European peg solitaire solution8/3/2023 ![]() 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. Begin to solve \(P_2 \,\square\, Q_1\) using Theorem 1 or Lemma 1. 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. Fibonacci Closure, Peano Integers, Concurrent pi, Concurrent Prime Sieve, Peg Solitaire Solver, Tree Comparison. ![]() 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. A valid move is to jump a peg orthogonally over an adjacent peg into a hole two positions away and then to remove the jumped peg. The objective is, making valid moves, to empty the entire board except for a solitary peg. Strictly k- solvable, if G is k-solvable but not l-solvable for any \(l ![]() K- solvable, if there is some \(v \in V\) such that the starting state \(S = \lbrace v \rbrace \) has an associated terminal state consisting of k vertices. Solvable, if there is some \(v \in V\) such that the starting state \(S = \lbrace v \rbrace \) has an associated terminal state consisting of a single vertex.įreely solvable, if for all \(v \in V\) the starting state \(S = \lbrace v \rbrace \) has an associated terminal state consisting of a single vertex. Therefore, we 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. AI Group Department of Computer Science University of York. In general, we begin with a starting state \(S \subset V\) of vertices that are empty (i.e., without pegs). Modelling and Solving English Peg Solitaire Chris Jefferson, Angela Miguel, Ian Miguel, Armagan Tarim. ![]()
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