# Haskell Quiz/Constraint Processing/Solution Jethr0

### From HaskellWiki

m (infixing "List") |
m (using an infix constructor for my list now) |
||

Line 37: | Line 37: | ||

<haskell> | <haskell> | ||

− | data List a = | + | data List a = a ::: (List a) | Empty deriving (Show) |

foldrL :: (a->b->b) -> b -> List a -> b | foldrL :: (a->b->b) -> b -> List a -> b | ||

foldrL _ start Empty = start | foldrL _ start Empty = start | ||

− | foldrL f start ( | + | foldrL f start (x ::: xs) = f x (foldrL f start xs) |

appendL :: List a -> List a -> List a | appendL :: List a -> List a -> List a | ||

− | appendL xs ys = foldrL | + | appendL xs ys = foldrL (:::) ys xs |

concatL :: List (List a) -> List a | concatL :: List (List a) -> List a | ||

Line 50: | Line 50: | ||

instance Functor List where | instance Functor List where | ||

− | fmap f = foldrL (\a b -> f a | + | fmap f = foldrL (\a b -> f a ::: b) Empty |

instance Monad List where | instance Monad List where | ||

− | return x = x | + | return x = x ::: Empty |

l >>= f = concatL . fmap f $ l | l >>= f = concatL . fmap f $ l | ||

Line 62: | Line 62: | ||

range :: (Integral a) => a -> a -> List a | range :: (Integral a) => a -> a -> List a | ||

range from to | from > to = Empty | range from to | from > to = Empty | ||

− | | from == to = to | + | | from == to = to ::: Empty |

− | | otherwise = from | + | | otherwise = from ::: range (from+1) to |

constr' = do a <- range 0 4 | constr' = do a <- range 0 4 |

## Revision as of 22:57, 20 December 2006

Basically there's nothing to be done in haskell for this quiz.

As the List Monad already provides non-deterministic evaluation with "guard" as a description of constraints, you really just have to write the problem in the List Monad and be done with it.

Of course one could write all kinds of wrapping hackery, but I think from the standpoint of usability and conciseness the built-in behaviour of haskell is already pretty optimal.

constr = do a <- [0..4] b <- [0..4] c <- [0..4] guard (a < b) guard (a + b == c) return ("a:",a,"b:",b,"c:",c) {- > constr [("a:",0,"b:",1,"c:",1) ,("a:",0,"b:",2,"c:",2) ,("a:",0,"b:",3,"c:",3) ,("a:",0,"b:",4,"c:",4) ,("a:",1,"b:",2,"c:",3) ,("a:",1,"b:",3,"c:",4)] -}

Obviously, solving a problem with built-in functionality always feels a little like cheating, because it's "pure chance" that the solution to the problem is already built-in.

So, for the sake of argument and because I was bored here's the same solution under the premise that haskell's List Monad didn't already exist.

Unfortunately I "had" to reimplement a fair bit of the Prelude, but that's what happens if you start creating unrealistic scenarios :). Of course, the "do" notation is also kind of a built-in, but representing it with bind (>>=) only made is less nice to look at.

data List a = a ::: (List a) | Empty deriving (Show) foldrL :: (a->b->b) -> b -> List a -> b foldrL _ start Empty = start foldrL f start (x ::: xs) = f x (foldrL f start xs) appendL :: List a -> List a -> List a appendL xs ys = foldrL (:::) ys xs concatL :: List (List a) -> List a concatL = foldrL appendL Empty instance Functor List where fmap f = foldrL (\a b -> f a ::: b) Empty instance Monad List where return x = x ::: Empty l >>= f = concatL . fmap f $ l instance MonadPlus List where mzero = Empty mplus = appendL range :: (Integral a) => a -> a -> List a range from to | from > to = Empty | from == to = to ::: Empty | otherwise = from ::: range (from+1) to constr' = do a <- range 0 4 b <- range 0 4 c <- range 0 4 guard (a < b) guard (a + b == c) return (a,b,c)