Liste des Groupes | Revenir à theory |
On 11/11/2024 5:06 AM, Mikko wrote:And not halting means never reaching that final state after an unbounded number of steps.On 2024-11-09 14:56:14 +0000, olcott said:Halt means reaching a final halt state to say otherwise
>On 11/9/2024 3:53 AM, Mikko wrote:>On 2024-11-08 14:39:20 +0000, olcott said:>
>On 11/8/2024 6:39 AM, Mikko wrote:>On 2024-11-07 16:39:57 +0000, olcott said:>
>On 11/7/2024 3:56 AM, Mikko wrote:>On 2024-11-06 15:26:06 +0000, olcott said:>
>On 11/6/2024 8:39 AM, Mikko wrote:>On 2024-11-05 13:18:43 +0000, olcott said:>
>On 11/5/2024 3:01 AM, Mikko wrote:>On 2024-11-03 15:13:56 +0000, olcott said:>
>On 11/3/2024 7:04 AM, Mikko wrote:>On 2024-11-02 12:24:29 +0000, olcott said:>
>HHH does compute the mapping from its input DDD>
to the actual behavior that DDD specifies and this
DOES INCLUDE HHH emulating itself emulating DDD.
Yes but not the particular mapping required by the halting problem.
Yes it is the particular mapping required by the halting problem.
The exact same process occurs in the Linz proof.
The halting probelm requires that every halt decider terminates.
If HHH(DDD) terminates so does DDD. The halting problmen requires
that if DDD terminates then HHH(DDD) accepts as halting.
void Infinite_Loop()
{
HERE: goto HERE;
return;
}
>
No that is false.
The measure is whether a C function can possibly
reach its "return" instruction final state.
Not in the original problem but the question whether a particular strictly
C function will ever reach its return instruction is equally hard. About
It has always been about whether or not a finite string input
specifies a computation that reaches its final state.
Not really. The original problem was not a halting problem but Turing's
Exactly. The actual Halting Problem was called that by Davis
in 1952. Not the same as Turing proof.
In early times there was variation in how things were presented and what
words were used. Post had studied the halting problem of his tag system
much earlier but didn't call it a machine. Many other problems were also
studied and later found to be more or less related to the halting
problem and its variants.
>*So we are back to The Halting Problem itself*>
>
has always been about whether or not a finite string input
specifies a computation that reaches its final state.
No, it has been a collection of related problems that includes that
particular one.
The halting problem has always been abuut halting
Nevertheless Turing's solution to his circularity problem is usually
regarded as the first solution to the halting problem.
>>As the problems are related and equally hard it does
not really matter which one you choose as long as you are clear
about your choice. To argue about the meaning of words id a clear
indcation of an intent to avoid an honest discussion.It is not the meaning of words it is the semantic>
property of the finite string pair HHH/DDD.
Above you have argued about the meanings of the words and
keep doing so below.
It is the meaning of the bytes of x86 code and
bytes of code are not words.
No, nothing you have said tells anuything about meanings of the bytes
of x86 code. (A pair of such bytes is sometimes called a "word").
You were just arguing about the meanings the verb "halt" and other
words.
>
is ignorant or dishonest.
No, it is an INCORRECT terminatation analyzer that INCORRECRTLY determines that DDD will not terminate._DDD()>>The halting problem has always been about whether a finite>
string input specifies a computation that will reach its
final halt state.
>
If you disagree then you must provide a complete and coherent
counter-example conclusively proving otherwise not merely
some vague reference to some other things somewhere else.
From https://www.tutorialspoint.com/automata_theory/ turing_machine_halting_problem.htm
>Turing Machine Halting Problem
Input − A Turing machine and an input string w.
Problem − Does the Turing machine finish computing of the string w in a finite number of steps? The answer must be either yes or no.
The computation specified by the finite string DDD
emulated by HHH cannot possibly reach its "return"
instruction final halt state.
It can and does if HHH is a decider and otherwise does not matter.
>
[00002172] 55 push ebp ; housekeeping
[00002173] 8bec mov ebp,esp ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404 add esp,+04
[00002182] 5d pop ebp
[00002183] c3 ret
Size in bytes:(0018) [00002183]
HHH is a correct termination analyzer that does correctly
determine that DDD emulated by HHH cannot possibly reach
its own [00002183] machine address.
The erroneous assumptions about halt deciders are anchoredNope, your ignorance is rooted in your false understanding of the properties of the system.
in the lack of a complete concrete example where the behavior
specified by the finite string can be unequivocally determined.
That people continue to stupidly try to get away with disagreeingNo, the fact that you IGNORE that the semantics of the x86 language says you can not "abort" your emulation and still necessarily get the FINAL behaivor of the code just shows your stupidity.
with the x86 language proves that anything less than the 100%
complete specification that the x86 language provides cannot
possibly overcome the strong indoctrination into an incoherent view.
And the "behavior" of their own inputs is DEFINED (for "programs") to be the results of the unbounded emulation of that program / the direct running of the program described, even if that isn't what the decider itself does.*It is the behavior of their own input that they must report on*The computation specified by the finite string DDD>
emulated by HHH1 IS NOT THE ACTUAL INPUT TO HHH.
HHH1 can take same inputs as HHH. These inputs specify some behaviour.
What they do with this input may differ.
>
It has always been ridiculously stupid for everyone here to
require HHH to ignore the actual behavior specified by its own
input and instead report on the behavior of the input to HHH1.
No, they must report if *THE* unbounded emulation of their input can possibly reach its own return instruction.Both HHH and HHH1 must report on whether or not their simulationHHH must compute the mapping FROM ITS INPUT TO THE>
BEHAVIOR THAT THIS INPUT SPECIFIES.
Not to full behaviour but to one feature of that behaviour.
Doesn't HHH1 need to?
>
of their own input can possibly reach its own "return" instruction
final halt state. They get different answers ONLY BECAUSE THE BEHAVIOR OF THEIR INPUT DDD IS DIFFERENT!
Les messages affichés proviennent d'usenet.