Do you want to see the list for those two cases? Because in either the recursive or iterative solution, there's only two cases.

The 2-disk moves are: [(a, c), (a, b), (c, b)]
The 3-disk moves are: [(a, b), (a, c), (b, c), (a, b), (c, a), (c, b), (a, b)]


On Sat, Feb 14, 2015 at 3:40 PM, YCH <dontdieych@gmail.com> wrote:

Would you mind show code for only two cases? Just think about two cases.

hanoi 2 a b c
hanoi 3 a b c

2015. 2. 15. 오전 6:35에 "Roelof Wobben" <r.wobben@home.nl>님이 작성:
I know.

Eveyone says there is a pattern but I do not see it at the moment.

Maybe I do the wrong first move with more then 2 disk.

You can put the first disk on the second or the thirth peg.

Roelof


Mike Meyer schreef op 14-2-2015 om 22:30:


On Sat, Feb 14, 2015 at 3:27 PM, Roelof Wobben <r.wobben@home.nl> wrote:
KC schreef op 14-2-2015 om 22:23:

Hint:

- think about what you need to do in each recursive step and in the base case

- there is also an interesting way of viewing the problem



Check if I can do another step or check if the end case all disk are on the last peg.

Im very curious what the interesting way is

I suspect it's the iterative version that was already mentioned here. You were close to discovering it when worked through the problem on paper. 


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