def check_eq(a, b, c, d, num, den):
if a % 10 == 0 and b % 10 == 0: return num, den
if a / gcd(a, b) == c / gcd(c, d) and b / gcd(a, b) == d / gcd(c, d):
num *= a
den *= b
return num, den
Given coprime positive integers m, n < m, with m - n odd, a pythagorean
triple is formed with a = m*m - n*n, b = 2*m*n, c = m*m + n*n
All primitive pythagorean triples are formed with this generator, so multiplying
each one by k in (1, x) while a+b+c < 1000 forms all pythagorean triples where
p is less than 1000.
I generate every pythagorean triple, and store the number of times each perimeter
is generated.
Runs in ~ .1 seconds
"""
from utils.misc import gcd
perimeters = dict([(x, 0) for x in xrange(12, 1001)])
for m in xrange(2, 21):
p = 12
n = m%2 + 1
while n < m and p <= 1000:
if gcd(m, n) > 1:
n += 2
continue
p = 2*m*m + 2*m*n
k = 1
while k*p <= 1000:
perimeters[k*p] += 1
k += 1
n += 2
print max(perimeters, key=lambda k: perimeters[k])
"""
from utils.misc import used_digits, gcd
from utils.primes import factor
num = 1
den = 1
def check_eq(a, b, c, d, num, den):
if a % 10 == 0 and b % 10 == 0: return num, den
if a / gcd(a, b) == c / gcd(c, d) and b / gcd(a, b) == d / gcd(c, d):
num *= a
den *= b
return num, den
for i in xrange(10, 100):
for j in xrange(i + 1, 100):
if len(used_digits(i) & used_digits(j)) > 0:
if i / 10 == j / 10:
num, den = check_eq(i, j, i % 10, j % 10, num, den)
elif i / 10 == j % 10:
num, den = check_eq(i, j, i % 10, j / 10, num, den)
elif i % 10 == j / 10:
num, den = check_eq(i, j, i / 10, j % 10, num, den)
else:
num, den = check_eq(i, j, i / 10, j / 10, num, den)
print den / gcd(num, den)
