On 6/22/2022 4:49 PM, Mr Flibble wrote:
> On Wed, 22 Jun 2022 16:41:43 -0500
> olcott <NoOne@NoWhere.com> wrote:
>
>> On 6/22/2022 4:20 PM, Mr Flibble wrote:
>>> On Wed, 22 Jun 2022 15:27:01 -0500
>>> olcott <NoOne@NoWhere.com> wrote:
>>>
>>>> On 6/22/2022 2:31 PM, Mr Flibble wrote:
>>>>> On Tue, 21 Jun 2022 21:38:56 -0500
>>>>> olcott <NoOne@NoWhere.com> wrote:
>>>>>
>>>>>> #include <stdint.h>
>>>>>> #define u32 uint32_t
>>>>>>
>>>>>> #include <stdint.h>
>>>>>> typedef void (*ptr)();
>>>>>>
>>>>>> void P(ptr x)
>>>>>> {
>>>>>> if (H(x, x))
>>>>>> HERE: goto HERE;
>>>>>> return;
>>>>>> }
>>>>>>
>>>>>> int main()
>>>>>> {
>>>>>> Output("Input_Halts = ", H(P, P));
>>>>>> }
>>>>>>
>>>>>> _P()
>>>>>> [000010d2](01) 55 push ebp
>>>>>> [000010d3](02) 8bec mov ebp,esp
>>>>>> [000010d5](03) 8b4508 mov eax,[ebp+08]
>>>>>> [000010d8](01) 50 push eax
>>>>>> [000010d9](03) 8b4d08 mov ecx,[ebp+08]
>>>>>> [000010dc](01) 51 push ecx
>>>>>> [000010dd](05) e820feffff call 00000f02
>>>>>> [000010e2](03) 83c408 add esp,+08
>>>>>> [000010e5](02) 85c0 test eax,eax
>>>>>> [000010e7](02) 7402 jz 000010eb
>>>>>> [000010e9](02) ebfe jmp 000010e9
>>>>>> [000010eb](01) 5d pop ebp
>>>>>> [000010ec](01) c3 ret
>>>>>> Size in bytes:(0027) [000010ec]
>>>>>>
>>>>>> Every sufficiently competent software engineer can easily verify
>>>>>> that the complete and correct x86 emulation of the input to
>>>>>> H(P,P) by H would never reach the "ret" instruction of P because
>>>>>> both H and P would remain stuck in infinitely recursive
>>>>>> emulation.
>>>>>>
>>>>>> If H does correctly determine that this is the case in a finite
>>>>>> number of steps then H could reject its input on this basis. Here
>>>>>> are the details of exactly how H does this in a finite number of
>>>>>> steps.
>>>>>>
>>>>>> typedef struct Decoded
>>>>>> {
>>>>>> u32 Address;
>>>>>> u32 ESP; // Current value of ESP
>>>>>> u32 TOS; // Current value of Top of Stack
>>>>>> u32 NumBytes;
>>>>>> u32 Simplified_Opcode;
>>>>>> u32 Decode_Target;
>>>>>> } Decoded_Line_Of_Code;
>>>>>>
>>>>>> machine stack stack machine assembly
>>>>>> address address data code language
>>>>>> ======== ======== ======== ========= =============
>>>>>> [000010d2][00211e8a][00211e8e] 55 push ebp
>>>>>> [000010d3][00211e8a][00211e8e] 8bec mov ebp,esp
>>>>>> [000010d5][00211e8a][00211e8e] 8b4508 mov eax,[ebp+08]
>>>>>> [000010d8][00211e86][000010d2] 50 push eax //
>>>>>> push P [000010d9][00211e86][000010d2] 8b4d08 mov ecx,[ebp+08]
>>>>>> [000010dc][00211e82][000010d2] 51 push ecx //
>>>>>> push P [000010dd][00211e7e][000010e2] e820feffff call 00000f02
>>>>>> // call H Infinitely Recursive Simulation Detected Simulation
>>>>>> Stopped
>>>>>>
>>>>>> // actual fully operational code in the x86utm operating system
>>>>>> u32 H(u32 P, u32 I)
>>>>>> {
>>>>>> HERE:
>>>>>> u32 End_Of_Code;
>>>>>> u32 Address_of_H; // 2022-06-17
>>>>>> u32 code_end = get_code_end(P);
>>>>>> Decoded_Line_Of_Code *decoded = (Decoded_Line_Of_Code*)
>>>>>> Allocate(sizeof(Decoded_Line_Of_Code));
>>>>>> Registers* master_state = (Registers*)
>>>>>> Allocate(sizeof(Registers));
>>>>>> Registers* slave_state = (Registers*)
>>>>>> Allocate(sizeof(Registers));
>>>>>> u32* slave_stack = Allocate(0x10000); // 64k;
>>>>>> u32 execution_trace =
>>>>>> (u32)Allocate(sizeof(Decoded_Line_Of_Code)
>>>>>> * 1000);
>>>>>>
>>>>>> __asm lea eax, HERE // 2022-06-18
>>>>>> __asm sub eax, 6 // 2022-06-18
>>>>>> __asm mov Address_of_H, eax // 2022-06-18
>>>>>> __asm mov eax, END_OF_CODE
>>>>>> __asm mov End_Of_Code, eax
>>>>>>
>>>>>> Output("Address_of_H:", Address_of_H); // 2022-06-11
>>>>>> Init_slave_state(P, I, End_Of_Code, slave_state,
>>>>>> slave_stack); Output("\nBegin Simulation Execution Trace
>>>>>> Stored at:", execution_trace);
>>>>>> if (Decide_Halting(&execution_trace, &decoded, code_end,
>>>>>> &master_state, &slave_state, &slave_stack, Address_of_H, P, I))
>>>>>> goto END_OF_CODE;
>>>>>> return 0; // Does not halt
>>>>>> END_OF_CODE:
>>>>>> return 1; // Input has normally terminated
>>>>>> }
>>>>>>
>>>>>> H knows its own machine address and on this basis it can easily
