Integer Programming Based Algorithmsfor Peg Solitaire Problems

Masashi KIYOMI1, Tomomi MATSUI11Department of Mathematical Engineering and Information Physics,

Graduate School of Engineering, University of Tokyo,7-3-1, Hongo, Bunkyo-ku, Tokyo 113-8656, Japan.{masashi, $\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{o}\mathrm{m}\mathrm{i}$ } $\copyright \mathrm{m}\mathrm{i}\mathrm{S}\mathrm{o}\mathrm{j}\mathrm{i}\mathrm{r}\mathrm{o}.\mathrm{t}.\mathfrak{U}-\mathrm{t}\mathrm{o}\mathrm{k}\mathrm{y}\mathrm{o}.\mathrm{a}\mathrm{c}.\mathrm{j}\mathrm{p}$

Abstract: Peg solitaire is a classical one player game. In this paper. we dealt with the pegsolitaire problem as an integer programming problem. We proposed algorithms based on thebacktrack search method and relaxation methods for integer programming problem.The algorithms first solve relaxed problems and get an upper bound of the number of jumpsfor $\mathrm{e}\mathrm{a}\mathrm{c}1_{1}$ jump position. This upper bound saves much time at the next stage of backtracksearching. While solving the relaxed $\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{b}\mathrm{l}\mathrm{e}\mathrm{m}\mathrm{S}_{i}$ we can prove many peg solitaire problemsare infeasible. We proposed two types of backtrack $\mathrm{s}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{g}_{j}$ forward-only searching andforward-backward searching. Our algorithnl can solve all the peg solitaire problem instanceswe tried and the total computational time is less than 20 minutes on an ordinary notebookpersonal computer.

Keywords: peg solitaire, integer programming, backtrack searching.

1 Introduction

Peg solitaire is a one player game using pegsand a board with some holes. In each configu-ration. each hole contains at most one peg (seeFigures 1. 2). A peg solitaire game has two spe-cial cohfigurations,$\cdot$ the starting configuration andthe finishing configuration. The aim of this gameis to get the $\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{i}_{\mathrm{S}}\mathrm{h}\mathrm{i}\mathrm{l}\mathrm{i}\mathrm{g}$ configuration from tlle start-ing configuration by moving and removing pegsas follows.

When there are (vertically or hori-zontally) consecutive three holes satis-fying that first and second holes containpegs and third hole is empty: the playercan remove two pegs from the consecu-tive holes and place a peg in the emptyhole (see Figure 3).

The move obeying the above rule is called a jump.In this paper, we propose an algorithm for peg

solitaire games based on integer programmingproblems. In Sections 2, 3, and 4, we considera peg solitaire problem defined below, which isa natural extension of the ordinary peg solitairegame. In Section 2, we formulate the peg solitaire

problem as an integer programming problem. InSection 3, we show a relation between the pagodafunction defined by Berlekamp. Conway and Guy[3] and our integer programming problem. Sec-tion 4 proposes an integer programming problemwhich is useful for pruning the backtrack searchdescribed in Section 6. We report computationalresults of our $\mathrm{a}\mathrm{l}\mathrm{g}\mathrm{o}\mathrm{r}\mathrm{i}\mathrm{t}_{\sigma \mathrm{h}}\mathrm{m}$ in Section 5.

There are many types of peg solitaire boardsand various pairs of starting and finishing config-urations. The most famous peg solitaire boardis that of Figure 1,. which is called the Englishboard.

A peg solitaire problem is defined by a peg soli-taire board and a pair of starting and finishingconfigurations. If there exists a sequence of $\mathrm{j}_{\mathrm{U}\ln_{\mathrm{P}}}\mathrm{S}$

which transforms the starting configuration intothe finishing configuration. we say that the givenpeg solitaire problem is feasiblei and the sequenceof jumps is a feasible sequence. The peg solitaireproblem finds a feasible sequence of jumps if it isfeasible; and answers “infeasible”, if the problemis not feasible.

