Re: D correctly simulated by pure function H cannot possibly reach its, own line 06 --- Linz

On 5/26/2024 12:48 PM, Richard Damon wrote:

On 5/26/24 1:26 PM, olcott wrote:

On 5/26/2024 12:16 PM, Richard Damon wrote:

On 5/26/24 1:01 PM, olcott wrote:

On 5/26/2024 11:31 AM, Richard Damon wrote:

On 5/26/24 10:13 AM, olcott wrote:

On 5/26/2024 6:43 AM, Richard Damon wrote:
>
typedef int (*ptr)();  // ptr is pointer to int function in C
00       int H(ptr p, ptr i);
01       int D(ptr p)
02       {
03         int Halt_Status = H(p, p);
04         if (Halt_Status)
05           HERE: goto HERE;
06         return Halt_Status;
07       }
08
09       int main()
10       {
11         H(D,D);
12         return 0;
13       }
>
>

Because, as I have said, the answer and reasoning changes depending on what you acknowledged are the implications of your stipulations. For instance, if your actual understanding of being a "Pure Function" is that the program is the equivalent of a Turing Machine, then we need to add a strictness to the definition that isn't normally used for just "Pure Functions", like accessing value of registers like the program counter or stack pointer might not be allowed in some cases. (which breaks you H).
>

>
Since we can see (and you already agreed) that D correctly simulated
by pure simulator H remains stuck in infinite recursive simulation then
we also know that D never reaches its own line 06 and halts in less
than an infinite number of correctly simulated steps.

>
But it depends on what H actually does, which you refuse to agree to.
>

>
When I specify that D is correctly simulated by pure function H, that
means that H is a pure simulator that stops simulating D after some
finite number of correctly simulated steps. There is absolutely nothing
else that needs to be known about H besides this.
>

>
And, do you accept that this exact description means that H is NOT the computational equivalent of a Turing Machine, and that this simulation doesn't prove that its input is not a non-halting program?
>

>
In other words you are saying that it is 100% totally impossible to
adapt an actual UTM so that it stops simulating after a finite number of
steps.

 Yep, at least and have it still be a UTM.
 

When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
When we see that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H in an
infinite number of steps cannot possibly reach its own simulated
final state of ⟨Ĥ.qn⟩ and halt then we correctly deduce that the
same thing applies when simulating halt decider embedded_H correctly
simulates less than an infinite number of steps of ⟨Ĥ⟩ ⟨Ĥ⟩.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer