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

From: William Hughes <wpihughes_at_hotmail.com>
Date: 20 Sep 2006 21:19:05 -0700

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
```>
>

> > > 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?
>
>
• William Hughes
Received on Thu Sep 21 2006 - 06:19:05 CEST

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