Misere and symmetry

Questions 1 to 5 of 8

1. 1. Match each position to who wins with best play.

   Choose the matching item for each row.

   Misere Nim, one pile of 1Choose…the player to move winsthe player to move losesthe player to move wins by taking the last stoneMisere Nim, two piles of 1Choose…the player to move winsthe player to move losesthe player to move wins by taking the last stoneNormal Nim, one pile of 1Choose…the player to move winsthe player to move losesthe player to move wins by taking the last stone
2. 2. Put the steps of the mirror argument for an even board in order.

   Use the arrows to put the steps in order.

   1. 1.A cell and its mirror never share a row or a column, so they never touch.↑↓
   2. 2.The mirror of a legal first move is legal after that move is placed.↑↓
   3. 3.The second player answers each move at its mirror, and the first player runs out of moves first.↑↓
   4. 4.No cell is its own mirror, because 2r = R - 1 has no whole-number solution when R is even.↑↓
3. 3. Which heap sizes are P (the player to move loses) in misere play of the subtraction game with moves 1, 2 and 5?

   Pick every answer that is true.

   Heap 0Heap 1Heap 4Heap 6
4. 4. Which position is an exception, where misere Nim and normal Nim give different answers?

   Piles 3 and 1, where the XOR is 2.Piles 2 and 2, where the XOR is 0.Piles 1, 1 and 1, where every pile has at most one stone.Piles 4, 4 and 4, where the XOR is 4.
5. 5. A teammate says: 'Grundy values work for misere too. A heap of 1 has value 1, so two heaps of 1 have XOR 0, which means the player to move loses.' What is wrong?

   Nothing, the XOR rule holds in misere play too.A heap of 1 has value 0 in misere play, so the sum is 0.Grundy values only work when every heap has at least 2 stones.Misere play changes the end of the game, so the XOR of values does not decide a sum. Two heaps of 1 are a win for the player to move.