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On 11/5/24 12:08 PM, olcott wrote:Well we did get somewhere on this so that is more progress.On 11/5/2024 6:03 AM, Richard Damon wrote:But the emulation by HHH isn't the correct measure of DDD reaching its return statement.On 11/4/24 10:15 PM, olcott wrote:>On 11/4/2024 8:42 PM, Richard Damon wrote:>On 11/4/24 8:32 PM, olcott wrote:>On 11/4/2024 6:21 PM, Richard Damon wrote:>On 11/4/24 7:48 AM, olcott wrote:>On 11/4/2024 6:07 AM, Richard Damon wrote:>On 11/3/24 11:03 PM, olcott wrote:>On 11/3/2024 9:57 PM, Richard Damon wrote:>On 11/3/24 10:19 PM, olcott wrote:>On 11/3/2024 7:46 PM, Richard Damon wrote:>On 11/3/24 8:38 PM, olcott wrote:>On 11/3/2024 7:26 PM, Richard Damon wrote:>On 11/3/24 8:21 PM, olcott wrote:>>>
What would an unbounded emulation do?
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Keep on emulating for an unbounded number of steps.
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Something you don't seem to understand as part of the requirements.
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Non-Halting isn't just did reach a final state in some finite number of steps, but that it will NEVER reach a final state even if you process an unbounded number of steps.
Would an unbounded emulation of DDD by HHH halt?
Not a valid question, as your HHH does not do an unbounded emulation, but aborts after a defined time.
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*Now you are contradicting yourself*
YOU JUST SAID THAT HHH NEED NOT DO AN UNBOUNDED
EMULATION TO PREDICT WHAT AN UNBOUNDED EMULATION WOULD DO.
Right. it doesn't NEED to do the operation, just report what an unbounded emulation would do.
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You asked about an "unbounded emulation of DDD by HHH" but that isn't possible, as HHH doesn't do that.
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On 11/3/2024 12:20 PM, Richard Damon wrote:
> On 11/3/24 9:39 AM, olcott wrote:
>>
>> The finite string input to HHH specifies that HHH
>> MUST EMULATE ITSELF emulating DDD.
>
> Right, and it must CORRECTLY determine what an unbounded
> emulation of that input would do, even if its own programming
> only lets it emulate a part of that.
>
>
*You JUST said that HHH does not need to do an unbounded emulation*
*You JUST said that HHH does not need to do an unbounded emulation*
*You JUST said that HHH does not need to do an unbounded emulation*
*You JUST said that HHH does not need to do an unbounded emulation*
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Right, it doesn't need to DO the unbounded emulatiohn just figure out what it would do.
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Just like we can compute:
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1 + 1/2 + 1/4 + 1/8 + ... + 1/2^n + ...
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Ether by adding the infinite number of terms, or we can notice something about it to say it will sum, in the infinite limit, to 2.
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In the same way, if HHH can see something in its simulation that tells it THIS this program can NEVER halt, it can report it.
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Anyone with sufficient technical competence can see that
the unbounded emulation of DDD emulated by HHH can never halt.
No, because the HHH that is given doesn't do that, and that is the only one that matters.
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On 11/3/2024 12:20 PM, Richard Damon wrote:
> On 11/3/24 9:39 AM, olcott wrote:
>>
>> The finite string input to HHH specifies that HHH
>> MUST EMULATE ITSELF emulating DDD.
>
> Right, and it must CORRECTLY determine what an unbounded
> emulation of that input would do, even if its own programming
> only lets it emulate a part of that.
>
>
If you are going to keep contradicting yourself
I am going to stop looking at anything you say.
And where is the contradiction?
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HHH doesn't need to do the unlimited emulation, just say what the unlimited emulation by the unlimited emulator (which WILL be a different program) will do.
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That is what I have been saying all along.
So, you agree that HHH1's emulation to the completion shows that the complete emulation of the input to HHH does halt, and thus the correct answer for HHH to give for *THIS* input, which has implicitly included *THIS* HHH as part of it, is that it halts.
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Nothing like this.
You continue to fail to understand that halting
requires reaching the "return" instruction final
halt state. DDD emulated by HHH never does this.
By the correct meaning of the statement, it is just false.ChatGPT explains why and how it <is> the correct measure
By your attempted meaning, it is just nonsense, and thus a lie to claim it to be true.*It is actually your words that are nonsense*
You are just trapped in your equivocation, unable to resolve it without admitting your error, but because of it, you are just showing your stupidity.It is ridiculously stupid to call the ACTUAL x86
>But it doesn't represent the program DDD.And thus, HHH is just wrong.>
>>>No contradiciton in that, unlike your claim that HHH can do a partial emulation to predict what it itself would do when it does a complete emulation of the exact same input using the exact same code.>
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I have never said that.
*ChatGPT explains that HHH does meet your model*
Nope, it admitted that it doesn't.
>>>
*Simplified Analogy*
Think of HHH as a "watchdog" that steps in during real
execution to stop DDD() from running forever. But when
HHH simulates DDD(), it's analyzing an "idealized" version
of DDD() where nothing stops the recursion. In the simulation,
DDD() is seen as endlessly recursive, so HHH concludes that
it would not halt without external intervention.
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https://chatgpt.com/share/67158ec6-3398-8011-98d1-41198baa29f2
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Except that your input is a LIE.
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It is very stupid to say that a byte string of x86 code is a lie.
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THAT is the lie.
That is not what the machine code of DDD that callsHow am I contradicting myself?It isn't analyzing an "idealized" version of its input, it is analyzing the input it is actually given, where DDD calls the existing HHH, not the idealized version of it.>
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You should know this, and thus all you have done is show that your logic is just based on LIES.
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The paragraph that you said that I keep quoting
is a breakthrough. that you keep contradicting
your own words seems quite dumb to me.
HHH doesn't need to actualy emulate DDD completely, just determine, like HHH1 does, that it will reach the return instruction.
That you called machine code a liar decimates your>Of the FULL program.Now, if you want to try to define some alternate model of computation>
No need to do this the x86 model of computation says it all.
(a) Finite string of x86 machine code DDD +
(b) The semantics of the x86 language +As show by the FULL emulation of the input/
(c) DDD is calling its own termination analyzerWhich is irrelevent to (b), and if we know that HHH is a termination analyser, we KNOW it will return an answer and thus DDD will halt
∴ HHH is correct to reject its input as non-haltingNope, and making the claim just shows that you are an ignorant liar.
A DDD does HALT this DDD IS NOT THE INPUT !!!>But the input halts, in both models, so your HHH is wrong both ways.
As far as the non-halting of the input to the conventional
HP proofs goes these two models are exactly the same.
-->where you can do this, go ahead and try. I suspect you will find it impossible to make such a system objectively consistant with inputs not matching themselves, or programs being able to include stuff that changes on them. And, until you DO actually define such a system, you can't use it as part of your arguement. Your don't even seem to have a real concept of such a system, but just think that is what is.>
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Sorry, you are just proving that you don't understand what you are talking about.
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