Abstract

In a 2011 paper, the game of peg solitaire is generalized to arbitrary boards, which are treated as graphs in the combinatorial sense. Of particular interest are graphs that are freely solvable, that is, graphs that can be solved from any starting position. In this paper we give several examples of freely solvable graphs including all such trees with ten vertices or less, numerous cycles with a subdivided chord, meshes, and generalizations of the wheel, helm, and web.

1. Introduction

Peg solitaire is a table game which traditionally begins with “pegs” in every space except for one which is left empty (i.e., a “hole”). If in some row or column two adjacent pegs are next to a hole (as in Figure 1), then the peg in can jump over the peg in into the hole in . The peg in is then removed. The goal is to remove every peg but one. If this is achieved, then the board is considered solved [1, 2].

Figure 1 

A typical jump in peg solitaire, .

In [3], this notion is generalized to graphs. A graph, , is a set of vertices, , and a set of edges, . Because of the restrictions of peg solitaire, we will assume that all graphs are finite, undirected, and connected graphs with no loops or multiple edges. For all undefined graph theory terminology, refer to West [4]. If there are pegs in vertices and and a hole in , then we allow the peg in to jump over the peg in into the hole in provided that ,  . The peg in is then removed. This jump is denoted by .

A graph is solvable if there exists some vertex so that, starting with a hole in , there exists an associated terminal state consisting of a single peg. A graph is freely solvable if this is true for all vertices. A graph is distance 2-solvable if there exists some vertex so that, starting with a hole in , there exists an associated terminal state consisting of two pegs that are distance 2 apart. In Figure 2, the left graph is not solvable. The middle graph in Figure 2 is solvable but not freely solvable. The right graph in this figure is freely solvable.

Figure 2 

An example of an unsolvable, a solvable, and a freely solvable graph.

In [5], it is shown that, of the nonisomorphic connected graphs on seven vertices or less, only are not freely solvable. However, determining whether a specific graph is freely solvable is NP-hard. Also [6] shows that, if is freely solvable and is obtained by appending a pendent vertex to any vertex of , then is (at worst) solvable. We are motivated by the above comments to give examples of freely solvable graphs in this paper. Known examples of freely solvable graphs include the Petersen graph, the platonic solids, the Archimedean solids, the complete graph, and the complete -partite graph. Other examples are given in [3, 5, 6].

Several useful results are included in [3]. The following theorem allows the completion of the game in reverse by exchanging the roles of pegs and holes.

Theorem 1 (see [3]). Suppose that is a starting state of with associated terminal state . Let and be the duals of and , respectively. It follows that is a starting state of with associated terminal state .

The following observation is also useful.

Observation 1 (see [3]). (i) If can be -solved with the initial hole in vertex and a jump is possible, then there is a first jump; say . Hence, if there are holes in and and pegs elsewhere, then can be -solved from this configuration. (ii) If is a -solvable spanning subgraph of , then is (at worst) -solvable.

Because of this observation, it is of particular interest to determine which trees are freely solvable. In Figure 3, we list all nonisomorphic freely solvable trees with ten vertices or less. The list of trees comes from the appendix to Harary [7]. An exhaustive computer search [8] was used to determine solvability.

Figure 3 

Freely solvable trees with .

2. The Cycle with a Subdivided Chord

The cycle with a subdivided chord, denoted by , is formed from a cycle on vertices, labeled in the usual way, by adding an edge from vertex to vertex to form a chord. This edge is then subdivided. The resulting vertex on the chord is labeled . Figure 4 shows a cycle with a subdivided chord. As usual, will denote the path on vertices with vertices labeled in the usual way.

Figure 4 

The cycle with a subdivided chord, .

In the following result, we show that is solvable with the initial hole in several vertices. However, these graphs may be solvable from additional vertices as well. In [3], it was shown that is solvable with the initial hole in vertex and the final peg in , where . If is odd, then is distance 2-solvable with the initial hole in and the final two pegs in and .

Lemma 2. The graph is solvable with the initial hole in vertex , , , , , , , , , , , , , , or . The graph is solvable with the initial hole in vertex , , , , , or .

Proof. Note that the vertices and are symmetric.
For , suppose the initial hole is in vertex ; jump . An even path subgraph is formed by , , with a hole in . This subgraph is solvable with the final peg in by [3]. Alternately, an even path is formed by , , ,   with a hole in . Solve this path with the final peg in . By Theorem 1, the graph may also be solved with the initial hole in or . A similar argument holds if the initial hole is in , 3, or .
Suppose that the initial hole is in 1. Jump . Solve the even cycle formed by the remaining pegs with a hole in . A similar argument holds when the initial hole is in , ,or .
Suppose that the initial hole is in 2. Jump . Solve the even path formed by the remaining pegs with a hole in . A similar argument holds for the case when the initial hole is in , , or .
Suppose that the initial hole in . Jump 1 over 0 into . Solve the even path formed by the remaining pegs with a hole in .
For , suppose the initial hole is in vertex . An even path is formed by , , with a hole in . Solve this path with the final peg in . Alternately, an even path is formed by , , , with a hole in . Solve this path with the final peg in . By Theorem 1, the graph may also be solved with the initial hole in or 2. A similar argument holds when the initial hole is in , , or .

