# Re: Idempotence and "Replication Insensitivity" are equivalent ?

Date: 20 Sep 2006 21:08:12 -0700

Message-ID: <1158811692.910726.12610_at_h48g2000cwc.googlegroups.com>

William Hughes wrote:

*> Marshall wrote:
**> > William Hughes wrote:
**> > >
*

> > > This is silly. I have a function f:A,A->A, but this is

*> > > too restrictive so I will change this to a function f:A,B->B,
**> > > but now I want to talk about idempotence so I will
**> > > let A=B, so I have a function f:A,A->A but this is
**> > > too restrictive so ...
**> >
**> > I have no idea what you're trying to say here.
**> >
**>
**> I will try again.
**>
**> The question is "can we talk about functions
**> being idempotent". There are two possibities.
**>
**>
**> 1. f:A,A->A -- we can talk about idempotent functions
**> but this restricts the functions we can get.
**>
**> 2. f:A,B->B -- we can not talk about idempotent functions,
**> but there is no restriction on the functions
**> we can get.
**>
**> Now you want to get the no restriction from case 2 and the
**> idempotent from case 1 by letting A=B. This does not work
**> If you let A=B then you are back to case 1.
*

> You only get the no restriction if B is allowed to

*> be something other than A.
*

And if you have a final expression that can use zero or more expressions of type 2 above.

> > Anyway, A, B -> B as the most general type of a

*> > the first argument to fold is not my formulation; it's
**> > been around for a long time.
**>
**> However, the term idempotent does not appy here
**> so I fail to see the relevence.
*

> > I would prove that with

*> > a Google search, but alas! Google throws away most
**> > punctuation, and the first hit for "A, B -> B" is "The
**> > Official BB King Website."
**> >
**> > > Have I got this straight.
**> > >
**> > > S can contain an arbitrary number of elements of A,
**> > > so f(a,S) takes an arbitrary number of elements of A, but
**> > > f is despite this a binary form?
**> >
**> > No; I wouldn't call it binary unless A = B. But is this question
**> > really important? Boy you are really hung up on nomenclature. :-)
**> >
**>
**> The question is only important if you make the claim that
**> you can talk about idempotence whenever you have a binary
**> function. If you agree that you can only talk about
**> idempotence when you have a function f:A^2->A, then
**> the question of what you call a binary function is
**> of very limited interest.
*

Marshall Received on Thu Sep 21 2006 - 06:08:12 CEST