Proposed by Toby Walsh

The problem is to put $n$ balls labelled ${1,…,n}$ into 3 boxes so that for any triple of balls $(x,y,z)$ with $x+y=z$, not all are in the same box. This has a solution iff $n < 14$. The problem can be formulated as an 0-1 problem using the variables, $M_{ij}$ for $i \in {1,…,n}, j \in {1,2,3}$ with $M_{ij}$ true iff ball $i$ is in box $j$. The constraints are that a ball must be in exactly one box, $M_{i1} + M_{i2} + M_{i3} = 1$ for all $i \in {1,…,n}$. And for each $x+y=z$ and $j \in {1,2,3}$, not $(M_{xj} \wedge M_{yj} \wedge M_{zj}$). This converts to, $(1-M_{xj}) + (1-M_{yj}) + (1-M_{zj}) \geq 1$ or, $M_{xj} + M_{yj} + M_{zj} \leq 2$.

One natural generalization is to consider partitioning into $k$ boxes (for $k>3$).

Ramsey numbers are closely related, and are described in .