Re: Every sufficiently competent C programmer knows --- Paraphrase of Sipser's agreement

On 3/15/2025 5:12 AM, Mikko wrote:

On 2025-03-14 14:39:30 +0000, olcott said:
 

On 3/14/2025 4:03 AM, Mikko wrote:

On 2025-03-13 20:56:22 +0000, olcott said:
>

On 3/13/2025 4:22 AM, Mikko wrote:

On 2025-03-13 00:36:04 +0000, olcott said:
>

>
void DDD()
{
   HHH(DDD);
   return;
}
>
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
When HHH correctly emulates N steps of the
above functions none of them can possibly reach
their own "return" instruction and terminate normally.

>
Nevertheless, assuming HHH is a decider, Infinite_Loop and Infinite_Recursion
specify a non-terminating behaviour, DDD specifies a terminating behaviour

>
_DDD()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov  ebp,esp  ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404     add  esp,+04
[00002182] 5d         pop  ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
>
What is the sequence of machine language
instructions of DDD emulated by HHH such that DDD
reaches its machine address 00002183?

>
Irrelevant off-topic distraction.

>
Proving that you don't have a clue that Rice's Theorem
is anchored in the behavior that its finite string input
specifies. The depth of your knowledge is memorizing
quotes from textbooks.

 Another irrelevant off-topic distraction, this time involving
a false claim.
 One can be a competent C programmer without knowing anyting about Rice's
Theorem.
 

YES.

Rice's Theorem is about semantic properties in general, not just behaviours.
The unsolvability of the halting problem is just a special case.
 

A property about Turing machines can be represented as the language of all Turing machines, encoded as strings, that satisfy that property.
http://kilby.stanford.edu/~rvg/154/handouts/Rice.html
Does THE INPUT TO simulating termination analyzer
HHH encode a C function that reaches its "return"
instruction [WHEN SIMULATED BY HHH] (The definition
of simulating termination analyzer) ???
<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>
<Accurate Paraphrase>
     If emulating termination analyzer H emulates its input
     finite string D of x86 machine language instructions
     according to the semantics of the x86 programming language
     until H correctly determines that this emulated D cannot
     possibly reach its own "ret" instruction in any finite
     number of correctly emulated steps then
     H can abort its emulation of input D and correctly report
     that D specifies a non-halting sequence of configurations.
</Accurate Paraphrase>

Memorizing quotes from textbooks is useful for practical purposes but
if it is too hard for you there are other ways.
 

The whole X represents TM X that halts on its input is inaccurate.
If you did not merely learn-by-rote you would already know this.
*Input Y to TM Z specifies a TM that halts when directly measured by Z*
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer