Peg Solitaire on Cartesian Products of Graphs

In 2011, Beeler and Hoilman generalized the game of peg solitaire to arbitrary connected graphs. In the same article, the authors proved some results on the solvability of Cartesian products, given solvable or distance 2-solvable graphs. We extend these results to Cartesian products of certain unsolvable graphs. In particular, we prove that ladders and grid graphs are solvable and, further, even the Cartesian product of two stars, which in a sense are the “most” unsolvable graphs.


Introduction
In [3], Beeler and Hoilman introduced the game of peg solitaire on graphs as a generalization of the classical peg solitaire game: For a connected, undirected graph G ¼ ðV; EÞ, we put pegs in some of the vertices of G. Given three vertices u, v, w with pegs in u and v and a hole in w such that uv; vw 2 E, we can jump with the peg from u over v into w, removing the peg in v (see Fig. 1). This jump will be denoted as u Á v Á w.
In general, we begin with a starting state S & V of vertices that are empty (i.e., without pegs). A terminal state T & V is a set of vertices that have pegs at the end of the game such that no more jumps are possible. A terminal state T is associated to a starting state S if T can be obtained from S by a series of jumps. We will always assume that the starting state S consists of a single vertex.
The goal of the original game is to remove all pegs but one. This is not possible for every graph. Therefore, we use the following notation. A graph G ¼ ðV; EÞ is called solvable, if there is some v 2 V such that the starting state S ¼ fvg has an associated terminal state consisting of a single vertex. freely solvable, if for all v 2 V the starting state S ¼ fvg has an associated terminal state consisting of a single vertex. k-solvable, if there is some v 2 V such that the starting state S ¼ fvg has an associated terminal state consisting of k vertices. strictly k-solvable, if G is k-solvable but not l-solvable for any l\k. In that case G has solitaire number PsðGÞ ¼ k.
Peg solitaire has been considered for quite a few classes of graphs, including paths, complete graphs, stars, double stars and caterpillars (for more results and variants see [1][2][3][4][5][6][7][8][9]). In this article we will consider the Cartesian product of two graphs. For graphs G and H, we denote the Cartesian product of G and H by G h H and use the (common) notation ðg; hÞ 2 VðG h HÞ for the vertex induced by g 2 VðGÞ and h 2 VðHÞ.
In [3], Beeler and Hoilman gave some results about Cartesian products of graphs, that are either solvable or distance 2-solvable (meaning that they are strictly 2-solvable such that the pegs in the terminal state have distance 2). We will extend these results to Cartesian products of an arbitrary graph and a solvable graph (with an extra condition). We also consider Cartesian products of stars (i.e., graphs that are furthest away from being solvable) and show that these Cartesian products are solvable, too. This gives rise to the conjecture, that all (non trivial) connected Cartesian products are solvable.

