For 1 the trick is in the first recursive call. When we subtract the current treasure from the target, we want to find solutions that include the current treasure. However inside that call, we don't know about the existence of that current treasure on the inner call - that fact is implicit in the fact that the target is smaller, and it will only get added back to the solution after the inner call returns.
in reply to Re^2: Mark Jason Dominus And Me - The Partition Problem
in thread Mark Jason Dominus And Me - The Partition Problem
For 2 the answer is, "Whenever we have a problem that we don't know how to solve, which can be somehow reduced to simpler versions of the same problem." For Fibonacci, that means smaller $n. In the case of the directory walk, that means farther down the directory tree. In the case of the partition problem, simpler means "fewer treasures".