Naively counting Pythagorean triples in Python and Julia

JuliaGlobals

total = 0
N = 300

start_time = time()

for a in 0:(N - 1)
	for b in 0:(N - 1)
		for c in 0:(N - 1)
			if a^2 + b^2 == c^2
				total = total + 1
			end
		end
	end
end

end_time = time()

end_time - start_time

# Repeat now that JIT has done its work

total = 0
N = 300

start_time = time()

for a in 0:(N - 1)
	for b in 0:(N - 1)
		for c in 0:(N - 1)
			if a^2 + b^2 == c^2
				total = total + 1
			end
		end
	end
end

end_time = time()

println(end_time - start_time)

JuliaLocals

function loop(N::Integer)
	total = 0

	start_time = time()

	for a in 0:(N - 1)
		for b in 0:(N - 1)
			for c in 0:(N - 1)
				if a^2 + b^2 == c^2
					total = total + 1
				end
			end
		end
	end

	end_time = time()

	return end_time - start_time
end

loop(300)

# Repeat now that JIT has done its work

println(loop(300))

Python

total = 0
N = 300

from time import time

start = time()

for a in range(N):
	for b in range(N):
		for c in range(N):
			if a**2 + b**2 == c**2:
				total = total + 1

end = time()

print(end - start)

Using cython via ipython notebook:
%load_ext cythonmagic

paste this into the next cell block:

%%cython
import numpy as np

cimport numpy as np
cimport cython

@cython.boundscheck(False)
@cython.wraparound(False)
cpdef calc(N=300):
    cdef:
        np.int total = 0
        Py_ssize_t i, j, k
    for i in xrange(N):
        for j in xrange(N):
            for k in xrange(N):
                if i**2 + j**2 == k**2:
                    total = total + 1
    return total

paste this into the next cell block:
%timeit -n3 calc()

3 loops, best of 3: 19.1 ms per loop

If N is defined as const in JuliaGlobals, the time changes from 660 ms to 34 ms on my machine