Re: H(D,D) cannot even be asked about the behavior of D(D) V3 ---IGNORING ALL OTHER REPLIES

On 6/15/24 5:56 PM, olcott wrote:

On 6/15/2024 11:33 AM, Richard Damon wrote:

On 6/15/24 12:22 PM, olcott wrote:

On 6/13/2024 8:24 PM, Richard Damon wrote:
 > On 6/13/24 11:32 AM, olcott wrote:
 >>
 >> It is contingent upon you to show the exact steps of how H computes
 >> the mapping from the x86 machine language finite string input to
 >> H(D,D) using the finite string transformation rules specified by
 >> the semantics of the x86 programming language that reaches the
 >> behavior of the directly executed D(D)
 >>
 >
 > Why? I don't claim it can.
>
The first six steps of this mapping are when instructions
at the machine address range of [00000cfc] to [00000d06]
are simulated/executed.
>
After that the behavior of D correctly simulated by H diverges
from the behavior of D(D) because the call to H(D,D) by D
correctly simulated by H cannot possibly return to D.

>
Nope, the steps of D correctly simulated by H will EXACTLY match the steps of D directly executed, until H just gives up and guesses.
>

 When we can see that D correctly simulated by H cannot possibly
reach its simulated final state at machine address [00000d1d]
after one recursive simulation and the same applies for 2,3,...N
recursive simulations then we can abort the simulated input and
correctly report that D correctly simulated by H DOES NOT HALT.

Nope. Because an aborted simulation doesn't say anything about Halting,
Remember, your analysis was done with the input CHANGING with H, which means any N used different than what your H actually uses, is INVALID for discussing what THIS D does.
IF you did it right, and asked what would an H that simulated farther THIS input, which means D calls the original H, then if that simulates father, it WOULD see the simulated ORIGINAL H abort and return to D and D halt.
But this H can't do that, so it can't see it.
So, you not only ask the wrong question, but you ask it of the wrong input.

 That you call this a guess seems disingenuous at least and dishonest
at most. it is as if you are denying mathematical induction.

But the D that Linz talks about has its own copy of H, as is needed to make it an actual computation, and thus doesn't change when you think about different machines simuating it.

 

By your defintions, the ONLY correct simulation of D by H MUST follow the call H into the instructions of H, and then continue there forever until H either gives up simulating, or it happens to simulate the return from H to D (which doesn't look like it will ever happen).
>

 Yes and when I say 2 + 3 = 5 you are not free to disagree and be
correct.

Just like you don't get to change the code of D, including the fact that it calls the ORIGINAL H, an d isn't a template anymore, and thatthe criteria *IS* does it halt when directly run, and not about "the correct simulation by H" which isn't even a valid criteria.

 

This is the ONLY thing that matches your definition of Correct Simulation.
>

 The x86 language defines the semantics of correct simulation that
you denigrate this to my opinion seems disingenuous at least and
dishonest at most.

Right, which means the ONLY "correct simulation" of D simulates the call to H and the code in H, so the correct simulation by H is EXACTLY the same as the direct execution up to the point when H gives up and stops, so you claim that they differ is just a LIE.

 

And that means that you definition of the "Input" is shown to be incorrect (or insufficient for H to do what it needs to do). The input CAN NOT be just those instructions shown below, but must include the contents of ALL the memory used by the program.
>

 It is the sequence of x86 instructions specified by the machine
language of D that is being examined. We cannot simply ignore
the recursive simulation effect of the call from D to H(D,D).
That D makes this call is what make D correctly simulated by H
fail to halt.

No, we done ignore them, we simulate through them.
Note, somehow you think thaT H only needs to simulate the instruction in the C function of D, but that is INCORRCT, and if you try to use that defintion, H must stop as soon as it gets to the call H and can not proceeed.
The problem here is that, to be the needed input, the input actually needs to include all the code that D uses, including H and everyting that H calls, or the input is just incorrect.

 

And this comes out naturally, because to ask about the program represented by the input, the input needs to represent a FULL PROGRAM, which includes ALL of the algorithm used by the code, so ALL of the program.
>

 That D calls H in recursive simulation is what makes D correctly
simulated by H never halt. H is merely watching what D does
until it see that D correctly simulated by H cannot possibly halt
on the basis of something just like mathematical induction.

Nope, That only holds if H is defined to never abort.
You are just proving you don't understand what you are talking about and are just being a pathological liar with a total and reckless disregard for the truth.

 We can directly see that 1,2,3...N recursive simulations do
not result in D correctly simulated by H reaching its simulated
final state of [00000d1d].

Nope, and in fact it seems that EVERYT simulation that you have done that tries to go past those first 7 instuctions violate you definition of a "Correct Simulation" and thus ALL You logic is just wrong.

 

Thus, the input "D", includes its copy of H that was DEFINED by Linz and Sipser to be part of the machine, and when you do your logic, you can't change that code.
>

 When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
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⟩
(d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
(e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(g) goto (d) with one more level of simulation
 When embedded_H is a UTM we can see that Ĥ ⟨Ĥ⟩ never halts.

And if embedded_H is a UTM< then so is H, and then H fails to be a decider.
So either H / embedded_H *IS* programmed to abort it simulation at some point, so that H (H^) (H^) can go to qn, and thus H^ (H^) will also have its embedded_H abort its simulation and go to qn and thus H^ (H^) Halts, or H (H^) (H^) just never answers.
Both are wrong.

 This by itself proves that the same thing applies when embedded_H
is based on a UTM that stops simulating and rejects its input at
any point after it sees

There is no sucn thing. You are effectively basing you argument on the puppy that *IS* a 15 story office building.

 two complete simulations show a pair of identical TMD's are
simulating a pair of identical inputs.  We can see this thus
proving recursive simulation.

But not infinite.
So, you are just wrong, and proving you are stupid as you don't understand what a UTM is.

 

If this means that you can't do your imagining of different Hs, then you can't do that operation.
>
Any H you imagine being given the same input as this H, must see the code for that original H, not the new hypothetical H you are imagining.
>
Which sort of blows you your whole argument.
>

>
_D()
[00000cfc](01) 55          push ebp
[00000cfd](02) 8bec        mov ebp,esp
[00000cff](03) 8b4508      mov eax,[ebp+08]
[00000d02](01) 50          push eax       ; push D
[00000d03](03) 8b4d08      mov ecx,[ebp+08]
[00000d06](01) 51          push ecx       ; push D
[00000d07](05) e800feffff  call 00000b0c  ; call H
[00000d0c](03) 83c408      add esp,+08
[00000d0f](02) 85c0        test eax,eax
[00000d11](02) 7404        jz 00000d17
[00000d13](02) 33c0        xor eax,eax
[00000d15](02) eb05        jmp 00000d1c
[00000d17](05) b801000000  mov eax,00000001
[00000d1c](01) 5d          pop ebp
[00000d1d](01) c3          ret
Size in bytes:(0034) [00000d1d]
>

>