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From: dbush <dbush.mobile@gmail.com>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1
Date: Sat, 29 Mar 2025 17:42:59 -0400
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On 3/29/2025 5:37 PM, olcott wrote:
> On 3/29/2025 3:11 PM, dbush wrote:
>> On 3/29/2025 3:46 PM, olcott wrote:
>>> On 3/29/2025 2:25 PM, dbush wrote:
>>>> On 3/29/2025 3:19 PM, olcott wrote:
>>>>> On 3/29/2025 2:01 PM, dbush wrote:
>>>>>> On 3/29/2025 2:58 PM, olcott wrote:
>>>>>>> On 3/29/2025 1:37 PM, dbush wrote:
>>>>>>>> On 3/29/2025 2:15 PM, olcott wrote:
>>>>>>>>> On 3/29/2025 4:31 AM, joes wrote:
>>>>>>>>>> Am Fri, 28 Mar 2025 14:27:36 -0500 schrieb olcott:
>>>>>>>>>>> On 3/28/2025 2:17 PM, dbush wrote:
>>>>>>>>>>>> On 3/28/2025 3:02 PM, olcott wrote:
>>>>>>>>>>>>> On 3/28/2025 1:12 PM, dbush wrote:
>>>>>>>>>>>>>> On 3/28/2025 1:57 PM, olcott wrote:
>>>>>>>>>>>>>>> On 3/27/2025 9:33 PM, dbush wrote:
>>>>>>>>>>>>>>>> On 3/27/2025 10:10 PM, olcott wrote:
>>>>>>>>>>>>>>>>> On 3/27/2025 8:24 PM, dbush wrote:
>>>>>>>>>>>>>>>>>> On 3/27/2025 9:21 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>> On 3/27/2025 8:09 PM, dbush wrote:
>>>>>>>>>>>>>>>>>>>> On 3/27/2025 9:07 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>> On 3/27/2025 7:38 PM, dbush wrote:
>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Good, because that's all that's required for a 
>>>>>>>>>>>>>>>>>> solution to the
>>>>>>>>>>>>>>>>>> halting problem:
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> There are sometimes when the behavior of TM Description D
>>>>>>>>>>>>>>>>> correctly simulated by UTM1 does not match the behavior 
>>>>>>>>>>>>>>>>> correctly
>>>>>>>>>>>>>>>>> simulated by UTM2.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Irrelevant, because to satisfy the requirements, the 
>>>>>>>>>>>>>>>> behavior of
>>>>>>>>>>>>>>>> the described machine when executed directly must be 
>>>>>>>>>>>>>>>> reported.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> I HAVE PROVED THAT THE REQUIREMENT IS WRONG NITWIT.
>>>>>>>>>> According to what? WE require it. YOU are answering a 
>>>>>>>>>> different question.
>>>>>>>>>>
>>>>>>>>>>>>>> Category error.
>>>>>>>>>>>>>> I want to know if any arbitrary algorithm X with input Y 
>>>>>>>>>>>>>> will halt
>>>>>>>>>>>>>> when executed directly.
>>>>>>>>>>>>>
>>>>>>>>>>>>> It is 100% impossible for any TM to take another executing 
>>>>>>>>>>>>> TM as its
>>>>>>>>>>>>> input.
>>>>>>>>>> Quit that.
>>>>>>>>>>
>>>>>>>>>>>> But it can take a complete description of a TM that
>>>>>>>>>>>
>>>>>>>>>>> Is not always a perfect proxy for the behavior of the direct 
>>>>>>>>>>> execution
>>>>>>>>>>> of the underlying machine.
>>>>>>>>>
>>>>>>>>>> Uh yes it is.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> That my proof that I am correct
>>>>>>>>> is over your head is less than
>>>>>>>>> no rebuttal what-so-ever.
>>>>>>>>
>>>>>>>> The fact that such TM description can be given to a UTM which 
>>>>>>>> will exactly replicate the behavior of the described TM when 
>>>>>>>> executed directly proves otherwise is apparently over your head.
>>>>>>>>
>>>>>>>
>>>>>>> One cannot correctly ignore the effect that a specified
>>>>>>> pathological relationship has between its simulator
>>>>>>> and its input on the behavior of this input.
>>>>>>>
>>>>>>
>>>>>> All it means is that HHH does not correctly map DDD to 1 as per 
>>>>>> the requirements:
>>>>>>
>>>>>
>>>>> int sum(int x, int y) { return x + y; }
>>>>> In the same way that sum(2,3) cannot be mapped to 7.
>>>>
>>>> It can, it just wouldn't meet the requirements of the mathematical 
>>>> "sum" function.
>>>>
>>>
>>> int DD()
>>> {
>>>    int Halt_Status = HHH(DD);
>>>    if (Halt_Status)
>>>      HERE: goto HERE;
>>>    return Halt_Status;
>>> }
>>>
>>> Likewise if HHH reported on the behavior of the
>>> directly executed DD 
>>
>> It would be behaving as required:
>>
> 
> Requiring a halt decider to not be able to return
> is an incorrect requirement. 


Which is why it's required to compute the halting function:


Given any algorithm (i.e. a fixed immutable sequence of instructions) X 
described as <X> with input Y:

A solution to the halting problem is an algorithm H that computes the 
following mapping:

(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly


And because the above can't be computed, halt deciders don't exist.