Nim

Matches set up in rows for a game of Nim. Players take turns to choose a row and remove any number of matches from it.
Genres	Mathematical game Abstract strategy game
Players	2
Chance	None

Nim is a mathematical game of strategy in which two players take turns removing (or "nimming") objects from distinct heaps or piles. On each turn, a player must remove at least one object, and may remove any number of objects provided they all come from the same heap or pile. Depending on the version being played, the goal of the game is either to avoid taking the last object or to take the last object.

Variants of nim have been played since ancient times.^[1] The game is said to have originated in China—it closely resembles the Chinese game of 捡石子 jiǎn-shízi, or "picking stones"^[2]—but the origin is uncertain; the earliest European references to nim are from the beginning of the 16th century. Its current name was coined by Charles L. Bouton of Harvard University, who also developed the complete theory of the game in 1901,^[3] but the origins of the name were never fully explained. The Oxford English Dictionary derives the name from the German verb nimm, meaning "take."

Nim is typically played as a misère game, in which the player to take the last object loses. Nim can also be played as a "normal play" game whereby the player taking the last object wins. In either normal play or a misère game, when there is exactly one heap with at least two objects, the player who takes next can easily win. If this removes either all or all but one objects from the heap that has two or more, then no heaps will have more than one object, so the players are forced to alternate removing exactly one object until the game ends. If the player leaves an even number of non-zero heaps (as the player would do in normal play), the player takes last; if the player leaves an odd number of heaps (as the player would do in misère play), then the other player takes last.

Normal-play nim (or more precisely the system of nimbers) is fundamental to the Sprague–Grundy theorem, which essentially says that in normal play every impartial game is equivalent to a nim heap that yields the same outcome when played in parallel with other normal play impartial games (see disjunctive sum).

While all normal-play impartial games can be assigned a nim value, that is not the case under the misère convention. Only tame games can be played using the same strategy as misère nim.

Nim is a special case of a poset game where the poset consists of disjoint chains (the heaps).

The evolution graph of the game of nim with three heaps is the same as three branches of the evolution graph of the Ulam-Warburton automaton.^[4]

At the 1939 New York World's Fair Westinghouse displayed a machine, the Nimatron, that played nim.^[5] From May 11, 1940, to October 27, 1940, only a few people were able to beat the machine in that six-month period; if they did, they were presented with a coin that said "Nim Champ."^[6][7][8] It was also one of the first-ever electronic computerized games. Ferranti built a nim-playing computer which was displayed at the Festival of Britain in 1951. In 1952 Herbert Koppel, Eugene Grant and Howard Bailer, engineers from the W. L. Maxon Corporation, developed a machine weighing 23 kilograms (50 lb) which played nim against a human opponent and regularly won.^[9] A nim playing machine has been described made from tinkertoys.^[10]

The game of nim was the subject of Martin Gardner's February 1958 Mathematical Games column in Scientific American. A version of nim is played—and has symbolic importance—in the French New Wave film Last Year at Marienbad (1961).^[11]

Game play and illustration

The normal game is between two players and is played with three heaps of any number of objects. The two players alternate taking any number of objects from any one of the heaps. The goal is to be the last to take an object. In misère play, the goal is instead to ensure that the opponent is forced to take the last remaining object.

The following example of a normal game is played between fictional players Bob and Alice, who start with heaps of three, four and five objects.

Winning positions

The practical strategy to win at the game of nim is for a player to get the other into one of the following positions, and every successive turn afterwards they should be able to make one of the smaller positions. Only the last move changes between misère and normal play.

* Only valid for normal play.

** Only valid for misère.

For the generalisations, n and m can be any value > 0, and they may be the same.

Mathematical theory

Nim has been mathematically solved for any number of initial heaps and objects, and there is an easily calculated way to determine which player will win and which winning moves are open to that player.

The key to the theory of the game is the binary digital sum of the heap sizes, i.e., the sum (in binary), neglecting all carries from one digit to another. This operation is also known as "bitwise xor" or "vector addition over GF(2)" (bitwise addition modulo 2). Within combinatorial game theory it is usually called the nim-sum, as it will be called here. The nim-sum of x and y is written x ⊕ y to distinguish it from the ordinary sum, x + y. An example of the calculation with heaps of size 3, 4, and 5 is as follows:

Binary Decimal
 
  0112    310    Heap A
  1002    410    Heap B
  1012    510    Heap C
  ---
  0102    210    The nim-sum of heaps A, B, and C, 3 ⊕ 4 ⊕ 5 = 2

An equivalent procedure, which is often easier to perform mentally, is to express the heap sizes as sums of distinct powers of 2, cancel pairs of equal powers, and then add what is left:

3 = 0 + 2 + 1 =     2   1      Heap A
4 = 4 + 0 + 0 = 4              Heap B
5 = 4 + 0 + 1 = 4       1      Heap C
--------------------------------------------------------------------
2 =                 2          What is left after canceling 1s and 4s

In normal play, the winning strategy is to finish every move with a nim-sum of 0. This is always possible if the nim-sum is not zero before the move. If the nim-sum is zero, then the next player will lose if the other player does not make a mistake. To find out which move to make, let X be the nim-sum of all the heap sizes. Find a heap where the nim-sum of X and heap-size is less than the heap-size; the winning strategy is to play in such a heap, reducing that heap to the nim-sum of its original size with X. In the example above, taking the nim-sum of the sizes is X = 3 ⊕ 4 ⊕ 5 = 2. The nim-sums of the heap sizes A=3, B=4, and C=5 with X=2 are

A ⊕ X = 3 ⊕ 2 = 1 [Since (011) ⊕ (010) = 001 ]

B ⊕ X = 4 ⊕ 2 = 6

C ⊕ X = 5 ⊕ 2 = 7

The only heap that is reduced is heap A, so the winning move is to reduce the size of heap A to 1 (by removing two objects).

As a particular simple case, if there are only two heaps left, the strategy is to reduce the number of objects in the bigger heap to make the heaps equal. After that, no matter what move your opponent makes, you can make the same move on the other heap, guaranteeing that you take the last object.

When played as a misère game, Nim strategy is different only when the normal play move would leave only heaps of size one. In that case, the correct move is to leave an odd number of heaps of size one (in normal play, the correct move would be to leave an even number of such heaps).

These strategies for normal play and a misère game are the same until the number of heaps with at least two objects is exactly equal to one. At that point, the next player removes either all objects (or all but one) from the heap that has two or more, so no heaps will have more than one object (in other words, so all remaining heaps have exactly one object each), so the players are forced to alternate removing exactly one object until the game ends. In normal play, the player leaves an even number of non-zero heaps, so the same player takes last; in misère play, the player leaves an odd number of non-zero heaps, so the other player takes last.

In a misère game with heaps of sizes three, four and five, the strategy would be applied like this:

A B C nim-sum 
 
3 4 5 0102=210   I take 2 from A, leaving a sum of 000, so I will win.
1 4 5 0002=010   You take 2 from C
1 4 3 1102=610   I take 2 from B
1 2 3 0002=010   You take 1 from C
1 2 2 0012=110   I take 1 from A
0 2 2 0002=010   You take 1 from C
0 2 1 0112=310   The normal play strategy would be to take 1 from B, leaving an even number (2)
                 heaps of size 1.  For misère play, I take the entire B heap, to leave an odd
                 number (1) of heaps of size 1.
0 0 1 0012=110   You take 1 from C, and lose.

Proof of the winning formula

The soundness of the optimal strategy described above was demonstrated by C. Bouton.

Theorem. In a normal nim game, the player making the first move has a winning strategy if and only if the nim-sum of the sizes of the heaps is not zero. Otherwise, the second player has a winning strategy.

Proof: Notice that the nim-sum (⊕) obeys the usual associative and commutative laws of addition (+) and also satisfies an additional property, x ⊕ x = 0.

Let x₁, ..., x_n be the sizes of the heaps before a move, and y₁, ..., y_n the corresponding sizes after a move. Let s = x₁ ⊕ ... ⊕ x_n and t = y₁ ⊕ ... ⊕ y_n. If the move was in heap k, we have x_i = y_i for all i ≠ k, and x_k > y_k. By the properties of ⊕ mentioned above, we have

    t = 0 ⊕ t
      = s ⊕ s ⊕ t
      = s ⊕ (x1 ⊕ ... ⊕ xn) ⊕ (y1 ⊕ ... ⊕ yn)
      = s ⊕ (x1 ⊕ y1) ⊕ ... ⊕ (xn ⊕ yn)
      = s ⊕ 0 ⊕ ... ⊕ 0 ⊕ (xk ⊕ yk) ⊕ 0 ⊕ ... ⊕ 0
      = s ⊕ xk ⊕ yk
 
(*) t = s ⊕ xk ⊕ yk.

The theorem follows by induction on the length of the game from these two lemmas.

