TITLE: Labeling a Square Array
QUESTION [7 upvotes]: Suppose that the $n^2$ cells of an $n\times n$ array are labeled with the integers $1, \dots, n^2$. Under the traditional left-to-right and top-to-bottom labeling, the labels of horizontally adjacent cells differ by $1$, and the labels of vertically adjacent cells differ by $n$.  Is it possible to relabel the array so that the labels of adjacent cells (horizontally or vertically) differ by less than $n$?
I suspect the answer is "no" but do not have a proof.  I made up this problem while contemplating a similar Putnam Competition problem (1981, A-2).  In this problem, adjacent cells are horizontal, vertical, or diagonal neighbors.

REPLY [10 votes]: Suppose such a labeling exists. Let $F_k$ the region formed by the cells labeled $\{1,2,\dots,k\}$. There is a minimum $k>1$ such that $F_k$ connects two opposite sides of the checkerboard (wlog suppose left and right side). Color in black the cells belonging to $F_{k-1}$, and in white the rest.
Then, $k$ is in a white cell touching a black cell, and in every other column there is at least one black and one white cell. So overall there are at least $n$ white cells touching black cells (one per column). One of them must contain a number $\geq k-1+n$, and (since it touches the black region) it touches a number $\leq k-1$. Contradiction!