Slowly compute pythagorean triples with pyDatalog

pythagorean_triples.py

from pyDatalog import pyDatalog as pyd
import sys
pyd.create_terms('L1, L2, I, triples')

def triples(maxleg):
    return (L1.in_(range(1,maxleg)) &
            L2.in_(range(1,maxleg)) &
            I.in_(range(1, 2 * maxleg)) &
            (I ** 2 == (L1 ** 2 + L2 ** 2))
            )
    # 2*maxleg is a nice integer upperbound for sqrt(2 maxleg^2)
    # also valid because of the triangular inequality

s = triples(int(sys.argv[1]))
print('\n'.join([str(x) for x in sorted(s) if x[0] < x[1]]))