On 8/11/24 8:40 AM, olcott wrote:
On 8/11/2024 6:08 AM, Richard Damon wrote:
On 8/10/24 10:38 PM, olcott wrote:
On 8/10/2024 9:21 PM, Richard Damon wrote:
On 8/10/24 9:43 PM, olcott wrote:
On 8/10/2024 8:13 PM, Richard Damon wrote:
On 8/10/24 8:51 PM, olcott wrote:
On 8/10/2024 7:20 PM, Richard Damon wrote:
On 8/10/24 7:52 PM, olcott wrote:
On 8/10/2024 5:47 PM, Richard Damon wrote:
On 8/10/24 6:41 PM, olcott wrote:
On 8/10/2024 4:53 PM, Richard Damon wrote:
On 8/10/24 5:37 PM, olcott wrote:
On 8/10/2024 4:33 PM, Richard Damon wrote:
On 8/10/24 5:18 PM, olcott wrote:
On 8/10/2024 3:58 PM, Richard Damon wrote:
On 8/10/24 4:36 PM, olcott wrote:
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As I have countlessly proven it only requires enough correctly
emulated steps to correctly infer that the input would never
reach is "return" instruction halt state.
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Except that HHH does't do that, since if HHH decides to abort and return, then the DDD that it is emulating WILL return, just after HHH has stopped its emulation.
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You just confuse the behavior of DDD with the PARTIAL emulation that HHH does, because you lie about your false "tautology".
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Denying a tautology seems to make you a liar. I only
say "seems to" because I know that I am fallible.
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Claiming a false statement is a tautology only make you a liar.
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In this case, you lie is that the HHH that you are talking about do the "correct emulation" you base you claim on.
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That is just a deception like the devil uses, has just a hint of truth, but the core is a lie.
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What I say is provably correct on the basis of the
semantics of the x86 language.
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Nope.
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The x86 language says DDD will Halt if HHH(DDD) returns a value.
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HHH is called by main() there is no directly executed DDD()
any where in the whole computation.
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Except in your requirements, and we can see what it does by adding a call to DDD from main, since nothing in your system calls main.
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All that you need to know is that there is not any
directly executed DDD() anywhere in the computation.
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But there ccould be, and the behavior of it is what matters.
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The key error of the halting problem proofs all of these
years has been the false assumption that a halt decider
must report on the behavior of the computation that itself
is contained within.
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But it isn't a false assemption, but an actual requirement.
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A Halt Decider must be able to correctly answer for ANY Turing Machine represented as its input.
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ANY includes those that are built from a copy of itself.
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So, a Halt Decider needs to be able to correctly answer about programs that include copies of itself, even with contrary behavior, which is what makes it impossible to compute.
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You seem to confuse non-computable with invalid, it seems in part because you don't understand the difference between knowledge and truth.
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Everyone has simply assumed that the behavior of the
input to a decider must exactly match the direct execution
of this input. They only did this because everyone rejected
simulation out-of-hand without review.
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Because that is the DEFINITION of what it is to decide on.
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You just don't understand what a requirement is.
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Since the DEFINITION of "Correct Simulation" that you are trying to use (from a UTM) means a machine the EXACTLY reproduces the behavior of the direct exectution of the machine described by the input, the correct simulation must exactly match the behavior of the direct execution.
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You can't get out of it by trying to lie about it being different.
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This caused them to never notice that the input simulated
according to its correct semantics does call its own decider
in recursive simulation thus cannot possibly return to its
caller. The Linz proof is sufficiently isomorphic so this equally
applies to the Linz TM proof.
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Nope, just shows you don't know what "Correct" means.
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Your proof is NOT "sufficiently isomorphic" since by your own claims it is clearly not even Turing Complete, so no where near isomorphic.
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If HHH were to report on the direct execution of DDD it would
be breaking the definition of a halt decider that only computes
the mapping from its input...
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Nope. Since the mapping that it is supposed to compute is DEFINED as based on the direct exectut
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No it never has been this. I has always been a mapping
from the behavior that the finite string specifies. It
has never been the behavior of the actual computation
that the decider is contained within.
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And thatg behavior is specified to be the behavior of the program the input represents. PERIOD.
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That has never been true. It is always the case that every
decider of any kind only computes the mapping from its input
finite string and never gives a rat's ass about anything else
anywhere else.
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No, you are confusing capability with requirements.
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A "Foo Decider" has ALWAYS been required to compute the "Foo" mapping, as that mapping is defined.
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The "Halting" mapping is defined as the behavior of the machine/input represented by the input, so the input needs to be a representation of the program and input and the decider tries to compute the mapping of that representation to the behavior that program represents.
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How that isn't the "mapping" of the input to a Halt Decider seems to put a big hole in your argument.
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So, the behavior of the program the input describes *IS* the mapping that HHH needs to try to compute to be a halt decider, as that is the mapping that Halting defines.
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Now, it is only CAPABLE of computing a computable mapping, like a finite length emulation of the input, and since it has been shown that Halting is NOT a computable mapping, there will be some inputs (like H^) that the decider WILL get wrong. That doesn't say the problem is wrong, or specifies the wrong mapping, just that the problem can't be done with a computation.
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Your HHH may be a decider, if it ALWAYS halts for any input give to it, but since the answer it gives doesn't always match the Halting mapping, it just isn't a Halt Decider.
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Ben should have known this. He didn't. People here that pretend
to know computer science should know that. That they don't proves
that they are (to some degree) fakers.
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What make you think they don't know it.
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After all, the mapping of the finite string that completely represents the program and input to be decided to whether that program and input that it represents will halt when run is a perfectly defined mapping.
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The semantics of my 100% fully specified concrete example
conclusively proves that the behavior of DDD correctly emulated
by HHH is a different sequence that the directly executed DDD().
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Nope, since you have never shown the requested output, you have no grounds to claim that you HHH does a correct x86 emulation of the input.
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Remember, that means the trace is of what HHH does, so starts with is emulating the beginnig of DDD(), and then following the call to HHH that it makes into HHH, all the way until the emulator decides to stop.
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Then you need to point out what instruction that it correctly emulated differed from what the actually directly executed machine would have done, and explain WHY its difference is correct.
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Even a pretty stupid person can see that each HHH does emulate
its input correctly even if it does this by wild guess. They
*merely have to bother to pay attention that the emulated lines*
*are the same lines as the x86 source code of DDD*
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No, each HHH CONDITIONALLY emulates the input it is given, and will abort its emulation and return to its caller if it decides the input isn't going to halt. As such, to prove infinite recursion you have to show that NONE of the emulated HHHs will abort their emulations and return, which, since they are the same HHH as deciding, means your top level HHH can't do that either, so it fails to be a decider.
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That after three years no one has bothered to do that can't seem
to have any plausible explanation besides playing sadistic trollish
head games.
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No, the fact that you can't prove that the "correct emulation" differs from the Direct Execution means your claim that they differ is a lie, and thus HHH is just wrong about it deciding that DDD will not halt, when it does.
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_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]
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Begin Local Halt Decider Simulation Execution Trace Stored at:1138cc
[00002172][001138bc][001138c0] 55 push ebp ; housekeeping
[00002173][001138bc][001138c0] 8bec mov ebp,esp ; housekeeping
[00002175][001138b8][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][001138b4][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
New slave_stack at:14e2ec
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Note, this is NOT a correct emulation, and is just your way to try to LIE.
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We need to see the instruction of HHH here.
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In other words you cannot see that the following code exactly
matches the x86 source-code of DDD thus proving that the second
HHH did emulate it input correctly?
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Your problem is that that is not the COMPLETE x86 source code of the PROGRAM DDD, as that needs the code for HHH included in it.
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A correct x86 emulation of DDD includes the correct emulation of HHH.
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It does do this yet mixing in the 200 pages of other code
makes it too difficult to see the execution trace of DDD.