Lemma 1. If s = 0, then t ≠ 0 no matter what move is made.

Proof: If there is no possible move, then the lemma is vacuously true (and the first player loses the normal play game by definition). Otherwise, any move in heap k will produce t = x_k ⊕ y_k from (*). This number is nonzero, since x_k ≠ y_k.

Lemma 2. If s ≠ 0, it is possible to make a move so that t = 0.

Proof: Let d be the position of the leftmost (most significant) nonzero bit in the binary representation of s, and choose k such that the dth bit of x_k is also nonzero. (Such a k must exist, since otherwise the dth bit of s would be 0.) Then letting y_k = s ⊕ x_k, we claim that y_k < x_k: all bits to the left of d are the same in x_k and y_k, bit d decreases from 1 to 0 (decreasing the value by 2^d), and any change in the remaining bits will amount to at most 2^d−1. The first player can thus make a move by taking x_k − y_k objects from heap k, then

t = s ⊕ xk ⊕ yk           (by (*))
  = s ⊕ xk ⊕ (s ⊕ xk)
  = 0.

The modification for misère play is demonstrated by noting that the modification first arises in a position that has only one heap of size 2 or more. Notice that in such a position s ≠ 0, and therefore this situation has to arise when it is the turn of the player following the winning strategy. The normal play strategy is for the player to reduce this to size 0 or 1, leaving an even number of heaps with size 1, and the misère strategy is to do the opposite. From that point on, all moves are forced.

Variations

The subtraction game

Interactive subtraction game: Players take turns removing 1, 2 or 3 objects from an initial pool of 21 objects. The player taking the last object wins.

In another game which is commonly known as nim (but is better called the subtraction game), an upper bound is imposed on the number of objects that can be removed in a turn. Instead of removing arbitrarily many objects, a player can only remove 1 or 2 or ... or k at a time. This game is commonly played in practice with only one heap.

Bouton's analysis carries over easily to the general multiple-heap version of this game. The only difference is that as a first step, before computing the nim-sums we must reduce the sizes of the heaps modulo k + 1. If this makes all the heaps of size zero (in misère play), the winning move is to take k objects from one of the heaps. In particular, in ideal play from a single heap of n objects, the second player can win if and only if

0 = n (mod k + 1) (in normal play), or

1 = n (mod k + 1) (in misère play).

This follows from calculating the nim-sequence of S(1,2,...,k),

${\displaystyle 0.123\ldots k0123\ldots k0123\dots ={\dot {0}}.123\ldots {\dot {k}},}$

from which the strategy above follows by the Sprague–Grundy theorem.

Player     Number
  1           1
  2           4
  1        5, 6 or 7
  2           8
  1       9, 10 or 11
  2          12
  1      13, 14 or 15
  2          16
  1      17, 18 or 19
  2          20
  1          21

. . . . . . . . . .

_ . . . . . . . _ _

_ . _ _ _ . . . _ _

_ . _ _ _ . . _ _ _

See also

References

Further reading

• W. W. Rouse Ball: Mathematical Recreations and Essays, The Macmillan Company, 1947.
• John D. Beasley: The Mathematics of Games, Oxford University Press, 1989.
• Elwyn R. Berlekamp, John H. Conway, and Richard K. Guy: Winning Ways for your Mathematical Plays, Academic Press, Inc., 1982.
• Manfred Eigen and Ruthild Winkler: Laws of the Game, Princeton University Press, 1981.
• Walter R. Fuchs: Computers: Information Theory and Cybernetics, Rupert Hart-Davis Educational Publications, 1971.
• G. H. Hardy and E. M. Wright: An Introduction to the Theory of Numbers, Oxford University Press, 1979.
• Edward Kasner and James Newman: Mathematics and the Imagination, Simon and Schuster, 1940.
• M. Kaitchik: Mathematical Recreations, W. W. Norton, 1942.
• Donald D. Spencer: Game Playing with Computers, Hayden Book Company, Inc., 1968.

External links

Wikimedia Commons has media related to Nim (game).

• 50-pound computer plays Nim- The New Yorker magazine "Talk of the Town" August, 1952(subscription required)
• The hot game of Nim – Nim theory and connections with other games at cut-the-knot
• Nim and 2-dimensional SuperNim at cut-the-knot
• Subtraction Game: a Subtraction Game illustration on Appstore.
• Classic Nim - Implementation of Nim for iOS.
• Matchstick Nim - Implementation of Nim for Android devices.
• NIM2 - Implementation of Nim for Android devices.