Liste des Groupes | Revenir à c theory |
On 4/26/2025 10:38 PM, Mike Terry wrote:You often say that you don't express something in a clear way, because "people get confused" when you say it that way. There is never any evidence whatsoever that anybody but you is confused about these points. What you really mean is that when clearly worded your confusions are more easily seen to be mistakes - which is not in your interest it seems.On 27/04/2025 04:07, Mike Terry wrote:People proved to fail to NOT comprehend it when it isOn 27/04/2025 01:22, olcott wrote:>On 4/26/2025 5:31 PM, dbush wrote:>On 4/26/2025 6:28 PM, olcott wrote:>On 4/26/2025 5:11 PM, dbush wrote:>On 4/26/2025 6:09 PM, olcott wrote:>On 4/26/2025 4:04 PM, Richard Damon wrote:>On 4/26/25 4:33 PM, olcott wrote:>On 4/26/2025 1:26 PM, Fred. Zwarts wrote:>Op 26.apr.2025 om 19:29 schreef olcott:>On 4/26/2025 12:16 PM, joes wrote:>Am Sat, 26 Apr 2025 11:22:42 -0500 schrieb olcott:>On 4/25/2025 5:09 PM, joes wrote:Which x86 semantics does a processor violate when deriving a haltingAm Fri, 25 Apr 2025 16:46:11 -0500 schrieb olcott:>On 4/25/2025 11:54 AM, Richard Damon wrote:>On 4/25/25 12:31 PM, olcott wrote:There are no finite string operations that can be applied to the inputOnce we understand that Turing computable functions are only allowedYouy have your words wrong. They are only ABLE to use finite
to derived their outputs by applying finite string operations to
their inputs then my claim about the behavior of DD that HHH must
report on is completely proven.
>
algorithms of finite string operations. The problem they need to
solve do not need to be based on that, but on just general mappings
of finite strings to finite strings that might not be described by a
finite algorithm.
The mapping is computable, *IF* we can find a finite algorith of
transformation steps to make that mapping.
>
to HHH(DD) that derive the behavior of of the directly executed DD
thus DD is forbidden from reporting on this behavior.Yes, there are, the operations that the processor executes. How did youWhen you try to actually show the actual steps instead of being stuck in
think it works?
>
utterly baseless rebuttal mode YOU FAIL!
state from the string description of DD?
>When any HHH emulates DD according to the finite string transformationYes, where is that line?
rules specified by the x86 language (the line of demarcation between
correct and incorrect emulation) no emulated DD can possibly reach its
final halt state and halt.
>
Everyone claims that HHH violates the rules
of the x86 language yet no one can point out
which rules are violated because they already
know that HHH does not violate any rules and
they are only playing trollish head games.
>
_DD()
[00002133] 55 push ebp ; housekeeping
[00002134] 8bec mov ebp,esp ; housekeeping
[00002136] 51 push ecx ; make space for local
[00002137] 6833210000 push 00002133 ; push DD
[0000213c] e882f4ffff call 000015c3 ; call HHH(DD)
[00002141] 83c404 add esp,+04
[00002144] 8945fc mov [ebp-04],eax
[00002147] 837dfc00 cmp dword [ebp-04],+00
[0000214b] 7402 jz 0000214f
[0000214d] ebfe jmp 0000214d
[0000214f] 8b45fc mov eax,[ebp-04]
[00002152] 8be5 mov esp,ebp
[00002154] 5d pop ebp
[00002155] c3 ret
Size in bytes:(0035) [00002155]
>
DD emulated by HHH according to the finite
string transformation rules of the x86 language
does emulate [00002133] through [0000213c] which
causes HHH to emulate itself emulating DD again
in recursive emulation repeating the cycle of
[00002133] through [0000213c].
>
Finite recursion,
Mathematical induction proves that DD emulated by
any HHH that applies finite string transformation
rules specified by the x86 language to its input
no DD can possibly reach its final halt state.
