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

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Sujet : Re: The philosophy of computation reformulates existing ideas on a new basis
De : mikko.levanto (at) *nospam* iki.fi (Mikko)
Groupes : comp.theory
Date : 07. Nov 2024, 10:56:54
Autres entêtes
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Message-ID : <vgi2t6$2js8i$1@dont-email.me>
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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
solution was so easily adapted to the halting problem that we can say
that Turing solved the halting problem before nobody had presented it.
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.
--
Mikko

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