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From: Richard Heathfield <rjh@cpax.org.uk>
Newsgroups: comp.theory
Subject: Re: Incorrect requirements --- Computing the mapping from the input
 to HHH(DD)
Date: Fri, 9 May 2025 22:40:53 +0100
Organization: Fix this later
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On 09/05/2025 21:15, olcott wrote:
> On 5/9/2025 3:07 PM, Richard Heathfield wrote:
>> On 09/05/2025 20:46, olcott wrote:
>>> We have not begun to get into any of those points.
>>> We are only asking can DDD correctly simulated
>>> by any HHH that can exist ever reach its own
>>> "return" instruction.
>>
>> DDD can't be correctly simulated by itself (which is 
>> effectively what you're trying to do when you fire up the 
>> simulation from inside DDD).
>>
> 
> How the Hell did you twist my words to say that?

I haven't touched your words. What I have done is to observe that 
DDD's /only/ action is to call a simulator. Since DDD isn't 
itself a simulator, there is nothing to simulate except a call to 
a simulator.

It's recursion without a base case - a rookie error.

HHH cannot successfully complete its task, because it never 
regains control after the first recursion. To return, it must 
abort the simulation, which means the simulation fails.

> 
> void DDD()
> {
>    HHH(DDD);
>    return;
> }
> 
> When 1 or more statements of DDD are correctly
> simulated by HHH then this correctly simulated
> DDD cannot possibly reach its own “return statement”.

On what grounds can you persuade an extraordinarily sceptical 
readership that HHH 'correctly simulated' DDD?

There are only two possibilities:

(a) HHH aborts the simulation prematurely, or
(b) it doesn't.

If (a), the simulation fails to 'play' DDD correctly.

If (b), the simulation fails to arrive at a decision.

-- 
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within