html趣味数学网页版,趣味数学网站

How to Win at Nim

One Heap

In a game of nim where you have one nim heap, and you are allowed

to remove from 1 to 4 objects at a time, you can win if you leave

the heap with a multiple of 5 objects after your

turn. Thereafter, no matter what your opponent

does, you can always restore the number of objects to a multiple of

5 when you take your turn. If you cannot leave

the heap with a multiple of 5 objects, then your opponent has the

advantage, and only an error by your opponent can put you in a

position to win the game.

More Than One Heap

In a game of nim that involves nim heaps where you can take as many

objects as you want from any one of the heaps during your turn, you

need to be able to compute a nim sum,

that characterizes the configuration of the game. Here's how to do

it:

Express the number of objects in each nim heap as a binary

number, with the only digits being 0 and 1.

Fill out the smaller binary numbers with '0's on the left, if

necessary, so that all the numbers have the same number of

digits.

Sum the binary numbers, but do not carry

Replace each digit in the sum with the remainder that results

when the digit is divided by 2.

This yields the nim sum.

To win at nim, always make a move, when possible, that leaves a

configuration with a nim sum of 0. If you cannot do this, your

opponent has the advantage, and you have to depend on his or her

committing an error in order to win.

Note that if the configuration you are given has a nim sum not

equal to 0, there always is a move that creates a new configuration

with a nim sum of 0. However, there are usually also moves that

will yield configurations that give nim sums not equal to 0, and

you need to avoid making these.

Also note that if you are given a configuration that has a nim

sum of 0, there is no move that will create a configuration that

also has a nim sum of 0.

Example of Computing a Nim Sum

Let's say you have three nim heaps, with 3, 7, and 11 matchsticks,

respectively.

As binary numbers, these quantities are 11, 111, and 1011,

respectively.

Filling out with '0's yields 0011, 0111, and 1011.

Summing with carrying, we get:

0011

+0111

+1011

----

1133

Taking the remainders after dividing each digit by 2

yields 1111, which is not equal

to 0. Therefore, this configuration is a winning position for the

person who is about to take a turn. It follows that you should

try not to create this

configuration when you take your turn. Now, you need to plan a move

that creates a configuration with a nim sum of 0. How about removing 7 matchsticks from the pile that now has 11,

leaving 4? We then have:

0011

+0111

+0100

----

0222

Taking the remainders after dividing by 2 gives

us 0000. This is a configuration

that we want to leave for our opponent.

Try the Nim

Sum Calculator Applet. You will need to have a

Java 2 plug-in to do this.

有意思的是，nim

游戏还有一个mirere的对等版本，也即拿走最后一块石子的人输；但是若对于只有两堆的情况，不管是哪个游戏版本，先动手的人必输无疑。