Key Statement:It is helpful to simulate some of it. A static analysis could achieve
--------------
"It is sufficient for an SHD to DETECT infinite recursion withoutto SIMULATE it."
having
the same thing by (for example) pretending to simulate it.
Op 21.mei.2025 om 17:54 schreef olcott:
On 5/21/2025 12:56 AM, Richard Heathfield wrote:Verifiable counter-factual.
On 21/05/2025 06:23, olcott wrote:A self-contradictory input and a proof by contradiction are not the
On 5/20/2025 9:15 PM, Richard Damon wrote:
On 5/20/25 3:10 PM, Mr Flibble wrote:
<snip>
Conclusion: ----------- Flibble sharpens his argument by clarifying >>>>>> that SHDs are not required to simulate infinite execution. They are >>>>>> expected to *detect* infinite behavior structurally and respond in >>>>>> finite time. This keeps them within the bounds of what a decider
must be and strengthens the philosophical coherence of his
redefinition of the Halting Problem.
But you can't "redefine" the Halting Problem and then say you have
answered the Halting Problem.
Do you mean like how ZFC resolved Russell's Paradox thus converting
"set theory" into "naive set theory"?
No, because there is no paradox in the Halting Problem. A proof by
contradiction is not a paradox.
same thing. A proof by contradiction would conclude that "this sentence
is not true" is true because it cannot be proved false.
ZFC shows how a whole way of examining a problem can be tossed out as
incorrect and replaced with a whole new way.
The HP proofs are based on defining a D that can actually do the
opposite of whatever value that H returns.
No such D can actually exist.
A better parallel would be Cantor's proof that there are uncountablyLikewise with Russell's Paradox it is assumed that there can be a set
many real numbers, or Euclid's proof that there is no largest prime.
Both of these proofs make a single assumption and then derive a
contradiction, thus showing that the assumption must be false. No
paradoxes need apply.
In the Halting Problem's case, the assumption is that a UNIVERSAL
algorithm exists for determining whether any arbitrary program halts
when applied to given arbitrary input. The argument derives a
contradiction showing the assumption to be false.
of all sets that do not contain themselves as members. This is
"resolved" as nonsense.
Whatever you think your HHH determines, we know from Turing that itvoid DDD()
doesn't determine it for arbitrary programs with arbitrary input. It
therefore has no bearing whatsoever on the Halting Problem.
{
HHH(DDD);
return;
}
DDD correctly simulated by HHH DOES NOT HALT.
The simulation of DDD does not reach a natural end only because HHH
prevents it to halt by a premature abort.
Due this premature abort HHH misses the part of the input that specifies
the conditional abort in Halt7.c, which specifies that the program
halts. If a simulator would not abort this input, it would reach the
natural end of the program. Proven by direct execution and world-class simulators. But HHH has a bug, which makes that it aborts before it can
see that the input halts, only because its programmer dreamed of an
infinite recursion, where there is only a finite recursion.
Come out of rebuttal mode, which makes that you do not pay enough
attention to this logic, but reject it immediately when it disturbs your dreams, after which you only repeat the clueless claims.
If the program runs on a machine with a finite number of states, this is actually possible; a repeated state implies that the program is in an infinite loop. The problem is that a Turing machine is not limited to a finite number of states (including the state of the tape), and such an analysis is not possible in the general case.
I think olcott thinks it is possible.
On 5/24/2025 10:12 AM, dbush wrote:
On 5/24/2025 11:04 AM, olcott wrote:You are a damned liar when you say that I said that HHH must report on
On 5/23/2025 8:09 PM, dbush wrote:
On 5/23/2025 9:07 PM, olcott wrote:int main()
On 5/23/2025 7:57 PM, dbush wrote:
On 5/23/2025 8:54 PM, olcott wrote:If you can't stay exactly on topic I am going to ignore everything
On 5/23/2025 7:44 PM, dbush wrote:What about this?
On 5/23/2025 8:08 PM, Mike Terry wrote:
I suppose Ben quoted PO saying this, because PO /uses/ it to >>>>>>>>> justify that a particular /halting/ computation will never halt, >>>>>>>>> PO's HHH simulates DDD (which halts) but before DDD halts it >>>>>>>>> spots a pattern in the simulation, and announces non-halting. >>>>>>>>> "Eh?" I hear you say! PO claims HHH has "correctly determined >>>>>>>>> that DDD would never halt" and so is correct to decide non-
halting. His "proof" that it is right to decide non-halting is >>>>>>>>> his "when-so- ever.." quote, which broadly matches the Sipser >>>>>>>>> quote.
