A Pi Calculation with DTM
A simple yet interesting use of DTM is the calculation of
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with a Monte Carlo approach. This approach is quite straightforward : if you randomly throw
n darts on a unit square, approximately
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will be inside a quadrant delimited by (0,1) and (1,0). Therefore, if a huge quantity of darts are thrown, one could estimate
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simply by computing the ratio between the number of darts inside and outside the quadrant. A comprehensive explanation of the algorithm can be found
here
Note
This example is intended to show a simple parallelization of an actual algorithm. It should not be taken as a good
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calculation algorithm (it is not).
A possible serial Python code reads as follow :
from random import random
from math import hypot
def test(tries):
# Each run of this function makes some tries
# and return the number of darts inside the quadrant (r < 1)
return sum(hypot(random(), random()) < 1 for i in xrange(tries))
def calcPi(n, t):
expr = (test(t) for i in range(n))
pi2 = 4. * sum(expr) / (n*t)
print("pi = " + str(pi2))
return pi2
piVal = calcPi(1000, 10000)
With DTM, you can now take advantage of the parallelization, and distribute the calls to the function test(). There are many ways to do so, but a mere one is to use repeat(), which repeats a function an arbitrary number of times, and returns a results list. In this case, the program may look like this :
from math import hypot
from random import random
from deap import dtm
def test(tries):
# Each run of this function makes some tries
# and return the number of darts inside the quadrant (r < 1)
return sum(hypot(random(), random()) < 1 for i in xrange(tries))
def calcPi(n, t):
expr = dtm.repeat(test, n, t)
pi2 = 4. * sum(expr) / (n*t)
print("pi = " + str(pi2))
return pi2
piVal = dtm.start(calcPi, 1000, 10000)
And so, without any major changes (and not at all in the test() function), this computation can be distributed.
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