Nim (Part 2) - harder

So, how can you figure out the solution to the more complex 'nim' game? (See my previous post for info on the game).

It's actually amazingly simple (well - somewhat simple) but needs a bit of background in binary numbers to appreciate.

First, a reminder that a binary number is composed of a 'ones' column (the right-most digit in the number), then a 'twos' column, then a 'fours' column, etc, like below:

So this is 1 'four' plus 0 'two' plus 1 'one' = 5 (in decimal)

In decimal of course, we have the 'ones' column, 'tens' column, etc, like below:

And this would be 1 'hundred' plus 1 'one' = 101 (in decimal)

Now for the solution. All you do is write out the number of toothpicks in each pile in binary, one below the other. Then add up the columns to see whether each column is even (0, 2, 4, etc) or odd. Finally, if all column sums are even then we call the overall sum 'even'. Everything else will be 'odd'.

Two things that I will note (but not prove):
1. If the overall sum is 'even', then if you take 1 or more toothpicks from any pile you will make the overall sum 'odd'.
2. If the overall sum is 'odd', then there is always a way to take away from one pile so that you make the overall sum 'even'.

So if you can make the sum 'even' on your turn, you can also always do that on your next turn since your opponent will always make it 'odd' on their turn.

a) If you follow this strategy you will win the game where the last person to take a toothpick wins.

Here's an example with three piles of 3, 5 and 8 toothpicks respectively:

Pile 1 (3)	0	1	1
Pile 2 (5)	1	0	1
Pile 3 (8)	1	0	0
	Even	Odd	Even

If you take away 2 from pile 1 you change it from binary 011 to binary 001 and now the sum of all the columns is even. If you keep this up, you will always leave toothpicks in at least 2 piles until you force your opponent to take the second-last pile and you then take the last one.

b) If you are playing the more common version of nim where the person to take the last toothpick loses, then play as above until only one heap has a size of 2 or more. Then remove all toothpicks from the larger heap and you win.

Finally, if you are playing a version of the game where you can only take away up to 'k' toothpicks at a time, then first (in your mind) reduce the size of each heap modulo k + 1 (that is, remove all multiples of k + 1 until you just have a remainder that is between 1 and k). Assuming that you are playing the 'last person to take a toothpick loses' version, then:
- If this makes all the heaps of size zero just take k objects from one of the heaps.
-Otherwise, follow the above strategy 'a'.

This is just an intro to nim, but should be enough that the explanations you read elsewhere make sense (e.g. the wikipedia nim entry)

Nim (Part 1) - easy

Nim is "a game in which two players alternately take one or more objects from one of a number of heaps, each trying to take, or to compel the other to take, the last remaining object".

(This post looks at an 'easy' version of Nim and how to solve it. My next post looks at the solution to the harder variants.)

My first experience with Nim was the simple version ('easy') where you start with a pile of (say) 15 toothpicks. You are allowed to take away 1, 2 or 3 toothpicks in your turn and then it is the other player's turn. You continue taking turns until someone has to take the last toothpick (and they lose).

This type of game is pretty easy to analyze. You can see that if you leave your opponent 1 toothpick you win. And if you leave them 1+4 = 5 toothpicks you also win (since if there are 5 toothpicks and they take 'x', you just take '4-x' on your turn and that leaves 1...giving you the win). So the winning strategy is to leave your opponent with 1 + (some multiple of 4) in the pile. If the pile starts with 15 and you get to go first, take 2 (leaving 13 = 1 + 4(3)) and you are already in a winning position.

Finally, if you are allowed to take 'k' toothpicks on each turn, then just leave 1 + some multiple of (k+1) toothpicks and you win. (And if the rules are that you need to take the last toothpick yourself to win, then just leave 1 + some multiple of k.)

The next variant I had heard of (call it 'hard') extended this game to 3 piles. The version I played had piles of 3, 5 and 8. During your turn you can take 1, 2, or 3 toothpicks from any one pile. Again, the objective of the game was to leave your opponent with the last toothpick.

I'll leave the analysis of this to be a simple case of the 'hardest' case:

• you have any number of piles
• each pile has some number of toothpicks
• during your turn you can take any number of toothpicks from any one pile

What is the winning strategy?

I'll leave this to my next post...but it's actually a fairly easy(!) solution - if you're good at converting numbers to binary and then adding up columns to see if they are even or odd.