So the problem is not so much the "when-so-ever.." words
themselves [or the words of Sipser's quote], but understanding >>>>>>>>> how PO is so thoroughly misinterpreting/misapplying them. How >>>>>>>>> can PO believe HHH has "correctly determined the DDD will never >>>>>>>>> halt" when DDD demonstrably halts?
PO is working in a different model than the rest of us, though he >>>>>>>> doesn't seem to understand that.
To him, when function H is deciding on something, the
implementation of H is allowed to vary. This results in
functions that call H to vary as a result. To him, "DDD" is the >>>>>>>> same computation *regardless of the implementation of HHH*, in >>>>>>>> cases where HHH is simulating DDD.
This is essentially the mapping he's operating with:
-----------------
For a function X with input Y and a function H which simulates X: >>>>>>>> POH(H,X,Y)==1 if and only if there exists an implementation of H >>>>>>>> that can simulate X(Y) to completion POH(H,X,Y)==0 if and only if >>>>>>>> there does not exist an implementation of H that can simulate
X(Y) to completion ----------------
And a "decider" in his case maps the following subset:
----------------
Hx is a PO-halt decider if and only if Hx(X,Y) == POH(Hx,X,Y)
----------------
So given his rules, HHH1(DDD) is deciding on a algorithm while >>>>>>>> HHH(DDD) is deciding on a C function whose subfunctions vary.
This of course has nothing to do with the halting problem but he >>>>>>>> doesn't get this. After having spent 22 years on this, he'll >>>>>>>> come up with any crazy justification to avoid admitting to
himself that he misunderstood the problem all this time. He once >>>>>>>> said (and I don't recall the exact wording) that "the directly >>>>>>>> executed D doesn't halt even though it appears to".
The problem is that people here are too stupid to notice that HHH >>>>>>> cannot report on the behavior of its caller.
int min()
{
DD(); // HHH cannot report on the behavior of its caller.
}
that you say.
HHH cannot report on the behavior of its caller AKA the direct
execution of DD().
In other words, you again agree with Linz and others that no H exists
that can perform the following mapping:
Given any algorithm (i.e. a fixed immutable sequence of instructions)
X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes the
following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed
directly
{
DD(); // The HHH called by DD cannot report on the behavior
} // of its caller. Is this OVER-YOUR-HEAD ?
Which means that no HHH exists that meets the below requirements, as
Linz and others proved and as you have *explicitly* agreed is correct:
the behavior of its caller.
No HHH can report on the behavior of its caller for the same reason that
no function can report on the value of the square-root of a dead cat.
olcott <polcott333@gmail.com> wrote:
On 5/23/2025 5:06 PM, Richard Heathfield wrote:
On 23/05/2025 22:06, olcott wrote:
On 5/23/2025 3:50 PM, Richard Heathfield wrote:
On 23/05/2025 21:24, olcott wrote:
<snip>
Liar
An unequivocal response, but it lacks persuasive power.
When I provide the exact detailed steps of exactly how people can
show that I am wrong and they refuse to show that I am wrong yet
claim that I am wrong this is the kind of reckless disregard for the
truth that loses defamation cases.
When your opponents point to the Turing proof that proves you're wrong
Without going through all of the detailed steps that I present that is
a reckless disregard for the truth that loses defamation cases.
There you are utterly wrong. The Halting Theorem has been proven, thus
is true. Anybody putting up a contrary argument must therefore be
wrong.
You might also put up a long winded argument why 2 + 2 = 5, and I would dismiss this likewise, without bothering to follow your exact detailed
steps.
You've also tried, without success, to dismiss one of the proofs of the Halting Therem as invalid.
Thus, as a general rule, it is sensible to dismiss as not worthy of
attention anything you assert about mathematical logic.
If somebody competent were to put up such a contrary argument, it might
well be worth looking at. But that's different - and it hasn't
happened.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
*“The Halting Theorem has been proven, thus is true.”*
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