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Clear - one liner option included solution in Clear category for Boolean Algebra by Pelmen323
OPERATION_NAMES = ("conjunction", "disjunction", "implication", "exclusive", "equivalence")
def boolean(x, y, operation):
if operation == 'conjunction' and (x == y == 1): return 1
if operation == 'disjunction' and (x or y == 1): return 1
if operation == 'implication' and (x == 0 or y == 1): return 1
if operation == 'exclusive' and (x != y and (x or y == 1)): return 1
if operation == 'equivalence' and x == y: return 1
return 0
# One liner option below (I don't like using it here since it lowers readability):
# return 1 if operation == 'conjunction' and (x == y == 1) or operation == 'disjunction' and (x or y == 1) or operation == 'implication' and (x== 0 or y==1) or operation == 'exclusive' and (x != y and (x or y == 1)) or operation == 'equivalence' and x == y else 0
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?"
print('All good! Go and check the mission.')
Aug. 30, 2021