Re: [Haskell-cafe] Type constraints for class instances

Actually, infinite trees wouldn't work, for a similar reason to above. You can't decide sort order on the infinite left branches, so you could never choose the correct right branch. Stephen 2008/3/21 Stephen Marsh :

There is a bug in the code:

*Main> ycmp [5,2] [2,5] :: ([Int], [Int]) ([2,2],[5,5])

I think it is impossible to define a working (YOrd a) => YOrd [a] instance. Consider:

let (a, b) = ycmp [[1..], [2..]] [[1..],[1..]]

head (b !! 1) -- would be nice if it was 2, but it is in fact _|_

We take forever to decide if [1..] is greater or less than [1..], so can never decide if [1..] or [2..] comes next.

However Ord a => YOrd [a] can be made to work, and that is absolutely awesome, esp. once you start thinking about things like Ord a => YOrd (InfiniteTree a). This really is very cool, Krzysztof.

Stephen

2008/3/20 Krzysztof Skrzętnicki :

...

Hello everyone,

I'm working on a small module for comparing things incomparable with Ord. More precisely I want to be able to compare equal infinite lists like [1..]. Obviously

(1) compare [1..] [1..] = _|_

It is perfectly reasonable for Ord to behave this way. Hovever, it doesn't have to be just this way. Consider this class

class YOrd a where ycmp :: a -> a -> (a,a)

In a way, it tells a limited version of ordering, since there is no way to get `==` out of this. Still it can be useful when Ord fails. Consider this code:

(2) sort [ [1..], [2..], [3..] ]

It is ok, because compare can decide between any elements in finite time. However, this one

(3) sort [ [1..], [1..] ]

will fail because of (1). Compare is simply unable to tell that two infinite list are equivalent. I solved this by producing partial results while comparing lists. If we compare lists (1:xs) (1:ys) we may not be able to tell xs < ys, but we do can tell that 1 will be the first element of both of smaller and greater one. You can see this idea in the code below.

--- cut here ---

{-# OPTIONS_GHC -O2 #-}

module Data.YOrd where

-- Well defined where Eq means equality, not only equivalence

class YOrd a where ycmp :: a -> a -> (a,a)

instance (YOrd a) => YOrd [a] where ycmp [] [] = ([],[]) ycmp xs [] = ([],xs) ycmp [] xs = ([],xs) ycmp xs'@(x:xs) ys'@(y:ys) = let (sm,gt) = x `ycmp` y in let (smS,gtS) = xs `ycmp` ys in (sm:smS, gt:gtS)

ycmpWrap x y = case x `compare` y of LT -> (x,y) GT -> (y,x) EQ -> (x,y) -- biased - but we have to make our minds!

-- ugly, see the problem below instance YOrd Int where ycmp = ycmpWrap instance YOrd Char where ycmp = ycmpWrap instance YOrd Integer where ycmp = ycmpWrap

-- ysort : sort of mergesort

ysort :: (YOrd a) => [a] -> [a]

ysort = head . mergeAll . wrap

wrap :: [a] -> [[a]] wrap xs = map (:[]) xs

mergeAll :: (YOrd a) => [[a]] -> [[a]] mergeAll [] = [] mergeAll [x] = [x] mergeAll (a:b:rest) = mergeAll ((merge a b) : (mergeAll rest))

merge :: (YOrd a) => [a] -> [a] -> [a] merge [] [] = [] merge xs [] = xs merge [] xs = xs merge (x:xs) (y:ys) = let (sm,gt) = x `ycmp` y in sm : (merge [gt] $ merge xs ys)

--- cut here ---

I'd like to write the following code:

instance (Ord a) => YOrd a where ycmp x y = case x `compare` y of LT -> (x,y) GT -> (y,x) EQ -> (x,y)

But i get an error "Undecidable instances" for any type [a]. Does anyone know the way to solve this?

Best regards

Christopher Skrzętnicki

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