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Naively counting Pythagorean triples in Python and Julia
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)
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))
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)
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johnmyleswhite May 10, 2013

On my MacBook Pro, the timings are:

  • Python - ~7000ms
  • Julia w/ Globals - ~500ms
  • Julia w/ Locals - ~10ms
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johnmyleswhite commented May 10, 2013

On my MacBook Pro, the timings are:

  • Python - ~7000ms
  • Julia w/ Globals - ~500ms
  • Julia w/ Locals - ~10ms
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carljv May 10, 2013

The array version is intentionally very compressed, but still, I'd rather be able to use arrays to solve this sort of problem instead of three nested loops. Does Julia perform as well on that front?

def pythag_triples_loops(n):
    "Count Pythagorean triples using loops"
    total = 0
    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
    return total

def pythag_triples(n):
    "Count Pythagorean triples for an array 0, ..., n-1"
    x = np.arange(n)
    return np.sum(np.in1d(reduce(lambda x, y: x**2 + y**2, np.meshgrid(x, x)), x**2))

%timeit -n3  pythag_triples_loops(300)
#  3 loops, best of 3: 4.57 s per loop

%timeit pythag_triples(300)
#10 loops, best of 3: 17.5 ms per loop

carljv commented May 10, 2013

The array version is intentionally very compressed, but still, I'd rather be able to use arrays to solve this sort of problem instead of three nested loops. Does Julia perform as well on that front?

def pythag_triples_loops(n):
    "Count Pythagorean triples using loops"
    total = 0
    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
    return total

def pythag_triples(n):
    "Count Pythagorean triples for an array 0, ..., n-1"
    x = np.arange(n)
    return np.sum(np.in1d(reduce(lambda x, y: x**2 + y**2, np.meshgrid(x, x)), x**2))

%timeit -n3  pythag_triples_loops(300)
#  3 loops, best of 3: 4.57 s per loop

%timeit pythag_triples(300)
#10 loops, best of 3: 17.5 ms per loop
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seanjtaylor May 10, 2013

Under PyPy I got a mean of 45ms over 100 runs, also on a recent Macbook Pro. 4.5x faster than the PyPy JIT is pretty awesome.

Under PyPy I got a mean of 45ms over 100 runs, also on a recent Macbook Pro. 4.5x faster than the PyPy JIT is pretty awesome.

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vgoklani May 11, 2013

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

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

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diegozea Jun 7, 2013

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

diegozea commented Jun 7, 2013

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

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