Re: How the requirements that Professor Sipser agreed to are exactly met

On 2025-05-13 13:54:15 +0000, olcott said:

On 5/13/2025 2:41 AM, Mikko wrote:

On 2025-05-12 18:17:37 +0000, olcott said:
 

Introduction to the Theory of Computation 3rd Edition
by Michael Sipser (Author)
4.4 out of 5 stars    568 rating
 https://www.amazon.com/Introduction-Theory-Computation-Michael-Sipser/ dp/113318779X
 int DD()
  {
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
  }
 DD correctly simulated by any pure simulator
named HHH cannot possibly terminate thus proving
that this criteria has been met:
 <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its
     input D until H correctly determines that its simulated D
     would never stop running unless aborted then

 This specifies two requirements:
1. H correctly simulates that part of the behaviour of D that starts
from the start of the execution and does not end before the second
requirement is satisfied.
2. H correctly determines that unsimulated part of the behaviour is
infinitely long.
 The second reuirement is not satisfied when HHH analyses the above
DD.

 In other words you believe that DD will halt
on its own without ever being aborted by HHH.
That is counter-factual.

I didn't say so. If HHH does not return zero then DD does not halt.
I said that your HHH, unlike some other partial halt deciders, fails
to analyze DD enough to correctly determine whether it halts.

_DD()
[00002133] 55         push ebp      ; housekeeping
[00002134] 8bec       mov ebp,esp   ; housekeeping
[00002136] 51         push ecx      ; make space for local
[00002137] 6833210000 push 00002133 ; push DD
[0000213c] e882f4ffff call 000015c3 ; call HHH(DD)
...
 HHH determines that DD correctly simulated by
HHH keeps calling HHH(DD)

which in reality does not happen as DD calls HHH only once.
--
Mikko