Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]

Groups > comp.software-eng > #3408 > unrolled thread

That P(P) of main() halts does not contradict H(P,P)==0 [ exception to the rule ]

Back to article view | Back to comp.software-eng

Contents

  That P(P) of main() halts does not contradict H(P,P)==0 [ exception to the rule ] olcott <NoOne@NoWhere.com> - 2021-08-28 11:47 -0500
    Re: That P(P) of main() halts does not contradict H(P,P)==0 [ air tight proof ] olcott <NoOne@NoWhere.com> - 2021-08-29 08:28 -0500
      Re: That P(P) of main() halts does not contradict H(P,P)==0 [ air tight proof ] olcott <NoOne@NoWhere.com> - 2021-08-30 23:00 -0500
        Re: That P(P) of main() halts does not contradict H(P,P)==0 [ air tight proof ] olcott <NoOne@NoWhere.com> - 2021-08-31 09:24 -0500
          Re: That P(P) of main() halts does not contradict H(P,P)==0 [ relative invocation order ] olcott <NoOne@NoWhere.com> - 2021-08-31 11:08 -0500
          Re: That P(P) of main() halts does not contradict H(P,P)==0 [ relative invocation order ] olcott <NoOne@NoWhere.com> - 2021-08-31 16:48 -0500
    Re: That P(P) of main() halts does not contradict H(P,P)==0 [ exception to the rule ] olcott <NoOne@NoWhere.com> - 2021-09-01 10:05 -0500
      Re: That P(P) of main() halts does not contradict H(P,P)==0 [ Pathological Input ] olcott <NoOne@NoWhere.com> - 2021-09-01 21:53 -0500
      Re: That P(P) of main() halts does not contradict H(P,P)==0 [ ignorance ] olcott <NoOne@NoWhere.com> - 2021-09-02 10:23 -0500
        Re: That P(P) of main() halts does not contradict H(P,P)==0 [ ignorance ] olcott <NoOne@NoWhere.com> - 2021-09-02 20:42 -0500
          Re: That P(P) of main() halts does not contradict H(P,P)==0 [ ignorance ] olcott <NoOne@NoWhere.com> - 2021-09-03 09:05 -0500
          Re: That P(P) of main() halts does not contradict H(P,P)==0 [ ignorance ] olcott <NoOne@NoWhere.com> - 2021-09-03 09:20 -0500
        Re: That P(P) of main() halts does not contradict H(P,P)==0 [ Linz H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ ] olcott <NoOne@NoWhere.com> - 2021-09-03 19:25 -0500
    Re: That P(P) of main() halts does not contradict H(P,P)==0 [ exception to the rule ] olcott <NoOne@NoWhere.com> - 2021-09-02 09:15 -0500
      Re: That P(P) of main() halts does not contradict H(P,P)==0 [ pathological inputs ] olcott <NoOne@NoWhere.com> - 2021-09-02 10:34 -0500
      Re: That P(P) of main() halts does not contradict H(P,P)==0 [ pathological inputs ] olcott <NoOne@NoWhere.com> - 2021-09-02 11:32 -0500

#3408 — That P(P) of main() halts does not contradict H(P,P)==0 [ exception to the rule ]

// Simplified Linz Ĥ (Linz:1990:319)
// Strachey(1965) CPL translated to C
void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
}

int main()
{
   Output("Input_Halts = ", H((u32)P, (u32)P));
   P((u32)P);
}


The H executed on line 1 of main() is the same function with the same 
input as The executed H called from  P(P) executed on line 2 of main. 
These two instances have identical behavior.

The simulated H called from the P simulated by the executed H on line 1 
of main() is the same function called with the same input as the 
executed H.

The simulated P called from the executed H on line 1 of main() is the 
same function called with the same input as the executed P on line 2 of 
main().

In this case we have a brand new exception to the rule that the same 
function called with the same input must have identical behavior.

When one of these function instances is directly executed and the other 
is simulated by a simulating halt decider their behavior need not be 
identical. The executed function can not be aborted and the simulated 
function can be aborted. This forms an exception to the rule that the 
same function called with the same input must have identical behavior.

The P executed on line 2 of main() is computationally distinct from the 
P simulated by H executed on the first line of main(). Because these are 
two different computations they can have different behavior without 
contradiction.

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation 


-- 
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre 
minds." Einstein

[toc] | [next] | [standalone]

#3409 — Re: That P(P) of main() halts does not contradict H(P,P)==0 [ air tight proof ]

On 8/29/2021 6:00 AM, Richard Damon wrote:
> On 8/29/21 12:00 AM, olcott wrote:
>> On 8/28/2021 7:07 PM, Richard Damon wrote:
>>> On 8/28/21 7:51 PM, olcott wrote:
>>>
>>>>
>>>> I will take this to mean that you already know that
>>>> That P(P) of main() halts does not contradict H(P,P)==0.
>>>>
>>>> The only way that I can know that my words are clear enough to escalate
>>>> to the next level of actual computer scientist review when I mostly only
>>>> have God damned liars for reviewers is that their "rebuttals" become
>>>> ridiculously lame.
>>>>
>>>
>>> WRONG.
>>
>>
>> void Infinite_Loop()
>> {
>>    HERE: goto HERE;
>> }
>>
>> int main()
>> {
>>    H0((u32)Infinite_Loop);
>>    Infinite_Loop();
>> }
>>
>> It is dead obvious that the first line of main() has different behavior
>> than the second line of main() even though the same function with the
>> same empty arguments is simulated / executed in both cases.
>>
>>  From this we can deduce that every input that never halts will have
>> different behavior when simulated by a simulating halt decider that
>> recognizes its infinite behavior pattern than it would when directly
>> executed.
>>
>> I am going to stop here to see if this much is understood.
>>
>>
> 
> And who ever said that the computation of H(P,I) was the same as the
> computation P(I). Only you seem to argue that strawmand.
> 

If any pair of computations are computationally distinct then that can 
have different behavior without contradiction.

The fact that the P(P) of main() halts does not contradict the fact that 
H(P,P) of main() correctly decides that its input never halts because 
these are two entirely different computations.

For six months people have been consistently pointing out a 
"contradiction" that does not exist. This non-existent "contradiction" 
has been their whole basis of rebuttal. While it seemed to be an actual 
contradiction it seemed to be an actual rebuttal. Now that it is known 
to not be a contradiction it is no longer any sort of rebuttal.

H1(P,P) of main() reports that its input does halt because it can see 
that H(P,P) of P aborts its input.

H(P,P) of main sees that unless it aborts its simulation of P(P) that 
P(P) will never halt. This is conclusively proved below to anyone that 
can understand the technical details and wants an honest dialogue.

> The point everyone is making is that the answer that H(P,I) needs to
> correspond to the actual behavor of P(I).
> 
> If P(I) Halts, then H(P,I) needs to return 1.
> If P(I) doesn't ever Halt, then H(P,I) needs to return 0.
> 
> Since Infinite_Loop() never Hals, H(Infinite_Loop, 0) should return 0.
> 
> Since H^(H^) DOES Halt, H(H^,H^) must return 1 to be right, but it
> returns 0.
> 
> Note, a PARTIAL simulation may be finite when the machine is being
> simulated is infinite, because a partial simulation (like used by a
> smulating halt decider) doesn't fully repoduce the machine like a PURE
> simulation needs to.
> 
> But, this does NOT mean that every simulation that a simulating halt
> decider aborts would be infinite if not aborted. The fact that this
> simulator aborted the simulation doesn't affact what the machine would
> do if run by itself.
> 
> I think the one key thing you keep on missing is when you start to argue
> about the H^ machine is that when you look at varying the definition of
> H, to see what answer is right, you need to decide what problem you are
> going to be looking at. If you want to see if the answer H gave was
> right for a given H/H^ pair, than when you change H, you must not change
> the version of H that H^ is using. Thus we can replace the top level H
> with a UTM, but leave the H in H^ to still have its abort in it, to see
> that this H^ does Halt. (Which it does as long as H(H^,H^) returns 0)
> 
> When you want to try to see if you can find an H that gives the right
> answer, then you do need to change the H that is inside H^, and you then
> need to compare the results of H(H^,H^) to what H^(H^) does.
> 
> You will find two major cases for H. There are H's that don't abort the
> simulation of H^, these Hn, fail to decide on Hn(Hn^,Hn^) so are wrong,
> and it doesn't matter that Hn^(Hn^) doesn't halt.
> 
> The second case are Ha's that do abort the simulation of Ha^ and say it
> is non-halting. For all of these Ha's, we find that Ha^(H^) will be
> halting, and thus Ha was wrong.
> 
> You keep on trying to do logic where you look at Hn(Ha^,Ha^) or
> Ha(Hn^,Hn^) and in both of these cases, the H can give the right answer,
> but the problem isn't in the right form to be a counter. You need to get
> the decider to decide correctly on the ^ construction of itself, not
> some other version of the decider.
> 

The bottom line is that this can be verified as correct entirely on the 
basis of the meaning of its words by anyone wanting an actual honest 
dialogue:

Simulating Halt Decider Theorem (Olcott 2020):
A simulating halt decider correctly decides that any input that never 
halts unless the simulating halt decider aborts its simulation of this 
input is an input that never halts.

Next we see if that criteria is met:

Simulating partial halt decider H correctly decides that P(P) never 
halts (V1)

// Simplified Linz Ĥ (Linz:1990:319)
// Strachey(1965) CPL translated to C
void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
}

int main()
{
   Output("Input_Halts = ", H((u32)P, (u32)P));
}

_P()
[00000c36](01)  55          push ebp
[00000c37](02)  8bec        mov ebp,esp
[00000c39](03)  8b4508      mov eax,[ebp+08] // 2nd Param
[00000c3c](01)  50          push eax
[00000c3d](03)  8b4d08      mov ecx,[ebp+08] // 1st Param
[00000c40](01)  51          push ecx
[00000c41](05)  e820fdffff  call 00000966    // call H
[00000c46](03)  83c408      add esp,+08
[00000c49](02)  85c0        test eax,eax
[00000c4b](02)  7402        jz 00000c4f
[00000c4d](02)  ebfe        jmp 00000c4d
[00000c4f](01)  5d          pop ebp
[00000c50](01)  c3          ret
Size in bytes:(0027) [00000c50]

_main()
[00000c56](01)  55          push ebp
[00000c57](02)  8bec        mov ebp,esp
[00000c59](05)  68360c0000  push 00000c36    // push P
[00000c5e](05)  68360c0000  push 00000c36    // push P
[00000c63](05)  e8fefcffff  call 00000966    // call H(P,P)
[00000c68](03)  83c408      add esp,+08
[00000c6b](01)  50          push eax
[00000c6c](05)  6857030000  push 00000357
[00000c71](05)  e810f7ffff  call 00000386
[00000c76](03)  83c408      add esp,+08
[00000c79](02)  33c0        xor eax,eax
[00000c7b](01)  5d          pop ebp
[00000c7c](01)  c3          ret
Size in bytes:(0039) [00000c7c]

  machine   stack     stack     machine    assembly
  address   address   data      code       language
  ========  ========  ========  =========  =============
[00000c56][0010172a][00000000] 55          push ebp
[00000c57][0010172a][00000000] 8bec        mov ebp,esp
[00000c59][00101726][00000c36] 68360c0000  push 00000c36 // push P
[00000c5e][00101722][00000c36] 68360c0000  push 00000c36 // push P
[00000c63][0010171e][00000c68] e8fefcffff  call 00000966 // call H(P,P)

Infinite recursion detection criteria:
If the execution trace of function X() called by function Y() shows:
(1) Function X() is called twice in sequence from the same machine 
address of Y().
(2) With the same parameters to X().
(3) With no conditional branch or indexed jump instructions in Y().
(4) With no function call returns from X().
then the function call from Y() to X() is infinitely recursive.

When we apply the above criteria to the following execution trace we see 
that it is met, conclusively proving that P never halts unless H aborts 
its simulation of P:

Begin Local Halt Decider Simulation at Machine Address:c36
[00000c36][002117ca][002117ce] 55          push ebp
[00000c37][002117ca][002117ce] 8bec        mov ebp,esp
[00000c39][002117ca][002117ce] 8b4508      mov eax,[ebp+08]
[00000c3c][002117c6][00000c36] 50          push eax       // push P
[00000c3d][002117c6][00000c36] 8b4d08      mov ecx,[ebp+08]
[00000c40][002117c2][00000c36] 51          push ecx       // push P
[00000c41][002117be][00000c46] e820fdffff  call 00000966  // call H(P,P)

[00000c36][0025c1f2][0025c1f6] 55          push ebp
[00000c37][0025c1f2][0025c1f6] 8bec        mov ebp,esp
[00000c39][0025c1f2][0025c1f6] 8b4508      mov eax,[ebp+08]
[00000c3c][0025c1ee][00000c36] 50          push eax       // push P
[00000c3d][0025c1ee][00000c36] 8b4d08      mov ecx,[ebp+08]
[00000c40][0025c1ea][00000c36] 51          push ecx       // push P
[00000c41][0025c1e6][00000c46] e820fdffff  call 00000966  // call H(P,P)
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

[00000c68][0010172a][00000000] 83c408      add esp,+08
[00000c6b][00101726][00000000] 50          push eax
[00000c6c][00101722][00000357] 6857030000  push 00000357
[00000c71][00101722][00000357] e810f7ffff  call 00000386
Input_Halts = 0
[00000c76][0010172a][00000000] 83c408      add esp,+08
[00000c79][0010172a][00000000] 33c0        xor eax,eax
[00000c7b][0010172e][00100000] 5d          pop ebp
[00000c7c][00101732][00000068] c3          ret
Number_of_User_Instructions(27)
Number of Instructions Executed(23721)

Thus we have X > Y and Y > Z therefore X > AKA air tight proof.


