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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: HHH(DDD)==0 is correct
Date: Fri, 4 Jul 2025 16:33:32 -0400
Organization: i2pn2 (i2pn.org)
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On 7/4/25 2:24 PM, olcott wrote:
> On 7/4/2025 12:48 PM, Richard Damon wrote:
>> On 7/4/25 9:32 AM, olcott wrote:
>>> On 7/4/2025 8:22 AM, joes wrote:
>>>> Am Fri, 04 Jul 2025 07:50:23 -0500 schrieb olcott:
>>>>> On 7/4/2025 2:35 AM, Mikko wrote:
>>>>>> On 2025-07-03 12:56:42 +0000, olcott said:
>>>>>>> On 7/3/2025 3:57 AM, Mikko wrote:
>>>>>>>> On 2025-07-03 02:50:40 +0000, olcott said:
>>>>>>>>> On 7/1/2025 11:37 AM, Mr Flibble wrote:
>>>>>>>>>> On Mon, 30 Jun 2025 21:12:48 -0400, Richard Damon wrote:
>>>>>>>>>>> On 6/30/25 2:30 PM, Mr Flibble wrote:
>>>>
>>>>>>>>>> No. A simulator does not have to run a simulation to 
>>>>>>>>>> completion if
>>>>>>>>>> it can determine that the input, A PROGRAM, never halts.
>>>
>>>> If. But here it confuses that with not being able to simulate past the
>>>> recursive call.
>>>>
>>>
>>> It is the correct simulation of the input that
>>> specifies the actual behavior of this input.
>>
>> Right, and that means correctly simulating EVERY instruction that 
>> makes up the PROGRAM, which must include the code of *THE* HHH, or you 
>> can't "correctly simulate" the call instruction.
>>
>> SIn
>>
>>>
>>> If this simulation cannot simulate past the recursive
>>> call then this correctly simulated input cannot possibly
>>> reach its own correctly simulated final halt state.
>>>
>>
>> But it can, as the call goes into HHH, and you then just simulate the 
>> code of HHH.
>>
> 
> Do you mean like this? (as I have been telling you for three years)
> https://liarparadox.org/HHH(DDD)_Full_Trace.pdf

Yes, so where in that trace does HHH's "correct simulation" differ from 
the exact same simulation generated by HHH1 before HHH aborts?


Note, that this trace PROVES that "the input" includes the code of HHH, 
as that is what HHH is looking at, and if you are simulating "the 
input", whatever you simulate must have been part of the input.

Note also, this trace isn't the "simulation done by HHH" but the trace 
of the execution of HHH as listed by x86utm.

This trace is also nearly identical to a correct simulation of the 
direct execution of DDD, it just starts at the addresses of main instead 
of DDD, but the source and object code of both main and DDD are 
identical (just relocated to different addresses).

If you compare the trace that HHH actually does, it will exactly match 
this to the point that HHH decides to abort.

If you disagree, point to the line where they differ, else you are just 
admitting that you are just a pathetic liar that has been caught in the 
trap.

> 
> I ignore the rest of your points because you
> prove that it is almost impossible for you
> to handle a single point.

No, you ignore them because you don't have an answer.

As I have pointed out, providing the trace doesn't show where it differs 
from the trace of the direct exectutuion of DDD.

> 
> When I go over the exact same point 500 times
> you never notice that I ever mentioned this point.
> 
> 

So, when are you going to answer the questions about that listing?