![]() ![]() ![]() In this project, we tried to create an algorithm to solve a game callled "Peg Solitaire" in it's classical ("English") version. The game, in it's english version, is a board game with 33 holes. It consists of a board with 33 holes, arranged in the shape of a cross - a 3×3 square with four 2×3 rectangles added. Standard Peg Solitaire, sometimes sold as Hi-Q, is the classic and most common version of the game. Initially, all the holes except the central hole are occupied by pegs. On this page I will describe some techniques that are used to analyse various peg solitaire puzzles. The goal of the game is to reverse the initial state: empty all the holes and leave a single peg in the center. Pegs (red circles in this implementation) are allowed to jump over adjacent (vertically or horizontally) pegs. The legal moves in the game are simple: a peg can "jump" over a neighbor peg, if there is an empty hole on the other side. Peg Solitaire (also known as Hi-Q) has very simple rules. The peg that was "jumped over" is taken off the board, and so on - until only one peg remains. We chose to two different methods - algorithms - to solve the problem. One of them is Bi-Directional Breadth-First Iterative Deepening A* search algorithm (BD-BFIDA*) with several heuristics, and the other is Constraint Satisfaction Problem (CSP) with backtracking search algorith. We tried to implement an improved version of the known IDA* algorith. The main reason to use a more sophisticated algorithm is run-time.Īs we know, breadth-first algorithm's run time increases as the algorithm advances. The basic game begins from a full board with one peg missing and nishes with one peg at a specied board. Bell Tech-X Corporation 5621 Arapahoe Ave, Suite A Boulder, CO 80303 USA Abstract We consider the one-person game of peg solitaire on a triangular board of arbitrary size. In out case, in the english version of peg solitaire, both the initial and final state are defined singularly - and we took advantage of that. Solving Triangular Peg Solitaire George I. The idea of the bidirectional search is to start simultaneously from the initial and final states, and expand both via BFIDA* search. algorithm, and another method that could forsee dead-ends - and alter the expansion in their direction. After each step (expansion), we checked if the two edges have reached one another - we checked if we found a solution in the middle, something that would save most of the running time of single-direction BFIDA* search.įurthermore, we used several heuristics to improve the algorithm, and another method that could forsee dead-ends - and alter the expansion in their direction. If(zetten.Propositions in our model - a list of the variables (holes) with either of the possible values - i.e., "in34", "out65" represent occupied hole in (3,4) and empty hole in (6,5) correspondingly.Domain: each hole can be either occupied by a peg, or empty.Ī state of the problem is an assignment of values from the domain to each of the variables.Variables: the holes on the game board (identified by their cartesian coordinates).In our model, the state is defined by the variables and the domain of the values for them: We created a simple model of the problem, that we could use in our algorithms. So first fillBoard() is called and then calculateMoves() Private boolean walkThroughList(ArrayList lijst) WalkThroughList(zettenLijst.get(zettenLijst.size()-1)) The problem is that I get a : heap space and I don't know why.Ġ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0This is the algorithm:įor(int i=0 i5 & i%1717 & i0 & i%1710)) & i>102 & i(zetten) ![]() Now, I'm trying to use backtracking for this: look for all possible moves (4 when starting) and then try one, look for all possible moves and then try one. Bitvector-compressed pattern databases result in log2 3 1.6 bits per state 14. The Solution Logic is as follows:1) Fill in Boxes 1,5,9 in with the digits 1-9 in any random order.2) Scan the entire grid and check for all possible values for each cell.3) If there is a cell with only 1 possible value, fill it in.4) If there is no such cell, pick the one with least possible values and pick one of the numbers at random.5) If a. the O in the middle) end ending up with the last peg in the center position. Peg Solitaire 41, Nine-Men-Morris 39, and Chineese Checkers 90,91. In either case, you are using a set of steps that can be generalized to solve all. People who solve Rubik's Cubes in ten seconds use algorithms, but so to people who use long division to solve 5246. Simply put, an algorithm is a set of steps that one can use to solve a problem. X x x x xThe goal of this game is to remove all pegs by jumping over another peg to an open field (e.g. Peg solitaire is a great way for kids to interact with algorithms. For examples of this algorithm applied to peg solitaire see and. I'm trying to work out an algorithm to solve the game "peg solitaire" for this structure: ![]()
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