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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Incorrect requirements --- Computing the mapping from the input
 to HHH(DD) --- DOUBLE DOG DARE YOU
Date: Fri, 9 May 2025 23:50:28 -0400
Organization: i2pn2 (i2pn.org)
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On 5/9/25 11:29 PM, olcott wrote:
> On 5/9/2025 10:17 PM, Richard Damon wrote:
>> On 5/9/25 10:27 PM, olcott wrote:
>>> On 5/9/2025 9:19 PM, Richard Heathfield wrote:
>>>> On 10/05/2025 02:50, olcott wrote:
>>>>> On 5/9/2025 8:32 PM, Richard Heathfield wrote:
>>>>>> On 10/05/2025 02:29, olcott wrote:
>>>>>>> On 5/9/2025 8:15 PM, Richard Heathfield wrote:
>>>>>>>> On 10/05/2025 01:51, olcott wrote:
>>>>>>>>> On 5/9/2025 7:29 PM, Richard Heathfield wrote:
>>>>>>>>>> On 10/05/2025 00:02, olcott wrote:
>>>>>>>>>>> Correctly emulating one or more instructions <is>
>>>>>>>>>>> the correct emulation of 1 or more instructions
>>>>>>>>>>> of DD. This is a truism.
>>>>>>>>>>
>>>>>>>>>> No, it's not. Correct emulation would entail accurately 
>>>>>>>>>> simulating the whole of DDD's behaviour.
>>>>>>>>>
>>>>>>>>> It is stupidly wrong to require the complete
>>>>>>>>> emulation of a non-terminating input.
>>>>>>>>
>>>>>>>> It is touchingly naive to think you can persuade people to 
>>>>>>>> accept incomplete emulation as 'correct'.
>>>>>>>>
>>>>>>>
>>>>>>> If one instruction is emulated correctly
>>>>>>> then is is dishonest to say that zero
>>>>>>> instructions were emulated correctly.
>>>>>>
>>>>>> Which instruction do you think is emulated correctly?
>>>>>>
>>>> <snip>
>>>>
>>>>> The entire sequence of the first four instructions
>>>>> of DDD is emulated correctly.
>>>>
>>>> Nope. Syntax errors don't count as correct.
>>>>
>>>
>>> Any HHH that emulates DDD according to the rules
>>> of the x86 language must emulate the first four
>>> instructions of DDD followed by HHH emulating
>>> itself emulated DDD and then the first three
>>> instructions of DDD when seven of the instructions
>>> of DDD are correctly emulated.
>>>
>>> This is axiomatic according to the rules of
>>> the x86 language applied to the input to HHH(DDD).
>>>
>>
>> Nope.
>>
>> I have shown an HHH that just by the rules of the x86 language can 
>> correctly emulate the code of DDD that references itself.
>>
> 
> _DDD()
> [00002172] 55         push ebp      ; housekeeping
> [00002173] 8bec       mov ebp,esp   ; housekeeping
> [00002175] 6872210000 push 00002172 ; push DDD
> [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
> [0000217f] 83c404     add esp,+04
> [00002182] 5d         pop ebp
> [00002183] c3         ret
> Size in bytes:(0018) [00002183]
> 
> You cannot possibly show any HHH that emulates this
> DDD according to the rules of the x86 language such
> that this DDD ever reaches its "ret" instruction
> final halt state.
> 
> 

Sure I have. I guess you are just admitting you are a blantant liar.

int HHH(ptr P) {
   static int flag = 0;
   if (flag) return 0;
   flag = 1;

   /* Now put your existing code for a correct simulator that never 
aborts */


Tell me why this simulator can't reach the final state of DDD when we 
call HHH(DDD)

it will simulate the introduction to DDD, then go into HHH and see that 
flag has been set to 1 by the initial call to HHH, and then immediately 
return 0 to DDD which will then halt.

THis has been explained several time TODAY, and thus your denial is just 
proof that you are just a stupid liar.

I guess you want to get the express ticket to that lake of fire.