Freely Solvable Graphs in Peg Solitaire

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.


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].
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.
In [5], it is shown that, of the 996 nonisomorphic connected graphs on seven vertices or less, only 54 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. 1 2 3 x y z x y z x y z  (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.

The Cycle with a Subdivided Chord
The cycle with a subdivided chord, denoted by CSC( , ), is formed from a cycle on vertices, labeled 0, 1, . . . , − 1 in the usual way, by adding an edge from vertex 0 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 0, 1, . . . , − 1 in the usual way.
In the following result, we show that CSC( , ) 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 2 is solvable with the initial hole in vertex 1 and the final peg in (2 − 2), where ≥ 2. If ≥ 5 is odd, then is distance 2-solvable with the initial hole in 1 and the final two pegs in ( − 3) and ( − 1).
Proof. Note that the vertices and + (mod ) are symmetric.
For CSC (2 , ), suppose the initial hole is in vertex 0; jump ⋅ ⃗ ⋅ 0. An even path subgraph is formed by − 1, , . . . , 2 − 1, 0, . . . , − 2 with a hole in . This subgraph is solvable with the final peg in −3 by [3]. Alternately, an even path is formed by + 1, , − 1, . . . , 0, 2 − 1, . . . , + 2 with a hole in . Solve this path with the final peg in + 3. By Theorem 1, the graph may also be solved with the initial hole in − 3 or + 3. A similar argument holds if the initial hole is in , 3, or 2 − 3.  Suppose that the initial hole is in 1. Jump ⋅ ⃗ 0 ⋅ 1. Solve the even cycle formed by the remaining pegs with a hole in 0. A similar argument holds when the initial hole is in − 1, + 1,or 2 − 1.
Suppose that the initial hole is in 2. Jump 0 ⋅ ⃗ 1 ⋅ 2. Solve the even path formed by the remaining pegs with a hole in 0. A similar argument holds for the case when the initial hole is in − 2, + 2, or 2 − 2.
Suppose that the initial hole in . Jump 1 over 0 into . Solve the even path formed by the remaining pegs with a hole in 0.
For CSC(2 + 1, ), suppose the initial hole is in vertex 0. An even path is formed by , 0, 1, . . . , 2 with a hole in 0. Solve this path with the final peg in 2 − 1. Alternately, an even path is formed by , 0, 2 , 2 − 1, . . . , 1 with a hole in 0. Solve this path with the final peg in 2. By Theorem 1, the graph may also be solved with the initial hole in 2 − 1 or 2. A similar argument holds when the initial hole is in , − 2, or + 2.
We now give some freely solvable examples of this type of graph.
Note that CSC(2 , 1) is isomorphic to the graph formed from 2 +1 by adding a chord between vertices 0 and 2. Thus, this provides partial progress on the open question from [5] as to whether all chorded odd cycles are freely solvable.
Suppose the initial hole is in . Jump 2 ⋅ ⃗ 3 ⋅ and 0 ⋅ ⃗ ⋅ 3. Solve the even path formed by the remaining pegs with a hole in 2.
It is currently unknown if CSC( , ) is freely solvable for other values of and .

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 ℎ ∈ ( ) by ( , ℎ).

