One Pile Nim with Arbitrary Move Function

Holshouser, Arthur; Reiter, Harold

doi:10.37236/1747

Cited by 3 publications

(2 citation statements)

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“…The proof of Theorem 2 is very similar to the proof of Theorem 1 and is left to the reader. [5] gives a complete proof of Theorem 2 when t = 1. Also, [5] gives several interesting applications of Theorem 2 for t = 1.…”

Section: Theorem 2 Supposementioning

confidence: 99%

Dynamic Single-Pile Nim Using Multiple Bases

Holshouser¹,

Reiter

2006

Electron. J. Combin.

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In the game $G_{0}$ two players alternate removing positive numbers of counters from a single pile and the winner is the player who removes the last counter. On the first move of the game, the player moving first can remove a maximum of $k$ counters, $k$ being specified in advance. On each subsequent move, a player can remove a maximum of $f(n,t) $ counters where $t$ was the number of counters removed by his opponent on the preceding move and $n$ is the preceding pile size, where $f:N\times N\rightarrow N$ is an arbitrary function satisfying the condition (1): $\exists t\in N$ such that for all $n,x\in N$, $f(n,x) =f(n+t,x) $. This note extends our earlier paper [E-JC, Vol 10, 2003, N7]. We first solve the game for functions $f:N\times N\rightarrow N$ that also satisfy the condition (2): $\forall n,x\in N$, $f(n,x+1) -f(n,x) \geq -1$. Then we state the solution when $f:N\times N\rightarrow N$ is restricted only by condition (1) and point out that the more general proof is almost the same as the simpler proof. The solutions when $t\geq 2$ use multiple bases.

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Section: Theorem 2 Supposementioning

confidence: 99%

Dynamic Single-Pile Nim Using Multiple Bases

Holshouser¹,

Reiter

2006

Electron. J. Combin.

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“…Such games have a vast literature (see the selected bibliography of Frankel [1]). Variants on the 1-pile version have included letting the number of stones a player can remove depend on how many stones are in the pile [5], letting the number of stones a player can remove depend on the player [2], allowing three players [7], viewing the stones as cookies that may spoil [6], and others. Grundy [4] and Sprague [8] showed how to analyze many-pile NIM games by analyzing the 1-pile NIM games that it consists of.…”

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confidence: 99%

NIM with Cash: A Concrete Approach

Chen¹,

Gasarch²

2019

Preprint

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Let A be a finite subset of N, and let n ∈ N. then NIM(A; n) is the two player game in which players alternate removing a ∈ A stones from a pile with n stones; the first player who cannot move loses. This game has been researched thoroughly.We discuss a variant of NIM in which Player 1 and Player 2 start with d and e dollars, respectively.When a player removes a ∈ A stones from the pile, he loses a dollars. The first player who cannot move loses, but this can now happen for two reasons:1. The number of stones remaining is less than min(A). The player has less than min(A) dollars.This game leads to much more interesting win conditions than regular NIM.We investigate general properties of this game. We then obtain and prove win conditions for the sets A = {1, L} and A = {1, L, L + 1}.

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