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# Chesslike Problem

Long ago, I invented a problem which solution looks like this:

The problem was to place a complete set of white chess pieces on the chessboard such that the number of bonds between them is maximal. Here you see that the king protects 8 neighbouring pieces and  the pawn in the top line covers no piece. The number of all protections is 53 and it can’t be done any better. (You may notice a pawn in the first line and that bishops are of the same color, but that’s fine.)

If you now add 40 more queens and rearrange all those 56 pieces, then there will be at most 338 bonds in total. We call that number B – the maximal bonding number for the set F. For all the white pieces it’s 53, for 2 knights only it’s 2, and so on.

We have a set F with bonding number B1. And a proper subset F’ with the bonding number B2. B1 is equal to B2. Which set is that?

In other words, for which set you can remove one or more pieces and the maximal bonding number remains the same as it was initially.

There is a trivial solution of any single piece and the empty set, but as always, we want a nontrivial solution.

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## 10 thoughts on “Chesslike Problem”

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2. thyjoking says:

32 bishops + 1 pawn?

• thyjoking says:

Or just 33 bishops actually

• This is one solution, yes. Congratulations.

• thyjoking says:

Well, 37 solutions to be exact, being various mixtures of pawns and bishops. It’s a lot easier to think of this problem as “is there a set of pieces were adding a piece won’t create any new bonds.” rather than the other way round.

• thyjoking says:

Sorry to reply so much (it’s late), but I think these are all the possible solutions due to it being impossible to create a non-full board where you can place a piece that a piece already on the board would bond to, except in the case of bishops and pawns using up all the possible diagonals, forcing the next piece to be on a different colour square and therefore have no bonds to already placed pieces.

• thyjoking says:

Aw crap, actually there’s only 33 solutions to the original problem. All possible sets of 33-x bishops and x pawns

3. > All possible sets of 33-x bishops and x pawns

A pawn can’t strike back, a bishop can. So you can’t mix those arbitrarily.

• thyjoking says:

Well, for 5 or less pawns, the maximum bonds will stay the same (4 on the bottom row and one on the other colour). For amounts of pawns above that, the maximum bonds will decrease but still won’t change when a pawn is removed from the set.