Re: D correctly simulated by H cannot possibly halt --- templates and infinite sets --- deciders

On 30/05/2024 16:21, olcott wrote:

On 5/30/2024 9:59 AM, Mike Terry wrote:

On 30/05/2024 15:43, olcott wrote:

On 5/28/2024 11:16 AM, olcott wrote:

>
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
*Formalizing the Linz Proof structure*
∃H  ∈ Turing_Machines
∀x  ∈ Turing_Machines_Descriptions
∀y  ∈ Finite_Strings
such that H(x,y) = Halts(x,y)
>

>
A decider computes the mapping from finite string inputs to
its own accept or reject state.
>
A decider does not and cannot compute the mapping from
Turing_Machine inputs to its own accept or reject state.
>
Halts(x,y) would report on the direct execution of x(y) thus ignores
the pathological behavior of x correctly simulated by pure function H.
This makes Halts(x,y) an incorrect measure of the correctness of H(x,y).
>
This is easier to see when we can see every single detail of all of
the steps as an x86 execution trace of D correctly simulated by pure
function H.
>
typedef int (*ptr)();  // ptr is pointer to int function in C
00       int HH(ptr p, ptr i);
01       int DD(ptr p)
02       {
03         int Halt_Status = HH(p, p);
04         if (Halt_Status)
05           HERE: goto HERE;
06         return Halt_Status;
07       }
08
09       int main()
10       {
11         HH(DD,DD);
12         return 0;
13       }
>
*Begin simulation of DD by HH*
Begin Local Halt Decider Simulation   Execution Trace Stored at:113075
[00001c22][00113061][00113065] 55         push ebp
[00001c23][00113061][00113065] 8bec       mov ebp,esp
[00001c25][0011305d][00103031] 51         push ecx
[00001c26][0011305d][00103031] 8b4508     mov eax,[ebp+08]
[00001c29][00113059][00001c22] 50         push eax         ; push DD
[00001c2a][00113059][00001c22] 8b4d08     mov ecx,[ebp+08]
[00001c2d][00113055][00001c22] 51         push ecx         ; push DD
[00001c2e][00113051][00001c33] e80ff7ffff call 00001342    ; call HH
New slave_stack at:14da95
[00001c22][0015da89][0015da8d] 55         push ebp
[00001c23][0015da89][0015da8d] 8bec       mov ebp,esp
[00001c25][0015da85][0014da59] 51         push ecx
[00001c26][0015da85][0014da59] 8b4508     mov eax,[ebp+08]
[00001c29][0015da81][00001c22] 50         push eax         ; push DD
[00001c2a][0015da81][00001c22] 8b4d08     mov ecx,[ebp+08]
[00001c2d][0015da7d][00001c22] 51         push ecx         ; push DD
[00001c2e][0015da79][00001c33] e80ff7ffff call 00001342    ; call HH
Local Halt Decider: Recursive Simulation Detected Simulation Stopped
>
DD correctly simulated by HH cannot possibly reach its own simulated
final state at line 06 in any number of steps including an infinite
number of steps because DD correctly simulated by HH remains stuck in
recursive simulation.
>

>
Your HH/DD above are nonsense functions - you have admitted that HH uses static variables deliberately to detect whether it is the outer (unsimulated) HH or an inner (simulated) HH.  In the event of the latter it branches into a completely different code branch from the outer HH, so the "simulated" behaviour of HH is /nothing like/ the behaviour of outer HH.
>
Any traces from such a HH/DD are completely worthless.
>
You know this, and yet you still claim to have a "fully operational" code etc..  So that is a LIE, just like when you claimed to have a fully operation TM implementing your ideas a few years back.
>
>
Mike.

 The fact that it uses static variables has no effect what-so-ever
on the fact that DD correctly simulated by HH cannot possibly
reach its own simulated final state at line 06.

But DD is NOT correctly simulated by HH.  HH uses static variables to modify its behaviour when simulated, so the "simulation" is rubbish.

 It is very easy to verify that DD correctly simulated by HH cannot
possibly reach its own simulated final state and halt on the basis
of the execution trace that I provided and this x86 source-code for DD.
 _DD()
[00001c22] 55         push ebp
[00001c23] 8bec       mov ebp,esp
[00001c25] 51         push ecx
[00001c26] 8b4508     mov eax,[ebp+08]
[00001c29] 50         push eax        ; push DD
[00001c2a] 8b4d08     mov ecx,[ebp+08]
[00001c2d] 51         push ecx        ; push DD
[00001c2e] e80ff7ffff call 00001342   ; call HH
[00001c33] 83c408     add esp,+08
[00001c36] 8945fc     mov [ebp-04],eax
[00001c39] 837dfc00   cmp dword [ebp-04],+00
[00001c3d] 7402       jz 00001c41
[00001c3f] ebfe       jmp 00001c3f
[00001c41] 8b45fc     mov eax,[ebp-04]
[00001c44] 8be5       mov esp,ebp
[00001c46] 5d         pop ebp
[00001c47] c3         ret
Size in bytes:(0038) [00001c47]
 You are correct that HH is not a pure function yet this has no effect
on the provided execution trace.

Nonsense - if HH is not correctly simulating DD+HH then the trace is just rubbish.
Mike.