D(C_n, -1)$, $\beta_n = D'(C_n, -1)$, and $\theta_n = D''(C_n, -1)$.
• Applied induction and modular arithmetic (mod 9) to analyze the $3$-adic valuation of coefficients in the recurrence.
• Used the fact that $D(G,x) = D(H,x)$ implies $G$ and $H$ are $k$-regular and have isomorphic components, leveraging Lemma 1 and Lemma 2.
• Combined derivative evaluations and modular constraints to eliminate the possibility of $G$ being a disjoint union of three cycles ($k=3$), proving $k=1$ is the only solution.

Experimental results

Research questions

• RQ1 Is every cycle $C_n$ uniquely determined by its domination polynomial, i.e., is $C_n$ ${\cal D}$-unique?
• RQ2 Can a graph other than $C_n$ have the same domination polynomial as $C_n$?
• RQ3 What invariants derived from the domination polynomial (e.g., evaluations at $x = -1$, $x = -3$, and derivatives) can distinguish $C_n$ from other graphs?
• RQ4 Does the ${\cal D}$-uniqueness of $C_n$ extend to related graphs like wheels $W_n = K_1 \vee C_{n-1}$?
• RQ5 What is the structure of the ${\cal D}$-equivalence class of $C_n$, and can it contain graphs other than $C_n$?

Key findings

• For all $n \geq 1$, the cycle $C_n$ is ${\cal D}$-unique, meaning no non-isomorphic graph shares its domination polynomial.
• The domination polynomial evaluation at $x = -1$ yields $D(C_n, -1) = 3$ if $n \equiv 0 \pmod{4}$, and $-1$ otherwise.
• The first derivative at $x = -1$ satisfies $D'(C_n, -1) = -n$ if $n \equiv 0 \pmod{4}$, $n$ if $n \equiv 1 \pmod{4}$, and $0$ otherwise.
• The second derivative at $x = -1$ is $n(n-4)/2$ if $n \equiv 0 \pmod{4}$, $-n(n-1)/2$ if $n \equiv 1 \pmod{4}$, $n(n+2)/4$ if $n \equiv 2 \pmod{4}$, and $0$ otherwise.
• The $3$-adic valuation ${\rm ord}_3(D(C_n, -3))$ is $\lceil n/3 \rceil$ if $n \equiv 2 \pmod{3}$, $\lceil n/3 \rceil + 1$ if $n \equiv 0 \pmod{3}$, and $\lceil n/3 \rceil$ or $\lceil n/3 \rceil + 1$ if $n \equiv 1 \pmod{3}$.
• The assumption that $G$ is a disjoint union of three cycles leads to a contradiction via derivative evaluations, proving $G$ must be a single cycle, so $k=1$ is the only possibility.