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On 6/3/2024 3:10 AM, Fred. Zwarts wrote:Nope.Op 03.jun.2024 om 05:20 schreef olcott:DD correctly simulated by HH meets this criteria:On 6/2/2024 10:13 PM, Richard Damon wrote:>On 6/2/24 10:54 PM, olcott wrote:>IT HAS ALWAYS BEEN ABOUT THE BEHAVIOR THAT THE INPUT SPECIFIES.>
That you did get confused by the Linz text proves that you do
get confused. Previously it looked just like willful deception.
Which is, for a Halt Decider, exactly and only the behavior of the Turing Machine the input describes.
>
PERIOD.
>
Anything else is just a LIE.
>>>You don't seem to understand that you can't just "redefine" the system to meet your desires.>
>
Deciders compute the mapping FROM THEIR INPUTS.
DD correctly simulated by HH specifies NON-HALTING.
No, Running DD(DD) and seeing that it will never, after an unbounded number of steps, indicate it is non-halting.
>
DEFINITION.
>>>
Deciders compute the mapping FROM THEIR INPUTS.
DD correctly simulated by HH specifies NON-HALTING.
Right, and the input is a representation of a Turing Machine and its input, whose behavior the decider is to decide on.
>>>
Deciders compute the mapping FROM THEIR INPUTS.
DD correctly simulated by HH specifies NON-HALTING.
>
And that is the machine the input describes.
>
ANYTHING ELSE IS JUST A LIE.
>You can't get away with implicitly saying that you>
just don't "believe in" UTMs.
I do, and a UTM is DEFINED as a machine that exactly reproduces the behavior of the machine described by its input.
>
*If that was true then you prove that this statement is false*
*We can see that the following DD cannot possibly halt when*
*correctly simulated by every HH that can possibly exist*
>
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 }
>
_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 1c22
[00001c2a] 8b4d08 mov ecx,[ebp+08]
[00001c2d] 51 push ecx ; push DD 1c22
[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]
>
It is clear (both from the C code as from the compiled code) that at the moment that the outer simulator reaches the place where DD calls HH then HH finds *itself* in a 'repeated state'. At that same moment DD is *not* (yet) in a repeated state. (DD may get in a repeated state if the called HH would start a new simulation.) So, the only conclusion is that it is HH that displays repetition. And because of that DD will repeat as well.
>
By using simulation, a new halting problem is introduced, that has nothing to do with the pathological part of DD n lines 04, 05 and 06. This pathological part is not even processed by the simulation.
On 10/13/2022 11:29:23 AM
MIT Professor Michael Sipser agreed this verbatim paragraph is correct
(He has neither reviewed nor agreed to anything else in this paper)
<Professor Sipser agreed>
If simulating halt decider H correctly simulates its input D until H
correctly determines that its simulated D would never stop running
unless aborted then
H can abort its simulation of D and correctly report that D specifies a
non-halting sequence of configurations.
</Professor Sipser agreed>
Therefore HH(DD,DD) == 0 is correct.
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