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How many diagonal rectangles are there in an 8x8 chessboard such

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File: valid.png (4KB, 292x295px) Image search: [Google]
valid.png
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How many diagonal rectangles are there in an 8x8 chessboard such that each square does not share a row or column with any other of the 3 squares?

The picture has some examples of valid rectangles.

I got 75 rectangles, but am curious what other people got and how you approached it. I will show my work if you would like.

Question modified by Elements of Discrete Mathematics, 3.35 - CL Liu. I purchased the book from a used book store and have been working through problems in my free time.
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>>8913281
Could you clarify what you mean by each square does not share a row or column with any other of the 3 squares?
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File: clarification.png (6KB, 295x295px) Image search: [Google]
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>>8913289
Sure, for example if you pick the filled in square in the picture, then the squares with lines in them cannot be picked to complete the diagonal rectangle
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>>8913299
Thanks. Can I pick say G8H7? Thats at the border, and the red lines that you drew for first picture would be out of bounds. Is that allowed?
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>>8913325
The red lines can go out of bounds, but you cannot pick a square out of bounds
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File: more clarification.png (7KB, 295x295px) Image search: [Google]
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>>8913332
For example, if you picked a corner square, you couldn't pick any of these squares
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>>8913332
Thanks. Let me try this.

Longest diagonal from A8 to H1 has 8 squares, which is itself a rectangle. Then there are two 7 squared rectangles, three 6 six squared, and so on. The total is 28 rectangles.

Moving on to the next diagonal B8 to H2, the longest rectangle is now only 7 squares long and we move the count "up" since there are no more eight squared rectangles. And so on we subtract by 7, yielding 21. For the next diagonal, we move it "up" again, and subtract by 6, yielding 15. And so on until the final diagonal G8H7. But B8 to H2 until H8 to H7 is mirrored on the other side of A8 to H1. So we get

28 + 2(21+15+10+6+3+1) = 140

Now we mirror the whole board in the other diagonal direction e.g. longest diagonal now A1 to H8. So I got 2x140 = 280.

t. non-math major
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>>8913340
Oh damn it. I didn't know if you picked one square the entire column and row is off limits for the rest of the choices. Ignore my post, which doesn't take that into account.>>8913359
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File: 149491094830383595611.jpg (2MB, 3264x2448px) Image search: [Google]
149491094830383595611.jpg
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>>8913281
I got 192. I may have made some mistakes though.

I basically brute-forced it, but I noticed symmetry between rectangles on dark squares and rectangles on the light squares, so I counted the light square rectangles and doubled that result.

To get a better look at it I rotated the board 45 degrees counter clockwise, and considered each 4-square as a square. So (after rotation and considering only the light colors) there is a stack of squares. See pic. Any rectangle must be made of these squares. Then I just counted the rectangles for each possible size.
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>>8913490
Just realized this is wrong. I counted illegal rectangles. After accounting for the rules, I got 136
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>>8913490
>>8913523

Did you figure out a general formula for an NxN board then?
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Thread posts: 12
Thread images: 5


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