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 Re: Proof of the following solution for div2-500/div1-250 (response to post by vasja_p) Yes, you are right.No matter what you do in a given position, you will always win or always lose it.This is because victory can be defined by the parity of the sum of the positions of the checkers, or rather the sum of the distances between their positions and the right-most cell. Yet it does not matter whether you use a normal move or a jump, since the jump jumps three cells, the parity of the distance can only be toggled and never maintained. And this realization translates into a very easy solution. Just about every other thread about this problem already explains this.
 Re: Proof of the following solution for div2-500/div1-250 (response to post by vasja_p) Yes, you are right.No matter what you do in a given position, you will always win or always lose it.This is because victory can be defined by the parity of the sum of the positions of the checkers, or rather the sum of the distances between their positions and the right-most cell. Yet it does not matter whether you use a normal move or a jump, since the jump jumps three cells, the parity of the distance can only be toggled and never maintained. Just about every other thread about this problem already explains this.