However, a series of consecutive jumps made at one time with a single piece can be regarded as a single move hence a player can aim not merely at solving a given puzzle, but also at finding the solution that requires the smallest number of moves. The number of jumps made in a game of solitaire equals the number of pegs removed. The player is required to finish the game with a single peg in the central cell. All cells are filled except the one in the center. "In the most widely known solitaire puzzle the 33-cell board is used. There are two kinds of boards used for peg solitaire.Īnother eliminates the corner cells, forming a cross-shaped pattern of cells (top right diagram). Jumps can be made only along the lattice lines (as shown in the diagram) a peg cannot jump diagonally." Following such a move, the peg over which the jump was made is removed from the board. There are many games and puzzles that can be devised for the solitaire board, but the method of making moves is common to all of them.Ī peg (assuming that this is the marker used) can be moved only by jumping it over a neighboring peg to a vacant space directly on the other side. His new invention, like the earlier game, used a board in which an array of holes - called cells - were bored as resting places for pegs or marbles. He was modifying an already existing game called Fox and Geese. "According to one old story, the game of peg solitaire often called simply solitaire, was invented in the 18th century by a French nobleman imprisoned in the Bastille, the grim fortress-prison in Paris. īy authors Pieter van Delft and Jack Botermans: The following quote is taken from the book Creative Puzzles Of The World. with many links to other internet areas which help supplement his wealth of information. This pruning technique dramatically reduces the size of the search tree and the algorithm runs within seconds (depending on the machine of course).For a great in-depth scientific notation on peg solitaire puzzles - and history, one can view my good friend, George Bell's website here. mirrored version of aforementioned stored states. Then at step #3, we restrict the recursive call to only those states that do not coincide with a rotated resp. Now, the idea is to store each occupation state for which the algorithm already determined no solution can exist. So, for each occupation of stones we have eight states that are equivalent w.r.t to solvability: rotate(0, 90, 180, 270) + mirror(rotate(0, 90, 180, 270)) To see this, by induction we easily can verify that the mirrored moves provide the same state in the mirrored game, either no movement available or solution (one in the center). Similarly, if you mirror the current state of occupation along the vertical center axis, again we would expect the answer if a solution still is possible to remain the same: O O O O O O O X X O O O O O O X O O O O O O O O O O O O O O O O O mirrored: O O O O O O O O O O X X O O O O O X O O O O O O O O O O O O O O O If you look at any occupation state of the stones and you now turn the game clockwise 90 degrees around its center, the answer if this state can lead to any solution remains the same. The idea is to use symmetries of the game. What we can do instead, is to use a very promising pruning technique. This number might be a little overestimated, but it already gives an impression that we can encounter problems.Īnd indeed, implementing the above approach takes hours to solve the problem (on my machine). If we estimate after each move there are two more possible moves available, then this amounts to 2^49 possible paths. In order to remove 49 of them, we need at least 49 moves. ![]() The problem though is, for exponential complexity the exponent is already a little too high. Its complexity clearly is exponential since most of the time there will be more than one move available. This algorithm is the typical approach of trying out all possible series of moves until the solution has been found. Each time with the resulting new state of occupations, go back to #1. If no possible moves are available, stop the search along this ‘path’. These are those that either vertically or horizontally let jump a stone over another into a non-occupied position and remove the other. ![]() Verify if the solution has been attained, that is, only at the center a stone is left over. ![]() We start with the initial state where every position except the center is occupied with a stone. The most basic approach is to solve the problem inductively.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |