I have played around with the puzzle for a little bit, but I haven't been able to complete it with one peg left but rather two. Is there only 1 unique solution for every starting configuration? Is there a definitive algorithm for completing the puzzle no matter what peg is left unfilled in the beginning? The other day, while watching La Carte aux trésors at my elderly neighbors house, I casually played peg solitaire on a board she has. Anyone who leaves a single peg elsewhere is an outstanding player. no matter what hole is left unfilled in the beginning? If so, are there solutions for each hole for every $n$-peg triangle? According to the game brochure (Milton Bradley Co., 1986), whoever succeeds in leaving the last peg in the center is a genius. Is there a solution for every starting configuration i.e. This paper concerns modelling and solving Peg Solitaire using constraint and mathematical programming techniques. The goal of the game is to jump the pegs in such a way that in the end, there is only one peg left. After the jump is completed, the peg that is jumped over is removed from the triangle. Our approach improves upon unidirectional BFIDA by usually avoiding the last iteration of search entirely, greatly speeding up search. The idea behind the game is that you fill 14 of the 15 holes with pegs, and the way to play is to "jump" another peg into an empty space adjacent to the peg that is being jumped over. We present a novel approach to bidirectional breadth-first IDA (BFIDA) and demonstrate its effectiveness in the domain of peg solitaire, a simple puzzle. Many of you may be familiar with a puzzle game that consists of a 15-peg triangle and is dubbed as "The Original IQ Test". 76 Share Save 27K views 13 years ago Peg Solitaire (44 pegs) solution.
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