# 99 questions/Solutions/46

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(**) Define predicates and/2, or/2, nand/2, nor/2, xor/2, impl/2 and equ/2 (for logical equivalence) which succeed or fail according to the result of their respective operations; e.g. and(A,B) will succeed, if and only if both A and B succeed.

A logical expression in two variables can then be written as in the following example: and(or(A,B),nand(A,B)).

Now, write a predicate table/3 which prints the truth table of a given logical expression in two variables.

The first step in this problem is to define the Boolean predicates:

-- Negate a Boolean argument
not' :: Bool -> Bool
not' True  = False
not' False = True

-- True if both a and b are True
and',or',nor',nand',xor',impl',equ' :: Bool -> Bool -> Bool
and' True True = True
and' _    _    = False

-- True if a or b or both are True
or' False False = False
or' _     _     = True

-- Negation of 'or'
nor'  a b = not' \$ or'  a b

-- Negation of 'and'
nand' a b = not' \$ and' a b

-- True if either a or b is true, but not if both are true
xor' True  False = True
xor' False True  = True
xor' _     _     = False

-- True if a implies b, equivalent to (not a) or (b)
impl' a b = (not' a) `or'` b

-- True if a and b are equal
equ' True  True  = True
equ' False False = True
equ' _     _     = False

The explicit implementation of logic functions above could be shortened using Haskell's builtin equivalents:

and'  a b = a && b
or'   a b = a || b
nand' a b = not (and' a b)
nor'  a b = not (or' a b)
xor'  a b = and' (or' a b) (nand' a b)
impl' a b = or' (not a) b
equ'  a b = a == b

Some could be reduced even further through Pointfree style:

and' = (&&)
or'  = (||)
equ' = (==)

Anyway, the only remaining difficulty is to generate the truth table. This approach accepts a Boolean function (Bool -> Bool -> Bool), then calls that function with all four combinations of two Boolean values:

table :: (Bool -> Bool -> Bool) -> IO ()
table f = mapM_ putStrLn [show a ++ " " ++ show b ++ " " ++ show (f a b)
| a <- [True, False], b <- [True, False]]

The table function in Lisp supposedly uses Lisp's symbol handling to substitute variables on the fly in the expression. I chose passing a binary function instead because parsing an expression would be more verbose in haskell than it is in Lisp. Template Haskell could also be used :)