is it possible to express the following Haskell program without FlexibleInstances, i.e. in pure Haskell2010?
{-# LANGUAGE FlexibleInstances #-}
class Funk a where truth :: a -> [Bool]
instance Funk [Bool] where truth = \x -> x
instance Funk Bool where truth = \x -> [x]
instance Funk b => Funk (Bool -> b) where
truth f = concat [truth (f True), truth (f False)]
This is inspired by the answer on How to write a Haskell function that takes a variadic function as an argument.
I suspect the problem is, that truth
returns something else than the function which it takes as an argument (which returns Bool
, not [Bool]
).
The purpose of this snippet is, to give a list of all evaluations of all possible configuration for a boolean function, i.e.
Main> truth (\x y -> x && y)
[True,False,False,False]
Main> truth (\x y -> x || y)
[True,True,True,False]
In the end, a truth-table is to be printed, like this (see boiler-plate at the end of this post to see the code which produces this):
Main> main
T T T | T
T T F | T
T F T | T
T F F | F
F T T | T
F T F | F
F F T | T
F F F | T
Here is some boiler-plate code for testing and visualizing, what the purpose of this function is:
class TruthTable a where
truthTable :: Funk f => f -> a
instance TruthTable [String] where
truthTable f = zipWith (++) (hCells (div (length t) 2)) (map showBool $ truth f)
where
showBool True = "| T"
showBool False = "| F"
hCells 1 = ["T ", "F "]
hCells n = ["T " ++ x | x <- hCells (div n 2)] ++ ["F " ++ x | x <- hCells (div n 2)]
instance TruthTable [Char] where
truthTable f = foldl1 join (truthTable f)
where join a b = a ++ "\n" ++ b
instance TruthTable (IO a) where
truthTable f = putStrLn (truthTable f) >> return undefined
main :: IO ()
main = truthTable (\x y z -> x `xor` y ==> z)
xor :: Bool -> Bool -> Bool
xor a b = not $ a == b
(==>) :: Bool -> Bool -> Bool
a ==> b = not $ a && not b