**Motivation.** Recently I was watching two people play a game that involved arranging sticks in a number of heaps and moving them around in certain allowed ways that I think I was able to infer from observation. (They told me the name of the game in their language, but I can't remember it.)


**Problem.** Let $\mathbb{N}\_0$ denote the set of non-negative integers and let $n\geq 4$ be an integer. For $a,b\in\mathbb{N}\_0^n$ say that $b$ *arises from a majority move* from $a$ if 


1. $\sum a\_i =\sum b\_i$, and
2. $|\{i: b\_i > a\_i\}| > |\{i: b\_i < a\_i\}|$. (This second condition motivates the "majority" part of nomenclature).


Let $\bf{1}\_n$ be the member of $\mathbb{N}\_0^n$ in which all entries are $1$. Let $S\_n$ be the set of members of $\mathbb{N}\_0^n$ that arise from some sequence of majority moves from ${\bf 1}\_n$. Let $M\_n$ be the maximal value of any entry of any member of $S\_n$.


What is the value of $M\_n$? Do we have $\lim M\_n/n = 1$? (Correct answers to either question will be accepted.)



Majority-driven manipulations of integer vectors
$M\_n$=1 and so $\lim M\_n/n=0\ne 1$.


(We assume your definitions mean that $\bf 1\_\bf n\in {\it S\_n}$.) 


We claim that $S\_n=\{\bf 1\_\bf n\}$. 


To derive a contradiction, let $(\lambda\_1,\dots,\lambda\_n)\in S\_n\setminus\{\bf 1\_\bf n\}$. WLOG, it is obtained by a majority move from $\bf 1\_\bf n$. Also, WLOG $\lambda\_1\ge\lambda\_2\ge\dots\ge\lambda\_l\ge2$, $\lambda\_{l+1}=\dots=\lambda\_{l+k}=1$, and $\lambda\_{k+l+1}=\dots=\lambda\_n=0$. Then $0\le k<n$, so either $\lambda\_1\ge2$ or $\lambda\_n=0$. If $\lambda\_1\le1$, then $\lambda\_1=1$ (since $\lambda\_1+\dots+\lambda\_n=n\ge4$); but then $\lambda\_n=0$, implying that $|\{i:\lambda\_i>1\}|=0$ and $|\{i:\lambda\_i<1\}|\ge1$, a contradiction. Thus $\lambda\_1\ge2$.


If $\lambda\_i\ge3$ for some $i\ge2$, then let $c\in\{1,\dots,n\}$ be the largest such $i$, so $(\lambda\_1+1,\lambda\_2,\dots,\lambda\_{c-1},\lambda\_c-1,\lambda\_{c+1},\dots,\lambda\_n)\in S\_n$ and is obtained by a majority move from $\bf 1\_\bf n$. Thus, WLOG, $\lambda\_2=\dots=\lambda\_l=2$. 


Now $l>n-(k+l)$, so $l>n-k-l$ and hence $2l>n-k$.


But $\lambda\_1=n-2(l-1)-k\ge2$, so $2\le n-2l+2-k\implies 0\le n-2l-k\implies 2l\le n-k$, a contradiction.