In [8], Uehara and Iwata dealt with the gener-alized Hi-Q problems which are equivalent to theabove peg solitaiie problems and showed the NP-

数理解析研究所講究録1185巻 2001年 100-108 100

Figure 3: an example of a jump

Figure 1: starting configuration example

Figure 2: finishing configuration example

$\bullet$ implies a hole with a peg. and $\mathrm{O}$ implies ahole with no peg.

completeness. In the well-known book ’‘Winningways for Mathematical Plays [3] $\text{ノ}.$

, Berlekamp.Conway and Guy discussed variations of prob-lems related to peg solitaire problems. Theyshowed the infeasibility of the peg solitaire prob-lem $‘:_{\mathrm{S}\mathrm{e}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{n}}\mathrm{g}$ scout 5 paces out into $\mathrm{d}\mathrm{e}\mathrm{s}\mathrm{e}\mathrm{r}\mathrm{t}^{i}$

’ by us-ing the pagoda function approach. In [7], Kannoproposed a linear programming based algorithmfor finding a pagoda function which guaranteesthe infeasibility of a given peg solitaire problem,if it exists. Recently, Avis and Deza [1] formu-lated a peg solitaire problem as a combinatorialoptimization problem and discussed the proper-ties of the feasible region called “a solitaire $\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{e}\text{ノ}’$ .

2 Integer Programming

In this section. we formulate the peg solitaireproblem as an integer $\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{g}\mathrm{l}\cdot \mathrm{a}\mathrm{m}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{g}$ problem.

We assume that all the holes on a given boardare indexed by integer numbers $\{1_{J}.2\ldots..n\}$ . Theboard of Figure 1 has 33 holes and so $n=33$ .We describe a state of certain configuration (pegsin the holes) by the $n$-dimensional 0-1 vector $p$

satisfying that the $i\mathrm{t}\mathrm{h}$ element of $p$ is 1 if andonly if the hole $i$ contains a peg. In the rest ofthis paper, we denote the starting configurationby $p_{s}$ and the finishing configuration by $p_{f}$ .

Let $J$ be the family of all the sequences of con-secutive three holes on a given board. Each ele-nlent in $J$ corresponds to a certain jump and sowe can denote a jump by a unit vector indexedby $J$ . In the rest of this paper. we assume thatall the $\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{t}_{\mathrm{S}}$ in $J$ are indexed by $\{$ 1. 2 $\ldots$ . . $n\mathrm{t}\}$ .

For example, the board of Figure 1 contains 76 se-quences of consecutive three holes and so $m=76$ .Given a peg solitaire board. we define $n\cross m$ ma-trix $\dot{A}=(a_{ij})’$. whose rows and columns are in-dexed by holes and jumps respectively: by

$a_{ij}=\{$

1 (a peg on the hole $i$ isremoved by the jump $j$ ),

$-1$ (a peg is placed on the hole $i$

by the jump $j$ ),$0$ (otherwise).

For any 0-1 vector $p,$ $\# p$ denotes the numberof ls in the elements of $p$ . We denote $\#\mathrm{P}_{S^{-}}\# pf$

by $l$ . If the given peg solitaire problem is fea-sible, any feasible sequence consists of $l$ jumps.For example, the peg solitaire problem definedby Figures 1 and 2 is feasible and there exists

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a feasible sequence whose length $l=32$ . Sinceeach jump corresponds to an m-d..imensiOnal unitvector, a feasible sequence corresponds to a se-quence of $l$ unit vectors $(x^{1}, x^{2}, \ldots, x^{l})$ such that$x^{k}=(x^{k}, x12’., x)^{\mathrm{T}}k..km$ for all $k\in\{1,2, \ldots, l\}$

and

$x_{j}^{k}=\{$1 (the $k\mathrm{t}\mathrm{h}$ move is the jump $j$ ).$0$ (the $k\mathrm{t}\mathrm{h}$ move is not the jump $j$ ).

