If \(v_k\) is contained in another path \(Q_j\) that has not been considered yet, continue with this path in the same way. If no such \(v_l\) exists in \(Q_i\), solve \(Q_i\) completely, ending with a hole in the subgraph \(P_2 \,\square\, \lbrace v_k \rbrace \) where we started with a hole (this can be done because of Lemma 1 and Theorem 1). Keep going with this process until no such vertex \(v_l\) exists. Then continue in this manner (stop solving \(Q_2\) and start solving \(Q_3\)). Let \(v_l \ne v_k\) be the vertex with largest index such that \(v_k\) lies in another path (if such a vertex exists). A soon as the subgraph \(P_2 \,\square\, \lbrace v_k \rbrace \) contains exactly one peg and one hole, stop solving \(Q_1\) and start solving \(Q_2\). Let \(v_k \in Q_1\) be the vertex with largest index such that \(v_k\) lies in another path. neighbor (horizontally or vertically) into an empty space opposite. A piece moves on this board by jumping over one of its immediate. Begin to solve \(P_2 \,\square\, Q_1\) using Theorem 1 or Lemma 1. 32 movable pegs (' ') are arranged on the following board such that. Use this to decompose T into paths \(Q_1,\ldots ,Q_m\). Choose a root vertex \(v_0\) and do a depth-first-search, enumerating the vertices in the order they occur. If T is a path this follows from Theorem 1. It suffices to show that \(P_2 \,\square\, T\) is solvable. \(P_2 \,\square\, G\) is freely solvable for any connected graph G. The first step in doing this is to show that Cartesian products are solvable if one of the components is the path \(P_2\). Using the super free solvability of ladders, we can prove a fairly general result about Cartesian products. In that case G has solitaire number \(\mathrm \cdot v\), we can finally solve \(G_1\) with terminal peg in t.Äue to symmetry, this covers all cases. Strictly k- solvable, if G is k-solvable but not l-solvable for any \(lwe use the following notation. The goal of the original game is to remove all pegs but one. We will always assume that the starting state S consists of a single vertex. A terminal state T is associated to a starting state S if T can be obtained from S by a series of jumps. A terminal state \(T \subset V\) is a set of vertices that have pegs at the end of the game such that no more jumps are possible. In general, we begin with a starting state \(S \subset V\) of vertices that are empty (i.e., without pegs).
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