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Somewhere down the message list here was a conversation about the TofH game. I had my Intermediate Algebra students discover the pattern; calculate the number of moves for 70 disks; and the time it would take to solve it making 1-trillion moves per second (a long time).
It got me thinking (but not too much yet), what if the rules were the same, but at any step you could move 2 disks. But why stop there? How many moves will it take to solve using "m" disks if you can move at most "n" at any time? Hmmmm.
I should probably do a search on this before pressing send, but it's more fun to ponder it awhile.
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The first few chapters of Knuth's Concrete Mathematics explores pretty much any variation of ToH that you might like.
Without much thought, being able to move two disks of a 70-disk stack seems like it would play much like a 35 disk stack without much change in complexity.
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Can you explain why allowing 2 disks to be moved in one step would have any effect other than halving the number of steps total?
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Almost simultaneous posts!
The way I thought the new rule worked was that you just can move two pieces per turn, hence you halve the number of moves.
I think the way you thought of it was that any two stacked pieces can be moved at once as a single piece, which acts like halving the number of pieces.
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Hi,
My first guess (a 2 seconds thinking) is that:
a ToH of 30 disk with a move of 2 disks at the time is the same as a ToH of 15 disks.
a ToH of 30 disk with a move of 3 disks at the time is the same as a ToH of 10 disks.
Patrice
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[bashful]Yup, like I said, didn't think about it much.[\bashful] Too much on my mind to have thought rationally for more than a second.
Edited: 18 Nov 2010, 8:40 p.m.
