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# johnmyleswhite/JuliaGlobals

Last active Aug 10, 2018
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)

### johnmyleswhite commented May 10, 2013

 On my MacBook Pro, the timings are: Python - ~7000ms Julia w/ Globals - ~500ms Julia w/ Locals - ~10ms

### 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 ``````

### seanjtaylor commented 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.

### vgoklani commented 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

### 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