No, to make a claim, you need to provide the actual proof.
Note, your "200 page" listing ISN'T the trace that HHH makes, is it begins in *main* not *ddd*, so this is a trace that x86utm is making of its runnning of HHH.
Note, in that trace, when HHH does a jmp to 000015E7, the trace of it stops, so HHH did NOT trace the behavior of HHH, but just the first couple of instructions of it, that didn't actually do anything important.
Note, those instruction ARE traces in the listing that x86UTM does, but NOT the sublisting of the instructions that HHH did.
So, you are caught in your lie.
https://liarparadox.org/HHH(DDD)_Full_Trace.pdf
I now have the first four instructions of HHH highlighted
in red proving that they were correctly emulated.
But this isn't the right emulation. This isn't the output of HHH emulating DDD but of x86UTM running main calling HHH.
And it shows that HHH stops emulating itself after doing only a couple of instructions.
The code trace for HHH emulating itself emulating DDD
has always been way beyond the capacity of the human mind
unless each instruction of all of the functions in 200
pages is highlighted in its own color.
Nope, maybe beyond YOUR mind, but that's because it is too small.
Maybe if you make the listing JUST the emulation that HHH does, and not the full trace of the program HHH making that trace, it would be more reasonable in length.
But, muy guess is you are just hacking at a program you don't really understand and can't figure out how to do that listing, but are just piggy backing on existing behavior that gives you stuff you don't actually want.
*THIS WAS NEVER EVER NEEDED IN THE LAST THREE YEARS*
Sure it has been. You have been makeing a FALSE claim all those years, trying to hide the lie behind the trace being too complicated.
It is dead obvious that HHH must have been emulated correctly
or it could not emulated DDD correctly. I show the trace of
each instance of DDD being emulated correctly.
But it didn't. Remember DDD includes HHH, which does a CONDITIONAL emulaiton of DDD, and the trace drops that conditionality, and thus leads you to an incorrect conclusion.
You have just lied to yourself, and made a fool of yourself to the whole world.
I do prove that each instance of HHH does derive the correct
execution trace of DDD, thus the trace is valid even if
each HHH does this by wild guess.
Nope, you forget that DDD includes HHH, and you don't emulation THAT correctly, so your whole trace is just incorrect.
One error spoils the whole proof.
HHH saying that HHH doesn't return is not a correct emulation of HHH, since HHH does return,
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You just don't understand what a PROGRAM is, or what CORRECT means.
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HHH and DDD are C functions. they have never been programs.
I already called you out on this and your ADD forgot.
But you build a program out of functions. Maybe part of your problem is that you don't understand that Computation Theory is about "programs", but programs that can be chained together, sort of like funcitons, but it means that a function includes everythinbg it calls as part of it.
HHH, when you include everything it calls is such a program.
DDD, when you inlcude everything it calls is also such a program, but without including HHH, it isn't.
I guess this is just another of your admissions of fundamentel errors in your understand that just blows up your whole proof.
When we are asked to decide on "DDD", we mean the DDD that includes ALL the code that it calls, including ALL of HHH.
Since HHH doesn't emulate all of the code of HHH that DDD calls, it isn't a correct emulation per the x86 language.
That people have not bothered to look at this for three years
is far too disingenuous to be called merely disingenuous.
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[00002172][0015e2e4][0015e2e8] 55 push ebp ; housekeeping
[00002173][0015e2e4][0015e2e8] 8bec mov ebp,esp ; housekeeping
[00002175][0015e2e0][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][0015e2dc][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
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And, HH may hve decided this, but that isn't a true determination.
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Again, more of your lies.
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Sorry, your "logic" just proves your stupidity, and the fact you can't see that just proves how stupid you are, and that is the worse kind of stupid.
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Who here is too stupid to know that DDD correctly simulated by HHH cannot possibly reach its own return instruction?
Re: Who here is too stupid to know that DDD correctly simulated,by HHH cannot possibly reach its own return instruction?