Sujet : Re: D simulated by H never halts no matter what H does V3
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theory sci.logicDate : 28. Apr 2024, 02:45:09
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <v0k66l$2djoe$9@i2pn2.org>
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User-Agent : Mozilla Thunderbird
On 4/27/24 8:21 PM, olcott wrote:
On 4/27/2024 6:44 PM, Richard Damon wrote:
On 4/27/24 7:29 PM, olcott wrote:
On 4/27/2024 6:09 PM, Richard Damon wrote:
On 4/27/24 6:56 PM, olcott wrote:
On 4/27/2024 5:45 PM, Richard Damon wrote:
On 4/27/24 6:29 PM, olcott wrote:
On 4/27/2024 5:19 PM, Richard Damon wrote:
On 4/27/24 6:02 PM, olcott wrote:
On 4/27/2024 4:45 PM, Richard Damon wrote:
On 4/27/24 5:36 PM, olcott wrote:
On 4/27/2024 4:24 PM, Richard Damon wrote:
On 4/27/24 3:48 PM, olcott wrote:
Simulating termination analyzer H determines whether or not
D(D) simulated by H can possibly reach its final state at its
own line 06 and halt whether or not H aborts its simulation.
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We can resolve exactly what I mean by this as an aspect of
staying on this one point. We cannot move on to the slightest
trace of any nuance of any other point until AFTER we have
100% complete mutual agreement on this point.
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(a) It is a verified fact that D(D) simulated by H cannot
possibly reach past line 03 of D(D) simulated by H whether H
aborts its simulation or not.
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When we have 100% perfect mutual agreement on that point
then we can move on to the next aspect of the point of the
paragraph.
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The problem is you don't seem to have a proper definition for a "program", as the input seems to change behavior as you analyize different options for what "different" H's might do.
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It seems that neither your D or your H actual meet the normal definition of what a "Program" is.
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I never even use the word "program"
*H and D are 100% completely specified right here*
https://github.com/plolcott/x86utm/blob/master/Halt7.c
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So, what is the defined "class" of the input to a Termination Analyzer.
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I am only talking about H and D. You always "read things in"
to what I say that I never said.
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So, what are "H" and "D", are they "Programs" per the standard definitions, or something else that you are stipulating?
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They are 100% completely defined in the complete source-code
that I just linked above.
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So, if H is "defined" by its source code, then it can only do one thing, and thus your criteria of talking about "whether it aborts its simulation or not" is a MEANINGLESS Statement.
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It <is> the mandatory prerequisite to proceeding to additional steps.
The end of these steps will show how it PRECISELY applies to the Linz
proof.
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You cannot truthfully say that I have not defined H and D sufficiently
when I provide the full source code to their fully operational system.
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Since there is now no "problem" left for H to be tested to see if it "solves" what can be agreed to?
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01 int D(ptr x) // ptr is pointer to int function
02 {
03 int Halt_Status = H(x, x);
04 if (Halt_Status)
05 HERE: goto HERE;
06 return Halt_Status;
07 }
08
09 void main()
10 {
11 H(D,D);
12 }
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Simulating termination analyzer H determines whether or not
D(D) simulated by H can possibly reach its final state at its
own line 06 and halt whether or not H aborts its simulation.
https://github.com/plolcott/x86utm/blob/master/Halt7.c
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But, since you have stipulated that there only exist ONE H, the H provided (since you won't define the class of object that H is) this statement reduces to:
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I cut-and-pasted the exact verbatim word-for-word post that
C experts found very easy to answer.
Did you actually explain to them the domain of the problem and the stipulation you have made? (Including that the options for the H and D have been restricted to EXACTLY the code listed)
I suspect not from the feedback we have gotten from Professor Sipser.
If you don't give me a straight answer I will assume that you
are dodging because you want to hide the fact that you are mostly
clueless about C.
The phrase "whether or not H aborts its simulation" is a meaningless phrase as there is EXACTLY one H, and it does abort its simulation.
So, your criteria is reduced to:
Simulating Termination Analyzer H determenes that D(D) Simulated and then Aborted by H will not reach its final state at its own line 06 because H aborts its simulation before the code can get there.
I.E., a toy problem.
Yes, your criteria has become the tautology that if H aborts its simulation then it can return 0 because if it aborts its simulation, its simulation will not reach the final state of the program.
THis answer is true INDEPENDENT of the input given to H, so H could have been equivalent defined as just a return 0 statement.
YOUR SYSTEM IS A TOY THAT DOESN'T PROVE ANYTHING.
I will point out that you are just proving yourself totally foolish and stupid, trying to ad-hominem attack me on my understanding of the C programming language when my qualification on that seem to exceed yours, including the fact that I am still gainfully employed where generating QUALIFIED code is part of my duties, and it sounds like I started programming YEARS before you (I was put into the position of being a teaching assistant for a programming course in 1971)