# PE 75 primitive Pythagorean triples

#### Arman777

Gold Member
Homework Statement
It turns out that 12 cm is the smallest length of wire that can be bent to form an integer sided right angle triangle in exactly one way, but there are many more examples.

12 cm: (3,4,5)
24 cm: (6,8,10)
30 cm: (5,12,13)
36 cm: (9,12,15)
40 cm: (8,15,17)
48 cm: (12,16,20)

In contrast, some lengths of wire, like 20 cm, cannot be bent to form an integer sided right angle triangle, and other lengths allow more than one solution to be found; for example, using 120 cm it is possible to form exactly three different integer sided right angle triangles.

120 cm: (30,40,50), (20,48,52), (24,45,51)

Given that L is the length of the wire, for how many values of L ≤ 1,500,000 can exactly one integer sided right angle triangle be formed?
Homework Equations
none
The question simply asks primitive pythagorean triples $(a,b,c)$ such that $S = a + b + c <15\times 10^{5}$
Python:
import time
import math

start = time.perf_counter()

pythagorean_triples = {(3, 4, 5) : 12}

for m in range(0, 10**3, 2):
for n in range(1, m, 2):
if math.gcd(m, n) == 1:
pythagorean_triples.update( {(m**2 - n**2, 2*m*n, m**2 + n**2): 2*(m*(m+n))} )

for m in range(1, 10**3, 2):
for n in range(0, m, 2):
if math.gcd(m, n) == 1:
pythagorean_triples.update( {(m**2 - n**2, 2*m*n, m**2 + n**2): 2*(m*(m+n))} )

pythagorean_triples_copy = pythagorean_triples.copy()
for key, value in pythagorean_triples.items():
if value >= 15 * 10 ** 5:
del pythagorean_triples_copy[key]

print(len(pythagorean_triples_copy))

end = time.perf_counter()

print(end - start, "sec")
According to the rule $gcd(m,n) = 1$ and if $m$ is even $n$ must be odd or vice verse to create primitive pythagorean triples. So here is my code however it seems something is wrong.

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#### Vanadium 50

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The question simply asks primitive pythagorean triples (a,b,c)(a,b,c) such that S=a+b+c<15×105
Does it? (3,4,5) and (5,12,13)

• Arman777

#### Arman777

Gold Member
oh yes sorry. I understand it now. Let me try like that.

#### Vanadium 50

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Why do you need primitive Pythagorean triples? As you wrote, 6-8-10 is a perfectly good solution.
Why is looping over m and n better than looping over a and b?
Why are you storing triples and not lengths?

#### jim mcnamara

Mentor
Hmm. Problem 75 looks a lot like your question: https://projecteuler.net/problem=75 ..... remarkably like your question.

#### Arman777

Gold Member
Why do you need primitive Pythagorean triples? As you wrote, 6-8-10 is a perfectly good solution.
Why is looping over m and n better than looping over a and b?
Why are you storing triples and not lengths?
I solved a similar problem (just the topic) like this on another site and I did not much pay attention to the PE part of the question. As you said ints unnecessery to write triples. I ll try to solve it and reply back as soon as I can.
Yes as you said 6-8-10 are also good solutions.
Well this method driectly gives us primatives and by multiplying each term with constant value we can get other solutions. Maybe we can do some sort of a sieve method.

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#### Vanadium 50

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Why is looping over m and n better than looping over a and b?

You seem to want to rush into writing code without thinking about what you want to write. This is likely to be as successful as building a house by cutting wood before drawing up plans.

• jim mcnamara

#### Vanadium 50

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For example, here is where I might start:

Code:
create array of ints from 12 to 1500000 called tally initialized to zero
loop over a from 3 to 459340 // largest a can be
loop over b from a+1 to 459340
c = sqrt(a^2 + b^2)
if c is an integer increment tally[a+b+c]
end loop over a and b

int n = 0
loop over i from 12 to 1500000
if tally[i] is 1, increment n
end loop over i

print n
The problem is that there are 100 billion square roots here. That may cause things to be slower than we would like. (PS there is also a bug - can you find it?)

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#### Arman777

Gold Member
Why is looping over m and n better than looping over a and b?
Well, initially I thought that we are just searching for primitive triples. In that case, using m and n is much better than the a and b ( I know that because I tried )

(PS there is also a bug - can you find it?)
probably here
loop over i from 12 to 1500000
if tally is 1, increment n

I am not sure I understand your code but you are referring tallyto tally and that's the bug? and why that should be equal to 1?

#### Arman777

Gold Member
The problem is that there are 100 billion square roots here. That may cause things to be slower than we would like.
To increase the speed of the code. As you said this approach cannot be good.

I thought some approaches and I would like to share them.

