![]() ![]() If \(v_k\) is contained in another path \(Q_j\) that has not been considered yet, continue with this path in the same way. The aim is to leave only one peg occupying the centre position. 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). Peg Solitaire is played on a board, which in its most usual version has the shape of a cross and 33 holes, with the central hole empty and the remaining holes occupied by pins or pegs. 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. Begin to solve \(P_2 \,\square\, Q_1\) using Theorem 1 or Lemma 1. ![]() A thorough analysis of the Peg solitaire game is known. The point is that user should always have a plan to remove peg which should not be separated completely. Use this to decompose T into paths \(Q_1,\ldots ,Q_m\). Solitaire marbles solution In order to find the peg solitaire solution the user should mainly focus on reachability of the marble or peg. 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. each having a different solution: Standard, Square, European, Asymmetric, Rhombus, German. Strictly k- solvable, if G is k-solvable but not l-solvable for any \(lgoal 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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