Sujet : Re: Every sufficiently competent C programmer knows --- Paraphrase of Sipser's agreement
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory comp.lang.c comp.lang.c++Suivi-à : comp.theoryDate : 15. Mar 2025, 22:27:00
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vr4rb6$bkso$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
User-Agent : Mozilla Thunderbird
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.htmlDoes 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; Geniushits a target no one else can see." Arthur Schopenhauer
Every sufficiently competent C programmer knows
Re: Every sufficiently competent C programmer knows