Re: Idempotence and "Replication Insensitivity" are equivalent ?

From: William Hughes <wpihughes_at_hotmail.com>
Date: 20 Sep 2006 20:55:26 -0700
Message-ID: <1158810926.196355.18580_at_b28g2000cwb.googlegroups.com>


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
> 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.

  • William Hughes
                                     -William Hughes
Received on Thu Sep 21 2006 - 05:55:26 CEST

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