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count only really distinct cuboids, for example the cuboids (A,B,C) (3,5,6) and (6,5,3) are considered equal.
key insight for me was that there are only 1.5 million pythagorean triplets that correspond to a cuboid with max size 400.000
if you use Euclid's formula (see for example https://en.wikipedia.org/wiki/Pythagorean_triple ) to generate pythagorean triplets and you are only interested in right triangles with maximum right side length M, then you can use 1.1*sqrt(M) as an upper bound for m.
If anybody is interested, I can write down my approach to solve this problem.
Curious to know how it works to compute million pythagorean triplets within the time limit. I've been using Euclid's formula, but that m += 1, n += 2 is supposedly killing it. It takes almost 5 seconds to generate triplets up to (with most of the time spent on finding the valid primitive triplets), let alone .
Perhaps it is because of the upper bound you mentioned? I don't know where that 1.1*sqrt(M) comes from, the bound I used was a while-loop with condition m**2 - (m-1)**2 <= M. When I look closely, this condition is unnecessarily loose, as it does not take into account the fact that either a >= b or b >= a should be true, while and .
EDIT: I tried using this upper bound, and got WA for Test Cases #3 through #9.
EDIT 2: Tried with sqrt(3*M) and it works. Even sqrt(2*M) did not work, where it was missing some pairs like m=917, n=218. Maybe the difference is due to the way I handled the triples, as I append whenever a >= b_plus_c / 2 + b_plus_c % 2 or b_plus_c >= a / 2 + a % 2.
I'm quite sure that this condition would fail again for much larger input. The naming here is quite confusing, as b_plus_c >= a / 2 + a % 2 in fact means b >= a_plus_c / 2 + a_plus_c % 2, but I just handle them all in one place.
EDIT 3: By the way, I find it interesting that no one mentioned another key insight involved in solving this problem, that is to visualise the unfolding of the 3D box. At the beginning, I was trying to solve the equation and find the derivative of it to look for a minimum, thinking about how this problem is supposed to be 35% difficulty.
Yes, the upper bound is crucial in finding all the Pythagorean triplets within the time limit.
Note that the M in my upper bound definition is not the same as the M in the problem statement. Sorry for the confusion about that. This might explain the difference in your findings about the upper bound.
The other insight you mention is crucial as well. I didn't want to spoil too much :-)
For me I needed to use m up to 1000 in the pythagorean triples 2mn, m^2-n^2, m^2+n^2 to generate the sides large enough to satisfy all the test cases that's sides bounded by M up to 400,000. The bound 1.1sqrt(M) is not enough. Because we need to satisfy max(2mn, m^2-n^2) < 800000 and min(2mn,m^2-n^2)<4000000. This can happen beyond the bound 1.1sqrt(M).
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Project Euler #86: Cuboid route
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Solved this problem in F#. Insights:
If anybody is interested, I can write down my approach to solve this problem.
Curious to know how it works to compute million pythagorean triplets within the time limit. I've been using Euclid's formula, but that
m += 1,n += 2is supposedly killing it. It takes almost 5 seconds to generate triplets up to (with most of the time spent on finding the valid primitive triplets), let alone .Perhaps it is because of the upper bound you mentioned? I don't know where that
1.1*sqrt(M)comes from, the bound I used was a while-loop with conditionm**2 - (m-1)**2 <= M. When I look closely, this condition is unnecessarily loose, as it does not take into account the fact that eithera >= borb >= ashould be true, while and .EDIT: I tried using this upper bound, and got WA for Test Cases #3 through #9.
EDIT 2: Tried with
sqrt(3*M)and it works. Evensqrt(2*M)did not work, where it was missing some pairs likem=917, n=218. Maybe the difference is due to the way I handled the triples, as I append whenevera >= b_plus_c / 2 + b_plus_c % 2orb_plus_c >= a / 2 + a % 2.I'm quite sure that this condition would fail again for much larger input. The naming here is quite confusing, as
b_plus_c >= a / 2 + a % 2in fact meansb >= a_plus_c / 2 + a_plus_c % 2, but I just handle them all in one place.EDIT 3: By the way, I find it interesting that no one mentioned another key insight involved in solving this problem, that is to visualise the unfolding of the 3D box. At the beginning, I was trying to solve the equation and find the derivative of it to look for a minimum, thinking about how this problem is supposed to be 35% difficulty.
Yes, the upper bound is crucial in finding all the Pythagorean triplets within the time limit.
Note that the M in my upper bound definition is not the same as the M in the problem statement. Sorry for the confusion about that. This might explain the difference in your findings about the upper bound.
The other insight you mention is crucial as well. I didn't want to spoil too much :-)
I hope this clears some things up.
For me I needed to use m up to 1000 in the pythagorean triples 2mn, m^2-n^2, m^2+n^2 to generate the sides large enough to satisfy all the test cases that's sides bounded by M up to 400,000. The bound 1.1sqrt(M) is not enough. Because we need to satisfy max(2mn, m^2-n^2) < 800000 and min(2mn,m^2-n^2)<4000000. This can happen beyond the bound 1.1sqrt(M).