### Sum Games

```.
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.
```