If a configuration $p’$ is obtained by applying thejump $j$ to a configuration $p,\cdot$ then $p’=p– Au$where $u$ is the $j\mathrm{t}\mathrm{h}$ unit vector in $\{0,1\}^{m}$ . Fromthe above discussion. we can formulate the pegsolitaire problem as the following integer pro-gramming problem;

$\mathrm{I}\mathrm{P}1$ : find $(x^{1}. X^{2}\ldots..x^{l})$

$\mathrm{s}$ . $\mathrm{t}$ . $\mathrm{A}(_{X^{1}+}x^{2l}+\cdots+x)=p_{S}-p_{f}$ .$0\leq p_{\mathrm{s}}-\mathrm{A}(x^{1}+x^{2}+\cdots+x^{\mathrm{k}})\leq 1$

$(\forall k\in\{1,2\ldots., l\})$ ,$x_{1}^{k}+x_{2m}^{k}+\mathrm{u}\cdot\cdot+X^{k\wedge}=1$

$(\forall k\in.\{.1.2\ldots.f\text{ノ}\}\})$ ,$x^{k}\in\{.0,1\}^{m}$

$(^{\forall}k\in\{1,2, \ldots, l\})$ .

The problem IP1 has a solution if and onlyif the given peg solitaire problem is feasible.Clearly, any solution $(x^{1..l}\text{ノ}.., x)$ of IP1 corre-sponds to a feasible sequence of jumps.

If we fornuulate the peg solitaire problem de-fined by Figures 1 and 2. then the number ofvariables is $m\cross l=76\cross 32=2_{i}4.32$ . the num-ber of equality constraints is $n+l=32+33=$$65$ . and the number of inequality constraints is2 $\mathrm{x}n\cross l=2\mathrm{x}33\cross 32=2.112$ . $\mathrm{T}\mathrm{h}\mathrm{u}\mathrm{s}_{\mathit{1}}$. the sizeof the integer programming problem is huge andso it is hard to solve the problem by commercialinteger programming software.

In Section 5. we propose an algorithm for pegsolitaire problem which is a combination of back-track search and pruning technique based on theabove integer programming problem.

3 Linear Relaxation andPagoda Function

In [3], Berlekamp. Conway and Guy proposed thepagoda function approach for showing the infea-sibility of some peg solitaire problems includingthe well-known problem $‘(\mathrm{s}\mathrm{e}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{n}\mathrm{g}$ scout 5 paces

Figure 4: an example of assignment of Pagodafunctions

out into desert’. In this paper., we show that thepagoda function approach is equivalent to the re-laxation approach for the integer programmingproblem.

A real valued function $\mathrm{p}\mathrm{a}\mathrm{g}:\{1_{i}2, \ldots, n\}arrow \mathrm{R}$

defined on the set of holes is called a pagodafunction when $\mathrm{p}\mathrm{a}\mathrm{g}(\cdot)$ satisfies the properties thatfor every (vertically or horizontally) consecutivethree holes $(i_{1}. i_{2_{i}}i3)$ . the pagoda function val-ues $\{\mathrm{p}\mathrm{a}\mathrm{g}(i_{1}). \mathrm{p}\mathrm{a}\mathrm{g}(i_{2})i\mathrm{p}\mathrm{a}\mathrm{g}(i3)\}$ satisfies $\mathrm{p}\mathrm{a}\mathrm{g}(i_{1})+$

$\mathrm{p}\mathrm{a}\mathrm{g}(i_{2})$ $\geq$ $\mathrm{p}\mathrm{a}\mathrm{g}(i_{3})$ . (Clearly. the sequence$(i_{3}. i_{2}, i_{1})$ is also a consecutive three holes. andso the inequality $\mathrm{p}\mathrm{a}\mathrm{g}(i_{3})+\mathrm{p}\mathrm{a}\mathrm{g}(i_{2})\geq \mathrm{p}\mathrm{a}\mathrm{g}(i_{1})$ alsoholds.) A pagoda function corresponds to an as-signment of real values to holes on the board sat-isfying the above properties. Figure 4 is an exam-ple of pagoda function defined on English board.

