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

On 2025-03-15 21:27:00 +0000, olcott said:

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

However, it is possible that for some property that language cannot
be presented.

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>

An attempt to deceive. Professor Sipser is does not say anuthing about
the qestion stated just before the that quote. Instead, is taking
about the halting problem. In particular, the clause "when simulated
by H" is not in the quoted text.

<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>

The "paraphrase" is not accurate. In particular, the phrase "in any
finite number of correctly emulated steps" is not in Sipsers words.

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.

Can't parse. What is the subject of "is inaccurate"?

If you did not merely learn-by-rote you would already know this.

Usually the word "know" is not used about false beliefs.

*Input Y to TM Z specifies a TM that halts when directly measured by Z*

Which TM halts, Y or Z?
--
Mikko