On 25/07/2024 14:56, olcott wrote:
On 7/24/2024 10:29 PM, Mike Terry wrote:
On 23/07/2024 14:31, olcott wrote:
On 7/23/2024 1:32 AM, 0 wrote:
On 2024-07-22 13:46:21 +0000, olcott said:
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On 7/22/2024 2:57 AM, Mikko wrote:
On 2024-07-21 13:34:40 +0000, olcott said:
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On 7/21/2024 4:34 AM, Mikko wrote:
On 2024-07-20 13:11:03 +0000, olcott said:
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On 7/20/2024 3:21 AM, Mikko wrote:
On 2024-07-19 14:08:24 +0000, olcott said:
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When we use your incorrect reasoning we would conclude
that Infinite_Loop() is not an infinite loop because it
only repeats until aborted and is aborted.
>
You and your HHH can reason or at least conclude correctly about
Infinite_Loop but not about DDD. Possibly because it prefers to
say "no", which is correct about Infinte_loop but not about DDD.
>
>
*Because this is true I don't understand how you are not simply lying*
int main
{
DDD();
}
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Calls HHH(DDD) that must abort the emulation of its input
or {HHH, emulated DDD and executed DDD} never stop running.
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You are the lying one.
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If HHH(DDD) abrots its simulation and returns true it is correct as a
halt decider for DDD really halts.
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>
(b) We know that a decider is not allowed to report on the behavior
computation that itself is contained within.
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No, we don't. There is no such prohibition.
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Turing machines never take actual Turing machines as inputs.
They only take finite strings as inputs and an actual executing
Turing machine is not itself a finite string.
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The definition of a Turing machine does not say that a Turing machine
is not a finite string. It is an abstract mathematical object without
a specification of its exact nature. It could be a set or a finite
string. Its exact nature is not relevant to the theory of computation,
which only cares about certain properties of Turing machines.
>
Therefore It is not allowed to report on its own behavior.
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Anyway, that does not follow. The theory of Turing machines does not
prohibit anything.
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Another different TM can take the TM description of this
machine and thus accurately report on its actual behavior.
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If a Turing machine can take a description of a TM as its input
or as a part of its input it can also take its own description.
Every Turing machine can be given its own description as input
but a Turing machine may interprete it as something else.
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In this case we have two x86utm machines that are identical
except that DDD calls HHH and DDD does not call HHH1.
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It is empirically proven that this changes their behavior
and the behavior of DDD.
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>
You say a lot about things that are "empirically proven" and without exception they are never "proven" at all.
>
It is empirically proven according to the semantics of the
x86 machine code of DDD that DDD correctly emulated by HHH
has different behavior than DDD correctly emulated by HHH1.
Perhaps your actual code does behave differently!
The questions are:
a) are HHH and HHH1 "identical copies", in the TM machine sense of incorporating
the algorithm of one TM inside another TM? (As Linz incorporates H
inside H^, meaning that the behaviours of H and embedded_H MUST be identical for any
input.)
[You claim HHH and HHH1 /are/ proper copies, and yet give different results for
input (D), which is impossible.]
b) If the two behaviours HHH/HHH1 are indeed different, WHAT precisely is the coding
difference that accounts for that different behaviour? (Like, with your H/H1 the
difference was that H used H's address as part of its algorithm, while H1 used H1's
address.)
