By my understanding of the article, they’re saying that they’ve classified every position of ten checkers as being a win for white, a win for black, and a draw; they’re assuming that if both sides play reasonably the game will not end up in any of the ten-piece positions which is a win for white or black. That’s probably true, but I’m curious what sort of formal argument one would use to show that one side couldn’t force the game to either end up in one of the ten-piece positions where he would be victorious, or for that matter to end up in an eleven- or twelve-piece position where he’d already won (the latter, of course, would seem pretty unlikely).
Maybe an easy way is to compare it to tic-tac-toe. Here there are less than 9*3 = 27 possible boards instead of something less than (5)^32.
You can easily list them out.
Intelligent play by both players will always lead to a draw.
I was thinking you could hold each board in a 16-byte structure. Use 4 bits to encode each square — red king, red pawn, black king, black pawn, empty square. Of course that takes only 3 bits, so you’d have 32 bits left over. Use one to say whose turn it is.
Then start building the game tree.