EDIT: Nope! Not far wrong, but definitely wrong.

My strategy of starting with the easy cases and building out to more complex cases was used successfully by other solvers. My analysis of the easy cases was correct, but there was a mistake in the way I built out. My diagram probably didn’t help; the one drawn by the winner didn’t distinguish between the larger and smaller piles, and I think that’s where my error crept in.

For example, when I considered (9, 5), I thought it was a losing position because I didn’t see that taking six coins from the larger pile produced (5, 3), since the larger pile became the smaller pile in that move. So the “simplification” I introduced, of specifying one pile as larger, while not wrong, made it easier for me to make a mistake. It helped to break my mental model by suggesting an identity for the piles (“larger” or “smaller”) that isn’t persistent, but depends on the progress of the game. Will try not to do that again!

EDIT 2: Revised the original spread sheet showing the rows that I had calculated wrongly, and producing a correct game state table.

Another week of being *fairly* confident about this week’s Riddler Classic. (Bolstered by getting exactly the right answer for the last one I did, if by a slightly inefficient method.)

The puzzle is a coin-picking game, with the usual aim of taking the last coins, but the unusual condition of having two coin stacks:

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