How to place 12 knights on a chess board so that all squares are either occupied or threatened?

Question

I was in my middle school chess club and the chess coach brought in a game and one of the problems I was stuck at was how to place the least amount of knights on a board so that all of the squares are either occupied or threatened. The hints were that it has rotational symmetry but not line symmetry and that it takes 12 knights. Could anyone please post their thoughts about this whether it is right or wrong to help me get an idea so I can try myself.

Answer 1

Your coach is too hard on you. I was unable to figure it out and I have an approximate rating of 2298 ELO. I came close on the first three solutions I thought of, but there was always one or two squares I couldn't cover.

Answer 2

Certainly this stipulation can be achieved, and in the following explanation you can see how it can logically be solved. If you only want a clue, then try first considering what to do about the corner squares, and read no further!

The knight is progressively negatively affected in scope by placing it further away from the central 16 squares of the board. Ultimately in the corner it attacks only 2 squares, whereas in the centre it attacks 8 (the rook is unaffected by placemant on an empty board, and always attacks 14!). The value of this problem is largely to do with realising this about the power of the knight.

So, do we place the first N in the corner, with a "value" of 3 (1 occupied+2 threatened), or, say, on c7, with a "value" of 7 (1 occupied+6 threatened)? Obviously c7 is the most sensible try, and using the clue about rotational symmetry, also N at g6, f2 and b3.

Next consider the 2 squares adjacent to each corner. There is only 1 position that a single N can threaten both of these, and as this try of most economy, place a N at c6, f6, f3 and c3.

The easiest way to see what remains to be done is to place, say, a pawn on each square yet to be occupied or attacked. Solving from this stage can be done by placing the remaining 4 N's at d6, f5, e3 and c4.

This kind of problem is excellent practice. Try placing 8 queens on the board, such that none attacks any other. This also can be solved, and is not just a frustrating "joke".