Chess Knight Distance

Question

On an $a×a$ chess-board knight takes $n$ jumps from the bottom left to the bottom right and $m$ jumps from the bottom left to the upper right.

For $a=7$ are 4 steps necessary.

Is $a=7$ the only number such that $n=m=4$?

Answer 1

Let the bottom left corner be $(1,1)$, so the upper right corner is $(a,a)$.

Each move of the knight changes the parity of exactly one coordinate. Hence, we always need an even number of moves to reach $(k,k)$, but when $k$ is even we need an odd number of moves to reach $(k,1)$.

It follows that when $a$ is even it is never possible to have $m=n$.

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When $a$ is odd, the total movement to reach $(a,1)$ is $(a-1,0)$. It can be reached with a combination of optimal moves $(4,0) = (2,1) + (2,-1)$, followed possibly by a move $(2,0) = (1,2)+(1,-2)$.

Hence, for $a - 1 = 4m$ we require $2m$ moves and for $a - 1 = 4m + 2$ we require $2m + 2$ moves. We can combine these as $n = 2\left\lceil (a-1)/4\right\rceil$ moves.

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On the other hand, each knight move increases the sum of coordinates by at most $3$, so we would require that $m$ be at least

$$\left\lceil \frac{2(a-1)}3\right\rceil.$$

It's easy to see hence that for $a$ large enough (say, $a\geqslant 13$) we have $m>n$.

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You can verify the cases of odd $a<13$ with $2\left\lceil (a-1)/4\right\rceil \geqslant \left\lceil 2(a-1)/3\right\rceil$ by hand (or computer?). There aren't many: only cases you'd need to check are $a = 3$ and $a=7$.

For $a=3$, we have $n=2$ and $m=4$. For $a=7$ we have $n=m=4$.

This shows more than what was asked: not only is $a=7$ the only board size with $m=n=4$, it is the only board size with $m=n$.