We now give some freely solvable examples of this type of graph.

Theorem 3. The graph is freely solvable.

Proof. If the initial hole is in vertex , then the graph is solvable by Lemma 2.
Suppose the initial hole is in , where is even and . An even path is formed by , , , with a hole in . Solve this path, ending in . Then jump , , and . Finally, solve the even path with a hole in , formed by , , .
Suppose the initial hole is in , where is odd and . An even path is formed by , with a hole in . Solve this path, ending in . Then jump , , , , and . Finally, solve the even path with a hole in , formed by , , , .

Theorem 4. The graph is freely solvable.

Proof. If the initial hole is in vertex , then the graph is solvable by Lemma 2.
Suppose the initial hole is in , where is even and . An even path is formed by , , with a hole in . Solve this path, ending in . Then jump . Finally, solve the even path with a hole in , formed by , .
Suppose the initial hole is in , where is odd and . An even path is formed by , , , with a hole in . Solve this path, ending in . Now jump . Finally, solve the even path with a hole in , formed by , .

Note that is isomorphic to the graph formed from by adding a chord between vertices and . Thus, this provides partial progress on the open question from [5] as to whether all chorded odd cycles are freely solvable.

Theorem 5. The graph is freely solvable.

Proof. If the initial hole is in vertex , , , , , or , then the graph is solvable by Lemma 2.
Suppose the initial hole is in , where is even and . An even path is formed by , with a hole in . Solve this path, ending in . Then jump and . Finally, solve the even path with a hole in , formed by , .
Suppose the initial hole is in , where is odd and . An even path is formed by , with a hole in . Solve this path, ending in . Then jump , , , and . Finally, solve the even path with a hole in , formed by .
Suppose the initial hole is in . Jump and . Solve the even path formed by the remaining pegs with a hole in .

It is currently unknown if is freely solvable for other values of and .

3. The Mesh and Related Graphs

In [3], it is shown that, if is solvable or distance 2-solvable and is solvable or distance 2-solvable, then the cartesian product is solvable. We now consider a specific case of when such graphs are freely solvable, namely, meshes . If or is even, then is Hamiltonian and has an even number of vertices. Hence, it is freely solvable by [3]. Thus, it suffices to consider the case where both and are odd. Label the vertex of induced by and by .

Lemma 6. For , there exists a peg solitaire solution to the mesh such that (i) the initial jump is and (ii) the final peg is in vertex .

Proof. Proceed by induction on . An exhaustive computer search [8] shows that the claim holds for . Assume the claim holds for , and consider . With the initial hole in vertex , jump , , and . The subgraph is solvable by hypothesis, ending in . Now jump , , , and to solve the graph.

Lemma 7. The mesh is freely solvable with the final peg in vertex .

Proof. Proceed by induction on . If , then use Lemma 6. Assume the claim holds for , and consider . Lemma 6 shows that the claim holds with the initial hole in vertex .
Case 1. Suppose the initial hole is in , where and . The subgraph is solvable by hypothesis, ending in . Now jump , , , , , and to solve the graph.
Case 2. Suppose the initial hole is in . Jump and . The subgraph is solvable by hypothesis, ending in . Finally, jump , , , , , and to solve the graph.
Case 3. Suppose the initial hole is in . Jump . The subgraph is solvable by Lemma 6, ending in . This reduces the graph to Case 1. Suppose the initial hole is in . Jump . This reduces the graph to the above case.
Case 4. Suppose the initial hole is in . Jump and . The subgraph is solvable by Lemma 6, ending in . This reduces the graph to Case 2. Suppose the initial hole is in . Jump and . This reduces the graph to Case 2. The case where the initial hole is in is analogous.

Theorem 8. The mesh is freely solvable.

Proof. We prove the stronger statement that is freely solvable with the final peg in vertex , which is symmetric to vertex . Proceed by induction on . If , then the claim holds by Lemma 7. Assume the claim holds for , and consider .
Case 1. Suppose the initial hole is in , where . The subgraph is solvable by hypothesis, ending in . Now jump and . The remaining pegs form an even path with a hole in . Solve this path with the final peg in .
Case 2. Suppose the initial hole is in . Jump . A hole is now in . Solve the subgraph by hypothesis, ending in .
If , then jump , , and . The remaining pegs form an even path with a hole in . Solve this path with the final peg in .
If , then jump . If , then jump . In either case, this reduces the graph to Case 1.
If , then jump and .
If , then, for , jump . Now jump , , , and . The remaining pegs form an even path with a hole in . Solve this path with the final peg in .
If , then, for , jump . If is odd, then the remaining pegs form an even path with a hole in . Solve this path with the final peg in . If is even, then , , form an even path with a hole in . Solve this path, ending in . Now jump , , and . The remaining pegs form an even path with a hole in . Solve this path with the final peg in .
Case 3. Suppose the initial hole is in . Jump or . In either case, this reduces the graph to Case 2.

