Sujet : Re: D simulated by H never halts no matter what H does V3
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory sci.logicDate : 27. Apr 2024, 16:32:04
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v0j295$dmbi$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33
User-Agent : Mozilla Thunderbird
On 4/27/2024 6:45 AM, Richard Damon wrote:
On 4/27/24 1:34 AM, olcott wrote:
On 4/26/2024 10:45 PM, Richard Damon wrote:
On 4/26/24 11:02 PM, olcott wrote:
On 4/26/2024 9:18 PM, Richard Damon wrote:
On 4/26/24 9:43 PM, olcott wrote:
On 4/26/2024 7:26 PM, Richard Damon wrote:
On 4/26/24 8:02 PM, olcott wrote:
On 4/26/2024 12:05 PM, olcott wrote:
On 4/26/2024 11:19 AM, Richard Damon wrote:
On 4/26/24 11:34 AM, olcott wrote:
On 4/26/2024 3:32 AM, Mikko wrote:
On 2024-04-25 14:15:20 +0000, olcott said:
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 D(D);
12 }
>
That H(D,D) must report on the behavior of its caller is the
one that is incorrect.
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What H(D,D) must report is independet of what procedure (if any)
calls it.
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Thus when H(D,D) correctly reports that its input D(D) cannot possibly
reach its own line 6 and halt no matter what H does then H can abort its
input and report that its input D(D) does not halt.
>
But since the program D(D) DOES reach its own line 6 when run, because H aborts its simulation and return 0 (since that is what you say this H will do), your statement is PROVEN TO BE A LIE, and you "logic" just a collection of contradictions.
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D simulated by H cannot possibly reach its own line 06 thus when we do
not use the strawman deception to refer to a different D then we know
that D simulated by H never halts.
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The fact that the D(D) executed in main does halt is none of H's
business because H is not allowed to report on the behavior of its
caller.
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In other words, H doesn't need to report on the Behavior of the Program described by its input because it isn't actually a Halt Decider, because you are just a LIAR.
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Anyone knowing the theory of computation knows that H is not allowed to
report on the behavior of its caller.
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In computability theory and computational complexity theory, an
undecidable problem is a decision problem for which it is proved to be
impossible to construct an algorithm that always leads to a correct yes-
or-no answer. https://en.wikipedia.org/wiki/Undecidable_problem
>
The behavior of the simulated D(D) before H aborts its simulation is
different than the behavior of the executed D(D) after H has aborted
its simulation.
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Every time that a simulated input would never stop running unless
aborted the simulating termination analyzer must abort this simulation
to prevent its own infinite execution.
>
H(D,D) is a case of this H1(D,D) is not a case of this even though
the only difference between H and H1 is that D calls H and D does
not call H1.
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D simulated by H would never stop running unless aborted and cannot
possibly reach its own line 06 and halt no matter what H does.
>
Thus whenever we do not use the strawman deception to refer to a
different D we know that D simulated by H specifies a non-halting
sequence of configurations to H.
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*This might be a more succinct way of summing that up*
When you understand that D simulated by H cannot possibly reach past its own line 03 (thus cannot possibly halt) no matter what H does and
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But since H does whatever H does, if H aborts and returns 0, the the direct execution of D, which is what actually matters, DOES get to that point.
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That is another much less useful way to make a universally correct
termination analyzer:
>
int H(ptr x, ptr y)
{
printf("The input program does whatever it does!\n");
return 777; // code for it does what it does
}
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But that doesn't make H answer the question.
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I guess you don't understand what I am saying.
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You said "no matter what H does", but that is a MEANINGLESS statement, because H will do what H is programmed to do, so we don't need to look at other behavior, but just the behavior that H ac
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It can be verified through ordinary software engineering that D(D)
simulated H cannot possibly reach past its own line 03.
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Yes, but that is just a lying RED HERRING, as the question isn't about what H's simulation of the input does, but what the program the input actually represents does when run.
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YOu are just effectively admitting that you are nothing but a stupid liar that doesn't know what he is talking about.
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It can be verified through computer science that this means that D(D) simulated H by never reaches its own final state and halts whether
H aborts its simulation or not.
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Which, since this H DOES abort its simulation is trying to introduce a red herring.
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D(D) simulated by H never halts
D(D) simulated by H never halts
D(D) simulated by H never halts
D(D) simulated by H never halts
D(D) simulated by H never halts
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STRAWMAN
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A method that I used very effectively on another forum and made much
progress with is utterly insisting that we cannot move on to any
additional point until we have 100% complete mutual agreement on
the current point.
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(1) If we (possibly falsely assume) that H(D,D) is supposed to report
on the behavior of D(D) simulated by H and
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(2) we understand that D(D) simulated by H cannot possibly reach past
its own line 3 thus cannot reach its own final state at line 6 and halt
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then (within this possibly false assumption) (1) and the understanding of (2) it <is> correct for H to abort its simulation and report halting.
>
Let go to actual DEFINITIONS shall we.
What do you define the Halting Problem to be?
THIS IS A STIPULATED DEFINITION (like a given" in Geometry)
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.
A stipulative definition is a type of definition in which a new or
currently existing term is given a new specific meaning for the purposes
of argument or discussion in a given context. When the term already
exists, this definition may, but does not necessarily, contradict the
dictionary (lexical) definition of the term.
https://en.wikipedia.org/wiki/Stipulative_definition-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer