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It can be done more efficiently than this. You shouldn't divide the work up evenly. Below are some special cases and the smallest solution sizes possible (I believe) for them.

If P is the number of pegs, D is the number of disks, and M is the minimum number of moves for which a solution exists, then:

for P >= 3 if D = P*(P-1)/2 then M = P*(P-2)*2+1 so P=3, D=3, M=7 P=4, D=6, M=17 P=5, D=10, M=31 P=6, D=15, M=49 ...

So I can beat your 35-move solution for the 5P10D case by 4 moves.

I may write up why this is so, but I suspect someone else is likely to beat me to that. And if you try to figure out why I picked these cases, you'll probably have a good hint for part of a good solution for the general problem. (:

- tye        


In reply to Re^2: Best possible counts for a few test cases (in readmore) (Hanoi Challenge) by tye
in thread Hanoi Challenge by tilly

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