What is the best way to generate Pythagorean triples?

问题

I have tried with that simple code when you just check all the combinations for a and b and then check if square root of c is an integer, but that code is really slow, then I have tried with Euclid's formula

a = d*(n^2 - m^2)
b = 2*n*m*d
c = d*(n^2 + m^2)

and I have written a code where you first find n with

trunc(sqrt(max_value))
//this is in pascal

and then you check every combination of 0 < m < n but I get duplicate results, like if n is 7, m is 5 and d is 1, and n is 6, m is 1 and d is 2 . In both cases you get 24, 70 and 74. So what is a good fast way to calculate the number of Pythagorean triples, I can't seem to find a way, also if I add all results to an array, and then check the array for duplicates, it just takes too much time... If anyone can help me with the code it can be pascal, c or python, I can understand all...

回答1:

I was curious so I decided to try this. I found that this algorithm was pretty easy to implement in Python and works pretty fast:

import math                                                                 

def pythagorean_triples(n):                                                 
    a, b, c = 1, 3, 0                                                       
    while c < n:                                                            
        a_ = (a * b) + a                                                    
        c = math.sqrt(a_**2 + b**2)                                         
        if c == int(c):                                                     
            yield b, a_, int(c)                                             
        a += 1                                                              
        b += 2                                                              

if __name__ == '__main__':                                                  
    import sys                                                              
    for pt in pythagorean_triples(int(sys.argv[1])):                        
        print(pt)

Try it by copying that script into pythagorean_triples.py and running python3 pythagorean_triples.py n where n is the maximum c you want it to generate. (You can use later Python2 if you like as well.)

回答2:

The Wikipedia page on Pythagorean triples gives us a hint:

The triple generated by Euclid's formula is primitive if and only if m and n are coprime and m − n is odd. If both m and n are odd, then a, b, and c will be even, and so the triple will not be primitive; however, dividing a, b, and c by 2 will yield a primitive triple if m and n are coprime

If you restrict m and n to coprime numbers and force m - n to be odd you will uiniquely generate all the primitive pythagorean triples. From this point on, you should be able to multiply these unique triples by factors of d to uniquely generate all triples.

In your example, allowing n=7 and m=5 was the problem, because their difference was even and the triple they generated was not primitive (you could divide all sides by 2 to get a smaller triple)

回答3:

Here is my solution:

import math
def pythagoreanTriplet(n):
  for b in range(n):
    for a in range(1, b):
        c = math.sqrt( a * a + b * b)
        if c % 1 == 0:
            print (a, b, int(c))

pythagoreanTriplet(12)

来源：https://stackoverflow.com/questions/22821210/what-is-the-best-way-to-generate-pythagorean-triples

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python

math

pascal