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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)
Date: Fri, 9 May 2025 07:12:27 -0400
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On 5/8/25 11:34 PM, olcott wrote:
> On 5/8/2025 10:14 PM, Mike Terry wrote:
>> On 09/05/2025 03:13, olcott wrote:
>>> On 5/8/2025 8:30 PM, Keith Thompson wrote:
>>>> olcott <polcott333@gmail.com> writes:
>>>>> On 5/8/2025 6:49 PM, Keith Thompson wrote:
>>>>>> olcott <polcott333@gmail.com> writes:
>>>>>> [...]
>>>>>>> void DDD()
>>>>>>> {
>>>>>>>     HHH(DDD);
>>>>>>>     return;
>>>>>>> }
>>>>>>>
>>>>>>> If you are a competent C programmer then you
>>>>>>> know that DDD correctly simulated by HHH cannot
>>>>>>> possibly each its own "return" instruction.
>>>>>> "cannot possibly each"?
>>>>>> I am a competent C programmer (and I don't believe you can make
>>>>>> the same claim).  I don't know what HHH is.  The name "HHH" tells
>>>>>> me nothing about what it's supposed to do.  Without knowing what
>>>>>> HHH is, I can't say much about your code (or is it pseudo-code?).
>>>>>>
>>>>>
>>>>> For the purpose of this discussion HHH is exactly
>>>>> what I said it is. It correctly simulates DDD.
>>>>
>>>> Does HHH correctly simulate DDD *and do nothing else*?
>>>>
>>>> Does HHH correctly simulate *every* function whose address is passed
>>>> to it?  Must the passed function be one that takes no arguments
>>>> and does not return a value?
>>>>
>>>> Can HHH just *call* the function whose address is passed to it?
>>>> If it's a correct simulation, there should be no difference between
>>>> calling the function and "correctly simulating" it.
>>>>
>>>> My knowledge of C tells me nothing about *how* HHH might simulate
>>>> DDD.
>>>>
>>>
>>> HHH can only simulate a function that take no arguments
>>> and has no return value. HHH also simulates the entire
>>> chain of functions that this function calls. These can
>>> take arguments or not and have return values or not.
>>>
>>> Thus HHH ends up simulating itself (and everything
>>> that HHH calls) simulating DDD in an infinite
>>> sequence of recursive emulation until OOM error.
>>>
>>>>> We need not know anything else about HHH to
>>>>> know that DDD correctly simulated by HHH cannot
>>>>> possibly REACH its own "return" instruction.
>>>>
>>>> Assuming that HHH(DDD) "correctly simulates" DDD, and assuming it
>>>> does nothing else, your code would be equivalent to this:
>>>>
>>>>      void DDD(void) {
>>>>          DDD();
>>>>          return;
>>>>      }
>>>>
>>>
>>> Exactly. None of these people on comp.theory could
>>> get that even after three years.
>>>
>>
>> PO is being quite deceptive here.
>>
> 
> I am only referring to the above hypothetical HHH/DDD pair.
> Everyone here has consistently denied that when DDD is
> correctly simulated by HHH that this DDD cannot possibly
> reach its "return" instruction (final halt state).

But that HHH isn't the required pure function that correctly emulate the 
input, as the input has been clearly defined to be JUST the code of the 
function DDD, and thus the call instruction can't be emulated as the 
instruction it references isn't in the input.

When you include the input, then DDD become a DDD/HHH pair, and thus the 
input for this hypotherical HHH is processing a different input than the 
input you are going to give to your actual HHH, and thus says nothing 
about it.

If you fix your statements so that it is clear that "the input" includes 
all of that memroy, we might be able to deal with THIS statement, but 
then you can't get to your next one.

And, if you remove the pure requirement, my version of HHH that uses the 
static flag proves your claim is just incorrect in the relaxed case.

> 
>> His simulation is in fact a single-stepped x86 instruction simulation, 
>> where the stepping of each x86 instruction is under the HHH's control. 
>> HHH can continue stepping the simulation until its target returns, in 
>> which case the situation is logically just like direct call, as you 
>> have described.  Or HHH could step just 3 x86 instructions (say) and 
>> then decide to return (aka "abort" its simulation).  Let's call that / 
>> partial/ simulation in contrast with /full/ simulation which you've 
>> been supposing.
>>
> 
> A full simulation of infinite recursion?
> I am only doing one tiny idea at a time here.

Sure, if that is what is there.

The fact that you HHH can't get by the call instruction and still meet 
the defintions, since you have been excluding that code as "part of the 
input" and just "in the system".

> 
>> Oh, did he forget to mention that?  Anyhow, in the general case with / 
>> partial/ simulation there is more to think about as it is obvioudly / 
>> not/ logically equivalent to direct execution.
>>
> 
> That is changing the subject away from
> DDD correctly simulated by HHH.

Which doesn't happen within your definitions.

> 
>>
>> Oh, and obviously everybody in comp.theory gets all this.  The problem 
>> is with PO and his inability to communicate his ideas properly and his 
>> inability to understand what other people understand or disagree with. 
>> He goes on for months/years claiming people don't understand things 
>> they agree with, but it's down to his duffer wording...
>>
>>>> Then the return statement (which is unnecessary anyway) will never be
>>>> reached. 
>>>
>>> It is only there to mark a final halt state.
>>>
>>>> In practice, the program will likely crash due to a stack
>>>> overflow, unless the compiler implements tail-call optimization, in
>>>> which case the program might just run forever -- which also means the
>>>> unnecessary return statement will never be reached.
>>>>
>>>
>>> Yes you totally have this correctly.
>>> None of the dozens of comp.theory people could
>>> ever achieve that level of understanding even
>>> after three years. That is why I needed to post
>>> on comp.lang.c.
>>
>> Everybody on comp.theory understands this much. 
> 
> No one here ever agreed that when 1 or more
> instructions of DDD are correctly simulated
> by HHH that DDD cannot possibly reach its
> own "return" instruction.

Because that is just a false phrase. Your HHH can not correctly emulate 
past the call HHH instruction (and still be the requried pure function) 
as the code there isn't part of what you have defined as the input.

> 
> Everyone has found one excuse or another to
> deny this.

Only by pointing out your errors.

> 
>> PO's plan is that when he goes elsewhere he can start with noobies and 
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