Ladders and Grids
In this section we will derive some general results. Our main result is the solvability of the Cartesian product of an even Hamiltonian graph and an arbitrary graph.
To deal with Cartesian products, we need the following new concept. A graph G is called super freely solvable if, for any (not necessarily distinct) v; w 2 V, terminal state fwg is reachable from starting state fvg.
The following (well known) lemma is easy to prove.  For m; n ! 2, we call P n h P 2 a ladder and P m h P n a grid. In particular, C 4 is both, a ladder and a grid. We will show that, apart from C 4 , all ladders and grids are super freely solvable.
Theorem 1 If n ! 3, the ladder P n h P 2 is super freely solvable. Moreover, for each i 2 f2; . . .; n À 1g, at some point in the solving process the subgraph fv i g h P 2 contains exactly one peg and one hole.
Proof We divide the proof into several cases, depending on the location of the starting hole and the terminal vertex. In each case it will be clear from the jumps that the second statement in the theorem is fulfilled. for 1 j b n 2 c À 1. If n is odd, we further jump ðn; 2Þ Á ðn; 1 ! Þ Á ðn À 1; 1Þ. This sequence of jumps will result in a state where we have pegs in (1, 1), (1, 2), (2j, 1) and ð2j þ 1; 2Þ for 1 j b n 2 c À 1 and, if n is odd, also in ðn À 1; 1Þ. Every other vertex is empty. Starting with the jump ð1; 1Þ Á ð1; 2 ! Þ Á ð2; 2Þ, we go up on the second component of the ladder and go down on the other side. This will end with a peg in (1, 1). If t ¼ ð1; 2Þ, we interchange every second component in the jumps above after the first jump. 3. Suppose s ¼ ð1; 1Þ and t ¼ ðk; iÞ with 1\k\n. Consider the ladder up to height k. We solve this using Case 1 with a peg in (k, i) (this is also possible if this ladder is a C 4 ). Now we are left with a ladder of height n À k þ 1 where t ¼ ð1; iÞ. If this is the C 4 , we can solve it since we have a peg in (k, i). If the remaining ladder is not the C 4 , solve it using Case 2. 4. Suppose that s ¼ ðk; 1Þ where 1\k\n and let t ¼ ðj; iÞ. W.l.o.g. assume j k.
Divide the ladder in L 1 and L 2 where L 1 contains the vertices (l, r) with l ! k and let L 2 be the ladder containing the vertices (l, r) with l k. Solve ladder L 1 , ending in ðk; i 0 Þ where i 0 6 ¼ i. This is possible if L 1 6 ¼ C 4 . Now solve L 2 ending in t. This is also possible if L 2 6 ¼ C 4 . The only cases left to consider are the cases L ¼ P 3 h P 2 and s ¼ ð2; 1Þ and t ¼ ð2; iÞ. It is easy to check that these are solvable. h Theorem 2 If n ! 3 or m ! 3, the grid P m h P n is super freely solvable.
Proof We prove the theorem by induction. If m ¼ 2 or n ¼ 2, the statement follows from Theorem 1. Hence, suppose that both m and n are at least 3. If G ¼ P 3 h P 3 , we find a ladder P 2 h P 3 in G containing a hole. Since this ladder is super freely solvable, it is easy to see that we can solve it and get a terminal peg (of the ladder) in a vertex such that we can solve the remaining grid with a peg in an arbitrary vertex t. Now, let m; n ! 3 be arbitrary with maxfm; ng ! 4 and let s ¼ ðj; kÞ be the starting hole.
1. If s ¼ ð1; 1Þ, let w.l.o.g. m ! 4 and divide the grid into the two graphs G 1 ffi P 2 h P n and G 2 ffi P mÀ2 h P n . If t 2 G 2 , solve the ladder G 1 with terminal peg next to G 2 . After two more jumps, we are left with the graph G 2 with exactly one hole. By induction hypothesis this is solvable with peg in t. If t 2 G 1 , start with the jump ð3; 1Þ Á ð2; 1 ! Þ Á ð1; 1Þ. Then, by induction hypothesis, we can solve G 2 , which is solvable with terminal peg in t. 2. If s ¼ ð1; kÞ with k 6 ¼ 1, we can proceed the same way, provided that m ! 4. If m ¼ 3, consider the two graphs G 1 and G 2 , where G 1 contains all vertices (r, s) such that s k and G 2 contains all vertices (r, If G 1 is a ladder, we can proceed analogously to the case when G ¼ P 3 h P 3 , solving G 2 first and dealing with the remaining 4 pegs later. If G 1 is not a ladder, solve G 2 with terminal peg in ð1; k þ 1Þ. Next do the jumps ð1; k À 2Þ Á ð1; k À 1 ! Þ Á ð1; kÞ and ð1; k þ 1Þ Á ð1; k ! Þ Á ð1; k À 1Þ. We are left with the grid G 1 which is freely solvable by induction hypothesis. 3. If s ¼ ðj; kÞ with 1\j\m and 1\k\n, divide the grid into two grids G 1 and G 2 that form a partition of G. Let w.l.o.g. s 2 G 1 . This can always be done in a way such that G 1 and G 2 are grids, neither of them is the grid C 4 , and s has distance 1 from its nearest vertex in G 2 . If t 2 G 2 we can proceed analogously to case 1. If After this jump, we solve G 2 with terminal peg in the unique vertex u 2 G 2 such that the subgraph induced by the vertices s, u, v, w is C 4 . After jumping u Á s ! Á v, we can finally solve G 1 with terminal peg in t.

Due to symmetry, this covers all cases. h
Using the super free solvability of ladders, we can prove a fairly general result about Cartesian products. The first step in doing this is to show that Cartesian products are solvable if one of the components is the path P 2 .
Theorem 3 P 2 h G is freely solvable for any connected graph G.
Proof Let T be a spanning tree of G. It suffices to show that P 2 h T is solvable. If T is a path this follows from Theorem 1. So suppose that T is not a path. Choose a root vertex v 0 and do a depth-first-search, enumerating the vertices in the order they occur. Use this to decompose T into paths Q 1 ; . . .; Q m . (cf. Fig. 2).
Start with a hole in v 0 . Begin to solve P 2 h Q 1 using Theorem 1 or Lemma 1. Let v k 2 Q 1 be the vertex with largest index such that v k lies in another path. A soon as the subgraph P 2 h fv k g contains exactly one peg and one hole, stop solving Q 1 and start solving Q 2 . Let v l 6 ¼ v k be the vertex with largest index such that v k lies in another path (if such a vertex exists). Then continue in this manner (stop solving Q 2 and start solving Q 3 ). Keep going with this process until no such vertex v l exists. If no such v l exists in Q i , solve Q i completely, ending with a hole in the subgraph P 2 h fv k g where we started with a hole (this can be done because of Lemma 1 and Theorem 1). 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 this is not the case, go all the way back and solve the ladders that have not been completely solved yet, always ending with a hole in the same subgraph in which we started with a hole. This process solves P 2 h T. Since v 0 was arbitrary, we have free solvability. h Theorem 4 Let G be a graph that has an even Hamiltonian path. Then for any connected graph H, G h H is solvable.

Cartesian products of stars
Keeping in mind the results from the last section, it seems natural to take a look at Cartesian products of unsolvable graphs. Since they are particularly far away from being solvable, we consider star graphs. Let K 1;n denote the star (graph) on n þ 1 vertices, where we denote the central vertex by 0 and the pendant vertices by 1; 2; . . .; n.
It is known that stars K 1;n are not solvable for n ! 3 [3]. However, the Cartesian product of any two stars on at least 2 vertices is solvable. We will prove several lemmas before turning our attention to this statement.
In Fig. 3, seven graphs (in the following often used as substructures in larger graphs) with a given (starting) state and a corresponding reachable (terminal) state are displayed. The notation EP i denotes the (so called) reduction (or elimination) process that is performed by several jumps to get from the left hand state to the right hand one. The reader is invited to check that all of these reductions are in fact possible. For the remainder of this chapter, we consider the Cartesian product K 1;n h K 1;m with n ! m ! 1. For even m (we will reduce everything to this case), we define A ¼ fðg; hÞ : g; h 2 ½mg 1 and A i ¼ fðg; hÞ 2 A : g À h 2 fÀ2i; À2i þ 1; m À 2i; m À 2i þ 1gg for i 2 1; m 2 À 1 Â Ã as well as A 0 ¼ fðg; hÞ 2 A : g À h 2 f1 À m; 0; 1gg: Since all possible differences are considered and appear in exactly one A i , we immediately get the following result. Proof For i ¼ 0 exactly two of the equations j À y ¼ a with a 2 f1 À m; 0; 1g have a solution in [m], to be specific, y ¼ j; a ¼ 0 and either y ¼ j À For i 2 ½ m 2 À 1, exactly two of the equations j À y ¼ a with a 2 fÀ2i; 1 À 2i; m À 2i; m À 2i þ 1g have a solution. We show this exemplarily for j\m À 2i, the other (three) cases follow in the same fashion. For a 2 fm À 2i; m À 2i þ 1g apparently no solution can be found since y is positive. The other two equations have the solutions j þ 2i and j þ 2i À 1, which lie in [m] since j\m À 2i. The arguments are similar for ðÁ; jÞ, therefore, part (i) is true.
For part (ii), we note, that x ðiÞ l;1 À x ðiÞ l;2 2 fÀ2i; 1 À 2ig (where the difference alternates) holds as long as x ðiÞ l;2 m, i.e. l 2m À 4i À 1. For l 0 ¼ 2m À 4i À 1, we have x ðiÞ l 0 ;1 À x ðiÞ l 0 ;2 ¼ À2i þ 1, and after changing the first second component to obtain x ðiÞ kþ1 , we have x ðiÞ l 0 þ1;1 À x ðiÞ l 0 þ1;2 ¼ m À 2i. Hence, for 2m À 4i l 2m À 2, we have x ðiÞ l;1 À x ðiÞ l;2 2 fm À 2i; m À 2i þ 1g (again the difference alternates). Since x ðiÞ 2mÀ1 ¼ ð1; 2iÞ and x ðiÞ 2m ¼ x ðiÞ 0 , the set X i containing the elements generated via (3) is a subset of A i . From (i) we immediately get jA i j ¼ 2m. Since the construction procedure of X i is periodic of length 2m, we obtain jX i j ¼ 2m. Therefore we have X i ¼ A i , which proves statement (ii). h Let us now consider the vertex set B ¼ fðg; hÞ : g; h 2 ½0; m; ðg; hÞ 6 ¼ ð0; 0Þ; g Á h ¼ 0g: The set A i [ B induces a cycle in K 1;n h K 1;m (with 2 j m) for each i 2 0; m 2 À 1 Â Ã . Furthermore, in this cycle an element of B is adjacent to two elements of A i and vice versa. Proof Due to symmetry it is sufficient to consider the case with holes in (i, 0) for i 2 ½2; m. First, we start by removing pegs from (0, i) for i 2 ½m and most of the vertices from A 0 (except (1, 1)). This can be done by jumping The basic idea in proving the following theorem is first eliminating pegs from (i, 0) and (0, i) for i 2 ½m. Next, the other pegs in the subgraph K 1;m h K 1;m will be removed. The last step is the elimination of the remaining pegs (i, j) for i 2 ½m þ 1; n (if n [ m). If at least one of n and m is odd, we have to be particularly thorough. The occurring problems will be dealt with by reducing these cases to even n and m.
Theorem 5 The Cartesian product K 1;n h K 1;m is solvable for all n; m 2 Z.
Proof If m ¼ 1 or n ¼ 1, the statement follows from Theorem 3. Thus, without loss of generality, we may assume m; n ! 2 and n ! m. Start with a hole in (0, 0). We distinguish four cases.

Conjecture 1
For any solvable graph G and any non trivial connected graph H, G h H is (freely) solvable.
In view of Theorem 5, because stars are, in a way, the most unsolvable graphs, we even believe the following much stronger statement to be true.
Conjecture 2 For any two connected non trivial graphs G and H, G h H is (freely) solvable.
We believe that this can be proven in a similar fashion as Theorem 3 via decomposing the graphs G and H into stars and paths and using the results we obtained about ladders, grids, and Cartesian products of stars. But the technical details seem very complicated.
Funding Open Access funding enabled and organized by Projekt DEAL.

Declaration
Conflict of interest The authors declare that they have no conflict of interest.
Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http:// creativecommons.org/licenses/by/4.0/.