-- 
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre 
minds." Einstein

[toc] | [prev] | [next] | [standalone]

#3415 — Re: That P(P) of main() halts does not contradict H(P,P)==0 [ air tight proof ]

On 8/30/2021 10:35 PM, André G. Isaak wrote:
> On 2021-08-30 21:20, olcott wrote:
>> On 8/30/2021 10:04 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/30/2021 9:17 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/29/2021 7:00 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 8/29/2021 11:19 AM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 8/29/2021 10:20 AM, Ben Bacarisse wrote:
>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>
>>>>>>>>>>>> The fact that the P(P) of main() halts does not contradict 
>>>>>>>>>>>> the fact
>>>>>>>>>>>> that H(P,P) of main() correctly decides that its input never 
>>>>>>>>>>>> halts
>>>>>>>>>>>> because these are two entirely different computations.
>>>>>>>>>>>
>>>>>>>>>>> What arguments must be passed to H for it to report on the 
>>>>>>>>>>> halting
>>>>>>>>>>> computation P(P) "of main()"?
>>>>>>>>>>
>>>>>>>>>> That is already answered in the part that you snipped.
>>>>>>>>>
>>>>>>>>> No it was not.  And if I am wrong, show the world that I am 
>>>>>>>>> wrong by
>>>>>>>>> writing them again here.  It can be no more that two simple C
>>>>>>>>> expressions.
>>>>>>>>
>>>>>>>> It is as I have said an airtight proof, notwithstanding false
>>>>>>>> assumptions to the contrary.
>>>>>>> No answer, as expected.
>>>>>>>
>>>>>>>> You can try to point to an actual error in the proof as its 
>>>>>>>> stands and
>>>>>>>> everyone that sufficiently understands the material will see your
>>>>>>>> mistake.
>>>>>>> No answer, as expected.
>>>>>>>
>>>>>>>> It is the case that H(P,P) is a computation that never halts unless
>>>>>>>> its simulation is aborted. It is the case that every computation 
>>>>>>>> that
>>>>>>>> never halts unless is simulation is aborted is a non halting
>>>>>>>> computation.
>>>>>>> No answer, as expected.
>>>>>>>
>>>>>>>> There is nothing besides carefully crafted double-talk that goes
>>>>>>>> against this.
>>>>>>> No answer, as expected.
>>>>>>
>>>>>> I will only answer questions that directly pertain to the points that
>>>>>> I made and will ignore dishonest dodges that attempt to change the
>>>>>> subject away from the points that I made.
>>>>> Dodge, dodge, dodge!  All the questions you have been asked pertain
>>>>> directly to the nonsense you post.
>>>>
>>>> I am only discussing that
>>>> H(P,P)==0 is correct is a necessary consequence of its two premises.
>>>
>>> You seem to forget that I agree.  H(P,P)==0 is correct given the
>>> criteria you have invented. 
>>
>> {H(P,P)==0 is correct} means that the input to H(P,P) is correctly 
>> decided as never halting.
> 
> H(P, P) is required to answer the question Does P(P) halt"
> 
> It is *specifically* being asked about the behaviour of int main() { 
> P(P); }
> 
> It is *not* being asked about the behaviour of P(P) "under the dominion 
> of a halt decider" (whatever that may mean) nor about any computation 
> other than int main() { P(P); }. This follows directly from how the 
> halting problem is *defined*.

    In computability theory, the halting problem is the
    problem of determining, from a description of an
    arbitrary computer program and an input, whether the
    program will finish running, or continue to run forever.
    https://en.wikipedia.org/wiki/Halting_problem

No that is not true. The halting problem is always about program 
descriptions not running programs. This means that it is always about 
the input to the halt decider not the direct execution of the program.

If the input to the halt decider never halts unless the halt decider 
aborts its simulation of this input then its input never halts.

