Coach
A 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, then shrink that row to the XORed size. Try each of the three rows.
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 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.
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 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 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 the question becomes: what can you leave that forces your opponent to take it?
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 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.
From a cold square every move lands on a "hot" square, so the player to move there is lost against perfect play. Check (3, 5) against the list.
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.
A case · you decide
Nima and Bela: making zero, and facing it
Nima & Bela — a player who computes the Nim-sum, and a friend who knows what it means to be handed a zero
The rows are 3, 5 and 7. Nima adds them in binary with no carrying and gets 1.
A Nim-sum that is not zero means the player to move can win.
Bela explains the goal: move so the sum becomes zero, and the opponent is the one who cannot fix it.
From 3, 5 and 7, which move does Nima make?
Now Bela is the one handed 3, 5 and 6. What is true about every move she has?
Nima's zero and Bela's zero are the same position, seen from opposite sides of the table.
A case · you decide
Mirren and Iver: copying, until the last token
Mirren & Iver — a player who wins by copying, and a friend who plays the version where taking the last token loses
Two rows of 4, misère rules. Mirren plans to copy every move Iver makes, the way it works in normal Nim.
Copying keeps the rows equal, and it works for a while.
Then Iver takes three from one row, leaving 4 and 1.
The rows are 4 and 1, and Mirren moves. What should Mirren do?
Copy while the rows are big. When they get small, count to the last token.