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 }
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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
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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.
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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.
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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
}
But that doesn't make H answer the question.
I guess you don't understand what I am saying.
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.
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.
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.
Which, since this H DOES abort its simulation is trying to introduce a red herring.
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
*I say that cats purr and you say no you are wrong dogs do bark*
D(D) simulated by H has different behavior than D(D) simulated
by H1 even though the ONLY difference between H and H1 is that
D calls H and D does not call H.
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This means that D(D) simulated by H unequivocally DOES NOT HALT!
Nope, prove you to be a stupid liar.
D(D) simulated by H never halts
D(D) simulated by H1 halts
D(D) simulated by H never halts
D(D) simulated by H1 halts
D(D) simulated by H never halts
D(D) simulated by H1 halts
Since an aborted simulationm tells you NOTHING about the actual behavior of the program being simulated, and your actual H does abort, then it is just a LIE to represent the behavior of a DIFFERENT PROGRAM that you DEzCEPTIVELY call H that doesn't abort as having ANYTHING to do
D simulated by H never reaches past its own line 03
whether H aborts its simulation or not AND YOU KNOW THIS!!!
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Universally everyone wants to use the strawman deception at this
point and refer to something else besides D(D) simulated by H.
Even these people might agree that D(D) simulated by H DOES NOT HALT.
But D(D) simulated by H IS THE STRAWMAN as the question is about the execution of the actual program describe by the input (which needs to FULLY describe the input, and thus includes the H it calls).
It may seem that way because 99% believe that yet prior to
Pythagoras 100% of the people "knew" the Earth was flat.
You are just PROVING you are just a totally ignorant pathological lying idiot to think you can just change the question and get away with it
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you understand that it is incorrect for H to report on the behavior of its caller: void main() { D(D); } then this necessitates
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But it MUST report on the program described to it, which is a call to D(D), and it doesn't matter if that is what calls H.
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So, your claim is just a STUPID LIE.
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Yes, you can't ask "What is the behavior of the program that called you?"
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Ah so you know the computer science of this, that is great!
Right, but that isn't the quest
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But you CAN ask what is the behavior of M(d) even if M(d) happens to call you.
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So you don't totally understand the computer science of this.
So, how does the input "D,D" mean "The program that called you"?
Go ahead, explain that.
You can't. because you know it is a DAMN LIE.
The execution trace of D(D) simulated by H1 is the same as
the execution trace of int main() { D(D); }
The execution trace of D(D) simulated by H is NOT the same
as The execution trace of D(D) simulated by H1 because the
behavior of the simulated D(D) before H has aborted its
simulation is different than behavior of the simulated D(D)
after H has aborted its simulation.
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It is always flat out incorrect for any computable function to
ever report on the behavior of its caller or even the behavior
of itself. The theory of computation DOES NOT ALLOW THAT!
REFERENCE?
Computable functions are the formalized analogue of the intuitive notion
of algorithms, in the sense that a function is computable if there
exists an algorithm that can do the job of the function, i.e. given an
input of the function domain it can return the corresponding output.
https://en.wikipedia.org/wiki/Computable_functionH must compute that mapping of the behavior of D(D) simulated
by H BEFORE H HAS ABORTED THIS SIMULATION.
You have been asked this before, so repeating the claim without a source is an admission that you don't have one and are just admitting it is a deceptive lie to throw the arguement off track.
Everyone that is really paying super close attention and really
really understands computable functions already knows this.
No one else ever spend 20,000 hours going over this material again and
again and again to see these details. I myself just came up with this
(in retrospect) obvious insight only in my 19th year of working on this.
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H(D,D) has different behavior than H1(D,D) even though the only
difference between H and H1 is that D calls H and does not call H1.
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DIFFERENT PROGRAMS ARE DIFFERENT
D(D) simulated by H has different behavior than D(D) simulated
by H1 even though the ONLY difference between H and H1 is that
D calls H and D does not call H.
The execution trace of D(D) simulated by H1 is the same as
the execution trace of int main() { D(D); }
The execution trace of D(D) simulated by H is NOT the same
as The execution trace of D(D) simulated by H1 because the
behavior of the simulated D(D) before H has aborted its
simulation is different than behavior of the simulated D(D)
after H has aborted its simulation.
Original Linz Turing Machine H
H.q0 ⟨M⟩ w ⊢* H.qy // M applied to w halts
H.q0 ⟨M⟩ w ⊢* Hqn // M applied to w does not halt
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Linz Turing Machine Ĥ
Ĥ.q0 ⟨M⟩ ⊢* embedded_H ⟨M⟩ ⟨M⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨M⟩ ⊢* embedded_H ⟨M⟩ ⟨M⟩ ⊢* Ĥ.qn
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This exact same reasoning applies to
embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ relative to Linz H ⟨Ĥ⟩ ⟨Ĥ⟩
Right, embedded_H, to meet the requirements, must go to H^.qy if H^(M) will halt.
embedded_H will also do exactly the same thing as H for the same inputs since they are copies of the exact same algorithm
Since H (H^) (H^) goes to Qn (by your claim of correctly saying non-halting) then H^.q0 (H^) => embedded_H (H^) (H^) => H^.qn also and we see that H^ (H^) will halt, and thus embedded_H TO BE CORRECT needed to have gone to H^.qy, but didn't, so embedded_H is wrong, and thus H is wrong.
And Peter Olcott is proved to be too dumb to understand that a wrong answer is not correct.
Which also makes him a pathological liar.
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You are just proving you total lack of understand of the nature of the problem.
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You can't show any error in my actual reasoning the best that
you can show is the conventional wisdom arrives at different
conclusions through different assumptions.
Sure I have, many times and you ignore it because you are too stupid and lack the understanding of the fundamenals.
The fundamentals that you refer to are simply a different set of
assumptions. I do know these assumptions and I do understand them.
I also understand their philosophical foundational error.
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H can abort its simulation of D and correctly report that D specifies a non-halting sequence of configurations.
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Nope, it CAN'T correctly say that a program that will halt when run is none halting.
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If D simulated by H is unequivocally non-halting and we are
only reporting on the behavior of D simulated by H then I am
definitely correct.
WRONG QUESTION so a STRAWMAN.
Within my assumptions (that you believe to be incorrect)
I AM CORRECT.
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If H is actually required to report on the behavior of its
caller then I am incorrect.
H is required to report on the behavior of the program whose description it was given (agree or admit you are just a liar).
You simply changed the subject.
If my assumptions are wrong yet my conclusion does correctly
follow from these incorrect assumptions then my conclusion is
correct within these incorrect assumptions.
That the program happens to call this decider is irrelevent.
> Thus, H *IS* actually required to report on the behavior of a program
that calls it, so you ARE incorrect.
No one that is really good at computer science would agreed to that.
Note, the thing you seem to be confusing is that there is no possible input that you can give to H that means, answer about whatever program happens to be calling this instance of H.
Those words are not very clear.
H is NOT ALLOWED to report on the behavior of the program that calls it
the same way that int sum(int x, int y) { return x + y; } when called
with sum(2,3) IS NOT ALLOWED TO REPORT THE SUM OF 4 + 5.
The fact that D calls H DOES CHANGE THE BEHAVIOR OF D(D)
simulated by H relative to THE BEHAVIOR OF D(D) simulated by H1.
You can ignore that into perpetuity yet the facts remain the same.
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer
Undecidability based on epistemological antinomies V2
Re: Undecidability based on epistemological antinomies V2