_DDD()
[00002177] 55 push ebp
[00002178] 8bec mov ebp,esp
[0000217a] 6877210000 push 00002177
[0000217f] e853f4ffff call 000015d7
[00002184] 83c404 add esp,+04
[00002187] 5d pop ebp
[00002188] c3 ret
Size in bytes:(0018) [00002188]
_main()
[00002197] 55 push ebp
[00002198] 8bec mov ebp,esp
[0000219a] 6877210000 push 00002177
[0000219f] e863f3ffff call 00001507
[000021a4] 83c404 add esp,+04
[000021a7] 33c0 xor eax,eax
[000021a9] 5d pop ebp
[000021aa] c3 ret
Size in bytes:(0020) [000021aa]
machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
[00002197][001037fb][00000000] 55 push ebp
[00002198][001037fb][00000000] 8bec mov ebp,esp
[0000219a][001037f7][00002177] 6877210000 push 00002177 ; push DDD
[0000219f][001037f3][000021a4] e863f3ffff call 00001507 ; call HHH1
New slave_stack at:10389f
Begin Local Halt Decider Simulation Execution Trace Stored at:1138a7
[00002177][00113897][0011389b] 55 push ebp
[00002178][00113897][0011389b] 8bec mov ebp,esp
[0000217a][00113893][00002177] 6877210000 push 00002177 ; push DDD
[0000217f][0011388f][00002184] e853f4ffff call 000015d7 ; call HHH
New slave_stack at:14e2c7
Begin Local Halt Decider Simulation Execution Trace Stored at:15e2cf
[00002177][0015e2bf][0015e2c3] 55 push ebp
[00002178][0015e2bf][0015e2c3] 8bec mov ebp,esp
[0000217a][0015e2bb][00002177] 6877210000 push 00002177 ; push DDD
[0000217f][0015e2b7][00002184] e853f4ffff call 000015d7 ; call HHH
New slave_stack at:198cef
[00002177][001a8ce7][001a8ceb] 55 push ebp
[00002178][001a8ce7][001a8ceb] 8bec mov ebp,esp
[0000217a][001a8ce3][00002177] 6877210000 push 00002177 ; push DDD
[0000217f][001a8cdf][00002184] e853f4ffff call 000015d7 ; call HHH
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
[00002184][00113897][0011389b] 83c404 add esp,+04
[00002187][0011389b][000015bc] 5d pop ebp
[00002188][0011389f][0003a980] c3 ret
[000021a4][001037fb][00000000] 83c404 add esp,+04
[000021a7][001037fb][00000000] 33c0 xor eax,eax
[000021a9][001037ff][00000018] 5d pop ebp
[000021aa][00103803][00000000] c3 ret
Number of Instructions Executed(352831) == 5266 Pages
that seems to be a (partial) trace of HHH1(D). It's not clear exactly what's happening - for example, who produces the message "Local Halt Decider: Infinite Recursion Detected Simulation Stopped"? It might be outer HHH1 or one of the inner HHH's. And what result did HHH1 report?
And more to the point what did HHH(D) do, and what is the difference?
AND WHAT CODING within HHH/HHH1 ACCOUNTS FOR THAT DIFFERENCE?
You previously claimed that H and H1 behaviours were different as evidence that "copies of routines" don't necessarily produce the same behaviour as the original routine, due to magical pathelogical relationships. But if the copies are done properly of course they will produce the same behaviour, because the x86 language is deterministic. I'm assuming you're not just cheating and using the mutable global data trick! or similar...
>
If you care study the code that I just provided
(and don't just ignore this and assume that I must be wrong)
you can see that when DDD is correctly emulated by HHH that
DDD does correctly have different behavior than DDD correctly
emulated by HHH1.
All I see above is a partial trace of HHH1(DDD). No behaviour is traced for HHH1, and there is no comparison with HHH(DDD).
Suggestion: make a post starting afresh with both traces, and a statement of how the behaviours differ, and ideally with your explanation of how the coding of HHH1/HHH accounts for that difference. [I suspect you don't actually know the answer to the last question, but you ought to be able to work it out.]
You could have seen this three years ago yet everyone simply
assumes that I must be wrong thus makes sure to ignore the
verified facts that N steps of DDD are correctly emulated by
HHH and M steps of DDD are correctly emulated by HHH1.
They guess that these steps are wrong entirely on the basis
of not bothering to check.
So what had you messed up with H/H1? H and H1 internally both used their /own/ addresses for comparisons with simulated code addresses. So
They do not do this.
H/H1 did that.
When HHH emulates itself emulating DDD is has no
idea that it is emulating itself. Likewise for when HHH1 emulates DDD
calling HHH(DDD) it does not know that it is not emulating itself.
Well that's great, you've not made the exact same mistake twice! :)
So by what you're saying, on the face of it we'd expect HHH/HHH1 behaviour to be the same - unless perhaps you've made some other mistake, of course...
they are not copies of each other - if they were, H1 would have used H's address (effectively this address is like a const global variable, and
That is the same as saying the identical twins have the exact same body
not two different bodies in two different locations.
HHH1 and HHH are essentially identical and the only reason why
DDD correctly emulated by HHH has different behavior than DDD
correctly emulated by HHH1 is that DDD does call HHH in recursive
emulation and DDD does not call HHH1 in recursive emulation.
No. In both cases we are discussing the (partial) simulation of DDD(). Since HHH and HHH1 are allegedly correct copies of each other, the simulation each performs should be identical, but for some reason it is not. Your supposed explanation simply does not address any possible cause of this. All you are saying is
1. the behaviour /is/ different, and
2. one difference in the setup is that DDD calls HHH, not HHH1,
3. therefore (2) causes the behaviour to be different.
(But (3) is not logic, and (2) would not explain why or how the behaviour could be different.)
[..snip silly OT stuff..]
When DDD calls HHH in recursive emulation
(as I have proven that it does**) and DDD does not call HHH1
in recursive simulation (as I have proven that it does not**)
then DDD will have different behavior.
No. In both cases we are talking of partial simulations of DDD(), and there is only one "DDD() behaviour" which is the FULL trace of computation DDD().
Different partial simulations may simulate more or less steps of DDD() according to their abort logic, but every single such simulation is simply calculating the steps [state transitions] of the one DDD() behaviour. And in the HHH/HHH1 scenario even that difference cannot occur, because HHH/HHH1 have /the same/ abort criteria, so must necessarily simulate THE SAME NUMBER OF STEPS.
**According to the semantics of the x86 machine code
provided above.
out to be just a case of naff programming on your part - nothing "magical" about pathelogical relationships between H/D which aren't present for H1/D; just naff coding. [H1 using a different algorithm meant it never matched H's abort decision, so effectively reverted to UTM behaviour that simulated (D,D) to completion, including the final return.]
>
Both HHH and HHH1 use the same algorithm. The input to HHH(DDD)
must be aborted to prevent the non-halting behavior of HHH, and DDD.
The input to HHH1(DDD) need not be aborted because the input to
HHH(DDD) already has been aborted.
The DDD of HHH(DDD) is in the different process state of needing
to be aborted than the DDD of HHH1(DDD) of *not* needing to be
aborted.
DDD needing to be aborted before any DDD has been aborted is
like being hungry before one has eaten.
DDD not needing to be aborted after some DDD has been aborted
is like *not* being hungry after one has eaten.
All this needing/not needing is rubbish - HHH and HHH1 do what they do because of how they are coded, not due to their "needs".
You claim they behave differently, so WHY (at the coding level) do they behave differently. The same code given the same input always behave the same, so you simply fail to understand something very basic here.
[..snip..]
It's your code so you should be the one to explain anomalies like this at the coding level. Why /exactly/ does HHH1 produce a different result to HHH?
Why does finite recursion halt and infinite recursion not halt?
I'm sure you know the answer to this. More to the point, both HHH and HHH1 are simulating the SAME computation DDD(), and so calculate the SAME computation steps. I.e. they are both simulating finite recursion, so your question is irrelevant.
HHH does see recursive emulation that will never stop unless aborted.
HHH1 does not see this.
HHH and HHH1 see exactly the same computation steps, and they have [according to you] the same logic evaluating those steps, and so should make the same halting decision.
Neither of them "sees" that "emulation will never stop unless aborted"! They don't have eyes, and can't see the future, and can't evaluate the alternative realities implied by your "unless".
In fact, what HHH and HHH1 both do is scan the same trace [of computation DDD()] looking for the same pattern of previous states, and if the pattern is found, they abort. So again the question is WHY DOES ONE OF THEM DECIDE TO ABORT AND NOT THE OTHER?
Don't answer this by explaining their "needs"; answer by explaining what code difference causes this. [Except you've claimed there /is/ no code difference, so you've got a challenge here!]
Mike.
DDD correctly emulated by HHH is correctly rejected as non-halting V2
Re: DDD correctly emulated by HHH is correctly rejected as non-halting V2