>>>>>> examine its stored execution_trace of P and determine:
>>>>>> (a) P is calling H with the same arguments that H was called
>>>>>> with. (b) No instructions in P could possibly escape this
>>>>>> otherwise infinitely recursive emulation.
>>>>>> (c) H aborts its emulation of P before its call to H is invoked.
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>> Technically competent software engineers may not know this
>>>>>> computer science:
>>>>>>
>>>>>> A halt decider must compute the mapping from its inputs to an
>>>>>> accept or reject state on the basis of the actual behavior that
>>>>>> is actually specified by these inputs.
>>>>>>
>>>>>> computation that halts … the Turing machine will halt whenever it
>>>>>> enters a final state. (Linz:1990:234)
>>>>>>
>>>>>> The "ret" instruction of P is its final state.
>>>>>>
>>>>>> Linz, Peter 1990. An Introduction to Formal Languages and
>>>>>> Automata. Lexington/Toronto: D. C. Heath and Company. (317-320)
>>>>>>
>>>>>
>>>>> void Px(u32 x)
>>>>> {
>>>>> H(x, x);
>>>>> return;
>>>>> }
>>>>>
>>>>> int main()
>>>>> {
>>>>> Output("Input_Halts = ", H((u32)Px, (u32)Px));
>>>>> }
>>>>>
>>>>> ...[000013e8][00102357][00000000] 83c408 add esp,+08
>>>>> ...[000013eb][00102353][00000000] 50 push eax
>>>>> ...[000013ec][0010234f][00000427] 6827040000 push 00000427
>>>>> ---[000013f1][0010234f][00000427] e880f0ffff call 00000476
>>>>> Input_Halts = 0
>>>>> ...[000013f6][00102357][00000000] 83c408 add esp,+08
>>>>> ...[000013f9][00102357][00000000] 33c0 xor eax,eax
>>>>> ...[000013fb][0010235b][00100000] 5d pop ebp
>>>>> ...[000013fc][0010235f][00000004] c3 ret
>>>>> Number of Instructions Executed(16120)
>>>>>
>>>>> It gets the answer wrong, i.e. input has not been decided
>>>>> correctly. QED.
>>>>>
>>>>> /Flibble
>>>>>
>>>>
>>>> You and Richard are insufficiently technically competent at
>>>> software engineering not meeting these specs:
>>>>
>>>> A software engineer must be an expert in: the C programming
>>>> language, the x86 programming language, exactly how C translates
>>>> into x86 and the ability to recognize infinite recursion at the
>>>> x86 assembly language level. No knowledge of the halting problem
>>>> is required.
>>>
>>> I cannot speak for Richard but I have 30+ years C++ experience; I
>>> also have C and x86 assembly experience (I once wrote a Zilog Z80A
>>> CPU emulator in 80286 assembly) and I can recognize an infinite
>>> recursion; the problem is that you cannot recognize the fact that
>>> the infinite recursion only manifests as part of your invalid
>>> simulation-based omnishambles:
>>
>> If you are competent then you already know this is true and lie about
>> it:
>>
>> Every sufficiently competent software engineer can easily verify that
>> the complete and correct x86 emulation of the input to H(Px,Px) by H
>> would never reach the "ret" instruction of P because both H and P
>> would remain stuck in infinitely recursive emulation.
>>
>> Otherwise you are incompetent.
>>
>>> the recursion simply isn't there for a "valid" halt
>>> decider, that being a halt decider that can return an answer in
>>> finite time to ALL invokers: H needs to return an answer to Px to be
>>> considered a valid halt decider.
>>>
>>> /Flibble
>
> Why did you ignore the second part? Again:
>
> The problem is that you cannot recognize the fact that the infinite
> recursion only manifests as part of your invalid simulation-based

It is easily provably correct. That you lack the technical competence to
verify that the x86 emulated behavior of the x86 emulation of the input
to H(P,P) by H precisely matches the behavior specified by P is far less
than no rebuttal at all.

I am not going to keep responding to your nonsense I don't really give a
rat's ass for the woefully incorrect opinion of incompetent people.

> omnishambles: the recursion simply isn't there for a "valid" halt
> decider, that being a halt decider that can return an answer in finite
> time to ALL invokers: H needs to return an answer to Px to be
> considered a valid halt decider.
>
> /Flibble
>

--
Copyright 2022 Pete Olcott

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