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@REPLYADDR Richard Damon
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@REPLYTO 2:5075/128 Richard Damon
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You keep on making the same mistakes,
On 8/25/23 2:48 PM, olcott wrote:
> A pair of C functions are defined such that D has the halting problem
> proof`s pathological relationship to simulating termination analyzer H.
> When it is understood that D correctly simulated by H (a) Is the
> behavior that H must report on and (b) Cannot possibly terminate
> normally then it is understood that D is correctly determined to be non-
> halting.
So, you are STIPULATING, that since D(x) calls H(x,x), that this means
askin H about the behavior of the program x given the input x, which
means that D(D) callihg H(D,D) means it is asking H about the behavior
of the program D(D)
>
> We can know that D correctly simulated by H must have different behavior
> than D(D) directly executed in main() because we can see (in its
> execution trace shown below) exactly how the pathological relationship
> between D and H changes the behavior of D relative to H.
Nope, that statement just PROVES that you don`t understand the meaning
of the words you are using, and are just a ignorant pathological lying
idiot.
>
> For any program H that might determine whether programs halt, a
> "pathological" program D, called with some input, can pass its own
> source and its input to H and then specifically do the opposite of what
> H predicts D will do. No H can exist that handles this case.
>
https://en.wikipedia.org/wiki/Halting_problem
So? what is wrong with the fact that no H can exist that handles this case?
>
> "A decision problem is a yes-or-no question on an infinite set of
> inputs"
https://en.wikipedia.org/wiki/Decision_problem#Definition
>
Right and a "X decider problem" means the decider must answer based on
the definition of problem X, and a "Halt Decider" must answer based on
whether to program described by its input will halt.
> Can D correctly simulated by H terminate normally?
Improper question. Have you stopped lying? (that one actually has an
answer, NO).
Since H doen`t actually correctly simulate its input, the question is
based on a false premise.
> The x86utm operating system:
https://github.com/plolcott/x86utm
> is based on an open source x86 emulator. x86utm enables one C function
> to execute another C function in debug step mode.
>
And your program shows that D(D) DOES HALT, and thus H`s answer is WRONG>
> // The following is written in C
> //
> 01 typedef int (*ptr)(); // pointer to int function
> 02 int H(ptr x, ptr y) // uses x86 emulator to simulate its input
> 03
> 04 int D(ptr x)
> 05 {
> 06 int Halt_Status = H(x, x);
> 07 if (Halt_Status)
> 08 HERE: goto HERE;
> 09 return Halt_Status;
> 10 }
> 11
> 12 void main()
> 13 {
> 14 H(D,D);
> 15 }
>
> *Execution Trace*
> Line 14: main() invokes H(D,D);
>
> *keeps repeating* (unless aborted)
LIE.
You H does no such thing, you don`t seem to understand that programs do
what they are actually programmed to do.
You are just proving you don`t understand the meaning of words like
"Program", "Correct" or even "Logic"
> Line 06: simulated D(D) invokes simulated H(D,D) that simulates D(D)
Right, your H is defined to ALWAYS abort its simulation of this input,
the, your statement of "keeps repeating" is just a LIE. You condition is
irrelevent.
>
> *Simulation invariant*
A meanilngless term (par for you) since in the problem we have a SINGLE
H, with a specified input, so there is nothing to "Vary", so we can`t
have an "invariant".
At best, you mean, if we look over all possible H`s that meet your
general definition, we can prove the following for all of them.
> D correctly simulated by H cannot possibly reach past its own line 06.
Which is a correct conclusion, it is IMPOSSIBLE for any H to be able to
simulate the program built by the template to that final state.
THe problem is that this just proves that NO H can, by your logic, can
show that its input is Halting. It says NOTHING about not it being
non-halting.
Remember, every H in this set is a different problem, and creates a
different input, so the results obtained about one input says nothing
about a different one.
>
> H correctly determines that D correctly simulated by H cannot possibly
> terminate normally on the basis that H recognizes a dynamic behavior
> pattern equivalent to infinite recursion. H returns 0 this basis.
Nope. UNSOUND LOGIC. Since any H that aborts its simulation doesn`t do a
"Correct SImulation", you don`t have anything to make your deduction on.
You are just proving your stupidity.
>
> *ADDENDUM*
>
> (1) The source-code of H and D conclusively proves that D correctly
> simulated by H cannot possibly terminate normally.
But ONLY if H DOES correctly simulate its input, at which point it never
aborts, and thus, as you proved above, it never answers.
>
> *THIS IS THE PART THAT EVERYONE LIES ABOUT*
> (2) The correct simulation of D by H must include the fact that
> D would continue to call H until stack overflow crash unless H
> aborts its simulation of D.
Nope, a correcrt simulaiton of D by ANYTHING (includeing H) must include
the ACTUAL behavior of H, not the behavior of some other program.
>
> (3) (2) Means that D is correctly simulated by H and this correctly
> simulated D is non-halting.
>
Nope, just shows you don`t understand what correct logic requires.
> (4) "A decision problem is a yes-or-no question on an infinite set
> of inputs"
https://en.wikipedia.org/wiki/Decision_problem#Definition
> This means that the behavior of non-inputs is not allowed to be
> considered.
>
Right, and it needs to give the correct answer to be correct.
You are just proving that you are an ignorant pathological lying idiot.
>
>
> *Termination Analyzer H is Not Fooled by Pathological Input D*
https://www.researchgate.net/publication/369971402_Termination_Analyzer_H_is_Not
_Fooled_by_Pathological_Input_D
>
>
Just shows your stupidity to the world.
Note, the fact that you keep on making the same DISPROVEN claims, shows
you are just a brain damaged insane moron.
You seem to be incapable of learning.
This seems to prove that you are just a pathological liar.
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