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On 6/27/2024 2:36 AM, Mikko wrote:That's right. But that question cannot be presented to the decider.On 2024-06-26 12:58:59 +0000, olcott said:In computability theory and computational complexity theory, a
On 6/26/2024 3:41 AM, Mikko wrote:Everyone knwos that H(P,P) is not asked anything.On 2024-06-26 02:29:59 +0000, olcott said:If this was true then everyone here would already know
On 6/25/2024 9:23 PM, Richard Damon wrote:Actually, it is the opposite. Everybody, or at least all computerOn 6/25/24 10:05 PM, olcott wrote:Yes and everyone knows that computer scientists are muchOn 6/25/2024 8:47 PM, Richard Damon wrote:Because I answered the actual question.On 6/25/24 1:45 PM, olcott wrote:If I ask you: What time is it?On 6/25/2024 9:46 AM, Alan Mackenzie wrote:But only if the function is, in fact, computable.Hi, Ben.<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
Ben Bacarisse <ben@bsb.me.uk> wrote:Alan Mackenzie <acm@muc.de> writes:In comp.theory olcott <polcott333@gmail.com> wrote:On 6/25/2024 4:22 AM, joes wrote:Am Sat, 22 Jun 2024 13:47:24 -0500 schrieb olcott:On 6/22/2024 1:39 PM, Fred. Zwarts wrote:Op 21.jun.2024 om 15:21 schreef olcott:When we stipulate that the only measure of a correct emulation is theYes. Which is wrong, because H0 should terminate.
semantics of the x86 programming language then we see that when DDD is
correctly emulated by H0 that its call to H0(DDD) cannot possibly
return.[ .... ]The call from DDD to H0(DDD) when DDD is correctly emulated
by H0 cannot possibly return.Until you acknowledge this is true, this is the
only thing that I am willing to talk to you about.I think you are talking at cross purposes. Joes's point is that H0
should terminate because it's a decider. You're saying that when H0 is
"correctly" emulating, it won't terminate. I don't recall seeing anybody
arguing against that.So you're saying, in effect, H0 is not a decider. I don't think anybody
else would argue against that, either.He's been making exactly the same nonsense argument for years. It
became crystal clear a little over three years ago when he made the
mistake of posting the pseudo-code for H -- a step by step simulator
that stopped simulating (famously on line 15) when some pattern was
detected. He declared false (not halting) to be the correct result for
the halting computation H(H_Hat(), H_Hat()) because of what H(H_Hat(),
H_Hat()) would do "if line 15 were commented out"!PO does occasionally make it clear what the shell game is.I think it's important for (relative) newcomers to the newsgroup to
become aware of this. Each one of them is trying to help PO improve his
level of learning. They will eventually give up, as you and I have
done, recognising (as Mike Terry, in particular, has done) that
enriching PO's intellect is a quite impossible task.
What's the betting he'll respond to this post with his usual short
sequence of x86 assembly code together with a demand to recognise
something or other as non-terminating?
--
Ben.
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>
On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
> I don't think that is the shell game. PO really /has/ an H
> (it's trivial to do for this one case) that correctly determines
> that P(P) *would* never stop running *unless* aborted.
>
> He knows and accepts that P(P) actually does stop. The
> wrong answer is justified by what would happen if H
> (and hence a different P) where not what they actually are.
>
Ben thinks that I tricked professor Sipser into agreeing
with something that he did not fully understand.
*The real issue is that no one here sufficiently understands*
*the highlighted portion of the following definition*
Computable functions are the formalized analogue of the
intuitive notion of algorithms, in the sense that a
function is computable if there exists an algorithm
that can do the job of the function, i.e.
*given an input of the function domain*
*it can return the corresponding output*
https://en.wikipedia.org/wiki/Computable_function
Since Halting isn't, you can't use that fact.
and you do not tell me the answer to the question hidden
in my mind "What did you have for dinner?" We cannot say
that you provided the wrong answer when you tell me what
time it is.
Just like the "Halt Decider" needs to answer the "Halt Decider Question" and not answer about POOP.
When we ask H to tell us whether its actual input haltsThen you are just admitting that it can't be a Halt Decider.
H can only answer that P correctly simulated by H will not halt.
H cannot answer the question hidden in your mind.
If it isn't what the definition requires, it just isn't one.
more infallible than God thus cannot possibly ever make
a definition that is incoherent in ways that these 100%
infallible computer scientists never noticed.
scientists and engineers, know that they, and all peaple, are fallible,
at least when making programs and when inferring about programs. Therefore
computer engineers demand that every program must be tested, and computer
scinetists demand that every claim is proven.
that H(P,P) is not even being asked about the behavior
of the directly executed P(P).
decision problem is a computational problem that can be posed as
a yes–no question of the input values.
https://en.wikipedia.org/wiki/Decision_problem
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