> since int main() { P(P); } halts, if your H(P, P) returns any answer 
> other than 'true', it is wrong by the very definition of the problem.
> 
> You can argue all you want that it is right about some *other* question, 
> or about some computation *other* than int main() { P(P); }, but then it 
> isn't answering the halting problem. It is answering something else, and 
> unless you can provide some reason why anyone should care about that 
> other problem I assure you no one will.
> 
> André
> 


-- 
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre 
minds." Einstein

[toc] | [prev] | [next] | [standalone]

#3416 — Re: That P(P) of main() halts does not contradict H(P,P)==0 [ air tight proof ]

On 8/30/2021 11:45 PM, André G. Isaak wrote:
> On 2021-08-30 22:31, olcott wrote:
>> On 8/30/2021 11:19 PM, André G. Isaak wrote:
>>> On 2021-08-30 22:00, olcott wrote:
>>>> On 8/30/2021 10:35 PM, André G. Isaak wrote:
>>>>> On 2021-08-30 21:20, olcott wrote:
>>>>>> On 8/30/2021 10:04 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 8/30/2021 9:17 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 8/29/2021 7:00 PM, Ben Bacarisse wrote:
>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>
>>>>>>>>>>>> On 8/29/2021 11:19 AM, Ben Bacarisse wrote:
>>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> On 8/29/2021 10:20 AM, Ben Bacarisse wrote:
>>>>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> The fact that the P(P) of main() halts does not 
>>>>>>>>>>>>>>>> contradict the fact
>>>>>>>>>>>>>>>> that H(P,P) of main() correctly decides that its input 
>>>>>>>>>>>>>>>> never halts
>>>>>>>>>>>>>>>> because these are two entirely different computations.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> What arguments must be passed to H for it to report on 
>>>>>>>>>>>>>>> the halting
>>>>>>>>>>>>>>> computation P(P) "of main()"?
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> That is already answered in the part that you snipped.
>>>>>>>>>>>>>
>>>>>>>>>>>>> No it was not.  And if I am wrong, show the world that I am 
>>>>>>>>>>>>> wrong by
>>>>>>>>>>>>> writing them again here.  It can be no more that two simple C
>>>>>>>>>>>>> expressions.
>>>>>>>>>>>>
>>>>>>>>>>>> It is as I have said an airtight proof, notwithstanding false
>>>>>>>>>>>> assumptions to the contrary.
>>>>>>>>>>> No answer, as expected.
>>>>>>>>>>>
>>>>>>>>>>>> You can try to point to an actual error in the proof as its 
>>>>>>>>>>>> stands and
>>>>>>>>>>>> everyone that sufficiently understands the material will see 
>>>>>>>>>>>> your
>>>>>>>>>>>> mistake.
>>>>>>>>>>> No answer, as expected.
>>>>>>>>>>>
>>>>>>>>>>>> It is the case that H(P,P) is a computation that never halts 
>>>>>>>>>>>> unless
>>>>>>>>>>>> its simulation is aborted. It is the case that every 
>>>>>>>>>>>> computation that
>>>>>>>>>>>> never halts unless is simulation is aborted is a non halting
>>>>>>>>>>>> computation.
>>>>>>>>>>> No answer, as expected.
>>>>>>>>>>>
>>>>>>>>>>>> There is nothing besides carefully crafted double-talk that 
>>>>>>>>>>>> goes
>>>>>>>>>>>> against this.
>>>>>>>>>>> No answer, as expected.
>>>>>>>>>>
>>>>>>>>>> I will only answer questions that directly pertain to the 
>>>>>>>>>> points that
>>>>>>>>>> I made and will ignore dishonest dodges that attempt to change 
>>>>>>>>>> the
>>>>>>>>>> subject away from the points that I made.
>>>>>>>>> Dodge, dodge, dodge!  All the questions you have been asked 
>>>>>>>>> pertain
>>>>>>>>> directly to the nonsense you post.
>>>>>>>>
>>>>>>>> I am only discussing that
>>>>>>>> H(P,P)==0 is correct is a necessary consequence of its two 
>>>>>>>> premises.
>>>>>>>
>>>>>>> You seem to forget that I agree.  H(P,P)==0 is correct given the
>>>>>>> criteria you have invented. 
>>>>>>
>>>>>> {H(P,P)==0 is correct} means that the input to H(P,P) is correctly 
>>>>>> decided as never halting.
>>>>>
>>>>> H(P, P) is required to answer the question Does P(P) halt"
>>>>>
>>>>> It is *specifically* being asked about the behaviour of int main() 
>>>>> { P(P); }
>>>>>
>>>>> It is *not* being asked about the behaviour of P(P) "under the 
>>>>> dominion of a halt decider" (whatever that may mean) nor about any 
>>>>> computation other than int main() { P(P); }. This follows directly 
>>>>> from how the halting problem is *defined*.
>>>>
>>>>     In computability theory, the halting problem is the
>>>>     problem of determining, from a description of an
>>>>     arbitrary computer program and an input, whether the
>>>>     program will finish running, or continue to run forever.
>>>>     https://en.wikipedia.org/wiki/Halting_problem
>>>>
>>>> No that is not true. The halting problem is always about program 
>>>> descriptions not running programs. This means that it is always 
>>>> about the input to the halt decider not the direct execution of the 
>>>> program.
>>>
>>> The *input* to the decider is a program description. But the question 
>>> the halt decider is expected to answer is about the *actual* program 
>>> which that description describes.
>>>
>>
>> In other words whether or not the the pure simulation of this 
>> description halts on its input.
> 
> No. The question is whether the *actual* computation described by the 
> input halts.
     In computability theory, the halting problem is the
    problem of determining, from a description of an
    arbitrary computer program and an input, whether the
    program will finish running, or continue to run forever.
    https://en.wikipedia.org/wiki/Halting_problem

The halting problem is always about program descriptions
not running programs. This means that it is always about
the input to the halt decider not the direct execution
of the program.

The most straight forward way to get the actual behavior of a program 
description is to simulate it.

> A pure simulation would have identical halting behaviour as the actual 
> computation, but the question isn't about simulations. It's about actual 
> computations. It isn't about what happens inside some decider; it's 
> about actual computations.
> 

When an "actual computation" has different behavior than the simulation 
of the input to the halt decider this proves that the actual computation 
is a different computation than the input to the halt decider thus not 
relevant to the halting problem.

> And since you claim that it has a *different* halting behaviour inside 
> your halt decider, you aren't dealing with a pure simulation but 
> something else. The correct answer to the question does P(P) halt is 
> determined by the *actual* computation and *nothing* else.
> 

In the case of the different between int main() { P(P); } and H(P,P) the 
different behavior is accounted for by many differences between the two 
computations.

We have this exact same issue of the difference between Ĥ applied to ⟨Ĥ⟩ 
and Ĥ.qx applied to ⟨Ĥ⟩ ⟨Ĥ⟩.

int main() { P(P); } halts only because H(P,P) correctly decides that 
its input never halts.

Ĥ applied to ⟨Ĥ⟩ halts only because Ĥ.qx applied to ⟨Ĥ⟩ ⟨Ĥ⟩ correctly 
decides that its input never halts.

There is a one way dependency relationship between
(a) int main() { P(P); } and H(P,P)
(b) Ĥ applied to ⟨Ĥ⟩ and Ĥ.qx applied to ⟨Ĥ⟩ ⟨Ĥ⟩

that is the key difference making the pairs of similar looking 
computations derive opposite results.

The key general principle is that distinct computations can have 
opposite behavior without contradiction.

These pairs of distinct computations can have opposite behavior without 
contradiction.
(a) int main() { P(P); } and H(P,P)
(b) Ĥ applied to ⟨Ĥ⟩ and Ĥ.qx applied to ⟨Ĥ⟩ ⟨Ĥ⟩

>>>> If the input to the halt decider never halts unless the halt decider 
>>>> aborts its simulation of this input then its input never halts.
>>>
>>> If P(P) halts, but the "simulation of P(P)" does not, then either 
>>> your simulator is broken or whatever is being simulated by the 
>>> decider *isn't* the the program which which was actually described by 
>>> its input. It is therefore answering about the *wrong* program.
>>>
>>> André
>>>
>>
>> Like I said H(P,P)==0 is correct as a logical consequence of its two 
>> premises and its two premises are true. There is no escape from this.
> 
> You still haven't given your two premises.
> 
> But it is entirely irrelevant anyways. As I said, the correct answer to 
> the question does P(P) halt is determined by the *actual* computation 
> and nothing else.
> 
> It is not possible to prove that H(P,P)==0 given that this doesn't match 
> the behaviour of the actual computation.
> 
> If you want to claim that it is the correct answer to the question 'does 
> P(P) 'under the dominion' of a simulating halt decider halt?' you can do 
> that to your heart's content.
> 

The actual computation must be at the same point in the execution trace 
as the input to the halt decider. The only way to do this is to simulate 
the input to the halt decider. Running the program at some other point 
in the execution trace does not count because it defines a different 
computation than the one under investigation.

int main() { P(P); }    halts because of what H(P,P) does later on.
int main() { H(P,P); }  halts because it aborts its simulation.

This proves that the above pair are different computations that can have 
opposite behavior without contradiction.

In the above two computations the relative order of P(P) to H(P,P) is 
reversed thus changing their relative placement in the execution trace 
which changes their behavior.

> But that *isn't* the question which the halting problem asks, nor is it 
> a question that anyone gives a damn about. It's basically just nonsense.
> 
> André
> 


-- 
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre 
minds." Einstein

[toc] | [prev] | [next] | [standalone]

#3417 — Re: That P(P) of main() halts does not contradict H(P,P)==0 [ relative invocation order ]

On 8/31/2021 10:07 AM, André G. Isaak wrote:
> On 2021-08-31 08:24, olcott wrote:
>> On 8/30/2021 11:45 PM, André G. Isaak wrote:
> 
>>      In computability theory, the halting problem is the
>>     problem of determining, from a description of an
>>     arbitrary computer program and an input, whether the
>>     program will finish running, or continue to run forever.
>>     https://en.wikipedia.org/wiki/Halting_problem
>>
>> The halting problem is always about program descriptions
>> not running programs. This means that it is always about
>> the input to the halt decider not the direct execution
>> of the program.
> 
> This is, of course, errant nonsense. The input to a halt decider is 
> simply data. Data doesn't *have* halting behaviour. Only the actual 
> computation described by that data can have halting behaviour. Your 
> inability to comprehend simple definitions is not the basis for a 
> coherent argument.
> 
> 
>> The most straight forward way to get the actual behavior of a program 
>> description is to simulate it.
> 
> No. The most straightforward way, and the ONLY 100% reliable way to get 
> the actual behaviour of a program description is to run the program it 
> describes.
> 
>>> A pure simulation would have identical halting behaviour as the 
>>> actual computation, but the question isn't about simulations. It's 
>>> about actual computations. It isn't about what happens inside some 
>>> decider; it's about actual computations.
>>>
>>
>> When an "actual computation" has different behavior than the 
>> simulation of the input to the halt decider this proves that the 
>> actual computation is a different computation than the input to the 
>> halt decider thus not relevant to the halting problem.
> 
> No, it proves that the 'simulation' performed by your halt decider is 
> not an accurate simulation.
> 
> André
> 

I think that I finally got it this time:

The difference in the relative invocation order of P(P) to H(P,P) 
accounts for the difference in behavior.

P(P) from main() invokes P(P) then H(P,P).

H(P,P) from main reverses this relative invocation order to H(P,P) then 
P(P).

Changes to the relative invocation order define distinctly different 
computations that can have different behavior without contradiction.


-- 
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre 
minds." Einstein

[toc] | [prev] | [next] | [standalone]

#3418 — Re: That P(P) of main() halts does not contradict H(P,P)==0 [ relative invocation order ]

On 8/31/2021 4:17 PM, Malcolm McLean wrote:
> On Tuesday, 31 August 2021 at 20:11:39 UTC+1, olcott wrote:
>> On 8/31/2021 1:40 PM, André G. Isaak wrote:
>>
>>>
>>> It makes no reference at all to the "relative invocation order" of H
>>> with respect to P. It makes no reference to H at all.
>>>
>>> If its answer doesn't correspond to the behaviour of P(P) which is a
>>> halting computation, then its answer is *wrong*.
>> H(P,P) specifies the invocation order of H(P,P) then P(P).
>>
>> If P did not have the pathological self-reference error the invocation
>> would not matter. Because P does have the pathological self-reference
>> error the invocation order does matter.
>>
>> It is very diffcult to correctly handle incorrect input.
>>
>> The key irrefutable point is that H(P,P) then P(P) is an entirely
>> different computation than P(P) then H(P,P)
>>
> So that's the central point. If H(P,P) from the root is a different computation
> to H(P,P),called from P, then indeed you can provide a counter-example to
> the Linz proof.
> 

// Simplified Linz Ĥ (Linz:1990:319)
// Strachey(1965) CPL translated to C
void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
}

// These are two different computations that can have
// opposite results without contradiction.

int main () { H(P,P); }  // invocation order H(P,P) then P(P)
int main () { P(P); }    // invocation order P(P) then H(P,P)

H(P,P) from anywhere has exactly the same behavior.

// H1 is an identical copy of H
int main () { H1(P,P); }  // reports that its input halts.

The different invocation order of H1 and H is how they can have 
identical code and different behavior. Halt deciders can only see the 
instructions that they (and their slave instances) simulate. H1 cannot 
see any of the instructions that H simulates.

H(P,P); P(P); and H1(P,P); are shown in sections V1,V2,V3 of this paper:

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation

-- 
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre 
minds." Einstein

[toc] | [prev] | [next] | [standalone]

