When I tried to solve this problem, I began by looking for some property invariant under moves, but after making no progress I discarded it when I noticed that:
Every move changes the parity of every pile. From there it was immediate that the single remaining chip at the end can only be in the pile that originally had a parity different from the other two, and that if they all have the same parity, it is not possible to reduce them to a single chip.
Yes, actually this turns out to be the same as the solution based on the invariant taking values in the Klein-four group: the kernel of the map f:Z3→(Z/2Z)2 given by (x1,x2,x3)→(x1+x2,x1+x3) is precisely the subspace of integer vectors with all entries having the same parity. It’s just that the invariant solution makes it look natural, while doing it this way around leads to having to express these parity conditions in a more elaborate way.
When I tried to solve this problem, I began by looking for some property invariant under moves, but after making no progress I discarded it when I noticed that:
Every move changes the parity of every pile. From there it was immediate that the single remaining chip at the end can only be in the pile that originally had a parity different from the other two, and that if they all have the same parity, it is not possible to reduce them to a single chip.
Yes, actually this turns out to be the same as the solution based on the invariant taking values in the Klein-four group: the kernel of the map f:Z3→(Z/2Z)2 given by (x1,x2,x3)→(x1+x2,x1+x3) is precisely the subspace of integer vectors with all entries having the same parity. It’s just that the invariant solution makes it look natural, while doing it this way around leads to having to express these parity conditions in a more elaborate way.
I wouldn’t call that ‘no progress’.