If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.
{20,48,52}, {24,45,51}, {30,40,50}
For which value of p < 1000, is the number of solutions maximised?

WARNING: CONTAINS MATHEMATICS

You may remember from school



(Difference of 2 squares)
let

and

so


and substituting these into
gives

Set and to get rid of that root and you have

Primitive triples are ones in lowest terms (if you know (3,4,5) you can get (6,8,10) etc by multiplying each side by some integer)
If p and q are coprime and different parity you just get primitive ones (i did not bother and just used set() to remove dupes from the list)

Here is my code

from __future__ import division
from collections import defaultdict
from time import time
     start = time()
counter = defaultdict(list)
    def triples():
    for q in range(1, 35):
        #35 is greater than sqrt(1000)
        for p in range(q+1, 35):
            a = p**2 - q**2
            b = 2 * p * q
            c = p**2 + q**2
            for m in range(1,int(1000/c)):
                #generate multiples till c is above 1000
                #triples with perimeter > 1000 are filtered later
                yield (m*min(a,b),m*max(a,b), m*c)
    for triple in triples():
        perimeter = sum(triple)
        if perimeter < 1001:
        counter[perimeter] += [triple]
<p>print set((counter[120]))
<p>maxlen = 1<br />
for i in counter:
    length = len(set(counter[i]))<br />
    if length > maxlen:<br />
        maxlen = length<br />
        print i<br />
end = time()</p>
<p>print "took ", end-start</p>
<p>

Which outputs:
set([(24, 45, 51), (20, 48, 52), (30, 40, 50)])
520
528
630
660
720
840
took 0.0140960216522

I tried brute forcing it today and knocked up the following

<br />
from collections import defaultdict<br />
counter = defaultdict(int)<br />
from time import time</p>
<p>def run():<br />
	start = time()<br />
	for perimeter in range(1,1001):<br />
		#print perimeter<br />
		for a in range(1,perimeter):<br />
			for b in range(1,perimeter-a):<br />
				c = perimeter - a - b<br />
				if a**2 + b**2 == c**2:<br />
					counter[a+b+c] += 1<br />
	mosttriples = 0<br />
	for i in counter:<br />
		if counter[i] > mosttriples:<br />
			mosttriples = counter[i]<br />
			print i<br />
	end = time()<br />
	print "TOOK", end-start</p>
<p>print "without psyco"<br />
run()</p>
<p>print "with"<br />
import psyco<br />
psyco.full()<br />
run()<br />

Which gives

without psyco
516
520
528
630
660
720
840
TOOK 134.663383007
with
516
520
528
630
660
720
840
TOOK 21.0988409519

So psyco makes it 6 times faster but a smarter algorithm is 2000 times faster still.

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