Re: DD simulated by HHH cannot possibly halt (Halting Problem)

On 4/5/2025 5:40 PM, olcott wrote:

On 4/5/2025 4:14 PM, dbush wrote:

On 4/5/2025 3:01 PM, olcott wrote:

On 4/5/2025 1:45 PM, Richard Heathfield wrote:

On 05/04/2025 19:11, olcott wrote:

On 4/5/2025 11:25 AM, dbush wrote:

>

On 4/5/2025 11:59 AM, olcott wrote:>>>>

Introduction to the Theory of Computation 3rd Edition
by Michael Sipser (Author) (best selling textbook)
>
<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
>
     H can abort its simulation of D and correctly report that D
     specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>

>
But not what you think he agreed to:
>
>

>
void DDD()
{
   HHH(DDD);
   return;
}
>

>
You have to show that by showing the details of how
what he agreed to is not accurately paraphrased by
*Simulating termination analyzer Principle*

>
No, you have to show firstly that your H determines anything at all about D's behaviour.

>
First of all it is the concrete DDD and the hypothetical HHH.

>
Category error.  The algorithm DDD is not fully specified if the code of the function HHH and everything it calls explicitly spelled out, as all of that is the code under test.

 DDD meets the spec of the
*Simulating termination analyzer Principle*
as long as HHH emulates enough steps of DDD
to see that it must stop simulating DDD.
 

And as such is unrelated to the halting problem, as the halting problem is about algorithms, and DDD as you've defined it is not an algorithm:
Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly