[Haskell-cafe] class MonoidBreak?

[Stefan O'Rear]

On Fri, Jun 08, 2007 at 07:24:09AM -0700, Alex Jacobson wrote:
> Dan Piponi wrote:
> >On 6/7/07, Alex Jacobson <alex at alexjacobson.com> wrote:
> >>Is there a standard class that looks something like this:
> >>
> >>class (Monoid m) => MonoidBreak m where
> >>     mbreak::a->m a->(m a,m a)
> >
> >I think you have some kind of kind issue going on here. If m is a
> >Monoid I'm not sure what m a means. Looks like you're trying to factor
> >elements of monoids in some way. Maybe you mean
> >
> >class (Monoid m) => MonoidBreak m where
> >   mbreak::a->m->(m,m)
> >
> >Though I'm not sure what the relationship between m and a is intended 
> >to be.
> 
> Ok how about this class:
> 
>   class (Monoid m) => MonoidBreak m where
>       mbreak::m->m->m
> 
> And the condition is
> 
>   mappend (mbreak y z) y == z

Consider baz x = mbreak x mempty

now:

baz x `mappend` x = mappend (mbreak x mempty) x = mempty

Thus, baz is a left-inverse operator, and (m, mappend, mempty, baz)
forms a group.

Going the other way using a hypothetical Group class:

instance Group m => MonoidBreak m where
    mbreak n p = p `mappend` negate n

satisfies your law.

Stefan