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Re^2: Hanoi Challenge

by Elgon (Curate)
on Oct 22, 2004 at 19:50 UTC ( #401660=note: print w/replies, xml ) Need Help??

in reply to Re: Hanoi Challenge
in thread Hanoi Challenge


A query; Surely it is possible only to solve the problem in exactly O(n) when there is one more peg than discs plus one. (It could be that I misunderstand the O(n), O(log n), O(n^2) notation...)

Example: In the quickest solution for three pegs and three discs, for example, the large disc moves once, the medium disc moves twice and the smallest disc moves four times.

Example 2: Where there are three rings and four pegs, each ring only moves twice.


UPDATE:Thanks to Thor for pointing out my error, below. As a general rule, I find that for n pegs and n discs the number of moves required is 2n + 1. If, however, there are n discs and n + 1 (or more) pegs, then 2n - 1 moves are required. Can someone with a better understanding of the formalisms tell me whether either of these is O(n) ???

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Re^3: Hanoi Challenge
by thor (Priest) on Oct 22, 2004 at 20:16 UTC
    Example 2: Where there are three rings and four pegs, each ring only moves twice.
    Does it? If we're trying to get ring C to peg 4
    A -> 1
    B -> 2
    C -> 4
    B -> 4
    A -> 4
    In the case of n disks and n+1 pegs, the last disk moves only once.


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