Developing an Efficient Algorithm for Nimber Calculation in Dots and Boxes by Andrew Brinker 

What is Dots and Boxes? 

Picture courtesy of Tiger66 < wiki/File:Dots-and-boxes.svg 

There are 4 levels of play 

You need chain control 

This is a chain 

There are actually 2 games 

Enter Nim 

NIM http://hanno-rein.de/images/main/2009_01/dsc00962.jpg 

http://hanno-rein.de/images/main/2009_01/dsc00962.jpg 

It all relies on the nim-sum 

And the Sprague–Grundy Theorem 

Win the 1st game, win the 2nd 

Presenting naive_nim() 

def mex(values): counter = 0 for value in values: if counter != value: return counter counter += 1 return values[len(values) - 1] + 1 ! def naive_nim(position): if all following moves for position are loony: return 0 return mex(all following positions to position) 

It’s O(n!) (really slow) 

Time for memoization (it’s a real word) 

It’s better in reverse 

Presenting fast_nim() 

function fast_nim(region): board = board with the same dimensions as region l = number of missing lines in region n = 1 saved = set of positions w/out 1 line and w/ value 0 while (n != l): G = set of antecedent positions w/out n lines for (each position in G): F = set of follower positions to position nim_values = [] for (each follower in F): {follower position will always be in saved} nim_values = saved[follower] saved[position] = mex(nim_values) return saved 

It’s O(n3) (much faster) 

It’s usable now! 

There’s still more to do 

Any questions?