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First solution in Clear category for Boolean Algebra by jeschlegel84
OP = ("conjunction", "disjunction", "implication", "exclusive", "equivalence")
def boolean(x, y, operation):
if operation == OP[0]: return (x and y)
elif operation == OP[1]: return (x or y)
elif operation == OP[2]:
if x: return y
else: return 1
elif operation == OP[3]: return (x+y)%2 #return x ^ y
elif operation == OP[4]: return x == y
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. 18, 2017