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.
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.
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?
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.
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
[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.