I thought 3 ways.

First, we can find all the primitive triples which its really easy by using $m$ and $n$ ( takes 1.6 seconds in python). And we know that any $S$, that is multiple of 2 primitive triples, will not satisfy the condition.

For instance, $(3,4,5) : 12$ is a primitive triple and $(5,12,13) : 30$ another one. So any multiple of 60 (since 60 is the first number which divisible by 12 and 30) should be excluded. So the problem reduces to catch all these multiple numbers are eliminate them.

Second, every triple can be found by an even $m$ and odd $n$ (or vice versa) and if $gcd(m, n) = 1$ then every, $(m^2 - n^2 , 2mn, m^2 + n^2)$ creates a primitive triple.

So

$$S = m^2 - n^2 + 2mn + m^2 + n^2 = 2m(m+n)$$

This implies 2 things. First every $S$ (primitive or not) should be divisible by $2$. Second, every $S$ should be divisible by $m$ and $(m+n)$. we also know that $m$ must be larger than $n$. By using this trick we might get somewhere. ( low probability? )

And also all non primitive triples are can be found like $S = C\times 2m(m+n)$ where C is some constant.

The third way, We can factor the $S$. If the $S$ contains at least two of the primitive triple $S$ ( For instance 60 contains 12 and 30 in its factors so it should be excluded). Or 120 contains ( 12, 30 and 40)

#### Vanadium 50

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In that case, using m and n is much better than the a and b ( I know that because I tried )
It is true that it is better, but why? This is what I mean by thinking and planning before jumping into the writing. If nothing else, it's inefficient to code things up N different ways and then picking.

The issue with using m and n instead of a, b, and c is you need three things.
1. It needs to be necessary. It produces only triples, with no non-triples.
2. It needs to be sufficient. It produces all the triples, with none left behind.
3. It needs to be unique. It produces each triple exactly once.
You need to understand why these are true - or more precisely, what you need to do in your code to ensure these are true.

I thought some approaches and I would like to share them.
Remember "profile your code"? I bet you are spending almost all of your time in math.gcd. (My code spends most of its time checking for coprimality.) So unless you can drastically speed up math.gcd or substantially reduce the number of times you call it, nothing will help much.

For what it's worth, my code runs at 10 ms. (The granularity of my timer) I didn't code up the algorithm I posted, but suspect it will take around an hour.

#### Arman777

Gold Member
It is true that it is better, but why? This is what I mean by thinking and planning before jumping into the writing. If nothing else, it's inefficient to code things up N different ways and then picking.
I did not code actually I am still in the thinking process. The code that I shared was code for another problem. I just modified a bit. But its totally useless for this problem.

Remember "profile your code"?
I am profiling when I need. For instance,

bet you are spending almost all of your time in math.gcd
In the original post if I loop over $a$ and $b$, float(c).is_integer() method takes much more time.

So it's better to use math.gcd(), $m's$ and $n's$ to save time. However, it finds only the primitive triples and we don't need them for this problem so its most likely useless.

As you said I ll loop over $a$ and $b$ and try to find a way an algorithm and then code.

In these days I am trying to code in C. I 'll take algorithm and data structure course next semester (which the course instructor will use C and I need to learn C). Since I have no idea about those things it might help me to come up with more clever solutions in general.

In the meantime I ll try to think about the problem and maybe I write my code in C instead of python :)

#### Vanadium 50

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The code that I shared was code for another problem. I just modified a bit. But its totally useless for this problem.
Then how can profiling it help you?

I'm not trying to give you a hard time. I am trying to get you to think logically and methodically.

it finds only the primitive triples and we don't need them for this problem so its most likely useless.
It's not true that it only finds primitive triples. It can be made to do so, but so can other generators. It's also true that primitive triples are not the answer, but does that mean it can't be part of an answer?

Make a plan. You need to make some decisions, so list the pros and cons of these decisions. Only then start writing code.

#### Arman777

Gold Member
Then how can profiling it help you?
Well it doesnt particularly for this problem however it shows that if I try to loop over a and b and try do float(c).is_integer() method its not going to work well. As you mentioned it takes time since there are lots of numbers. I can loop over a and b maybe or not ( I am not sure ) but in any case I need something clever.
but does that mean it can't be part of an answer?
It can be I think yes.
Make a plan. You need to make some decisions, so list the pros and cons of these decisions. Only then start writing code.
Yes I ll try to make an algorithm and decide to do loop over which variable or how to shorten things.. and then try to work on it. Maybe I mix things up Idk.

I'm not trying to give you a hard time. I am trying to get you to think logically and methodically.
Yes I know. Thanks for helping me out.

