Coach
A real case · you decide
Nima and the Nim-sum
Nima — a player who keeps losing Nim and cannot see the pattern
Three rows face you: 3, 4 and 5 tokens. It is your move, and you may take any number of tokens from a single row. Somewhere here is a move that wins — if you can find the number hiding in the position.
The Nim-sum is the bitwise XOR of the row sizes: 3 ⊕ 4 ⊕ 5. In binary that is 011 ⊕ 100 ⊕ 101 = 010 = 2. Bouton's theorem says the player to move WINS exactly when the Nim-sum is non-zero — and the winning move is any that makes it zero.
To zero a Nim-sum of 2, find a row whose size DROPS when you XOR it with 2. Here 5 (101) ⊕ 2 (010) = 7 — bigger, no good — but 4 (100) ⊕ 2 = 6 — bigger — while 3 (011) ⊕ 2 = 1 — smaller. So take the row of 3 down to 1.
The Nim-sum of (3, 4, 5) is 2. Which move hands your opponent a losing position?
Nim is not about tokens; it is about one number. Zero it on every turn and you cannot lose.
A real case · you decide
Bela reads a losing square
Bela — a player who wants to know, before moving, whether the position is already lost
Two rows of 2 tokens sit in front of you: (2, 2). It is your turn. Something feels wrong, and it is: this is a position where every move you make loses against perfect play.
2 ⊕ 2 = 0. A zero Nim-sum is a P-position — "Previous player wins" — because ANY move you make breaks the balance and hands a non-zero (winning) position to the opponent, who simply restores the zero.
You cannot escape a P-position by playing cleverly; you can only hope your opponent errs. The lesson is to recognise it and — when it is YOUR move that creates a P-position for THEM — to reach for it deliberately.
It is your move at (2, 2). What is the honest read?
A P-position is not a trap you spring — it is the shape you leave BEHIND you, move after move, until your opponent runs out of tokens.
A real case · you decide
Mirren mirrors
Mirren — a player who wins two-row games without ever computing an XOR
Someone hands you the SECOND move in a game with two equal rows: (5, 5). No binary, no theorem needed — there is a strategy so simple it feels like cheating.
Two equal rows have Nim-sum zero, so the player to move (your opponent) is already lost. Your plan: whatever they take from one row, take the SAME amount from the other, restoring equality.
Every turn you re-create two equal rows — a P-position — so your opponent always faces the losing side. Symmetry is just the Nim-sum theorem you can see with your eyes.
Your opponent takes 3 from the first row: (2, 5). What do you play?
Symmetry is the Nim-sum for the naked eye: keep the two piles equal and the game keeps itself.
A real case · you decide
Iver and the last-token twist
Iver — a player who mastered normal Nim and then lost every misère game the same way
Same game, one flipped rule: whoever takes the LAST token now LOSES. You reach a single row of 3 tokens, and it is your move. Your normal-Nim instinct — "take them all and win" — is about to betray you.
Misère Nim is played EXACTLY like normal Nim right up until your move would leave only single-token rows; there, and only there, the rule inverts. In normal Nim a lone row of 3 is a trivial win: take all three and you took the last token.
In MISÈRE, taking the last token loses — so from a lone row of 3 you must NOT clear it. Take 2 and leave a single token: your opponent has no choice but to take it, and taking the last token, they lose.
Misère (last token loses), your move at a single row of 3. What wins?
Misère is normal Nim wearing a disguise — the same theorem, with one careful step at the very end.
A real case · you decide
Wyn and the golden squares
Wyn — a player who found an irrational number hiding inside a two-heap game
Wythoff's game: two heaps, and a move takes any amount from one heap OR the same amount from BOTH. You are shown the position (3, 5) and asked the impossible-sounding question — is it already lost for the player to move?
Wythoff's losing (cold) positions are (⌊nφ⌋, ⌊nφ²⌋) for n = 0, 1, 2, …: (0,0), (1,2), (3,5), (4,7), (6,10)… The two coordinates always differ by exactly n, and their ratio marches toward φ ≈ 1.618.
(3, 5) is the n = 2 cold position — a P-position. From a cold square every move lands on a "hot" square, so the player to move there is lost against perfect play.
Is (3, 5) a win or a loss for the player to move in Wythoff's game?
Wythoff's cold squares are where a children's counting game and an irrational number turn out to be the same thing.