My First, Recursive speedy solution for Pearls in the Box by d_rabenko - python coding challenges

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My First, Recursive solution in Speedy category for Pearls in the Box by d_rabenko

from decimal import *


def prob(marbles, step):
    p_w = marbles.count('w') / len(marbles)
    if step == 1:
        return p_w
    else:
        return (1 - p_w) * prob(marbles.replace('b', 'w', 1), step - 1) + \
               p_w * prob(marbles.replace('w', 'b', 1), step - 1)


def checkio(marbles, step):
    temp = Decimal(prob(marbles, step) * 100).quantize(Decimal()) / 100
    return float(temp)

#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
    assert checkio('bbw', 3) == 0.48, "1st example"
    assert checkio('wwb', 3) == 0.52, "2nd example"
    assert checkio('www', 3) == 0.56, "3rd example"
    assert checkio('bbbb', 1) == 0, "4th example"
    assert checkio('wwbb', 4) == 0.5, "5th example"
    assert checkio('bwbwbwb', 5) == 0.48, "6th example"

Oct. 9, 2017

Comments:

ZweiStein on May 28, 2019, 1:10 p.m.
<p>This will fit better under speedy if you exchange string replace with arithmatics and do the count only once.</p>

d_rabenko on May 28, 2019, 7:53 p.m.
<p>thank you for your advise!</p>

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