1 Mathematical Games Great Theoretical Ideas In Computer Science V. Adamchik D. Sleator CS 15-251 Spring 2010 Lecture 10 Feb. 11, 2010 Carnegie Mellon University Plan Introduction to Impartial Combinatorial Games References Game Theory, by T. Ferguson Download from: http://www.math.ucla.edu/~tom/Game_Theory/Contents.html Related courses 15-859, (21-801) - Mathematical Games Look for it in Spring ‘112 21 chips Two Players: 1 and 2 A move consists of removing one, two, or three chips from the pile Players alternate moves, with Player 1 starting Player that removes the last chip wins A Take-Away Game Which player would you rather be? Try Small Examples! If there are 1, 2, or 3 only, player who moves next wins If there are 4 chips left, player who moves next must leave 1, 2 or 3 chips, and his opponent will win With 5, 6 or 7 chips left, the player who moves next can win by leaving 4 chips 0, 4, 8, 12, 16, … are target positions; if a player moves to that position, they can win the game Therefore, with 21 chips, Player 1 can win! 21 chips What if the last player to move loses? If there is 1 chip, the player who moves next loses If there are 2,3, or 4 chips left, the player who moves next can win by leaving only 1 In this case, 1, 5, 9, 13, … are a win for the second player Combinatorial Games • A set of positions • Two players • Rules specify for each player and for each position which moves to other positions are legal moves • The players alternate moving • A terminal position in one in which there are no moves • The game ends when a player has no moves • The game must end in a finite number of moves • (No draws!) Normal Versus Misère Normal Play Rule: The last player to move wins Misère Play Rule: The last player to move loses A Terminal Position is one where neither player can move anymore3 What is Omitted No random moves No hidden state No draws in a finite number of moves (This rules out games like poker) (This rules out games like battleship) (This rules out tic-tac-toe) Impartial Versus Partizan A combinatorial game is impartial if the same set of moves is available to both players in any position. A combinatorial game is partizan if the move sets may differ for the two players In this lecture we'll study impartial games. Partizan games will not be discussed P-Positions and N-Positions P-Position: Positions that are winning for the Previous player (the player who just moved) (Sometimes called LOSING positions) N-Position: Positions that are winning for the Next player (the player who is about to move) (Sometimes called WINNING positions) 0, 4, 8, 12, 16, … are P-positions; if a player moves to that position, they can win the game 21 chips is an N-position 21 chips What’s a P-Position? “Positions that are winning for the Previous player (the player who just moved)” That means: For any move that N makes There exists a move for P such that For any move that N makes There exists a move for P such that … There exists a move for P such that There are no possible moves for N P-positions and N-positions can be defined recursively by the following: (1) All terminal positions are P-positions (normal winning condition) (2) From every N-position, there is at least one move to a P-position (3) From every P-position, every move is to an N-position4 Chomp! Two-player game, where each move consists of taking a square and removing it and all squares to the right and above. Player who takes position (1,1) loses Show That This is a P-position N-Positions! Show That This is an N-position P-position! Let’s Play! I’m player 1 No matter what you do, I can mirror it! Mirroring is an extremely important strategy in combinatorial games!5 Theorem: Player 1 can win in any square starting position of Chomp Proof: The winning strategy for player 1 is to chomp on (2,2), leaving only an “L” shaped position Then, for any move that Player 2 takes, Player 1 can simply mirror it on the flip side of the “L” Theorem: Every rectangle is a N-position Proof: Consider this position: This is either a P or an N-position. If it’s a P-position, then the original rectangle was N. If it’s an N-position, then there exists a move from it to a P-position X. But by the geometry of the situation, X is also available as a move from the starting rectangle. It follows that the original rectangle is an N-position. The Game of Nim x y z Two players take turns moving A move consists of selecting a pile and removing chips from it (you can take as many as you want, but you have to at least take one) In one move, you cannot remove chips from more than one pile Winner is the last player to remove chips x y z We use (x,y,z) to denote this position (0,0,0) is a: P-position Analyzing Simple Positions One-Pile Nim What happens in positions of the form (x,0,0)? The first player can just take the entire pile, so (x,0,0) is an N-position Two-Pile Nim P-positions are those for which the two piles have an equal number of chips If it is the opponent’s turn to move from such a position, he must change to a position in which the two piles have different number of chips From a position with an unequal number of chips, you can easily go to one with an equal number of chips (perhaps the terminal position)6 Nim-Sum The nim-sum of two non-negative integers is their addition without carry in base 2 We will use ⊕ to denote the nim-sum 2 ⊕ 3 = 5 ⊕ 3 = 7 ⊕ 4 = (10)2 ⊕ (11)2 = (01)2 = 1 (101)2 ⊕ (011)2 = (110)2 = 6 (111)2 ⊕ (100)2 = (011)2 = 3 ⊕ is associative: (a ⊕ b) ⊕ c = a ⊕ (b ⊕ c) ⊕ is commutative: a ⊕ b = b ⊕ a For any non-negative integer x, x ⊕ x = 0 Cancellation Property Holds If x ⊕ y = x ⊕ z Then x ⊕ x ⊕ y = x ⊕ x ⊕ z So y = z Bouton’s Theorem: A position (x,y,z) in Nim is a P-position if and only if x ⊕ y ⊕ z = 0 Let Z denote the set of Nim positions with nim-sum zero Proof: Let NZ denote the set of Nim positions with non-zero nim-sum We prove the theorem by proving that Z and NZ satisfy the three conditions of P-positions and N-positions (1) All terminal positions are in Z (2) From each position in NZ, there is a move to a position in Z The only terminal position is (0,0,0) 001010001 100010111 111010000 010010110 ⊕ 001010001 100010111 101000110 000000000 ⊕ Look at leftmost column with an odd # of 1s Rig any of the numbers with a 1 in
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