# $6$ bishops and a knight on an infinite chessboard

Player $$A$$ places $$6$$ bishops wherever he/she wants on the chessboard with infinite number of rows and columns. Player $$B$$ places one knight wherever he/she wants.

Then $$A$$ makes a move, then $$B$$, and so on...

The goal of $$A$$ is to checkmate $$B$$, that is, to attack knight of $$B$$ with bishop in such a way that the knight of $$B$$ cannot move anymore because all the squares at which he can move are under attack by the bishops.

$$B$$ wins if he has a winning strategy, that is, if he moves in such a way that $$A$$ will never checkmate him and, of course, $$B$$ always plays the best possible strategy.

We tried in a chat-room to prove that $$6$$ bishops are always enough for $$A$$ to win, but, we still do not have a proof.

Is $$6$$ the minimum number of bishops needed to secure $$A$$ a winning strategy?

There are two versions:

1) A knight is allowed to capture bishops.

2) A knight is not allowed to capture bishops.

I would like to see the solution of at least one version.

• Do you know if 7+ bishops is enough? – TomGrubb Jun 24 at 22:23
• @TomGrubb There is in a chat-room a strategy for 8 bishops. – Grešnik Jun 24 at 23:05
• Very important question: if the knight can't move into a free square, is it forced to go into the attacked squared and captured? Or it's getting stalemated and w fail to checkmate it? If it's forced to get captured, then it's easy to show that we can kill it with 6 bishops, but if we can stalemate it, then my idea fails and the knight is getting stalemated. – Kusavil Jun 24 at 23:49
• @Kusavil There can be a stalemate, of course, the knight can, if in danger, try to construct a stalemate-strategy. So, stalemate is allowed. – Grešnik Jun 24 at 23:51
• chat.stackexchange.com/rooms/95333/… – Grešnik Jun 25 at 0:20

The strategy I will describe builds upon the answer here. The knight will be allowed to capture bishops, but we will make sure that he will never have the opportunity.

1. Place 3 black-square and 3 white-square bishops far enough apart so that the knight cannot immediately fork two of them.

2. Next, form a "diagonal wall" far enough away from the knight by putting three bishops next to each other on one row. Do the same thing with the other three bishops on the other side of the knight.

Note that the walls will be of different "parity", one consists of two black-square bishops and one white-square and the other with the colours switched.

In the answer linked above, one wall had four bishops. This enabled $$A$$ to steadily move this wall forward while keeping a wall of width three intact. As the knight cannot cross a wall of width three, $$B$$ eventually runs out of space.

The wall here cannot move on its own without creating gaps. To still ensure that the knight can never jump the wall, we will make use of the fact that the knight can only ever jump on a square of the dual colour of his current square. We describe the sequence for the wall of two white-square and one black-square bishop(s). In all the following pictures, white is to move. It is important that the knight is on a black square. The green arrows will represent the wall.

The exact position of the knight is only an example here. White first moves the middle bishop. EDIT: Moving the middle bishop might runs into problems. As ArsenBerk pointed out, one should move the bishop that is furthest back first (and only then the middle one. This is only problematic as long as the knight is on a black square in the diagonal directly infront of the wall. In that case, move the other wall.

Right now, there is a gap in the wall, but as the knight was on a black sqaure, he could not take advantage of this. Next white moves the back bishop to build the new face of the wall.

The same sequence just mirrored will enable the three bishops to get back to the starting position of three bishops in a row.

If the knight ever comes close to a group of bishops, we just move them far away along the wall without moving the wall.

1. With the sequence of moves described above, we force the knight into corridor of width two illustrated here:

The bishops are some distance away here and not in the picture. The highlighted squares are the remaining squares of the knight. He can only move up or down the corridor. With 7 bishops, $$A$$ would be able to made the knight by closing up the wall. With only 6, the bishops will have to get a little cosy with the knight now. We can assume that white is to play here.

1. Mate as follows:

We will drag one bishop down the wall to challenge the knight.

For the sake of illustration, we assume he moves down. (The board is shifted to the up and right now).

We will now block the road down for the knight, but to avoid stalemate, we leave one square for the knight to move to.

We immediately challenge the knight, leaving only one square open.

This is the final mating position. Observe that no bishop is ever under attack in this sequence and that (execept for the initial up or down choice) all moves are forced for black.

• In second picture, knight can move to h4 to get out of the wall (after white plays). If there are $2$ bishops in white squares and $1$ in black squares, B's best strategy is to prepare a jump to black diagonal that is not being attacked whenever A tries to move the wall. – ArsenBerk Jun 25 at 14:40