In sorting situations where the final destination of each item is known, it is natural to repeatedly choose items and place them where they belong, allowing the intervening items to shift by one to make room. However, it is not obvious that this algorithm necessarily terminates. We show that in fact the algorithm terminates after at most $2^{n-1}-1$ steps in the worst case, and that there are super-exponentially many permutations for which this exact bound can be achieved. The proof involves a curious symmetrical binary representation.

This is joint work with Peter Winkler.