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Naively counting Pythagorean triples in Python and Julia
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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) |
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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)) |
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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) |
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
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.
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
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On my MacBook Pro, the timings are: