Newsportal USENET - Re: Incorrect requirements --- Computing the mapping from the input to HHH(DD)

On 5/10/2025 9:09 PM, wij wrote:

On Sat, 2025-05-10 at 20:56 -0500, olcott wrote:
On 5/10/2025 8:44 PM, wij wrote:
On Sat, 2025-05-10 at 20:26 -0500, olcott wrote:
On 5/10/2025 8:17 PM, wij wrote:
On Sat, 2025-05-10 at 17:03 -0500, olcott wrote:
On 5/10/2025 4:44 PM, wij wrote:
On Sat, 2025-05-10 at 14:29 -0500, olcott wrote:
On 5/10/2025 2:02 PM, wij wrote:
>
>
You don't know the counter example in the HP proof, your D is not the case what HP says.
>
>
Sure I do this is it! (as correctly encoded in C)
>
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);
}
>
>
>
Try to convert it to TM language to know you know nothing.
>
>
I spent 22 years on this. I started with the Linz text
>
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
    or
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
(a) Ĥ copies its input ⟨Ĥ⟩
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩ ...
>
Thus ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H
cannot possibly reach its simulated final halt state
⟨Ĥ.qn⟩
>
To refute the HP, you need to understand what it exactly means in TM.
>
I have known this for 22 years.
A working TM. Build it explicitly from transition function, then explain
your derivation. You know nothing.

That would be like examining how an operating system
works entirely from its machine code.
We only have to actually know one detail:
Every counter-example input encoded in any model
of computation always specifies recursive simulation
that never halts to its corresponding simulating
termination analyzer.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

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26 Mar 26	…

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