# All the King's Horses and All the King's Men

What is the largest N by N square chessboard where it's possible to construct a position such that 2 knights and N pawns either occupy or threaten every square?

• pawn can threat both sides since no side part is given? ibb.co/gzYji9 these pawn threads everywhere? – Oray Aug 15 '18 at 20:06
• I assume pawns can only move a single square, right? Also, as Oray said, is there a specific orientation of the chess board? – EKons Aug 15 '18 at 20:18
• @ΈρικΚωνσταντόπουλος and @ Oray, a pawn threatens two squares diagonally in the direction of its movement. I believe the pawn's double move rules are irrelevant here, since a pawn never threatens any squares it can move into. – Bass Aug 15 '18 at 21:01
• @Bass But a pawn doesn't threaten all four squares diagonally adjacent to it if there's an orientation. – EKons Aug 15 '18 at 21:02
• I love the title! :D $\color{darkorange}{\bigstar}$ – Mr Pie Aug 16 '18 at 1:22

5x5 solution which seems to be optimal.

Here is

It is impossible to have

7x7 squares whatsoever, because a knight can thread at most 8 squares and with himself, it will be 9 squares thread+himself, and as totaly it is 18 squares. Pawn can thread 2 squares and with themselves, it will be 3 squares. and there are 49 squares for 7x7.

So with simple calculation we can conclude that

$18+7\times3=39$ squares is maximum number of thread + pieces and it is smaller than $49$. So even though theoretically 6x6 is possible, there is no such case exists.

Here is why

6x6 is impossible too.

With the previous calculation, we can find that

with 2 knights and 6 pawns, the maximum occupation and thread will be $36$, which is the same as the available empty squares. So there should not be any conflict square whatsoever.

so

there are only a few squares where you can put two knights to thread and occupy 18 squares and only combination where they thread distinct squares is only possible when they are positioned in the middle and next to each other. Other combination would not cover 18 squares (except symmetries).

like below:

As you see in the image above,

You cannot put your pawns where they occupy and thread 3 squares wherever you put them. Actually u need much more than 6 pawns.

• It looks like a pawn only threatens 2 squares. – EKons Aug 15 '18 at 21:04
• @ΈρικΚωνσταντόπουλος damn, need to edit the answer then :) – Oray Aug 15 '18 at 21:10
• Starting to look really good! A pawn cannot exist on the final rank, where it would be promoted, but that shouldn't be too hard to fix. More difficult thing might be the bit where you should calculate 6*3, not 6*2. – Bass Aug 15 '18 at 21:37
• @Bass fixed that part too. – Oray Aug 15 '18 at 21:39
• I edited my previous comment just as you replied to it: there's a small but crucial mistake in the math – Bass Aug 15 '18 at 21:45