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Logic Level 5

What is the maximum number of bishops that can be placed on a 8 × 8 8 \times 8 chessboard such that at most three bishops lie on any diagonal?

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The answer is 38.

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1 solution

If the chessboard is colored black and white as usual, then any diagonal is a solid color, so we may consider bishops on black and white squares separately. In one direction, the lengths of the black diagonals are 2 , 4 , 6 , 8 , 6 , 4 2, \, 4, \, 6, \, 8, \, 6, \, 4 and 2 2 . . Each of these can have at most three bishops, except the first and last which can have at most two, giving a total of at most 2 + 3 + 3 + 3 + 3 + 3 + 2 = 19 2+3+3+3+3+3+2 = 19 bishops on black squares. Likewise there can be at most 19 19 bishops on white squares for a total of at most 38 38 bishops.

I had to do it practically and my board was more or less same as yours in the picture, just the bishop at h8 in your board was placed at b2 and the bishop at a8 was placed at b7 in mine. By the way, your logic is great and a great problem and a great solution too. (+1)

Yatin Khanna - 4 years, 12 months ago

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Thanks! ¨ \ddot\smile

Samara Simha Reddy - 4 years, 12 months ago

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Also, congrats for reaching 200000 points !!

Yatin Khanna - 4 years, 12 months ago

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@Yatin Khanna Thanks Dude! ¨ \Large \ddot\smile

Samara Simha Reddy - 4 years, 12 months ago

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