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

From: Marshall <marshall.spight_at_gmail.com>
Date: 20 Sep 2006 21:08:12 -0700

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.

> > 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
> > 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

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