# Sushi sharing problem

The problem was first brought up by Sam McCauley when I was still in Stony Brook.

Two germophobic friends are sharing sushi. There are two kind of sushi, \(0\) and \(1\). The sushi are lied out on a \(n\times m\) matrix. There are \(a\) sushi of type \(0\) and \(b\) sushi of type \(1\), both are even numbers and \(a+b = nm\). One can pick up a sushi with a chopstick by picking it up horizontally, or vertically. It will touch the adjacent sushis horizontally or vertically, respectively.

Each person want to eat exactly \(a/2\) sushi of type \(0\) and \(b/2\) sushi of type \(1\), and no one want to eat a sushi touched by someone else. Is this always possible? Yes it is!

First, it's easy to see each person can pick up a entire row or column, and collapse the remaining matrix. All the remaining sushi in the matrix are clean. We will assume each person pick up a entire row/column.

We prove this by induction. Let \(f(i)\) be the number of \(1\)'s in the \(i\)th column, and \(f(S)=\{f(i)|i\in S\}\). As long as both friends pick up the same number of each sushi before collapse the matrix, we can go to a smaller case.

Before the case by case analysis, we note of two conditions:

- Even sum condition
\(\sum_{i\in [1..m]} f(i)=b\), it must be even.

- Even side condition
Because \(nm\) is even, either \(n\) or \(m\) is even.

\(n>m\), rotate the matrix.

\(n+1<m\), by pigeonhole principle, \(f(i)=f(j)\) for some \(i\) and \(j\). One person pick \(i\)th column and the other pick the \(j\)th column.

- \(n+1\geq m\) and \(m\geq \frac{1 + \sqrt{1+16n}}{2}\). If there is no \(2\) columns with the same number of \(0\)'s and \(1\)'s, then let \(f([1..m])=S\) to be a set of \(m\) elements. Now, consider we pick 4 distinct elements \(a,b,c,d\) from \(S\) and if \(a+b=c+d\), then we can let one person pick columns \(f^{-1}(\{a,b\})\) and the other pick columns \(f^{-1}(\{c,d\})\). Note if \(a+b=c+d\), and we know that \(a,b\) and \(c,d\) are distinct, and \(\{a,b\}\neq \{c,d\}\), then \(a,b,c,d\) must be all distinct. \(a+b\) can take \(2n-1\) different values, between \(1\) and \(2n-1\). There are \({m \choose 2}\) pairwise sums. This shows there must be at least \(2\) pairs that both have the same difference by pigeonhole principle when \({m\choose 2}\geq 2n\), which is true when \(m\geq \frac{1 + \sqrt{5+16n}}{2}\). To translate this,
- If \(n+1=m\), then this is always true when \(n\geq 3\).
- If \(n=m\), then this is always true when \(n\geq 5\).

\(n\leq 2\) and \(n+1=m\). Since both \(0+1\) and \(0+1+2\) is odd, we must have two columns have the same value.

\(n=m=2\). If there is no \(2\) columns with the same number of \(0\)'s and \(1\)'s, then \(f([1..2])=\{0,2\}\) in order to satisfy the even condition. But each row would have the same number of different kind of sushi. Let each friend take a row.

\(n=m=3\). This is not possible with even side condition.

\(n=m=4\). This is the last case. If no \(2\) columns have same number of \(0\)'s and \(1\)'s, then \(f([1..4])=\{a_1,a_2,a_3,a_4\}\) where \(a_1<a_2<a_3<a_4\) and has to equal to one of the following due to the even sum condition. In all cases, \(a_1+a_4=a_2+a_3\).

- \(0,1,2,3\)
- \(0,1,3,4\)
- \(1,2,3,4\)