Okay. Here is a further exposition of the solution. It took me a while
to remember where I had used it before.
i = n
j = n
do while not (i = j)
i = A(A((i))
j = A(j)
end
/* Now a=b can be reached at either 2k or k steps from n, */
/* where k is some integer between 1 and n. */
i = n
do while not (i = j)
i = A(i)
j = A(j)
end
print a
Dave
On Jan 16, 11:57 am, juver++ <avpostni...@gmail.com> wrote:
> @Dave
> Cycle finding algo (Floyd's, Brent's) can be applied only for the iterated
> function values.
> This means: f(x0) = x1; f(x1) = x2 and etc.
> Suppose we have the following array: 1 2 3 2 4. Value 2 have two different
> transitions.
>
> Clarify your proposed method if it needs additional observations.
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