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Topic: When triangles are squares (Read 2028 times) |
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towr
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When triangles are squares
« on: Oct 20th, 2002, 1:17pm » |
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problem: Make an algorithm that runs in linear time which can find all pairs x,y (both integer > 0) so that x(x+1)/2 == y^2
It's a problem arising from looking at triangles and squares (made from circles) and trying to find when they're equal in surfacearea (measured in circles)
It's an easy problem.. Though finding a linear solution makes it sufficiently harder to make it interesting =)
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| « Last Edit: Nov 7th, 2002, 10:29am by towr » |
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Pietro K.C.
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Re: NEW PROBLEM: When triangles are squares
« Reply #1 on: Oct 20th, 2002, 2:51pm » |
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I don't get it... linear in terms of what? What is the input parameter? If there are an infinite number of such pairs, an algorithm to find ALL of them would run forever; if there is only a finite number, the algorithm would take constant time! 
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| « Last Edit: Oct 20th, 2002, 4:44pm by Pietro K.C. » |
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TimMann
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Re: NEW PROBLEM: When triangles are squares
« Reply #2 on: Oct 20th, 2002, 7:30pm » |
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I would think it's supposed to be either linear in the size of x or linear in the number of pairs found. That is, either finding all pairs with x < N should be O(N), or finding the first n pairs should be O(n). Which is it, towr?
p.s. I have an algorithm that's O(N1/2). That's either better than requested or not good enough...
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| « Last Edit: Oct 20th, 2002, 7:36pm by TimMann » |
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towr
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Re: NEW PROBLEM: When triangles are squares
« Reply #3 on: Oct 21st, 2002, 1:20am » |
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It's linear in terms of number of answers..
So 1000 pairs takes about 10 times as long as 100 pairs.. (actually if you find x then y is trivial to find, so you don't even need to find a whole pair)
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S. Owen
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Re: NEW PROBLEM: When triangles are squares
« Reply #4 on: Oct 21st, 2002, 2:19pm » |
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I have a result and algorithm that I spotted purely by inspection and luck, though I don't know why it works. Anyone care to tell me why?
The first two pairs are of course (1,1) and (8,6).
You can compute y in the nth pair (call it yn) from the previous two y's. yn = (yn-12-1) / yn-2. I don't see why just yet, but it appears to hold:
(1,1)
(8,6)
(49,35)
(288,204)
(1681,1189)
(9800,6930)
...
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TimMann
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Re: NEW PROBLEM: When triangles are squares
« Reply #5 on: Oct 22nd, 2002, 1:30am » |
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Nice, that's a lot farther than I got.
I managed to do enough analysis to know that either x or x+1 must be an odd square, which forms the basis for an O(N1/2) algorithm. That's fast enough to quickly compute some example values, but I didn't spot the pattern SOwen did, and I don't yet see why the pattern holds either.
I read something about Diophantine equations of degree greater than 1 recently, but I can't remember where...
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towr
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Re: NEW PROBLEM: When triangles are squares
« Reply #6 on: Oct 22nd, 2002, 6:21am » |
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S. Owen did indeed find my answer.. [edit](well almost anyway, I just use yn-1^2/yn-2 and an extra constant. It also works for x but with another constant, but same factor. dunno which is better)[/edit] And I also don't know why it works.. But, it does seem to work in some variant for all ax^2+bx+c = dy^2 + ey +d
I found the square-triangle solution half a dozen years ago or so.. And more recently I was working with a hexagonal field and found a similar relation (one solution is a side 8 hexagon and a side 13 square and of course 1,1).. I have no idea how to prove it is mathematicly sound though..
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| « Last Edit: Oct 22nd, 2002, 6:33am by towr » |
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