Related to open problem #1: Don't fight each other

Logic Level 2

What is the maximum number of bishops you can place on this 5 × 5 5 \times 5 chessboard such that none of them can attack each other in one move?

7 8 4 5 6 9

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1 solution

Munem Shahriar
Oct 12, 2017

8 \boxed{8} bishops is maximum.


We have shown that 8 can be archived. Now we are going to show why we can't place 9 or more bishops.

A bishop move like this:

Meaning that it can only attack in this way.

We will not place any bishops at a5 to e5 and a1 to e1. Here is a possibility:

Which is only 5 bishops.

Now, we will place bishops at center, up and down. Note that if we place bishops at center, we cannot place it at corner. Here is a possibility:

Which is only 7 bishops.

Now, we will place bishops at corner, up and down. Here is a possibility:

Which is only 6 bishops. Now, we have only one strong choice. Which is to place the bishops at only at upside and downside.

I agree the answer is 8, but as I see it, this solution is incomplete. It doesn't proof why it's impossible to place 9 or more bishops on the board. That part was the trickiest for me.

Stefan van der Waal - 3 years, 6 months ago

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Well, let me think about that.

Munem Shahriar - 3 years, 6 months ago

Is that okay?

Munem Shahriar - 3 years, 6 months ago

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Writing good proofs is very hard and there are many ways to do it.

I approached this starting from a basic fact about bishops: a bishop can only attack squares of the same color it's on. So the problem can get rephrased like this: maximum amount of bishops = maximum of black bishops + maximum amount of white bishops.

By just looking at the black squares, it quickly becomes clear it's impossible to place 5 or more black bishops, and when looking at the white squares, it quickly becomes clear it's impossible to place 5 or more white bishops (I know this step in my logic isn't very rigorous). So the maximum amount of bishops = 4 + 4 = 8.

Stefan van der Waal - 3 years, 6 months ago

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@Stefan van der Waal Ok. What is your opinion about my solution?

Munem Shahriar - 3 years, 6 months ago

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@Munem Shahriar I think your current solution looks fine.

Stefan van der Waal - 3 years, 6 months ago

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@Stefan van der Waal Thanks for sharing your thoughts.

Munem Shahriar - 3 years, 6 months ago

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