>> I understood (apparently correctly) that Brett *knows* the inverse of
>> functions he is interested in.
Exactly.
>> The problem is that in Haskell it is not possible to distinguish
>> between functions, they do not belong to the Eq class
Yes, I believe this is the heart of the problem. A
Brett A. Letner precises his wishes concerning the inverse function.
Somebody objected :
> >> that is not possible in the general case.
> >> Not to forget that even if it was, not all functions has inverses.
I understood (apparently correctly) that Brett *knows* the inverse
[EMAIL PROTECTED] writes
>> Do you want to have a function
>> inverse :: (a -> b) -> (b -> a)
>> so that (inverse f) . f = id
>> ?
>> []
>> that is not possible in the general case.
>> Not to forget that even if it was, not all functions has inverses.
> Perhaps I should elaborate. The specif
>> Do you want to have a function
>> inverse :: (a -> b) -> (b -> a)
>> so that (inverse f) . f = id
>> ?
Yes.
>> that is not possible in the general case.
>> Not to forget that even if it was, not all functions has inverses.
I would like to do this for 3 specific cases, where I would supply th
Wed Dec 29 1999, [EMAIL PROTECTED] ->
> Is it possible to define an inverse function such that given the functions
> f,g, and h and their inverses f',g', and h', the inverse function would
> return the appropriate inverse?
> e.g.
> inverse(f)=f'
> inve
[EMAIL PROTECTED] writes
> Is it possible to define an inverse function such that given the functions
> f,g, and h and their inverses f',g', and h', the inverse function would
> return the appropriate inverse?
> e.g.
> inverse(f)=f'
> inverse(g)=g'
Haskell group-
Is it possible to define an inverse function such that given the functions
f,g, and h and their inverses f',g', and h', the inverse function would
return the appropriate inverse?
e.g.
inverse(f)=f'
inverse(g)=g'
inverse(h)=h'
I'm a newbie to