For any configuration $p\in\{0,1\}^{n_{J}}$. we denotethe sum total $\sum_{i=1}^{n}\mathrm{p}\mathrm{a}\mathrm{g}(i)\cross p_{i}$ by $\mathrm{p}\mathrm{a}\mathrm{g}(p)$ . Thedefinition of the pagoda functions implies that ifa configuration $p’$ is obtained by applying ajumpto a configuration $p$ , then $\mathrm{p}\mathrm{a}\mathrm{g}(p)\geq \mathrm{p}\mathrm{a}\mathrm{g}(p)’$ .Thus, if a given peg solitaire problem is feasi-ble, then the inequality $\mathrm{p}\mathrm{a}\mathrm{g}(p_{S})\geq \mathrm{p}\mathrm{a}\mathrm{g}(p_{f})$ holdsfor any pagoda function $\mathrm{p}\mathrm{a}\mathrm{g}(\cdot)$ . So, the existenceof a pagoda function $\mathrm{p}\mathrm{a}\mathrm{g}(\cdot)$ satisfying $\mathrm{p}\mathrm{a}\mathrm{g}(p_{S})<$

$\mathrm{p}\mathrm{a}\mathrm{g}(p_{f})$ shows that the given peg solitaire prob-lem is infeasible.

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In $[\overline{(}]i$ Kanno showed that there exists a pagodafunction which guarantees the infeasibility of thegiven peg solitaire problem if and only if the op-timal value of the following linear programmingproblem is negative.

any pagoda function showing the infeasibility. Al-though, the pagoda function approach was a pow-erful tool for proving the infeasibility of the prob-lem “sending scout 5 paces out into desert”, it isnot so useful for peg solitaire problems de.fined onEnglish board.

PAG-D: $\min$ . $(p_{S^{-}}p_{f})^{\mathrm{T}}y$

$\mathrm{s}$ . $\mathrm{t}$ . $A^{\mathrm{T}}y\geq 0$ .

It is easy to see that for any feasible solu-tion $y$ of $\mathrm{P}\mathrm{A}\mathrm{G}- \mathrm{D}_{l}$. the function $\mathrm{p}\mathrm{a}\mathrm{g}(\cdot)$ defined by$\mathrm{p}\mathrm{a}\mathrm{g}(i)=y_{i}$ for all $i\in\{1_{J}.2\ldots., n\}$ is a pagodafunction. Thus. it is clear that if the optimalvalue of PAG-D is negative. then the given pegsolitaire problem is infeasible. Unfortunately, theinverse implication does not hold: that is. thereexists an infeasible peg solitaire problem instancesuch that the optimal value of the correspondinglinear programming problem (PAG-D) is equal to$0$ (see Kanno [7] for example).

The dual of the above linear programmingproblem is

$\max\{0^{\mathrm{T}}x|Ax=p_{s}-p_{f}, x\geq 0\}$ .

Since the objective function $0^{\mathrm{T}}x$ is always $0$ , theabove problem is equivalent to the following prob-$1\mathrm{e}\mathrm{m}_{j}$

.

PAG-P: find $x$

$\mathrm{s}$ . $\mathrm{t}$ . $Ax=p_{s}-p_{f}$ , $x\geq 0$ .

Thus. there exists a pagoda function which showsthe infeasibility of the given peg solitaire problemif and only if the above linear inequality systemPAG-P is infeasible.

In the rest of this section, we show that theproblem PAG-P is obtained by relaxing the in-teger programming problem $\mathrm{I}\mathrm{P}1$ . First, we intro-duce a new variable $x$ satisfying $x–X^{1}+\cdots+xl$ .Next, we relax the first constraint in IP1 by$Ax=p_{s}-p_{f}$ , remove second and third con-straints, and relax the last 0-1 constraints of origi-nal variables by nonnegativity constraints of arti-ficial variables $x$ . Then the problem IP1 is trans-formed into PAG-P.

