On 6/22/2022 7:32 PM, Richard Damon wrote:
> On 6/22/22 4:27 PM, olcott 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.
>>
>>
>
> Except that I am probably at least as much of an expert on those fields
> as you are. My one problem is that I do have knowledge of the halting
> problem, that seems to make me resistant to your lies.
>
> You might have more experiance with x86, but only because I work on a
> number of different architectures, but your argument really isn't x86
> specific, that is just the case you are working in.
>
> (I was producing 'production' assembly routines 50 years ago, so I have
> seen a number of different types of machines)

First you agree that my words are perfectly correct within their
specified context

The software engineering aspect of this original post
On 6/21/2022 9:38 PM, olcott wrote:

then after this we can proceed with your objection that they do not meet
the definitions.

It is required that we have incremental closure on sub-points or full
closure will be impossible to achieve.

--
Copyright 2022 Pete Olcott

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