99 questions/Solutions/46

From HaskellWiki
< 99 questions‎ | Solutions
Revision as of 04:07, 18 July 2010 by Wapcaplet (talk | contribs) (more cleanup and commenting)

Jump to: navigation, search

(**) 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:

-- NOT negates a single Boolean argument
not' :: Bool -> Bool
not' True  = False
not' False = True

-- Type signature for remaining logic functions
and',or',nor',nand',xor',impl',equ' :: Bool -> Bool -> Bool

-- AND is True if both a and b are True
and' True True = True
and' _    _    = False

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

-- NOR is the negation of 'or'
nor'  a b = not' $ or'  a b

-- NAND is the negation of 'and'
nand' a b = not' $ and' a b

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

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

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

The above implementations build each logic function from scratch; they 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 using Pointfree style:

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

The only remaining task is to generate the truth table; most of the complexity here comes from the string conversion and IO. The approach used here accepts a Boolean function (Bool -> Bool -> Bool), then calls that function with all four combinations of two Boolean values, and converts the resulting values into a list of space-separated strings. Finally, the strings are printed out by mapping putStrLn across the list of strings:

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 :)