[Paper Review] Misere Hackenbush is NP-Hard

Fraser Stewart|arXiv (Cornell University)|Feb 17, 2012

Artificial Intelligence in Games3 references3 citations

TL;DR

This paper proves that Red-Blue-Green Misère Hackenbush is NP-hard by reducing it from normal-play Red-Blue Hackenbush, which is already known to be NP-hard. While Red-Blue Misère Hackenbush is solvable in polynomial time via a simple count of grounded edges, the addition of green edges—removable by both players—introduces combinatorial complexity that elevates the problem to NP-hardness under misère play rules.

ABSTRACT

Hackenbush is a two player game, played on a graph with coloured edges where players take it in turns to remove edges of their own colour. It has been shown that under normal play rules Red-Blue Hackenbush (all edges are coloured either red or blue) is NP-hard. We will show that Red-Blue Hackenbush is in P, but that Red-Blue-Green Hackenbush is NP-Hard, when played under misère rules.

Motivation & Objective

• To determine the computational complexity of Red-Blue-Green Misère Hackenbush under misère play rules.
• To contrast the complexity of Red-Blue Misère Hackenbush—solvable in polynomial time—with the more complex case involving green edges.
• To establish that the inclusion of green edges, removable by both players, significantly increases problem difficulty, leading to NP-hardness.
• To investigate whether restricted forms of the game (e.g., only grounded green edges) remain NP-hard.

Proposed method

• A reduction from normal-play Red-Blue Hackenbush to misère-play Red-Blue-Green Hackenbush is constructed.
• The transformation involves replacing the ground with a single vertex and attaching the resulting graph to a single grounded green edge.
• The structure ensures that neither player will cut the green edge early, as doing so would lose the game under misère rules.
• The outcome of the transformed game $ G_m $ mirrors the outcome of the original $ G' $ under normal play rules.
• The proof relies on the fact that the player who wins $ G' $ under normal play will win $ G_m $ under misère play.
• The reduction preserves NP-hardness, proving that solving $ G_m $ is at least as hard as solving the original NP-hard problem.

Experimental results

Research questions

• RQ1 Is Red-Blue-Green Misère Hackenbush NP-hard, despite Red-Blue Misère Hackenbush being solvable in polynomial time?
• RQ2 Can the NP-hardness of normal-play Red-Blue Hackenbush be transferred to the misère version with green edges?
• RQ3 Does the presence of green edges, removable by both players, fundamentally alter the game's computational complexity?
• RQ4 Are there restricted subclasses of Red-Blue-Green Misère Hackenbush (e.g., only grounded green edges) that remain NP-hard?

Key findings

• Red-Blue Misère Hackenbush is solvable in polynomial time, with the outcome determined solely by comparing the number of grounded blue and red edges.
• If the number of grounded blue edges exceeds red edges, Left wins; if red exceeds blue, Right wins; if equal, the first player wins.
• The addition of green edges in Red-Blue-Green Misère Hackenbush increases the problem's complexity to NP-hardness.
• The NP-hardness is established via a reduction from normal-play Red-Blue Hackenbush, which is already known to be NP-hard.
• The reduction constructs a game $ G_m $ where the outcome under misère play mirrors the outcome of $ G' $ under normal play.
• The proof shows that neither player will cut the single grounded green edge in $ G_m $, as doing so would result in an immediate win for the opponent under misère rules.