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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: Defining a correct simulating halt decider
Date: Sun, 8 Sep 2024 12:46:58 -0000 (UTC)
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Am Sat, 07 Sep 2024 08:56:02 -0500 schrieb olcott:
> On 9/7/2024 3:27 AM, Mikko wrote:
>> On 2024-09-06 11:42:48 +0000, olcott said:
>>> On 9/6/2024 6:19 AM, Mikko wrote:
>>>> On 2024-09-05 13:24:20 +0000, olcott said:
>>>>> On 9/5/2024 2:34 AM, Mikko wrote:
>>>>>> On 2024-09-03 13:00:50 +0000, olcott said:
>>>>>>> On 9/3/2024 5:25 AM, Mikko wrote:
>>>>>>>> On 2024-09-02 16:38:03 +0000, olcott said:
>>>>>>>>
>>>>>>>>> A halt decider is a Turing machine that computes the mapping
>>>>>>>>> from its finite string input to the behavior that this finite
>>>>>>>>> string specifies.
>>>>>>>> A halt decider needn't compute the full behaviour, only whether
>>>>>>>> that behaviour is finite or infinite.
Like Sipser said.

>>>>>>> New slave_stack at:1038c4 Begin Local Halt Decider Simulation  
>>>>>>> Local Halt Decider: Infinite Recursion Detected Simulation Stopped
>>>>>>> Hence  HHH(DDD)==0 is correct
>>>>>>
>>>>>> Nice to see that you don't disagree with what said.
>>>>>> Unvortunately I can't agree with what you say.
>>>>>> HHH terminates, so DDD obviously terminates, too.
>>>>>
>>>>> DDD emulated by HHH never reaches it final halt state.
>>>> If that iis true it means that HHH called by DDD does not return and
>>>> therefore is not a ceicder.
>>> The directly executed HHH is a decider.
>> 
>> If the called HHH behaves differently from the direcly executed HHH
>> then the DDD is not relevant to classic proofs of the impossibility of
>> a halting decider.
>> If you can't show encoding rules that permit the encoidng of the
>> behaviour of the directly executed DDD to HHH then HHH is not a halting
>> decider.
> I SHOW THE ACTUAL EXECUTION TRACE AND EVERYONE DISAGREES WITH IT.
Your implementation is buggy.

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.