On the English board the equivalent alternative games are to start with a hole and end with a peg at the same position. There is no solution to the European board with the initial hole centrally located, if only orthogonal moves are permitted.
This is easily seen as follows, by an argument from Hans Zantema. Divide the positions of the board into A, B and C positions as follows:. Initially with only the central position free, the number of covered A positions is 12, the number of covered B positions is 12, and also the number of covered C positions is After every move the number of covered A positions increases or decreases by one, and the same for the number of covered B positions and the number of covered C positions.
Hence after an even number of moves all these three numbers are even, and after an odd number of moves all these three numbers are odd. Hence a final position with only one peg can not be reached: then one of these numbers is one the position of the peg, one is odd , while the other two numbers are zero, hence even. There are, however, several other configurations where a single initial hole can be reduced to a single peg.
A tactic that can be used is to divide the board into packages of three and to purge remove them entirely using one extra peg, the catalyst, that jumps out and then jumps back again. Other alternate games include starting with two empty holes and finishing with two pegs in those holes.
Also starting with one hole here and ending with one peg there. On an English board, the hole can be anywhere and the final peg can only end up where multiples of three permit. Thus a hole at a can only leave a single peg at a , p , O or C. They introduced a notion called pagoda function which is a strong tool to show the infeasibility of a given generalized peg solitaire problem. A problem for finding a pagoda function which concludes the infeasibility of a given problem is formulated as a linear programming problem and solvable in polynomial time see Kiyomi and Matsui Uehara and Iwata dealt with the generalized Hi-Q problems which are equivalent to the peg solitaire problems and showed their NP-completeness.
Avis and Deza formulated a peg solitaire problem as a combinatorial optimization problem and discussed the properties of the feasible region called 'a solitaire cone'. Kiyomi and Matsui proposed an efficient method for solving peg solitaire problems. One consequence of this analysis is to put a lower bound on the size of possible 'inverted position' problems, in which the cells initially occupied are left empty and vice versa.
Any solution to such a problem must contain a minimum of 11 moves, irrespective of the exact details of the problem. It can be proved using abstract algebra that there are only 5 fixed board positions where the game can successfully end with one peg. The shortest solution to the standard English game involves 18 moves, counting multiple jumps as single moves:.
The puzzle in Figure 2 can be solved in six moves. You can make the moves yourself by clicking and dragging on the figure, or click through the steps here in sequence:. If things go awry, or you want to start over, click the reset button next to the board. Play to a single counter. Click and drag on the figure above to make moves, or click the links in the walkthough to see the solution. The port and sweep moves can take some getting used to especially if you're used to ordinary peg solitaire.
Hopefully you won't get stuck for too long on any of them, but they illustrate some typical maneuvers. Each move has source squares the squares you're taking counters away from , and a target square the square you're adding to. To clarify the rules a little further: The source for a port move must have 2 counters, and the target must have 0 or 1 so that there's room to add one more.
The "in-between" square for a port is completely unaffected by the move, and may have any number of counters. Some boards Figure 7 have gaps missing squares , and you can port across a gap in the board.
All three source squares for a sweep must have counters, but they can be 1s or 2s, and they don't all have to be the same. Try both possibilities for the first move in the upper-right puzzle. One solution uses two sweeps; the other uses just one. PaSS puzzles typically have a large number of possible solutions, and looking for a solution that ends with a long sequence of ports can be an interesting extra challenge. Play to a single counter in the center. The four puzzles in Figure 4 all finish with a single counter in the center of the board, though they exhibit a variety of symmetries in the initial arrangement of their counters.
Despite the modest size, I think you'll find them satisfying to solve! PaSS works well on a variety of board shapes. Figure 5 poses a few problems on the square diamond.
They don't all finish in the center, and that can make it harder to visualize a solution. That's where mathematics can help out: with some simple arithmetic, it is easy to deduce where the final counter could end up in these puzzles, even before you solve them. More about the calculations later, though! To the Main Page "Mathematische Basteleien". Generally you play this game without a strategy, so that you can't repeat the moves later. If you like to tell a solution, you have to establish a notation scheme.
If a peg is in the middle red , jumps in all directions are possible. Two jumps can be made from the blue field and one jump from the green one. There are 76 jumps altogether. You can make another classification of the holes of the field.
There are four classes 4 colours.
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