Re: D simulated by H never halts no matter what H does V3 ---

Am Sun, 05 May 2024 09:30:20 -0500 schrieb olcott:

On 5/5/2024 5:36 AM, Fred. Zwarts wrote:

Op 05.mei.2024 om 05:17 schreef olcott:

On 5/4/2024 9:49 PM, Richard Damon wrote:

On 5/4/24 9:30 PM, olcott wrote:

On 5/4/2024 8:05 PM, Richard Damon wrote:

On 5/4/24 8:49 PM, olcott wrote:

On 5/4/2024 7:44 PM, Richard Damon wrote:

On 5/4/24 8:20 PM, olcott wrote:

On 5/4/2024 7:07 PM, Richard Damon wrote:

On 5/4/24 7:51 PM, olcott wrote:

On 5/4/2024 6:32 PM, Richard Damon wrote:

On 5/4/24 7:01 PM, olcott wrote:

On 5/4/2024 5:36 PM, Richard Damon wrote:

On 5/4/24 6:08 PM, olcott wrote:

On 5/4/2024 4:43 PM, Richard Damon wrote:

On 5/4/24 5:18 PM, olcott wrote:

On 5/4/2024 3:40 PM, Richard Damon wrote:

On 5/4/24 2:46 PM, olcott wrote:

On 5/4/2024 12:15 PM, Richard Damon wrote:

On 5/4/24 12:31 PM, olcott wrote:

On 5/4/2024 10:52 AM, Richard Damon wrote:

On 5/4/24 10:48 AM, olcott wrote:

On 5/4/2024 9:39 AM, Alan Mackenzie wrote:

olcott <polcott333@gmail.com> wrote:

On 5/4/2024 5:56 AM, Alan Mackenzie wrote:

[ Followup-To: set ]

>

In comp.theory olcott <polcott333@gmail.com>
wrote:

>

[please snip your replies a little]

Note, you CAN'T just "Stipulate" that a given machine IS a UTM except
by defining that it works just like a UTM, which means, for one
thing, it can NEVER abort its simulation, not even after determining
that it will simulate this input forever.
>

None-the-less a TM that correctly simulates N steps cannot be said to
have simulated those N steps incorrectly on the basis that it could
have simulated N+1 steps.
>

Those N steps were simulated correctly, but the fact that it stops
after N steps make it an incorrect simulation.

 
In other words a decider is wrong unless it never stops simulating an
non-halting input?

Correct. Simulating here means producing the exact same behaviour.
If it is correct up to a point, it might still make a mistake later.
The only way to know is to keep simulating.

--
joes