# Nim-like game with splitting heaps

The motivation for this question came from a game introduced in a math(s) class, and so I thought it would be interesting to see if one could develop a winning strategy.

Let $$S$$ be a multiset $$[a_1,a_2,\dots,a_n]$$, with $$a_i\in\mathbb{N}=\{0,1,2,3,\dots\}$$. A game is played on $$S$$ where two players take turns to play single moves where a move consists of the following:

Replace an element $$a\in S$$ with $$b,c\in\mathbb{N}$$, such that $$b+c.

The last player to (be able to) make a (legal) move loses the game, i.e: s/he leaves each element of $$S$$ as zero.

This can be though of (and indeed was originally presented to me) as a misère Nim-like game where the elements of $$S$$ indicate the inital number of objects in $$n$$ heaps, and, where in normal Nim a move consists of removing some number of objects from the end of a heap, instead in this game you can remove a contiguous run of objects from anywhere in the heap, potentially separating the heap into two separate, non-empty heaps.

Given the simplicity, I imagine there exists a winning strategy, for the first or second player depending on the initial $$S$$. However, I have never studied games mathematically, and don't really how to go about development of a winning strategy - this is my first point of enquiry. The second (or maybe zeroth) being: is this equivalent to an existing (well-understood?) mathematical game?

After some intial analysis of my own, I determined $$[1,1,1]$$, $$[2,2]$$, and $$[1,1,2,2]$$ is won by the second player, and thus $$[1,2,5]$$ is won by the first player.

• This would be the octal game $0.\dot7$. Surprisingly, a first glance through the literature doesn't reveal any information about it. I can't imagine why there wouldn't be, as it's simple enough that you would think it would be named after someone by now.
– user694818
Dec 16 '19 at 11:58
• $[2,2]$ (like any "symmetric" position) should be a win for the second player by simply stealing the first player's strategy. So $[1,2,X]$ is a win for the first player whenever $X>3$ by reducing the position to $[1,1,2,2]$.
– user694818
Dec 16 '19 at 12:07
• Yes - sorry I mis-interpreted my notes: let me correct Dec 16 '19 at 12:09
• @MatthewDaly you expect a solution... though none has been found? Dec 16 '19 at 12:13
• Some octal games have complete solutions and others with equally clear rules don't. I wouldn't venture a guess in this case.
– user694818
Dec 16 '19 at 12:22