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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: Anyone with sufficient knowledge of C knows that DD specifies non-terminating behavior to HHH
Date: Sat, 8 Feb 2025 12:25:07 +0200
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On 2025-02-07 23:13:04 +0000, olcott said:

> Experts in the C programming language will know that DD
> correctly simulated by HHH cannot possibly reach its own
> "if" statement.

Wrong, they understand that nothing below exludes the possibility that
HHH is a program that can correctly simulate DD to its "if" statement.
The code of HHH might exlude that but that is not sohwn below.

> The finite string DD specifies non-terminating recursive
> simulation to simulating termination analyzer HHH.

No, it does not. DD as quoted below pecifies nothing about the behaviour
of HHH, only its argument types and return type.

> This makes HHH necessarily correct to reject its input as
> non-halting.

No, it does not. It is not correct to reject the input to HHH as non-halting
unless it really is non-halting.

> typedef void (*ptr)();
> int HHH(ptr P);
> 
> int DD()
> {
>    int Halt_Status = HHH(DD);
>    if (Halt_Status)
>      HERE: goto HERE;
>    return Halt_Status;
> }
> 
> int main()
> {
>    HHH(DD);
> }

> https://github.com/plolcott/x86utm/blob/master/Halt7.c
> has fully operational HHH and DD

No, it has not. There is no DD there.

> The halting problem has always been a mathematical mapping
> from finite strings to behaviors.

The problem is not a mappling. It requires that its solution is.

-- 
Mikko