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

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.

Assembly (or C) also work, but you need to understand more details and
be able to map every assembly instruction or C expressions to TM language.
The form of the DD above (in some books maybe) is for layman to understand,
which is not exactly the case that the HP provides. Don't be silly.
 

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer