# Numerical methods challenge: Day 27

During October (2017) I will write a program per day for some well-known numerical methods in both Python and Julia. It is intended to be an exercise then don't expect the code to be good enough for real use. Also, I should mention that I have almost no experience with Julia, so it probably won't be idiomatic Julia but more Python-like Julia.

## Monte Carlo integration

Today we have Monte Carlo integration. Where we use random sampling to numerically compute an integral. This method is important when we want to evaluate higher-dimensional integrals since common quadrature techniques imply an exponential growing in the number of sampling points.

The method computes an integral

where \(\Omega\) has volume \(V\).

The integral is approximated as

where the \(x_i\) distribute uniform over \(\Omega\).

Following are the codes

### Python

from __future__ import division, print_function import numpy as np def monte_carlo_int(fun, N, low, high, args=()): ndims = len(low) pts = np.random.uniform(low=low, high=high, size=(N, ndims)) V = np.prod(np.asarray(high) - np.asarray(low)) return V*np.sum(fun(pts, *args))/N def circ(x, rad): return 0.5*(1 - np.sign(x[:, 0]**2 + x[:, 1]**2 - rad**2)) N = 1000000 low = [-1, -1] high = [1, 1] rad = 1 inte = monte_carlo_int(circ, N, low, high, args=(rad,))

### Julia

using Distributions function monte_carlo_int(fun, N, low, high; args=()) ndims = length(low) pts = rand(Uniform(0, 1), N, ndims) for cont = 1:ndims pts[:, cont] = pts[:, cont]*(high[cont] - low[cont]) + low[cont] end V = prod(high - low) return V*sum(fun(pts, args...))/N end function circ(x, rad) return 0.5*(1 - sign.(x[:, 1].^2 + x[:, 2].^2 - rad^2)) end N = 1000000 low = [-1, -1] high = [1, 1] rad = 1 inte = monte_carlo_int(circ, N, low, high, args=(rad,))

One of the most common examples is the computation of \(\pi\), this is illustrated in the following animation.

### Comparison Python/Julia

Regarding number of lines we have: 20 in Python and 24 in Julia. The comparison
in execution time is done with `%timeit` magic command in IPython and
`@benchmark` in Julia.

For Python:

%timeit monte_carlo_int(circ, N, low, high, args=(rad,))

with result

10 loops, best of 3: 53.2 ms per loop

For Julia:

@benchmark monte_carlo_int(circ, N, low, high, args=(rad,))

with result

BenchmarkTools.Trial: memory estimate: 129.70 MiB allocs estimate: 46 -------------- minimum time: 60.131 ms (5.15% GC) median time: 164.018 ms (55.64% GC) mean time: 204.366 ms (49.50% GC) maximum time: 357.749 ms (64.04% GC) -------------- samples: 25 evals/sample: 1

In this case, we can say that the Python code is roughly 3 times faster than Julia code.

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