We applied pagoda function approach to prob-lems defined on English board and found manyinfeasible problem instances which do not have

4 Upper Bound of the Numberof Jumps

In this $\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}_{\mathrm{i}}$ we propose a method for findingan upper bound of the number of jumps for each(fixed) jump $j$ contained in a feasible sequence.And additionaly, this method is proved to be avery strong tool to check the feasibilities of thegiven problems. In the next section. We proposea pruning technique for backtrack search usingthe upper bound described below.

We consider the following integer programmingproblem for each jump $j_{\text{ノ}}$.

$\mathrm{U}\mathrm{B}j:\max$ . $x_{j}^{1}+x_{j}^{2}\cdots+x_{j}^{l}$

$\mathrm{s}$ . $\mathrm{t}$ . $A(x+\cdots+1lX)=p_{s}-p_{f}$ ,$0\leq p\mathrm{S}-\mathrm{A}(_{X+\cdots+}1x^{\mathrm{k}})\leq 1$

$(\forall k\in\{1,2_{l}\ldots., l\})$ ,$x_{1^{+X}2m}^{k}kk+\cdots+x=1$

$(\forall k\in.\{1_{/}. \cdot 2_{j}\ldots li\})$ .$x^{k}\in\{0.1\}^{m}$

$(^{\forall}k\in\{1_{i}2\ldots., l\})$ .

Since the set of constraints of $\mathrm{U}\mathrm{B}j$ is equivalentto that of $\mathrm{I}\mathrm{P}1_{\iota}$. the given peg solitaire problem isfeasible, if and only if $\mathrm{U}\mathrm{B}j$ has an optimal solu-tion. We denote the optimal value of $\mathrm{U}\mathrm{B}j$ by $z_{j}^{*}$ .if it exists. It is clear that any feasible sequenceof the given problem contains the jump $j$ at most$z_{j}^{*}$ times. $\mathrm{H}_{\mathrm{o}\mathrm{W}\mathrm{e}\mathrm{V}\mathrm{e}}\mathrm{r}\text{ノ}$. the size of the above problemis equivalent to the original problem IP1 and $\mathrm{s}\mathrm{o}_{J}$.it is hard to solve. In the following, we relax theabove problem to a well-solvable problem.

To decrease the number of $\mathrm{v}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{b}\mathrm{l}\mathrm{e}\mathrm{S}_{i}$ we con-sider only the pair of starting and finishing config-urations and ignore intermediate configurations.The above relaxation corresponds to the replace-ment of the variables $x^{1}+\cdots+x^{l}$ by $x$ . Wedecrease the number of constraints by droppingsecond and third constraints of $\mathrm{U}\mathrm{B}j$ . Then wehave the following relaxed problem of $\mathrm{U}\mathrm{B}j$

)

103

Figure 5: an example of peg solitaire problem

RU.B$j$ : $\max$ . $x_{j}$

$\mathrm{s}$ . $\mathrm{t}$ . $Ax=p_{S}-p_{fi}$

$x_{1},$ $x_{2,.\prime},..x_{m}$ arenon-negative integers.

If a problem RUBj is infeasible for an index$j\in\{1.2\text{ノ}’\ldots : m\}$ , the original peg solitaire prob-lem is infeasible and the problem RUBj is also in-feasible for each index $j$ . Since the above problemis a relaxed problem of $\mathrm{U}\mathrm{B}j$ , the optimal value isan upper bound of the optimal value of $\mathrm{U}\mathrm{B}j$ . Ifwe deal with the problem defined by Figures 1and 2, the problem RUBj has $m=76$ integervariables and $n=33$ equality constraints. Therelaxed problems RUBj of the problem defined byFigure 7 are infeasible and so the original problenlis also infeasible.

