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Most legal plays

Posted By: Timothy Chow
Date: Friday, 14 January 2011, at 12:51 a.m.

In Response To: Most legal plays (Ray Kershaw)

You don't need to do any programming yourself because you can just put the position into GNU and ask for a hint; it will display 2226 options for you.

If I had to work it out by hand, I'd proceed as follows. Let A(n), B(n), C(n), D(n) denote the number of different ways to play 1, 2, 3, 4 aces respectively when you have a checker on the 3pt and n further pairs of checkers stretched out behind it as Zare recommends. Then we want to show that D(7) = 2226. I think it should not be too hard to work out by hand how to write formulas for A(n), B(n), C(n), and D(n) in terms of A(n-1), B(n-1), C(n-1), D(n-1). For example, to compute D(n) you split into several cases: either you don't touch the back two checkers and move all four aces with the remaining checkers (there are D(n-1) ways to do this), or you move one ace with the back checkers and move three aces with the remaining checkers (there are 2*C(n-1) ways to do this), or you move two aces with the back checkers and two with the remaining checkers, etc. As you pointed out, one has to be careful not to double-count here, so one needs to make a decision whether a move like 23/19 is treated as 4, 3, or 2 aces with the back checkers. Any choice will work as long as you apply it consistently.

Then you can work out A(1), B(1), C(1), D(1) by hand and iterate the formulas to find D(7).

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