#### Vanadium 50

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The code is so fast now I am having a hard time timing it. As I said, the granularity is 10 ms, so "10 ms" can mean anything between 5 and 15. By running it 100x, it looks like it was a hair under 15 and with my latest attempts to speed it up, it's a hair under 12. So about a 300,000x improvement over the original hour.

I still want you to code it up yourself, but once you have done it (and not before), there are three lines of my code I want you to think about. Why would I write this?

C++:
 return (a<b) ? coprime(b,a) : !(a%b) ? (b==1) : coprime(b, a%b);
Code:
for(n=1+(m%2); n < m; n+=2)
Code:
halflength=msquared+m*n;

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• Arman777

#### Arman777

Gold Member
I could not come up with a good algorithm. I tried to think about some approaches however they are too complex or useless. Meanwhile I am also trying to solve problems in CodeAbbey (which have great questions for proggraming).

Thats really fast, impressive.

I dont know what to do. I ll try to think again but I dont have much hope. Maybe you can give some hints ? Otherwise I can try own my own but it will take time.

#### Vanadium 50

Staff Emeritus
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I could not come up with a good algorithm.
Then start with a bad one.

#### Arman777

Gold Member
Then start with a bad one.
Okay then I ll write some code and share by tomorrow.

#### Vanadium 50

Staff Emeritus
Science Advisor
Education Advisor
Better still...write a design for some code.

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#### Arman777

Gold Member
I thought something like this

Code:
Create an array (primitive_triples) that contains the sum of the all primitive triple values under 1.5 million.
Take a number from the primitive_triple and multiply it by c (for c in range(2, 1.5 * 10**6/num)) and put these numbers in the array (num_array)
Take the factors of each element in the num_array and then check if two of the factors are in the primitive_triple array.
If it contains, remove all of its multiples from the num_array.
Take the lenght of the num_array
Repeat this process for every number in the primitive_triples.
Sum the length's of each num_array
Its terrible I guess. The problem is I cannot think any mathematical trick to simplify the process

#### Vanadium 50

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OK, now let's go through your algorithm by hand not for n = 1,500,000 but for n = 60.

1. Primitive_triples is 12, 30, 40, 56. How did you get that, you say? That's a sign that you need more detail in step one.

2. "Take a number from the primitive_triple..." okay, I'll pick 30, "and multiply it by c (for c in range 2..60)." I get 30, 60, 90, 120...up to 1800.

3. Take the factors of each element in num_array. {2,3,5,6,10,15,30}, {2,3,4,5,6,10,12,15,30,60}... and remove the ones where two factor are in the primitive triple array.

By this point, even doing this by hand should show you a couple of things.
1. You need an algorithm to generate Pythagorean triples.
2. You don't need to be multiplying by c up to 60.
3. It will take forever to get a list of factors of your elements, especially if that list is too long.
4. The thing you are working with, the list of factors of num_array, is not an array.
5. It's not yet obvious that this gets the right answer.
So, why don't you work out by hand the answer for n = 60. I believe the answer is 7. (60 itself is doubled) Then try to write down the algorithm you used.

#### Arman777

Gold Member
I guess I wrote my algorithm wrong. Let me explain it again by using n = 60.

primitive_triples_sum = [12, 30, 40 ,56]
create arrays for each number such that
num_array_1 = [12, 24,36,48]
num_array_2 = [30, 60]
num_array_3 = 
num_array_4 = 

So, why don't you work out by hand the answer for n = 60. I believe the answer is 7. (60 itself is doubled) Then try to write down the algorithm you used.
Actually it was better for trying by hand

Python:
import math

max_sum = 15 * 10**5
pythagorean_triples = []

for m in range(0, 10**3, 2):
for n in range(1, m, 2):
if math.gcd(m, n) == 1 and 2*(m*(m+n)) <= max_sum:
pythagorean_triples.append (2*(m*(m+n)))

for m in range(1, 10**3, 2):
for n in range(0, m, 2):
if math.gcd(m, n) == 1 and 2*(m*(m+n)) <= max_sum:
pythagorean_triples.append (2*(m*(m+n)))

pythagorean_triples.sort()
pythagorean_triples.remove(2) # removing 2 from the list

num_0 = pythagorean_triples[-1]
num_set_0 = {num_0*c for c in range(1, max_sum // num_0 + 1)}

for num in pythagorean_triples[:-1]:
num_set = {num*c for c in range(1, max_sum // num + 1)}
num_set_0 = (num_set_0 | num_set) - (num_set & num_set_0)

print(len(num_set_0))
Maybe you can check for small max_num and if the code is correct then we might try to improve the time..?

Creating arrays are bad .....

#### Vanadium 50

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Creating arrays are bad .....
Why? Or do you mean creating so many named arrays is bad?

Maybe you can check for small max_num a
Maybe you could?

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