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
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