No, it doesn't, as you can't have an infinte series of a function that has been defined to be a specific instance.
>
One recursive emulation of HHH emulating itself emulating
DD after DD has already been emulated by DD once conclusively
proves that
>
simulated DD would never stop running unless aborted
>
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its input D
until H correctly determines that its *simulated D would never*
*stop running unless aborted* then
>
H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>
And again you lie by implying that Sipser agrees with you when it has been proven that he doesn't:
>
>
On Monday, March 6, 2023 at 2:41:27 PM UTC-5, Ben Bacarisse wrote:
> I exchanged emails with him about this. He does not agree with anything
> substantive that PO has written. I won't quote him, as I don't have
> permission, but he was, let's say... forthright, in his reply to me.
>
That professor Sipser did not have the time to
understand the significance of what he agreed to
does not entail that he did not agree with my
meanings of what he agreed to.
>
Professor Sipser did not even have the time to
understand the notion of recursive emulation.
Without this it is impossible to see the significance
of my work.
In other words, he did not you agree what you think he agreed to, and your posting the above to imply that he did is a form of lying.
>
*He agreed to MY meaning of these words*
He most certainly did not! He presumably agreed to what he /thought/ you meant by the words.
>
Since there is a natural interpretation of those words which would be correct, and relevant to a discussion concerning a simulating HD, my GUESS would be that he thought that was what you were saying: basically, the D in the quote below is clearly intended to represent *one* *specific* input whose halt status is being determined, namely the input D.
>
There is talk of "would never stop running if not aborted", which is saying that if H were replaced by a UTM (which never aborts its input) THEN UTM(D) WOULD RUN FOREVER. That amounts to the same thing as saying that H has determined [through examination of its simulation steps] that D does not halt [when run directly/natively]. Of course if H has determined that D does not halt, there's no point in simulating further, and H can just decide "non-halting" straight away.
>
NOTE: I said UTM( *D* ), not UTM(UTM) or UTM(D2) where D2 is some modified version of D that reflects changes to the embedded copy of modified H internal to D. The role of D in all this is /data/ viz the string representing the particular D being discussed. The role of H is /code/, H being the halt decider deciding the input D. D does not change when applying the "simulated D would never stop unless aborted", or imagining whatever hypothetical changes to H you are thinking of - only code of H is being (hypothetically) changed.
I suppose I should have made this clear, as you get confused by this point - The TM description D which is not changing, includes the [TM description of the] embedded copy of [original] H. I.e. H without any of your hypothetical imagined changes.
>
Much better still, stop imagining hypothetical changes to things and phrase things by introducing new objects with new names when required, so that a given name always means the same thing....
>
For example: rather than saying "DDD emulated by HHH cannot return whatever number of steps HHH simulates from 1 to oo" or whatever, say:
>
"suppose HHH_n is a SHD which simulates its corresponding input DDD_n for
n steps before aborting. Then for each n, HHH_n does not simulate DDD_n as far as
DDD_n's return."
>
That way it's clear that the DDD_n are all different programs, none of which is simulated to its return statement. The way you say it above sounds like you have ONE DDD whose simulation never returns however far it is simulated! That would be a non-halting DDD, but that's not the situation at all. You don't want to invite deliberate confusion, do you? (Each DDD_n is different, and each /would/ return if simulated further, e.g. perhaps DDD_2 simulated by HHH_100 simulates as far as its final ret and so on.)
>
Mike.
>
said that way. They get stuck in a shell game.
*This is a better way to say that*There are a few problems with that wording. :
∄HHH ∈ X86 emulators that emulate 0 to ∞ steps of DD |
(DD emulated by HHH reaches its own final halt state)
The category of correct emulator HHH where DD is emulated
by HHH and the emulated DD reaches its final halt state
DOES NOT EXIST.
Les messages affichés proviennent d'usenet.