Figure 6 shows the optinlal values of RUBj foieach jump $j\in\{1_{j}2\ldots., m\}$ of the peg solitaireproblem defined by Figure 5. Since RUBj is arelaxation of $\mathrm{U}\mathrm{B}j$ . feasibility of RUBj does notguarantee the feasibility of the given peg solitaireproblem. For example, all the relaxed problemsRUBj $(j\in\{1,2_{i}\ldots.m\})$ have optimal solutionsas shown in Figure 6 and the given peg solitaireproblem defined by Figure 5 is infeasible.

If the problem PAG-P is infeasible. then theproblem RUBj is also infeasible for each $j\in$

$\{1,2, \ldots, m\}$ . However, the inverse implicationdoes not hold. For example. PAG-P defined byFigure 7 is feasible, and RUBj is infeasible foreach $j\in\{1,2, \ldots., m\}$ .

Kanno [7] gave 10 infeasible peg solitaire prob-lem instances such that the pagoda function ap-proach failed to show the infeasibility. Our infea-sibility check method can show infeasibility of all10 examples given by Kanno. From the above,

Figure 6: maximal numbers of jumps of Figure 5

Figure 7: a problem which can be proved to beinfeasible

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our infeasibility check method compares severelywith the pagoda function approach.

Here we note an important comment related tothe next section. See Figure 6, we can see thatthere are many jumps not to be done at all andnot to be done twice and so on. So, we can prunemany and many branches during the backtracksearching which we deal with in the next section.

Clearly. there are variations of relaxation prob-lems of $\mathrm{U}\mathrm{B}j$ . We have examined many relaxationproblems and chose the above problem RUBjby considering the trade-off between the requiredcomputational efforts and the tightness of the ob-tained upper bound. However. the choice de-pends on the available softwares and hardwaresfor solving integer programming problems. InSection 6. we will describe the environment and

the conf iguration ‘ ‘ end’ ’ ;

if (search ( $\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{r}\mathrm{t}$ , end, upper-bound,rest) $!=$ true) $\{$

print ‘ ‘ infeasible. ’ $j$ ;exit;

$\}$

$\}$

boolsearch ( $\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}\mathrm{i}\mathrm{g}\mathrm{u}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}$ start,

configuration end,$\mathrm{t}\mathrm{a}\mathrm{b}\mathrm{l}\mathrm{e}_{-}\mathrm{o}\mathrm{f}-\mathrm{a}\mathrm{l}1_{-}\mathrm{j}\mathrm{u}\mathrm{m}_{\mathrm{P}^{\mathrm{S}}}$ upper-bound,int rest) $\{$

if (rest $<=0$ ) $\{$

if (start $==$ end)

results of our $\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{l}\mathrm{P}^{\mathrm{U}}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{a}1$experiences in detail.

5 Backtrack Search

return true;else

return false;$\}$

First, we propose forward-only backtrack searchalgorithm for solving peg solitaire problems. Be-fore executing the backtrack $\mathrm{s}\mathrm{e}\mathrm{a}\mathrm{r}\overline{\mathrm{c}}\mathrm{h}$, we solvedthe problem RUBj for each jump position $j\in$

$\{1.2.‘\ldots, m\}\text{ノ}$ and found an upper bound of thenumber of jumps. We can effectively prune thebacktrack search by using the obtained upperbound of the number of jumps. The algorithmis described below:

$\mathrm{v}\mathrm{o}$ idsolve ( $\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}\mathrm{i}\mathrm{g}\mathrm{u}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}$ start,

configuration end) $\{$

$\mathrm{t}\mathrm{a}\mathrm{b}\mathrm{l}\mathrm{e}_{-^{\mathrm{O}}--\mathrm{j}\mathrm{P}\mathrm{s}}\mathrm{f}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{u}\mathrm{m}$ upper-bound;int rest;

for (each jump j) $\{$

solve RUBj ;if (RUBj is infeasible) $\{$

print ‘ ‘ inf easible. ’ ’ ;exit;

$\}$

set the optimal value of RUBjupper-bound $[\mathrm{j}]$ ;

$\}$

rest $=$ the number of pegs ofthe conf iguration ‘ ‘ start’ ’

- the number of pegs of

for (all possible jumps) $\{$

if (upper-bound of the jump $<=0$ )

continue;

upper-bound of the jump $=$

upper-bound of the$\mathrm{j}\mathrm{u}\mathrm{m}\mathrm{p}\mathrm{b}^{\mathrm{T}}t‘ 1$

;update the conf iguration start’ $f$

by applying the$\mathrm{j}\mathrm{u}\mathrm{m}\dot{.}\mathrm{p}\prime \mathrm{Y}$

operation... $’\sim$

if (search ( $\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{r}\mathrm{t}$ , end, upper-bound,rest - 1) $==$ true) $\{$

display the configuration‘ ‘ start $y$ ’ ;

restore the$\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}\mathrm{i}\mathrm{g}\mathrm{u}‘ \mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\iota \mathrm{S}\mathrm{t}\mathrm{a}\mathrm{r}\mathrm{t}$

”

by applying the reversejump operation;

return true;

$\}$ else $\{$

upper-bound of the jump $=$

upper-bound of the jump $+1$ ;restore the

$\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}\mathrm{i}\mathrm{g}‘ \mathrm{u}_{\mathrm{S}\mathrm{t}}‘ \mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{a}\mathrm{r}\mathrm{t}$

”

by applying the reversejump operation;

$\}$

105

$\}$

return false;$\}$

To avoid searching the same configurationsmore than twice, we used a hash table with$2.097’.169$ entries for maintaining all the scannedconfigurations. If the backtrack search algorithmfinds that the present configuration is containedin the hash table. we can stop searching thepresent configuration. We used the hash tablewhose size is the maximum prime number cur-rently available at our computer.

Here we point out the importance of the hashtechnique for solving peg solitaire problem. Forexample. if we use both IP and hash method. wecan solve the problem defined by Figure 8 in 2seconds on the notebook perspnal $\mathrm{C}\mathrm{o}\mathrm{m}\mathrm{p}_{\mathrm{U}\mathrm{t}}\mathrm{e}1$ wewill use in the next section. But if we use onlyIP upper bound or use only hash technique. theprogram doesn’t stop in ten minutes.

Figure 8: an example problem to which hash isefficient

Second, we propose forward-backward back-track search algorithm. The second algorithmuses the properties of a peg solitaire problemthat the number of jumps required to solve theproblem is known and solving the problem in for-ward direction is essentially equivalent to solvingthe problem in backward direction. Here, solv-ing problem in backward $\mathrm{d}\mathrm{i}\dot{\mathrm{r}}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}$ means that westart from the finishing configuration, repeat the‘reverse jump’ operation and aim to get the start-ing configuration. Our second algorithm executesbacktrack searching from the finishing configu-ration to the haif depth of the search tree andmaintains all the obtained configurations by the

Figure 9: jump upper bounds of the problenl de-fined by Fig. 8

hash table. Next, the algorithm begins backtracksearching from the starting configuration to thehalf depth of the search tree. When an obtainedconfiguration is in the hash table, the originalsolitaire problem is feasible. If all the scannedconfigurations are not in the hash table, the orig-inal problem is infeasible. The idea of the abovesecond algorithm is very simple and effective forsome problems but not all. When we appliedthe second algorithm, some problems require avery large hash table whose size is greater than2,097,169.

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6 Computational Results

In this section, we deal with peg solitaire prob-lems defined on the English board. We used anotebook personal computer with MMX-Pentium$233\mathrm{M}\mathrm{H}\mathrm{z}$ CPU, $64\mathrm{M}\mathrm{B}$ memory, and Linux $\mathrm{O}\mathrm{S}$ . Wesolved the relaxed problem RUBj for each $j_{\mathit{1}}$. bythe software $lp$-solve 3.0. This lp-solve version isreleased under the LGPL license. One can findthe latest version of $lp$ -solve at the following ftpsite.

