Re: Ben Bacarisse fails understand that deciders compute the mapping from inputs

On 2024-08-27 13:04:26 +0000, olcott said:
 

On 8/27/2024 12:45 AM, joes wrote:

Am Mon, 26 Aug 2024 18:03:41 -0500 schrieb olcott:

On 8/26/2024 7:42 AM, Ben Bacarisse wrote:

Mike Terry <news.dead.person.stones@darjeeling.plus.com> writes:

On 23/08/2024 22:07, Ben Bacarisse wrote:

>

We don't really know what context Sipser was given.  I got in touch
at the time so I do know he had enough context to know that PO's
ideas were "wacky" and that had agreed to what he considered a "minor
remark". Since PO considers his words finely crafted and key to his
so-called work I think it's clear that Sipser did not take the "minor
remark" he agreed to to mean what PO takes it to mean!  My own take
if that he (Sipser) read it as a general remark about how to
determine some cases, i.e. that D names an input that H can partially
simulate to determine it's halting or otherwise.  We all know or
could construct some such cases.

>
Exactly my reading.  It makes Sipser's agreement natural, because it
is both correct [with sensible interpretation of terms], and moreover
describes an obvious strategy that a partial decider might use that
can decide halting for some specific cases.  No need for Sipser to be
deceptive or misleading here, when the truth suffices.  (In particular
no need to employ "tricksy" vacuous truth get out clauses just to get
PO off his back as some have suggested.)

>
Yes, and it fits with his thinking it a "trivial remark".

>

That aside, it's such an odd way to present an argument: "I managed to
trick X into saying 'yes' to something vague".  In any reasonable
collegiate exchange you'd go back and check: "So even when D is
constructed from H, H can return based on what /would/ happen if H did
not stop simulating so that H(D,D) == false is correct even though D(D)
halts?".  Just imagine what Sipser would say to that!

Is this an accurate phrasing, pete?

>
Deciders never compute the mapping of the computation
that they themselves are contained within.

 Why not? A decider always either accepts or rejects its input.

If the
input specifies a computation then it maps the computation to either
accept or reject. Whether the computation contains the decider does
not affect that. If the "decider" neither accpets nor rejects it is
not a decider.