Complexity of Bondage and Reinforcement

Let $G=(V,E)$ be a graph. A subset $D\subseteq V$ is a dominating set if every vertex not in $D$ is adjacent to a vertex in $D$. A dominating set $D$ is called a total dominating set if every vertex in $D$ is adjacent to a vertex in $D$. The domination (resp. total domination) number of $G$ is the smallest cardinality of a dominating (resp. total dominating) set of $G$. The bondage (resp. total bondage) number of a nonempty graph $G$ is the smallest number of edges whose removal from $G$ results in a graph with larger domination (resp. total domination) number of $G$. The reinforcement number of $G$ is the smallest number of edges whose addition to $G$ results in a graph with smaller domination number. This paper shows that the decision problems for bondage, total bondage and reinforcement are all NP-hard.

💡 Research Summary

The paper investigates the computational complexity of three graph parameters that measure how the domination number changes under edge deletions or additions: the bondage number b(G), the total bondage number bₜ(G), and the reinforcement number r(G). A dominating set D of a graph G = (V, E) is a subset of vertices such that every vertex not in D has a neighbor in D; a total dominating set requires every vertex (including those in D) to have a neighbor in D. The domination number γ(G) is the size of a smallest dominating set, and γₜ(G) is defined analogously for total domination. The bondage number b(G) is the smallest number of edges whose removal strictly increases γ(G); total bondage bₜ(G) is defined with respect to γₜ(G); reinforcement r(G) is the smallest number of edges whose addition strictly decreases γ(G).

Although it is known that computing γ(G) (or γₜ(G)) is NP‑complete, the complexity of b(G), bₜ(G), and r(G) had not been settled for general graphs. The authors prove that each of the three decision problems—“Is b(G) ≤ k?”, “Is bₜ(G) ≤ k?”, and “Is r(G) ≤ k?”—is NP‑hard by giving polynomial‑time many‑one reductions from the classic NP‑complete problem 3‑SAT.

Reduction for the bondage problem.
Given a 3‑SAT instance with variables U = {u₁,…,uₙ} and clauses C = {C₁,…,Cₘ} (each clause contains exactly three literals), the authors construct a graph G as follows:

1. For each variable uᵢ, create a triangle Tᵢ with vertices {uᵢ, ¬uᵢ, vᵢ}.
2. For each clause Cⱼ = {ℓ₁, ℓ₂, ℓ₃}, introduce a clause vertex cⱼ and connect it to the three literal vertices representing ℓ₁, ℓ₂, ℓ₃.
3. Add a path P = s₁–s₂–s₃ and connect both s₁ and s₃ to every clause vertex cⱼ.
4. Set k = 1.

The authors prove four claims: (i) γ(G) ≥ n + 1 and any minimum dominating set of size n + 1 must contain exactly one vertex from each triangle and s₂; (ii) γ(G) = n + 1 iff the original 3‑SAT formula is satisfiable; (iii) for any edge e, γ(G – e) ≤ n + 2; (iv) γ(G) = n + 1 iff b(G) = 1. Consequently, the formula is satisfiable exactly when b(G) ≤ 1, establishing NP‑hardness of the bondage decision problem.

Reduction for the total bondage problem.
A similar construction is used, but each variable gadget Hᵢ now consists of four vertices {uᵢ, ¬uᵢ, vᵢ, vᵢ′} with edges forming a small “diamond” that forces any total dominating set to include vᵢ and one of uᵢ or ¬uᵢ. Clause vertices are connected as before. An additional gadget H with five vertices {s₁,…,s₅} and a specific edge pattern is attached, and s₁ and s₃ are linked to every clause vertex. Again k = 1. The authors show: (i) γₜ(G) ≥ 2n + 2, and a minimum total dominating set of that size must contain vᵢ for each i and exactly two vertices from H (s₂ and s₄); (ii) γₜ(G) = 2n + 2 iff the 3‑SAT instance is satisfiable; (iii) γₜ(G – e) ≤ 2n + 2 for any edge e; (iv) γₜ(G) = 2n + 2 iff bₜ(G) = 1. Hence, deciding whether bₜ(G) ≤ 1 is NP‑hard.

Reduction for the reinforcement problem.
The reinforcement reduction mirrors the bondage construction but works in the opposite direction: the authors identify a single edge whose addition reduces the domination number precisely when the original formula is satisfiable. By constructing a graph where γ(G) = n + 2 and adding a specific edge lowers it to n + 1 if and only if a satisfying assignment exists, they prove that determining whether r(G) ≤ 1 is NP‑hard.

Throughout, the size of the constructed graph is linear in the size of the 3‑SAT instance, guaranteeing that the reductions are polynomial‑time. The paper’s methodology hinges on a careful analysis of how dominating (or total dominating) sets must be structured in the gadgets, ensuring that any deviation from the intended configuration forces the domination number to increase, thereby linking the graph‑theoretic parameters directly to the logical satisfiability of the original formula.

In conclusion, the authors successfully demonstrate that the decision versions of bondage, total bondage, and reinforcement are all NP‑hard. This result closes a gap in the literature concerning the algorithmic difficulty of these parameters and suggests that, unless P = NP, no polynomial‑time algorithms exist for computing them on arbitrary graphs. The work also provides a template for future reductions involving domination‑related graph invariants.