8-liner: cleanest recursive clear solution for Probably Dice by przemyslaw.daniel - python coding challenges

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8-liner: cleanest recursive solution in Clear category for Probably Dice by przemyslaw.daniel

from functools import lru_cache

@lru_cache(maxsize=None)
def probability(dice_number, sides, target):
    if dice_number == 1:
        return (1 <= target <= sides**dice_number)/sides
    return sum([probability(dice_number-1, sides, target-x)
                for x in range(1, sides+1)])/sides

Feb. 23, 2018

Comments:

ymdhtt on Dec. 26, 2018, 2 a.m.
看看

juestr on April 12, 2019, 12:07 a.m.
Very nice. In the if-branch **dice_number can be removed since it is known to be 1.

yuanwxu on Jan. 5, 2020, 5:10 p.m.
The `@lrn_cache` decorator is the key

Kirill_Morhun-Pospolitaki on Jan. 17, 2020, 8:09 p.m.
Thanks a lot, very cool solution. Especially when you realize that `target-x for x in range(1, sides+1)])/sides` imitates dice which was subtracted here `dice_number-1`

CDG.Axel on Nov. 8, 2021, 5:17 p.m.
it's better replace dicenumber == 1 --> dicenumber == 0 <pre class='brush: python'>return sum(probability(dice_number-1, sides, target+~i) for i in range(sides)) / sides if dice_number else not target </pre> <a href="https://py.checkio.org/mission/probably-dice/publications/CDG.Axel/python-3/one-line-recursive-with-lru_cache/">https://py.checkio.org/mission/probably-dice/publications/CDG.Axel/python-3/one-line-recursive-with-lru_cache/</a>

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