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

Date: 20 Sep 2006 21:19:05 -0700

Message-ID: <1158812345.690596.72300_at_h48g2000cwc.googlegroups.com>

Marshall wrote:

*> 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.
*

>

> Well, sure. The set of functions that belong to case 1

*> is a subset of the functions that belong to case 2.*

*> I have no idea what you mean by "this does not work."*

> >

> > 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 understood your first paragraph to mean that you

*> thought I had made up the A, B -> B type on the*

*> spot to escape some recently-uncovered "flaw" in the*

*> A, A -> A form, when in fact the use of both forms*

*> are probably decades old.*

> >

> > > 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.*

>

> I would say the question of what *I* call a binary function

*> is of very limited interest, regardless, even to me. Definitions*

*> are somewhat boring. In any event, I was under the impression*

*> that we both agree that idempotence only applies to functions*

*> of type A, A -> A, in which case, what are we discussing again?*

>

>

The question posed by the title of this thread "are idempotence and replication insensitivity equivalent?". Since idempotence only applies to functions of the type A,A->A we get that idempotence only applies to a restricted set of functions on M(A). Since this restricted set does not include all the functions we would like to apply the term "replication insensitive" to, I would conclude that the terms "idempotent" and "replication insensitive" are not equivalent.

- William Hughes