As it only affected two cells, we used the 2-periodic coloring (white and black) or equivalently the 2-periodic labeling with. If we try to use the same language for the dominos, then there we had the rule. The hard question now is where to look for properties like we have in the dominos (parity, and the white\black counting) which are easier to compute, and are enough to show that some board are unsolvable. For example, while is solvable, the state is not.
#Peg solitaire rules how to
It already becomes quite complicated for 5 pegs, and it is not clear how to describe the solvable states. With this equivalence, the only solvable boards with at most 5 pegs are: So for example, from we can go both to and, and both of these are equivalent (by moving 3 places at the same direction). Also, for simplicity, we will consider two boards which are the same up to a translation to the right or left equivalent. Just to see some examples of solvable games, instead of starting with an initial state and eating pegs, lets start with the final one peg state and create new ones by reversing the eating rule, namely the new rules are and. Let us start by thinking about a board with a single line, and ask which initial states of pegs can we solve in the sense that we can transform it using our rule above to a state with a single peg.Įven in this one dimensional board, and even with a single initial state, this is not a trivial manner. Here the game is more complicated, and when dealing with a complicated problem, you should always look for the simplest variant of it which you can solve, and hope to get some intuition for the more general problem. Also, we could easily analyze it for rectangular games, which gave us some intuition about how to continue to general games. In the domino game in the previous post, our lives were much easier – we could add a domino piece in every place that had two subsequent “holes”. Our question now is, given a variant of this game, which can be one of the variants above, or more generally any given board with given initial and final state of pegs, can this game be won, and if not, is there an easy way to show it?
So for example, if the initial hole is on the bottom right, then the last peg should be on the top left. In the European version, the last peg should be in the place which is 180 degrees rotation of the initial hole. In the English version, the initial hole and the final peg are usually at the same place (and in particular, you start with a hole in the middle and try to end with a peg in the middle). The goal of this game is remove more and more pegs and eventually remain with a single peg, where the location of the initial hole and the final peg is determined by the variant of the game. Or, in a more formal way, if symbolizes a peg and a hole, then we can move from to (these can appear in the up, down, left or right directions).
Then, at each turn, we need to choose 3 subsequent places where the first two have pegs and the third one doesn’t, and we can use the first peg, to jump over and “eat” the second peg, in order to reach the third empty place.
#Peg solitaire rules full
In this game we usually start with a board full with pegs except for one place. The two main variants are the English version and the European version. The Peg Solitaire is quite an old game, which according to wikipedia was probably played by Louis XIV at the year 1697. In this post, we will deal with a new game – the Peg Solitaire – and see if and how we can find such invariants for this game. These are “elements” that do not change as the game progresses, so if the initial and final state have different invariants, then we could not reach from one to the other. We wanted to find a way such that if a board couldn’t be covered, then we could prove it easily, and this led us to the definition of the game invariants. In the previous post we considered a simple domino game where we needed to cover a board with dominos.
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