## Heap Games, Numeration Systems and Sequences (1998)

Venue: | Ann. of Combinatorics |

Citations: | 5 - 3 self |

### BibTeX

@ARTICLE{Fraenkel98heapgames,,

author = {Aviezri Fraenkel},

title = {Heap Games, Numeration Systems and Sequences},

journal = {Ann. of Combinatorics},

year = {1998},

volume = {2},

pages = {197--210}

}

### OpenURL

### Abstract

. We propose and analyze a 2-parameter family of 2-player games on two heaps of tokens, and present a strategy based on a class of sequences. The strategy looks easy, but it is actually hard. A class of exotic numeration systems is then used, which enables us to decide whether the family has an efficient strategy or not. We introduce yet another class of sequences and demonstrate its equivalence with the class of sequences defined for the strategy of our games. Keywords: heap games, numeration systems, sequences 1. Example Given a 2-player game played on two heaps (piles) of finitely many tokens. There are two types of moves: (I) Take any positive number of tokens from one heap, possibly the entire heap. (II) Take from both heaps, k from one and l from the other, with, say, k l. Then the move is constrained by the condition 0 ! k l ! 2k + 2, which is equivalent to 0 l \Gamma k ! k +2; k ? 0. The player making the last move (after which both heaps are empty) wins, and the opponent ...

### Citations

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Citation Context ...dd number of 0s. b. For every (An, Bn) ∈ P, the representation of Bn is the “left shift” of the representation of An. Thus (1, 4) of Table 1 has representation (1, 10), and (6, 22) has representation =-=(12, 120)-=-: 10 is the “left shift” of 1, 120 the left shift of 12. We remark that the second part of a is not independent; it follows from its first part, since A and B are complementary. In the next section we... |

265 |
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Citation Context ...r can win, i.e., the player moving from q. The position (0, 0) (two empty heaps) is a P-position, since the first player cannot even make a move, so the second wins by default. The next P-position is =-=(1, 4)-=-: if Jean takes an entire heap, then Gill takes the other and wins. If Jean takes any part of the larger heap, Gill can take the balance of both heaps. Lastly, Jean cannot remove both heaps, and if sh... |

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Citation Context ...e Bn have representations ending in an odd number of 0s. b. For every (An, Bn) ∈ P, the representation of Bn is the “left shift” of the representation of An. Thus (1, 4) of Table 1 has representation =-=(1, 10)-=-, and (6, 22) has representation (12, 120): 10 is the “left shift” of 1, 120 the left shift of 12. We remark that the second part of a is not independent; it follows from its first part, since A and B... |

18 | The golden section, phyllotaxis and Wythoff’s game - Coxeter - 1953 |

15 | Fraenkel, A linear algorithm for nonhomogeneous spectra of numbers - Boshernitzan, S - 1984 |

12 |
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Citation Context ...resentations ending in an odd number of 0s. b. For every (An, Bn) ∈ P, the representation of Bn is the “left shift” of the representation of An. Thus (1, 4) of Table 1 has representation (1, 10), and =-=(6, 22)-=- has representation (12, 120): 10 is the “left shift” of 1, 120 the left shift of 12. We remark that the second part of a is not independent; it follows from its first part, since A and B are compleme... |

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9 |
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Citation Context ...r can win, i.e., the player moving from q. The position (0, 0) (two empty heaps) is a P-position, since the first player cannot even make a move, so the second wins by default. The next P-position is =-=(1, 4)-=-: if Jean takes an entire heap, then Gill takes the other and wins. If Jean takes any part of the larger heap, Gill can take the balance of both heaps. Lastly, Jean cannot remove both heaps, and if sh... |

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8 |
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Citation Context ...of Lemma 1. Then Vk +1 is old, but Vk +2 is clearly evil, since R(Vk+2) ends in 1, so Vk+1 = Vk+2. Then Vk+1 = 1+(d2k+1+ 1)u2k+1+ ∑n i=2k+2 diui, so R(Vk+1) = u0+(d2k+1+1)u2k+1+ ∑n i=2k+2 diui. Hence =-=(5)-=- LR(Vk+1) − sR(Vk+1) = (u1 − su0) + (d2k+1 + 1)(u2k+2 − su2k+1) n∑ n∑ + di(ui+1 − sui) = t + u2k+2 − su2k+1 + di(ui+1 − sui). i=2k+2 For case (ii) of Lemma 1 we have by (3), i=2k+1 tk = r(u1 − su0) + ... |

7 | Nonhomogeneous spectra of numbers - Boshernitzan, Fraenkel - 1981 |

3 | Zeckendorf [1972], Représentation des nombres naturels par une somme de nombres de Fibonacci ou de nombres de - unknown authors |

2 |
Fraenkel [1984], A linear algorithm for nonhomogeneous spectra of numbers
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Citation Context ...s remainder, yields this unique representation. The case b1 = b2 = 1 gives a binary representation known as the Zeckendorf representation [Zec72]. Example. We consider the case u−1 = 1 2 , (b1, b2) = =-=(3, 2)-=-. Then u1 = 4, u2 = 14, u3 = 50, u4 = 178, . . . . The representations of the integers 1 to 60 in this numeration system are displayed in Table 2. A question we just might ask at this point is whether... |

1 |
Boshernitzan and A.S. Fraenkel [1981], Nonhomogeneous spectra of numbers, Discr
- unknown authors
(Show Context)
Citation Context ... which is equivalent to 0 ≤ l − k < (s − 1)k + l, k ∈ Z + . The example presented in §1 is the special case s = t = 2. Denote by P the set of all P-positions. Theorem 1. P = ⋃ ∞ i=0 {(Ai, Bi)}, where =-=(2)-=- An = mex{Ai, Bi : 0 ≤ i < n}, Bn = sAn + tn (n ∈ Z 0 ). Proof. Let A = ⋃∞ n=1 An, B = ⋃∞ n=1 Bn. Then A, B are complementary with respect to Z +: A ∪ B = Z + follows from the mex property. Suppose th... |