Case 3. Suppose the initial hole is in
It follows from Theorem 8 that the cylinder ◻ is freely solvable. The generalized web is obtained from ◻ by appending a pendant vertex to ( , − 1) for = 0, . . . , − 1. Similarly, the generalized wheel is obtained from ◻ by adding a new vertex such that is adjacent to ( , 0) for = 0, . . . , − 1. Finally, the generalized helm is obtained from the generalized wheel by appending a pendant vertex to ( , − 1) for = 0, . . . , − 1.
Corollary 9. The generalized web, generalized wheel, and the generalized helm are freely solvable for ≥ 3 and ≥ 2.
For the generalized wheel, if the hole is in ( , ), then solve the mesh subgraph with the final peg in (0, 0). Now jump ⋅ ⃗ (0, 0) ⋅ (1, 0). If the hole is in , then jump (0, 0) ⋅ ⃗ (1, 0) ⋅ . This reduces the graph to the case of the mesh after the first jump.
Case 1. If 0 = 1 = 2 = 1, then the graph is 6 , which is freely solvable by [3]. If 0 = 1 = 2 ≥ 2, then it suffices to show solutions with the initial hole in vertices . . , 0 , to eliminate the 0 − 1 remaining pegs in 0 , 1 , and 2 . This solves the graph with the final peg in 0 . The final jump could also be 2, 1 ⋅ ⃗ 2 ⋅ 1,1 . Thus, by Theorem 1, the graph can also be solved with the initial hole in 1,1 and the final peg in 0 .
Suppose the initial hole is in 0 . Solve the Fp(3; 1 , 1 , 1 ) subgraph, ending in 0 , as in Case 1. Now use the method described in Configuration A to eliminate the 0 − 1 remaining pegs in 0 . If 0 − 1 is even, then the final peg is in 0 . However, the final jump could also be 0, 0 ⋅ ⃗ 1 ⋅ 0,1 . If 0 − 1 is odd, then the final peg is in 1 . However the final jump could also be 0, 0 ⋅ ⃗ 0 ⋅ 0,1 . In either case, the graph can be solved with the initial hole in 0,1 by Theorem 1.
Suppose the initial hole is in 0 . Solve the Fp(3; 2 , 2 , 2 ) subgraph, ending in 0 , as in Case 1. Now use the method described in Configuration B to eliminate the remaining 0 − Suppose the initial hole is in vertex 2,1 . First jump 0, 0 ⋅ ⃗ 0 ⋅ 2,1 . If 0 = 2, then this reduces the graph to Case 1. Assume that 0 ≥ 3. Jump 1 ⋅ ⃗ 0,1 ⋅ 0 , 0,2 ⋅ ⃗ 0 ⋅ 0, 0 , 2 ⋅ ⃗ 1,1 ⋅ 1 , and 1 ⋅ ⃗ 1,2 ⋅ 2 . Now use the method described in Configuration A to eliminate the 2 pegs in 2 . If 2 is even, then the ending peg is in 2 . If 2 is odd, then the ending peg is in 0 . Next use the method described in Configuration B to eliminate the remaining 0 − 2 pegs from 0 and 1 . If 0 − 2 and 2 have the same parity, then the final peg is in 2 . Otherwise, the final peg is in 0 .
Case 4. If 0 > 1 > 2 ≥ 1, then the graph may be reduced to a solvable configuration with the initial hole in each vertex.
(i) Suppose the initial hole is in 0 . Use the method described in Configuration A to eliminate 0 − 1 pegs from 0 . If 0 − 1 is even, then the ending peg is in 1 . If 0 − 1 is odd, then the ending peg is in 0 . This reduces the graph to Case 3. The case where the initial hole is in 1 is analogous. (ii) Suppose the initial hole is in 2 . Use the method described in Configuration A to eliminate 1 − 2 pegs from 1 . This reduces the graph to Case 2. (iii) Suppose the initial hole is in 0,1 . Jump 0, 0 ⋅ ⃗ 0 ⋅ 0,1 . This reduces the graph to Case 3 or 4(i). The cases where the initial hole is in 1,1 or 2,1 are analogous.

Open Problems
We end this paper by listing a few open problems as possible avenues for future research. (i) What other graphs are freely solvable? (ii) Is the corona of a (freely) solvable graph likewise (freely) solvable? (iii) What is the solvability of a cycle with a chord that is subdivided multiple times? (iv) For what graphs is it possible to start with a specific initial jump and end with another specific jump? (v) Suppose that we want to start with our initial hole in any vertex and end with our final peg in any vertex . For what graphs is this possible? (vi) What is the solvability of other fat polygons?