AE1APS Algorithmic Problem Solving John Drake.

 Previously we introduced matchstick games, using one pile of matches  It is possible to have more than one pile of matches  First a pile must be chosen, then matches may be removed from the chosen pile according to some rule (which may differ from pile to pile)  The game is thus a combination of two games  Combinations of games is such a way are known as sum games

 We have two games, each with its own rules  Let us call the two games, the left and right games  A position in the sum game is the combination of the position in the left game and the position in the right game  A move in the sum game is a move in one of the left or right games

 A graph represents a game.  The positions are represented by nodes.  The moves are represented by edges.  Imagine a coin placed on a node.

 A graph represents a game.  The positions are represented by nodes.  The moves are represented by edges.  Imagine a coin placed on a node.

 A graph represents a game.  The positions are represented by nodes.  The moves are represented by edges.  Imagine a coin placed on a node.

 A graph represents a game.  The positions are represented by nodes.  The moves are represented by edges.  Imagine a coin placed on a node.

 In the sum game, two coins are used  One coin being placed over a node in each of the two graphs  A move is then to choose one of the coins, and move it along an edge to another node  A move changes the position of one of the coins

 The two games are unstructured, thus we have to use brute force to determine if they are winning or losing positions  The left game has 15 positions, the right game has 11 positions  How many positions are there in the sum game?

 There are 15*11 positions (165)  Brute force is not a desirable approach  We now look at how to compute a strategy for the sum of two games  We find that the computation effort needed to find winning and losing states is not the product, but the sum of the computational effort for the individual games.

 Suppose there are two piles of matches  A move is to choose a pile of matches, and remove at least one match from that pile  The game is lost when a player cannot remove any matches.

 Given a pile of matches, remove at least one  The winning positions are the positions where there is at least one match  The winning strategy is to remove all the matches  The losing position contains no matches  Draw the state transition diagram

 The single game strategy cannot be applied directly to the sum game  Why not?

 Symmetry between left and right allows us to identify a winning strategy.  Let m, n represent the number of matches in the two piles.  In the end m=n=0.  This suggests m=n identifies losing positions.  Choose the larger pile and restore the property that m=n. i.e. maintain the invariant

 The property n!=m is the precondition for the winning strategy to be applied  Equivalently m<n or n<m.  If m<n, we infer 1 <= n-m <=n  So n-m matches can be removed from the pile with n matches.  After the assignment n:=n-(n-m), the property m=n will hold.

 Suppose K matches can be removed (K is fixed)  This restriction disallows some winning moves  We are not allowed to remove m-n matches when K<m-n  The property m=n no longer characterises losing positions.

 Example K=1  Lets set K=1  (2, 0).  A player is forced to move to (1, 0) and the opponent can then win the game.

 It seems that the strategy of restoring symmetry no longer applies  Worse, if the number of matches we are allowed to remove from the two piles differs

 Even worse, the left and right games may be completely different!!!

 With one pile of matches, where M matches at most can be removed,  The winning strategy is to continually restore the property that the remainder after dividing the number of matches by M+1 is 0 (i.e. m mod (M+1) == 0)  The number, m mod (M+1) == 0, determines if the position is winning or not.

 The one player game suggests the symmetry as:m mod (M+1) = n mod (N+1)  We will prove this in later lectures.  In the final position (0,0), this property is satisfied.