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In From knights to kings and From Knights to Kings on a rectangle we tried to help knights be next to their friends after they were promoted to kings. What happens if we have kings who are deposed and become knights?

We start off with an $N\times M$ board filled with kings. A king is friends with all the kings on squares that he could move to, i.e all adjacent kings. After they are deposed, they want to have all their friends still be in squares that they could get to in a single move. So two kings A and B:

A B .
. . .

would want to be like this after becoming knights:

A . .
. . B

Unfortunately A's two other friends (originally in the two spaces below him) cannot be put anywhere that allows A to still reach them. So a $3\times 2$ board doesn't work.

A $1\times 1$ board is trivially solved - no friends either before or after the change.

Any $n\times 1$ board (where $n>1$) is obviously impossible - each king has one friend if he is at the end, or two friends if he isn't. After they become knights, none of them can move so none of them can be a knight's move away from their friends.

Are there any board sizes that allow the kings to be a knight's move away from all of their friends?

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  • $\begingroup$ How many kings? Any limit on number of moves? $\endgroup$ – Josh Caswell Feb 11 '15 at 20:53
  • $\begingroup$ @JoshCaswell full board, they want to be a single move away from their friends. I was thinking about asking a version of this or the knights to kings problem where the board wasn't full, but I figured I'd ask one at a time. $\endgroup$ – Rob Watts Feb 11 '15 at 21:42
  • $\begingroup$ Oh, but they're "magically transported" to their new locations? I thought they needed to move as a knight to get there, so that your 2x3 board was unsolvable. $\endgroup$ – Josh Caswell Feb 11 '15 at 21:46
  • $\begingroup$ @JoshCaswell you pick up all the pieces and put them in their new places. You just need to figure out where they need to go to let them stay in touch with their friends. $\endgroup$ – Rob Watts Feb 11 '15 at 21:55
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The answer for a full n x m board of kings is:

No. The reason is that you can't position three knights such that each is friends with the other two. You've already eliminated n x 1. If we have n x 2, then each King has three friends who are all friends with each other, and we can't even handle two of them.

If we're allowed less than a full board:

then it becomes a friendly-king-packing problem. The limit there is at least 2/3 (for some board sizes). For example:
123 _ _ _ 153
xxx _ > _ xxx
456 _ _ _ 426
Here the numbers are kings, and this is two 3x3 boards. But maybe I'm getting ahead of things...

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