Am Wed, 17 Jul 2024 08:14:43 -0500 schrieb olcott:
On 7/17/2024 2:08 AM, Mikko wrote:
On 2024-07-16 14:46:40 +0000, olcott said:
On 7/16/2024 2:18 AM, Mikko wrote:
On 2024-07-15 13:32:27 +0000, olcott said:
On 7/15/2024 2:57 AM, Mikko wrote:
On 2024-07-14 14:48:05 +0000, olcott said:
On 7/14/2024 3:49 AM, Mikko wrote:
On 2024-07-13 12:18:27 +0000, olcott said:
Any input that must be aborted to prevent the non termination of >>>>>>> HHH necessarily specifies non-halting behavior or it would never >>>>>>> need to be aborted.
Disagreeing with the above is analogous to disagreeing with
arithmetic.
A lame analogy. A better one is: 2 + 3 = 5 is a proven theorem just >>>>>> like the uncomputability of halting is.
The uncomputability of halting is only proven when the problem is
framed this way: HHH is required to report on the behavior of an
input that was defined to do exactly the opposite of whatever DDD
reports.
No, it is proven about the halting problem as that problem is.
Which is simply a logical impossibility
Yes, a halting decider is a logical impossibility, as can be and has
been proven.
If it is a logical impossibility then it places no actual limit on computation otherwise we would have "the CAD problem" of the logical impossibility of making a CAD system that correctly draws a square
circle.
It is impossible for a program to return the halting status of every
program, although we would all like it very much. That IS a limit.
thus no actual limit to computation more that this logical
impossibility: What time is it (yes or no)?
In contrast to this example, the halting problem does have a welldefined answer.
As construction of a halting decider is already known to be impossible
why would anyone care whether there is other limitations about it?
And of course the impossibility of halting decider prevents any
applicaions of it, for example as a tool to solve other problems.
Only because we have framed the problem as a logical impossibility. When
we re-frame the problem so that it is not a logical impossibility then
the practical applications can still be derived.
You may not care about the halting decider counterexample. That has
nothing to do with the fact that termination can be confirmed for some
programs (like those with static loop limits and no recursion).
*This is isomorphic the HP decider/input pair*
Can Carol correctly answer “no” to this (yes/no) question?
(Hehner:2018:2)
And the answer is no, Carol cannot, but others can.
Perhaps you can use the isomorphism to proove that Carol can't.
But that should be faily easy anyway.
Carol's question is isomorphic to the halting problem decider/input pair showing that the halting problem is simply a cheap trick.
It is certainly a neat turn on itself. That makes it all the more
valuable.
--
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.
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