It follows from Theorem 8 that the cylinder is freely solvable. The generalized web is obtained from by appending a pendant vertex to for . Similarly, the generalized wheel is obtained from by adding a new vertex such that is adjacent to for . Finally, the generalized helm is obtained from the generalized wheel by appending a pendant vertex to for .

Corollary 9. The generalized web, generalized wheel, and the generalized helm are freely solvable for and .

Proof. We first show that a mesh of even order may be solved with the final peg in an outermost vertex. For , suppose the initial hole is in vertex . Without loss of generality, assume . Solve the subgraph ending in . Jump , , and . The remaining pegs form an even path with a hole in . Solve this path with the final peg in .
For , suppose the initial hole is in vertex . Without loss of generality, assume that . Solve the subgraph, ending in . The remaining pegs form an even path with a hole in . Solve this path, ending in .
For the generalized web, if the hole is in , then solve the mesh subgraph with the final peg in . Now, jump for . If the hole is in , then jump . Next, jump for . Finally, jump . Ignoring the pendants, the graph is a mesh after the first jump.
For the generalized wheel, if the hole is in , then solve the mesh subgraph with the final peg in . Now jump . If the hole is in , then jump . This reduces the graph to the case of the mesh after the first jump.
For the generalized helm, if the hole is in or , then solve the generalized web subgraph with the final peg in . Now jump . If the initial hole is in , then solve the wheel subgraph, ending in . Now, jump for .

4. The Fat Triangle

We now construct the fat polygon, denoted by . Label the vertices of , with , . Replace the edge between and with edges, where . For , place a new vertex, , on the th edge between and . So and . Further, let . In this section, we show that (i.e., the fat triangle) is freely solvable (see Figure 5). Without loss of generality, we assume .

Figure 5 

Fat triangle .

Theorem 10. The fat triangle is freely solvable for all .

Proof. For convenience of exposition, we define and solve the following configurations.
Configuration A. Suppose there are pegs in vertices , , and a hole in . If , then jump , ending in . If is even, say , then, for , jump and , ending in . If is odd, then make the additional jump , ending in .
Configuration B. Suppose there are pegs in vertices , , and , and holes in and . If , then jump and , ending in . If is even, say , then, for , jump , , , and , ending in . If is odd, then perform the additional jumps and , ending in .
Case 1. If , then the graph is , which is freely solvable by [3]. If , then it suffices to show solutions with the initial hole in vertices and .
Suppose the initial hole is in . Jump , , , and . Now jump , , and , for , to eliminate the remaining pegs in , , and . This solves the graph with the final peg in . The final jump could also be . Thus, by Theorem 1, the graph can also be solved with the initial hole in and the final peg in .
Case 2. If , then it suffices to show solutions with the initial hole in vertices , , , and .
Suppose the initial hole is in . Solve the subgraph, ending in , as in Case 1. Now use the method described in Configuration A to eliminate the remaining pegs in . If is even, then the final peg is in . However, the final jump could also be . If is odd, then the final peg is in . However the final jump could also be . In either case, the graph can be solved with the initial hole in by Theorem 1.
Suppose the initial hole is in . Jump , , , and . Use the method of Configuration A to eliminate the remaining pegs in . If , then this solves the graph. If is even, then the final peg is in . If is odd, then the final peg is in . If , then use the method described in Configuration B to eliminate the remaining pegs in and . If and have the same parity, then the final peg is in . Otherwise, the final peg is in .
Suppose the initial hole is in . Jump . If , then this reduces the graph to Case 1. If , then this reduces the graph to the case of the initial hole in .
Case 3. If , then it suffices to show solutions with the initial hole in vertices , , , and .
Suppose the initial hole is in . Solve the subgraph, ending in , as in Case 1. Now use the method described in Configuration B to eliminate the remaining pegs in and . As in Case 2, the graph may also be solved with the initial hole in .
Suppose the initial hole is in vertex . First jump . If , then this reduces the graph to Case 1. Assume that . Jump , , , and . Now use the method described in Configuration A to eliminate the pegs in . If is even, then the ending peg is in . If is odd, then the ending peg is in . Next use the method described in Configuration B to eliminate the remaining pegs from and . If and have the same parity, then the final peg is in . Otherwise, the final peg is in .
Suppose the initial hole is in . Jump , , , and . If necessary, use the method described in Configuration A to eliminate the remaining pegs in . If is even, then the ending peg is in . If is odd, then the ending peg is in . Next use the method described in Configuration B to eliminate the remaining pegs from and . If and have the same parity, then the final peg is in . Otherwise, the final peg is in .
Case 4. If , then the graph may be reduced to a solvable configuration with the initial hole in each vertex.(i)Suppose the initial hole is in . Use the method described in Configuration A to eliminate pegs from . If is even, then the ending peg is in . If is odd, then the ending peg is in . This reduces the graph to Case 3. The case where the initial hole is in is analogous.(ii)Suppose the initial hole is in . Use the method described in Configuration A to eliminate pegs from . This reduces the graph to Case 2.(iii)Suppose the initial hole is in . Jump . This reduces the graph to Case 3 or 4(i). The cases where the initial hole is in or are analogous.