Good point. By fold/unfold transformation you get the following: contains = flip elem [Eureka] = contains xs e = flip elem xs e [Expose data structures] = contains [] e = False contains (x:xs) e = flip elem (x:xs) e [Instantiate] = contains [] e = False contains (x:xs) e = elem e x:[] || flip elem xs e [Unfold flip one step] = contains [] e = False contains (x:xs) e = elem e x:[] || contains xs e [Fold back to original defintion] = contains [] e = False contains (x:xs) = e==x || contains xs e [Substitute]
Apparently, the fold/unfold transformation law will always yield an equally or more efficient computation. So this begs the question... contains [] e = False contains (x:xs) = e==x || contains xs e OR contains = flip elem Neil Mitchell wrote: > > Hi > >> > contains :: Eq a => [a]->a->Bool >> > contains [] e = False >> > contains (x:xs) e = if x==e then True else contains xs e >> >> contains = flip elem > > And even if not using the elem function, the expression: > > if x==e then True else contains xs e > > can be written as: > > x==e || contains xs e > > Thanks > > Neil > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > > -- View this message in context: http://www.nabble.com/Knowledge-tp14423007p14577605.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe