[Haskell-cafe] Haskell function help

[Carsten Schultz]

Am 01.02.11 14:05, schrieb Houdini:
> 
> that is really.....complicated...hmm

Maybe you find


assign :: (Atom,Bool)->Formula->Formula

assign _ [] = []
assign (a,b) (c:cs)
       | (b,a) `elem` c = cs'
       | otherwise = filter ((/= a).snd) c : cs'
       where cs' = assign (a,b) cs


easier to read.  With


example :: Formula
example = [[p,q,r],[n p,q,n r],[p,n q]]
    where p = l "P"
          q = l "Q"
          r = l "R"
          l a = (True, a)
          n (b,a) = (not b, a)


we get

*Algorithm> example
[[(True,"P"),(True,"Q"),(True,"R")],[(False,"P"),(True,"Q"),(False,"R")],[(True,"P"),(False,"Q")]]
*Algorithm> assign ("P", True) example
[[(True,"Q"),(False,"R")]]
*Algorithm> assign ("P", False) example
[[(True,"Q"),(True,"R")],[(False,"Q")]]

But note that we also get

*Algorithm> assign ("P", False)
[[(True,"P"),(True,"Q"),(True,"R")],[(False,"P"),(True,"Q"),(False,"R")],[(True,"P")]]
[[(True,"Q"),(True,"R")],[]]

So this is not reduced to [[]].

hth C.