The Seven-Pebble Strategy

Solving Pebbles from seven pebbles produces not a move but a policy — an answer prepared for every reply the opponent can make.

Extend the table upward. Six left: taking one leaves five, a winning count for the opponent; taking two leaves four — also winning for them. Six is poisoned. Seven left: take one, and the opponent faces the poisoned six — so seven is a win, and the winning move is forced. But a move is not yet a strategy; the strategy must survive both replies. You take one, leaving six. If the opponent takes one, five remain: take two, leaving the poisoned three. If the opponent takes two, four remain: take one, leaving the poisoned three again. Either way the opponent now faces three; whatever they take, one or two pebbles remain, and you take everything left. Written as one sentence, the policy is: leave your opponent a multiple of three, forever. Check it against the table: three, six — the poisoned counts — are exactly the multiples of three, and every winning line above passes through them. That is what solving a game means: not finding a good move, but holding an answer for every branch the striped tree can grow — the same shape of object a delivery route's contingency plan or a tournament preparation sheet is, with the arithmetic doing all of the preparing. And notice the economy: the full seven-pebble tree holds dozens of branches, but the policy compresses them into one test — is the count a multiple of three? — which a player can run faster than the opponent can move.