How does this work? Weird Towers of Hanoi Solution

Question

I was lost on the internet when I discovered this unusual, iterative solution to the towers of Hanoi:

for (int x = 1; x         


        

Answer 1

antti.huima's solution is essentially correct, but I wanted something more rigorous, and it was too big to fit in a comment. Here goes:

First note: at the middle step x = 2^N-1 of this algorithm, the "from" peg is 0, and the "to" peg is 2^N % 3. This leaves 2^(N-1) % 3 for the "spare" peg. This is also true for the last step of the algorithm, so we see that actually the authors' algorithm is a slight "cheat": they're moving the disks from peg 0 to peg 2^N % 3, rather than a fixed, pre-specified "to" peg. This could be changed with not much work.

The original Hanoi algorithm is:

hanoi(from, to, spare, N):
    hanoi(from, spare, to, N-1)
    move(from, to)
    hanoi(spare, to, from, N-1)

Plugging in "from" = 0, "to" = 2^N % 3, "spare" = 2^N-1 % 3, we get (suppressing the %3's):

hanoi(0, 2**N, 2**(N-1), N):
(a)   hanoi(0, 2**(N-1), 2**N, N-1)
(b)   move(0, 2**N)
(c)   hanoi(2**(N-1), 2**N, 0, N-1)

The fundamental observation here is: In line (c), the pegs are exactly the pegs of hanoi(0, 2^N-1, 2^N, N-1) shifted by 2^N-1 % 3, i.e. they are exactly the pegs of line (a) with this amount added to them.

I claim that it follows that when we run line (c), the "from" and "to" pegs are the corresponding pegs of line (a) shifted by 2^N-1 % 3. This follows from the easy, more general lemma that in hanoi(a+x, b+x, c+x, N), the "from and "to" pegs are shifted exactly x from in hanoi(a, b, c, N).

Now consider the functions
f(x) = (x & (x-1)) % 3
g(x) = (x | (x-1)) + 1 % 3

To prove that the given algorithm works, we only have to show that:

• f(2^N-1) == 0 and g(2^N-1) == 2^N
• for 0 < i < 2^N, we have f(2^N - i) == f(2^N + i) + 2^N % 3, and g(2^N - i) == g(2^N + i) + 2^N % 3.

Both of these are easy to show.