Nim

From Wikipedia, the free encyclopedia

For other uses, see Nim (disambiguation).

Nim is a two-player mathematical game of strategy in which players take turns removing objects from distinct heaps. 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.

Variants of Nim have been played since ancient times. The game is said to have originated in China (it closely resembles the Chinese game of "Jianshizi", or "picking stones"), 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, but the origins of the name were never fully explained. The name is probably derived from German nimm! meaning "take!", or the obsolete English verb nim of the same meaning. It should also be noted that rotating the word NIM by 180 degrees results in WIN (see Ambigram).

Nim is usually 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, which means that the person who makes the last move (i.e., who takes the last object) wins. This is called normal play because most games follow this convention, even though Nim usually does not.

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).

It is important to note that 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.

A version of Nim is played–and has symbolic importance–in the French New Wave film Last Year at Marienbad (1961).

[edit] Illustration

A normal play game may start with heaps of 3, 4 and 5 objects:

In order to win always leave an even total number of 1's, 2's, and 4's.

Sizes of heaps  Moves
A B C
 
3 4 5           I take 2 from A
1 4 5           You take 3 from C
1 4 2           I take 1 from B
1 3 2           You take 1 from B
1 2 2           I take entire A heap leaving two 2's.
0 2 2           You take 1 from B
0 1 2           I take 1 from C leaving two 1's. (In misère play I would take 2 from C leaving (0, 1, 0).) 
0 1 1           You take 1 from B
0 0 1           I take entire C heap and win.

[edit] Mathematical theory

Nim has been mathematically solved for any number of initial heaps and objects; that is, there is an easily-calculated way to determine which player will win and what winning moves are open to that player. In a game that starts with heaps of 3, 4, and 5, the first player will win with optimal play, whether the misère or normal play convention is followed.

The key to the theory of the game is the binary digital sum of the heap sizes, that is, the sum (in binary) neglecting all carries from one digit to another. This operation is also known as exclusive or (xor) or vector addition over GF(2). Within combinatorial game theory it is usually called the nim-sum, as will be done 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's 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's left after cancelling 1s and 4s

In normal play, the winning strategy is to finish every move with a Nim-sum of 0, which 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. Take the Nim-sum of each of the heap sizes with X, and find a heap whose size decreases. 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

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 no heap of size 2 or larger. In that case, the correct move is to leave an odd number of heaps of size 1 (in normal play, the correct move would be to leave an even number of such heaps).

In a misère game with heaps of sizes 3, 4 and 5, 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.

The previous strategy for a misère game can be easily implemented in Python.

def nim_misere(heaps):
	"""Computes next move for Nim in a misère game, returns tuple (chosen_heap, nb_remove)"""
	X = reduce(lambda x,y: x^y, heaps)
	if X == 0: # Will lose unless all non-empty heaps have size one
		for i in range(len(heaps)):
			if heaps[i] > 0: # Empty any (non-empty) heap
				chosen_heap, nb_remove = i, heaps[i]
				break
	else:
		sums = [t^X < t for t in heaps]
		chosen_heap = sums.index(True)
		nb_remove = heaps[chosen_heap] - (heaps[chosen_heap]^X)
		heaps_twomore = 0
		for i in range(len(heaps)):
			n = heaps[i]-nb_remove if chosen_heap == i else heaps[i]
			if n>1: heaps_twomore += 1
		# If move leaves no heap of size 2 or larger, leave an odd number of heaps of size 1
		if heaps_twomore == 0: 
			chosen_heap = heaps.index(max(heaps))
			heaps_one = sum([t==1 for t in heaps])
			# even? make it odd; odd? keep it odd
			nb_remove = heaps[chosen_heap]-1 if heaps_one%2==0 else heaps[chosen_heap]
	return chosen_heap, nb_remove

[edit] 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 first player has a winning strategy if and only if the nim-sum of the sizes of the heaps is nonzero. 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 (technically speaking, the nonnegative integers under ⊕ form an Abelian group of exponent 2).

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 ⊕ ... ⊕ x_n) ⊕ (y₁ ⊕ ... ⊕ y_n)
      = s ⊕ (x₁ ⊕ y₁) ⊕ ... ⊕ (x_n ⊕ y_n)
      = s ⊕ 0 ⊕ ... ⊕ 0 ⊕ (x_k ⊕ y_k) ⊕ 0 ⊕ ... ⊕ 0
      = s ⊕ x_k ⊕ y_k
 
(*) t = s ⊕ x_k ⊕ y_k.

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

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 ⊕ x_k ⊕ y_k           (by (*))
  = s ⊕ x_k ⊕ (s ⊕ x_k)
  = 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. 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.

[edit] Other variations of Nim

[edit] The subtraction game S(1,2,...,k)

In another game which is commonly known as Nim (but is better called the subtraction game S(1,2,...,k)), 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 (for instance with k = 3 in the game Thai 21 on Survivor: Thailand, where it appeared as an Immunity Challenge).

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 a play from a single heap of n objects, the second player can win iff

n ≡ 0 (mod k+1) (in normal play), or

n ≡ 1 (mod k+1) (in misère play).

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

$0.123...k0123...k0123... = \dot0.123\ldots\dot{k}$ ,

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

[edit] The 21 game

The game "21" is played as a misère game with any number of players who take turns saying a number. The first player says "1" and each player in turn increases the number by 1, 2, or 3, but may not exceed 21; the player forced to say "21" loses. This can be modeled as a subtraction game with a heap of 21–n objects.

[edit] A multiple-heap rule

In another variation of Nim, besides removing any number of objects from a single heap, one is permitted to remove the same number of objects from each heap.

[edit] Circular Nim

Yet another variation of Nim is 'Circular Nim', where any number of objects are placed in a circle, and two players alternately remove 1, 2 or 3 adjacent objects. For example, starting with a circle of ten objects,

. . . . . . . . . .

three objects be taken in the first move

_ . . . . . . . _ _

then another three

_ . _ _ _ . . . _ _

then one

_ . _ _ _ . . _ _ _

but then three objects cannot be taken out in one move.

[edit] Grundy's game

In Grundy's game, another variation of Nim, a number of objects are placed in an initial heap, and two players alternately divide a heap into two nonempty heaps of different sizes. Thus, 6 objects may be divided into piles of 5+1 or 4+2, but not 3+3. Grundy's game can be played as either misère or normal play.

[edit] See also

[edit] References

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.
C. L. Bouton: Nim, a game with a complete mathematical theory, Annals of Mathematics 3 (1901-02), 35-39.
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.
Donal D. Spencer: Game Playing with Computers, Hayden Book Company, Inc., 1968.

[edit] External links

Nim

From Wikipedia, the free encyclopedia

Contents

[edit] Illustration

[edit] Mathematical theory

[edit] Proof of the winning formula

[edit] Other variations of Nim

[edit] The subtraction game S(1,2,...,k)

[edit] The 21 game

[edit] A multiple-heap rule

[edit] Circular Nim

[edit] Grundy's game

[edit] See also

[edit] References

[edit] External links

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