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On 11/9/2024 3:53 AM, Mikko wrote:But to have meaning, you need all the bytes, which include that of HHH, and thus you can't change those bytes in your arguement.On 2024-11-08 14:39:20 +0000, olcott said:It is the meaning of the bytes of x86 code and
>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.
>
bytes of code are not words.
Which computation are you meaning?The computation specified by the finite string DDDThe 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.
emulated by HHH cannot possibly reach its "return"
instruction final halt state.
The computation specified by the finite string DDDSure it is, if you mean the actual computation that string (when extended to include the whole computation) is repesenting.
emulated by HHH1 IS NOT THE ACTUAL INPUT TO HHH.
HHH must compute the mapping FROM ITS INPUT TO THERight, and that mapping is the full behavior that the full input represent, which means the UNBOUNDED emulation of that input, which HHH1 does, but HHH doesn't, so HHH can't look to its own emulation as that isn't the property that the behavior of the input is defined to use.
BEHAVIOR THAT THIS INPUT SPECIFIES.
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