The Game of Nim is played with several piles of objects. That the game starts with more than one pile is important. Since there is no limitation on how many objects can be removed on a move, Nim, on a single pile, is a bland, one move win for a first player. The situation changes when the rules of the games introduce limitations on available moves. For example, Scoring and the Subtraction games may be meaningfully played on a single pile. Scoring limits the size of a move from above. In the Subtraction games, the move is restricted to a finite set of alternatives.

One Pile is the most direct generalization of Scoring and the simplest of the Subtraction games. On each move a player is permitted to remove any number of objects bounded both from above and below. (In the applet, a move is performed by pressing one of the buttons located on the perimeter of the drawing area. The Min and Max attributes can be modified by clicking a little off their central line.)

The Grundy numbers for the various sizes of the pile are easily found with the Mex rule. For example, for Min = 3 and Max = 5, we get

The P-positions correspond to the piles whose size S falls into the range 0 S mod (Min + Max) < Min.

There is a small body of literature that delves into the variations of One Pile in which Min = 1 while Max depends on the previous move. These are usually referred to as the Take-Away games. On the first move a player is allowed to remove any number of objects, but the whole pile. Assuming x is the size of a move -- the number of the removed objects -- let f(x) stand for the maximum number of objects that could be legally removed on the subsequent move. Different choices of the function f lead, in principle, to different games. The following applet implements three prototypical games corresponding to f(x) = x (No more than ...), f(x) = 2x-1 (Less than twice ...), and f(x) = 2x (No more than twice ...). (A. J. Schwenk gave a complete solution to the games with f nondecreasing and satisfying f(x) x.)

Schwenk makes a case for the importance of the losing starting positions, i.e. the positions in which, all other factors equal, the second player has a winning strategy. In all games, the smallest losing position holds a single object. (The second player wins without lifting a finger. See, e.g. Russell.)

Let then H₁ = 1 and define the sequence H_k+1 = H_k + H_m, where m = min {j: f(H_j) H_k}. Since, in the very least, f(H_k) H_k, the definition does make sense.

The fact that the sequence {H_i} consists of losing starting positions stems from two assertions.

(1) leads to a game specific binary representation of the size of the pile - position in a game.

Such a move reduces the number of units /N/ and also insures that the next move could not accomplish the same feat. Thus, if the game is played right, unit reduction is only possible on every other move. Unit reduction is the key to the ultimate success because the only integer N for which /N/ = 0 is 0, the desired outcome of the final and winning move. (For more details, see Take-Away Games.)

Let's now determine the sequence {H_j} for the three games. By definition, {H_j} is always increasing.

f(x) = x

H₁ = 1, H₂ = H₁ + H₁ = 2, H₃ = H₂ + H₂ = 4, and, since f(H_j-1) = H_j-1 < H_j, H_j+1 = H_j + H_j, in general. H_i = 2ⁱ.

f(x) = 2x - 1

H₁ = 1, H₂ = H₁ + H₁ = 2, H₃ = H₂ + H₂. By induction, if H_j = H_j-1 + H_j-1, then f(H_j-1) = H_j - 1 < H_j, so that H_j+1 = H_j + H_j.

A surprise! Here too, H_i = 2ⁱ.

f(x) = 2x

H₁ = 1, H₂ = H₁ + H₁ = 2, and, since 2x x + 1, for x 1, H₃ = H₂ + H₁, H₄ = H₃ + H₂. Now, by induction, if H_j = H_j-1 + H_j-2, then, on one hand, f(H_j-2) = 2H_j-2 < H_j-1 + H_j-2 = H_j. On the other hand, f(H_j-1) = 2H_j-1 > H_j-1 + H_j-2 = H_j. Therefore, f(H_j+1) = H_j + H_j-1, by definition.

Daykin showed that Zeckendorf's theorem gives in fact a characterization of the Fibonacci sequence: if {f_i} is a sequence such that any positive integer has a unique representation as the sum of f_i's with no two successive terms in the expansion, then {f_i} is necessarily increasing and f_i = F_i, for i > 0.

Similarly, if we assume that a sequence {h_i} is increasing and every positive integer has a unique representation (1) as the sum h_j1 + h_j2 + h_j3 + ... + h_jk, where, for i = 1, ..., k-1, f(h_ji) < h_ji+1, for a nondecreasing function f with f(x) x, then necessarily h_i = H_i, with H_i's defined as above.

The proof is left to the reader as a playful exercise.

References

1. D. E. Daykin, Representation of Natural Numbers As Sums of Generalized Fibonacci Numbers, J London Math Soc, 35 (1960), 143-160
2. B. Russell, In Praise of Idleness, Routledge, 1996
3. A. J. Schwenk, Take-Away Games, The Fibonacci Quarterly, v 8, no 3 (1970), 225-234

Fibonacci Numbers

1. Ceva's Theorem: A Matter of Appreciation
2. When the Counting Gets Tough, the Tough Count on Mathematics
3. I. Sharygin's Problem of Criminal Ministers
4. Single Pile Games
5. Take-Away Games
6. Number 8 Is Interesting
7. Curry's Paradox
8. A Problem in Checker-Jumping
9. Fibonacci's Quickies

Golden Ratio

1. Golden Ratio in Geometry
2. Golden Ratio in an Irregular Pentagon
3. Golden Ratio in a Irregular Pentagon II
4. Inflection Points of Fourth Degree Polynomials
5. Wythoff's Nim
6. Inscribing a regular pentagon in a circle – and proving it
7. Cosine of 36 degrees
8. Continued Fractions

Copyright © 1996-2009 Alexander Bogomolny

Generalized Converse of Zeckendorf's Theorem

We are given that every positive integer N can be uniquely represented as a sum h_j1 + h_j2 + h_j3 + ... + h_jn, where, for I = 1, ..., n-1, f(h_ji) < h_ji+1, for a nondecreasing function f with f(x) x.

We are to prove that, for k = 1, 2, ..., h_k+1 = h_k + h_m, where m = min{j: f(h_j) h_k}.

If h_k+1 = h_k + h_m is not true, then either h_k+1 > h_k + h_m or h_k+1 < h_k + h_m.

The first of the inequalities is impossible, because then, taking N = h_k + h_m, we have h_k < N < h_k+1. Since the expansion of N is bound to end with h_k: N = h_k + h_j1 + ... + h_jn, we get h_m = h_j1 + ... + h_jn, in violation of the uniqueness of the expansion.

Assume then h_k+1 < h_k + h_m. Then h_k+1 > h_k + h_m-1. For, otherwise, h_k + h_m-1 would have an expansion that ends with h_k+1. Furthermore, from the choice of m, f(h_m-1) < h_k. We thus have

As before, the expansion of h_k+1 - h_k is bound to include h_m-1:

where f(h_ji) < h_ji+1, I = 1, ..., n-1, and h_jn = h_{m - 1}. We therefore get the expansion (for f(h_m-1) < h_k)

which, too, contradicts the uniqueness assumption.

As we see, h_k+1 = h_k + h_m is the only possibility.

Copyright © 1996-2009 Alexander Bogomolny