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2 Comments

Anyway, I'm coming to 127. I figure it as follows:

For a 1 disk problem: 1 disk move.

For a 2 disk problem:

Move the smallest disk to an empty

Move the bottom to another empty

Move the smallest onto the other

3 moves

For a 3 disk problem:

Move the top two to an empty:3 moves

Move the bottom to an empty

Move the two onto the bottom:3 moves

7 moves

So the number of moves for n disks looks like the number of moves for a stack one less times 2, plus one:

moves for stack of n=2*(moves for stack of n-1)+1

So we have:

1 - 1

2 - 3

3 - 7

4 - 15

5 - 31

6 - 63

7 - 127

That's almost the same you have. Are we defining the problem the same? Are we counting exactly the same thing? What do you think?

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