Liste des Groupes | Revenir à c theory |
On 2025-07-05 15:46:21 +0000, olcott said:*That is provably incorrect and no one ever noticed in 90 years*
On 7/5/2025 3:32 AM, Mikko wrote:No, it does not. The domain is specified in the statement of theOn 2025-07-04 12:34:39 +0000, olcott said:>
>On 7/4/2025 2:25 AM, Mikko wrote:>>>
What HHH correctly or otherwise simulates is merely an implementation
detail.
It is a detail that defines a partial halt decider
that makes the "do the opposite" code unreachable.
No, it does. The proof that a counter-example can be constructed
does not refer to any implementation details, so it applies to
every implementation that is does not violate the requirements
so obviously that the proof is not needed.
>>What matters is the beahviour DD specifies.>
The behavior that an input specifies is only correctly
measured by correctly simulating this input.
Wrong. It is correctly measured by a direct execution.
Since no Turing machine can possibly take another directly
executing Turing machine as an input this makes all directly
executed Turing machines outside of the domain of every Turing
machine based decider.
halting problem which includes every Turing machine.
Turing machines (by definition) can only take finite string inputs.The requirement that a halt decider report on the behaviorThe magic word "bogus" has no effect on the problem statement.
of things outside of its domain has always been bogus.
Instead of this deciders must report on the behavior thatIt does not work that way. Instead, the solution of the problem must
their input actually specifies.
have, in addittion to the decider, a user's manual that specifies
what the input must be in order to specify the actual computation.
For example, if the computation asked about is DDD(), what must the
input be so that the decider says "it halts"?
Les messages affichés proviennent d'usenet.