
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
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"
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
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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