Hi,
Are there 1120 different rectangles that can be situated in a 5x7 grid?
Thanks!
Problem 147
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See also the topics:
Don't post any spoilers
Comments, questions and clarifications about PE problems.
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axelbrz
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uws8505
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Re: Problem 147
I get 420 for a 5*7 grid, but I'm not so sure about my answer.
Math and Programming are complements
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axelbrz
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Re: Problem 147
Oh, I mean vertical + horizontal + cross-hatched rectangles, but I've verified it using brute force.
Thanks for the time!
Thanks for the time!
Last edited by axelbrz on Mon Nov 10, 2008 8:14 am, edited 1 time in total.
"think(O(n))+O(n) sometimes is better than think(O(1))+O(1)"
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uws8505
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Re: Problem 147
I'm trying to solve it but I haven't found any algorithm to find the number of cross hatched ones yet 
Math and Programming are complements
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axelbrz
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Re: Problem 147
Lol, try to see the cross-hatched grid as a normal matrix 
Good luck!
Good luck!
"think(O(n))+O(n) sometimes is better than think(O(1))+O(1)"
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rayfil
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Re: Problem 147
The question is a little ambiguous. Do you mean only in a strictly 5x7 grid or in that size grid and ALL smaller ones as required in the problem???axelbrz wrote:Are there 1120 different rectangles that can be situated in a 5x7 grid?
When you assume something, you risk being wrong half the time.
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axelbrz
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Re: Problem 147
No, no, I was referring in a strictly 5x7 grid, but I've resolved the problem, I had a trivial mistake in the program when I asked that.
So, thanks too!
So, thanks too!
"think(O(n))+O(n) sometimes is better than think(O(1))+O(1)"
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lg5293
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Problem 147
Hi,
Problem 147 (View Problem)
Can anyone confirm if these rectangles are correct. This is assuming just strictly that rectangle, so 3x2 is just 37.
3x3 = 87
3x4 = 56
3x5 = 209
4x4 = 264
4x5 = 395
2x12 = 343
2x13 = 392
1x20 = 229
10x9 = 7660
10x10 = 9502
Thanks for your help.
Problem 147 (View Problem)
Can anyone confirm if these rectangles are correct. This is assuming just strictly that rectangle, so 3x2 is just 37.
3x3 = 87
3x4 = 56
3x5 = 209
4x4 = 264
4x5 = 395
2x12 = 343
2x13 = 392
1x20 = 229
10x9 = 7660
10x10 = 9502
Thanks for your help.
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jaap
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Re: Problem 147
3x3 is correct, the rest is not.
Here something is obviously wrong, as the latter contains at least all the rectangles of the former.lg5293 wrote:3x3 = 87
3x4 = 56
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Schu-ism
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Re: Problem 147
I hope it's not too much to emphasize that the result for a 5 by 7 grid is NOT 1120.
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rustleg
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Re: Problem 147
I have difficulty understanding the conditions in this question.
First when considering the second set of 35 rectangles contained within the horizontal-vertical rectangles, surely you are double counting, not only the smaller grids themselves, but also their contents, so they aren't different.
Also for example the 2x2 grid contains 4 1x1 horizontal-vertical rectangles as well as 2x1's and 1x2's which themselves can be re-cross-hatched and double counted as above to contain yet more rectangles. Where do you stop? Does the question imply a limit to the reduction of contained horizontal-vertical rectangles?
First when considering the second set of 35 rectangles contained within the horizontal-vertical rectangles, surely you are double counting, not only the smaller grids themselves, but also their contents, so they aren't different.
Also for example the 2x2 grid contains 4 1x1 horizontal-vertical rectangles as well as 2x1's and 1x2's which themselves can be re-cross-hatched and double counted as above to contain yet more rectangles. Where do you stop? Does the question imply a limit to the reduction of contained horizontal-vertical rectangles?
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jaap
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Re: Problem 147
They are not sub-grids of each other. The question just asks you to examine one grid of each possible size up to some maximum size limit, and add their results together.
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rustleg
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Re: Problem 147
Not sub-grids, ok. I now realise how I misread the question. Thanks for the clarification.