$\mathrm{f}\mathrm{t}\mathrm{p}://\mathrm{f}\mathrm{t}\mathrm{p}$ . ics.ele. $\mathrm{t}\mathrm{u}\mathrm{e}.\mathrm{n}1/\mathrm{p}\mathrm{u}\mathrm{b}/1_{\mathrm{P}-\mathrm{S}}01_{\mathrm{V}}\mathrm{e}/$

In the following. we discuss the problems RUBj$(j\in\{1_{;}2. . . ., m\})\mathrm{j}$ which are called relaxed prob-lems in the following. It took about 16 min-utes to solve all the 76 relaxed problems definedb.v Figures 1 and 2. (Since the pair of start-ing and finishing configurations are symmetrical.we actually need to solve only 12 relaxed prob-lems. However, our computer program does notuse the information depending on the symmetric-ity.) We tried more than 20 peg solitaire probleminstances, and the computational time requiredfor solving 76 relaxed problems for each instanceis less than 16 minutes. Here we note that thereare some problems which took only 10 seconds tosolve whole 76 relaxed problems.

We compared the forward-only backtracksearching and the forward-backward searchmethod. The forward-only backtrack searchmethod solves the peg solitaire $\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{b}\mathrm{l}\mathrm{e}\ln$ definedby Figures 1 and 2 in 1 second. However. ifwe apply the forward-backward search method.the hash overflows after 3. $5\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{s}$ . (Here wenote that the program concludes that the hashhas overflowed when the 80 percent of the hashis used). Since this problem is symmetric. theforward-only search method finds a feasible se-quence easily. However. the symmetricity in-creases the upperbound of the number of jumpsand so the backward search generates many con-figurations and the hash overflows.

We also tried to solve the symmetrical peg soli-taire problem shown by Figure 10. The forward-only search method finds a feasible sequence in 37minutes. However, the forward-backward searchmethod solves the problem in 1.4 seconds. Wethink that the performance of two methods de-pends not only on the symmetricity of the prob-lem but also on the the number of jumps required.

Figure 10: a peg solitaire problem fit for back-$\backslash \mathrm{v}\mathrm{a}\mathrm{r}\mathrm{d}$ and forward searching

The length of the feasible sequence of the prob-lem defined by Figures 1 and 2 is 31 and that ofthe problem defined by Figure 10 is 13. Since thedepth of search tree of the latter problem is small.the hash does not overflow during the algorithm.

We solved more than 20 problems includingthe instances given in Kanno [7]. Either forward-only search method or forward-backward searchmethod solves each instances we tried in at most20 minutes. If we select faster algorithm for eachinstance we solved, the total computational timeof our algorithm is less than 20 minutes.

References

[1] Avis. D.. and Deza. A.: Solitaire Cones.Technical Report No. SOCS-96.8. 1996.

[2] Beasley J. D.: Some notes on Solitaire. Eu-reka. 25 (1962). 13-18.

[3] Berlekamp, E. $\mathrm{R}_{i}$. Conway, J. H.. and Guy,R. K.: Winning Ways for MathematicalPlays. Academic $\mathrm{P}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}_{i}$ London,. 1982.

[4] de Bruijn N. G.: A Solitaire Game and ItsRelation to a Finite Field. Journal of Recre-ational Mathematics, 5 (1972), 133-137.

[5] Cross D. C.: Square Solitaire and varia-tions. Journal of Recreational Mathematics,1 (1968), 121-123.

[6] Gardner M.: Scientific American, 206$\# 6$ (June 1962), 156-166; 214 $\# 2(\mathrm{F}\mathrm{e}\mathrm{b}.$1966),112-113; 214 $\# 5$ (May 1966), 127.

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[7] Kanno, E.: Linear Programming Algorithmfor Peg Solitaire Problems, Bachelor thesis,Department of Mathematical Engineering,Faculty of Engineering, University of Tokyo,1997 (in Japanese).

[8] Uehara, R., Iwata, S.: Generalized Hi-Q is$\mathrm{N}\mathrm{P}$-complete, Trans. IEICE, 73 (1990), 270-273.

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