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On 6/2/2024 12:55 PM, joes wrote:Similarly:Am Sun, 02 Jun 2024 10:02:54 -0500 schrieb olcott:Introduction to the Theory of Computation, by Michael SipserOn 6/2/2024 4:36 AM, joes wrote:The simulation is incorrect if it stops at that point and the simulatedAm Sat, 01 Jun 2024 17:37:28 -0500 schrieb olcott:>On 6/1/2024 5:30 PM, Richard Damon wrote:>On 6/1/24 5:27 PM, olcott wrote:On 6/1/2024 4:15 PM, Richard Damon wrote:On 6/1/24 4:35 PM, olcott wrote:On 6/1/2024 3:29 PM, Richard Damon wrote:On 6/1/24 12:46 PM, olcott wrote:On 6/1/2024 11:33 AM, Richard Damon wrote:On 6/1/24 12:18 PM, olcott wrote:On 6/1/2024 11:08 AM, Richard Damon wrote:On 6/1/24 11:58 AM, olcott wrote:On 6/1/2024 10:46 AM, Richard Damon wrote:On 6/1/24 10:00 AM, olcott wrote:Not simulating an infinite number of steps of infinite recursion is>Every DD correctly simulated by any HH of the infinite set of HH/DD>
pairs that match the above template never reaches past its own
simulated line 03 in 1 to ∞ steps of correct simulation of DD by HH.
But since the simulation was aborted,
*The above never mentions anything about any simulation being aborted*
incorrect. You always forget this requirement: the simulation must be
complete.
When HH correctly simulates N steps of DD it is incorrect to say that
these N steps were incorrectly simulated.
machine is not in a final state, even if it was correct up to that point.
It also matters what steps where simulated, not only if each was correct.
>
https://www.amazon.com/Introduction-Theory-Computation-Michael-Sipser/dp/113318779X/
On 10/13/2022 11:29:23 AM
MIT Professor Michael Sipser agreed that this verbatim paragraph is correct
(He has neither reviewed nor agreed to anything else in this paper)
<Professor Sipser agreed>
If simulating halt decider H correctly simulates its input D until H correctly determines that its simulated D would never stop running unless aborted then
H can abort its simulation of D and correctly report that D specifies a non-halting sequence of configurations.
</Professor Sipser agreed>
There are an infinite number of different HH/DD pairs specified by thatI see only a single DD. All H that stop simulating D before it reaches aWhat are the sets of HH and DD? I thought they were concrete machines.The infinite set of every HH/DD pair where HH correctly simulates 1 or
>
more steps of DD is the infinite set that I am referring to.
final state are already wrong.
>
template. One class of them simulates N steps and one class of them
simulates ∞ steps. Some of them play a game of bingo before simulating
any steps. Some of them play a game of chess after simulating N steps.
*IN NONE OF THESE CASES DOES DD CORRECTLY SIMULATED BY HH HALT*
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