Recent content by acheong87

Background This started out as a question, but answered itself in the process of being written up. I thought I'd post it anyway, in case it might help clarify something for someone else. It is an alternative derivation of the solution to a (similar) question asked in Six Not-So-Easy Pieces...

Recall the problem. G(n) = n - G(G(n-1)) G(0) = 0 Try it! G(0) = 0 G(1) = 1 - G(G(0)) = 1 - G(0) = 1 - 0 = 1 G(2) = 2 - G(G(1)) = 2 - G(1) = 2 - 1 = 1 G(3) = 3 - G(G(2)) = 3 - G(2) = 3 - 1 = 2 G(4) = 4 - G(G(3)) = 4 - G(2) = 4 - 1 = 3 G(5) = 5 - G(G(4)) = 5 - G(3) = 5 - 2 = 3...