very clear lambdas clear solution for Boolean Algebra by fishsouprecipe - python coding challenges

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very clear lambdas solution in Clear category for Boolean Algebra by fishsouprecipe

OPERATION_NAMES = {
    "conjunction": lambda a, b: a & b,
    "disjunction": lambda a, b: a | b,
    "implication": lambda a, b: a <= b,
    "exclusive": lambda a, b: a ^ b,
    "equivalence": lambda a, b: a == b
}


def boolean(x, y, operation):
    return OPERATION_NAMES[operation](x, y)

March 8, 2020

Comments:

veky on March 15, 2020, 3:13 p.m.
It can be even clearer <a href="https://py.checkio.org/mission/boolean-algebra/publications/veky/python-3/an-operator-by-any-other-name/">without them</a>. :-]

fishsouprecipe on March 15, 2020, 3:15 p.m.
haha, that's nice. Please, check it out <a href="https://py.checkio.org/mission/four-to-the-floor/publications/fishsouprecipe/python-3/method-monte-carlo/">PY.CHECKIO.ORG</a>

veky on March 15, 2020, 3:22 p.m.
Sorry, haven't solved that one yet. Is it numerical approximation or exact calculation? ^_^

fishsouprecipe on March 15, 2020, 3:25 p.m.
I solved that task with <a href="https://en.wikipedia.org/wiki/Monte_Carlo_method">Method Monte Carlo</a>. The whole code there is just <pre class='brush: python'>import random import math def is_covered(room, sensors): for _ in range(5000000): random_x, random_y = (random.random() * c for c in room) if not any(sensor_r >= math.sqrt(math.pow(random_x - sensor_x, 2) + math.pow(random_y - sensor_y, 2)) for sensor_x, sensor_y, sensor_r in sensors): return False return True </pre>

veky on March 15, 2020, 3:28 p.m.
Ah, nice. You don't even seed your random generator, just counting on luck? :-) I thought they'd probably give some tests where a tiny bit of space is left so it isn't hit by Montecarlo, but obviously I overestimated the task authors. B-)

max-belichenko on April 8, 2020, 2:15 p.m.
Quite clear and compendious

Look up in the Dict()

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