First clear solution for Boolean Algebra by lukasz.bogaczynski - python coding challenges

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First solution in Clear category for Boolean Algebra by lukasz.bogaczynski

def boolean(first_operand, second_operand, operation):
    return {
        "conjunction": conjunction,
        "disjunction": disjunction,
        "implication": implication,
        "exclusive": exclusive,
        "equivalence": equivalence,
    }[operation](first_operand, second_operand)


def conjunction(first_operand, second_operand):
    return int(bool(first_operand) and bool(second_operand))


def disjunction(first_operand, second_operand):
    return int(bool(first_operand) or bool(second_operand))


def implication(first_operand, second_operand):
    return 0 if first_operand == 1 and second_operand == 0 else 1


def exclusive(first_operand, second_operand):
    return int(bool(first_operand != second_operand))


def equivalence(first_operand, second_operand):
    return int(first_operand == second_operand)


if __name__ == '__main__':
    #These "asserts" using only for self-checking and not necessary for auto-testing
    assert boolean(1, 0, "conjunction") == 0, "and"
    assert boolean(1, 0, "disjunction") == 1, "or"
    assert boolean(1, 1, "implication") == 1, "material"
    assert boolean(0, 1, "exclusive") == 1, "xor"
    assert boolean(0, 1, "equivalence") == 0, "same?"

Oct. 21, 2017

Comments:

fenryxo on Feb. 5, 2019, 6:56 p.m.
<p>I like the usage of a dict instead of "if-else" hell but there is too much unnecessary int/bool casting. It is enough to cast only the final result to int. 1 behaves the same as True, 0 as False. </p>

Look up in the Dict()

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