Re: The philosophy of computation reformulates existing ideas on a new basis

On 11/7/24 11:39 AM, olcott wrote:

On 11/7/2024 3:56 AM, Mikko wrote:

On 2024-11-06 15:26:06 +0000, olcott said:
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On 11/6/2024 8:39 AM, Mikko wrote:

On 2024-11-05 13:18:43 +0000, olcott said:
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On 11/5/2024 3:01 AM, Mikko wrote:

On 2024-11-03 15:13:56 +0000, olcott said:
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On 11/3/2024 7:04 AM, Mikko wrote:

On 2024-11-02 12:24:29 +0000, olcott said:
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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.

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Yes but not the particular mapping required by the halting problem.

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Yes it is the particular mapping required by the halting problem.
The exact same process occurs in the Linz proof.

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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.

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void Infinite_Loop()
{
   HERE: goto HERE;
   return;
}
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No that is false.
The measure is whether a C function can possibly
reach its "return" instruction final state.

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Not in the original problem but the question whether a particular strictly
C function will ever reach its return instruction is equally hard. About

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It has always been about whether or not a finite string input
specifies a computation that reaches its final state.

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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.
 *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 always been about trying to make a computation that given a finite string representation of a program and input, decide if the program will halt on that input.
It should be noted that the problem STARTS with a program, which gets represented with a finite string, and that string might be different for different deciders, as the problem doesn't define a specific encoding method.
Your insistance that the problem starts with a finite-string just shows your ignorance.
Try to show a reliable source that defines it as the string is the DEFINITION of what is being asked about, as opposed to being a representation of the program being asked about.
Go ahead, TRY to do it.

DDD specifies a non-halting computation to HHH because
DDD calls HHH in recursive simulation.

No, because the HHH that DDD calls is programmed to break that recursive simulation, and thus make the results finite.
If you change HHH to not abort, then DDD does become non-halting, but HHH doesn't give the right answer. That is a DIFFERENT HHH, and thus a DIFFERENT DDD (as DDD to be a program includes ALL the code it uses, so it includes the code of HHH, which you changed)

 DDD specifies a halting computation to HHH1 because
DDD DOES NOT CALL HHH1 in recursive simulation.

And since it is the SAME PROGRAM described, the OBJECTIVE behavior of that program is the same.

 *Ignoring these key differences is ridiculously foolish*

But it isn't a difference in the behavior of DDD, only the inability of HHH to get the right answer because you didn't program it right.
All you are doing is proving you don't understand the diffference between an OBJECTIVE specification and a SUBJECTIVE one.
The Halting Problem is specified as an OBJECTIVE specification, as the behavior of the program. While that behavior is dependent on the behavior of HHH, since HHH was a DEFINED PROGRAM, that behaivor is FIXED and unchanging, and thus DDD has fixed behavior.
You just are showing you don't understand anything about what you are talking about, especially what a program is, or even the definition of the problem.

 

solution was so easily adapted to the halting problem that we can say
that Turing solved the halting problem before nobody had presented it.
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Turings original problem was to find a method to determine whether the
given Turing machine with given input ceases to write unerasable symbols.
Modern Turing machines don't even start as any symbol can be erased or
overwritten.
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