#3420

On 9/1/2021 9:44 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
> 
>> Ĥ.qx applied to ⟨Ĥ⟩ ⟨Ĥ⟩ never halts [UNLESS] it aborts its simulation,
>> not very hard at all for people that care about truth as opposed to
>> and contrast with winning an argument.
> 
> (correction added from your own follow-up)
> 
> Ĥ.qx applied to ⟨Ĥ⟩ ⟨Ĥ⟩ halts.  It halts in rejecting state Ĥ.qn.  There
> is no dispute about this fact from you or anyone else.  The /reason/ it
> halts is interesting to you, but /not/ to anyone else.
> 
> The facts remain: ⟨Ĥ⟩ ⟨Ĥ⟩ encodes a halting computation and you were
> flat-out wrong to say that is does not.  And H (the machine embedded in
> Ĥ at Ĥ.qx) is wrong to reject the string for that reason.  You will
> never admit either mistake.
> 
> That you are wrong is so blinding obvious that any paper you write about
> the theorem will go in the editor's bin in seconds.  (Unless he or she
> decides it's worth pinning on the staff room notice board for fun.)
> 

The reason that I created the x86utm operating system was to enable 
every single detail of the halting problem to be specified at the high 
level of abstraction of C/x86 so that people don't merely imagine 
details that are not true.

When we examine the x86 execution trace of the simulation of the input 
to H(P,P) we can determine that unless H aborts its simulation that its 
input never halts.

Those that take the time to analyze this realize that it is a reasonable 
criterion measure of never halting. Those only interested in winning an 
argument that don't give a rat's ass for truth never notice this.

These same people reject that the simulation of P(P) meets the following 
criteria because they simply don't bother to check, their entire focus 
is on rebuttal thus verifying that I am correct is off the table.

The infinite recursion detection criteria are met by the above execution 
trace:
(a) P calls H twice in sequence from the same machine address.
(b) With the same parameters: (P,P) to H.
(c) With no conditional branch or indexed jump instructions in the 
execution trace of P.
(d) We know that there are no return instructions in H because we know 
that H is in pure simulation mode.




-- 
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre 
minds." Einstein

[toc] | [prev] | [next] | [standalone]

#3421 — Re: That P(P) of main() halts does not contradict H(P,P)==0 [ Pathological Input ]

On 9/1/2021 9:34 PM, Richard Damon wrote:
> On 9/1/21 10:18 PM, olcott wrote:
>> On 9/1/2021 8:42 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 9/1/2021 7:58 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 9/1/2021 9:44 AM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> Ĥ.qx applied to ⟨Ĥ⟩ ⟨Ĥ⟩ never halts [UNLESS] it aborts its
>>>>>>>> simulation,
>>>>>>>> not very hard at all for people that care about truth as opposed to
>>>>>>>> and contrast with winning an argument.
>>>>>>>
>>>>>>> (correction added from your own follow-up)
>>>>>>> Ĥ.qx applied to ⟨Ĥ⟩ ⟨Ĥ⟩ halts.  It halts in rejecting state Ĥ.qn.
>>>>>>> There
>>>>>>> is no dispute about this fact from you or anyone else.  The
>>>>>>> /reason/ it
>>>>>>> halts is interesting to you, but /not/ to anyone else.
>>>>>>>
>>>>>>> The facts remain: ⟨Ĥ⟩ ⟨Ĥ⟩ encodes a halting computation and you were
>>>>>>> flat-out wrong to say that is does not.  And H (the machine
>>>>>>> embedded in
>>>>>>> Ĥ at Ĥ.qx) is wrong to reject the string for that reason.  You will
>>>>>>> never admit either mistake.
>>>>>>> That you are wrong is so blinding obvious that any paper you write
>>>>>>> about
>>>>>>> the theorem will go in the editor's bin in seconds.  (Unless he or
>>>>>>> she
>>>>>>> decides it's worth pinning on the staff room notice board for fun.)
>>>>>>
>>>>>> The reason that I created the x86utm operating system was to enable
>>>>>> every single detail of the halting problem to be specified at the high
>>>>>> level of abstraction of C/x86 so that people don't merely imagine
>>>>>> details that are not true.
>>>>>
>>>>> You created it to distract from the massive lies you told in Dec 2018:
>>>>>
>>>>>      "I now have an actual H that decides actual halting for an
>>>>> actual (Ĥ,
>>>>>      Ĥ) input pair.  I have to write the UTM to execute this code, that
>>>>>      should not take very long.  The key thing is the H and Ĥ are 100%
>>>>>      fully encoded as actual Turing machines."
>>>>>      "Everyone has claimed that H on input pair (Ĥ, Ĥ) meeting the Linz
>>>>>      specs does not exist. I now have a fully encoded pair of Turing
>>>>>      Machines H / Ĥ proving them wrong."
>>>>>
>>>>> You need to concentrate the steaming pile of x86 code you are hiding
>>>>> rather than on the TM you lied about having.  And try to avoid saying
>>>>> anything clearly, because every time you do you get burned.  Your ⟨Ĥ⟩
>>>>> ⟨Ĥ⟩ encodes a halting computation and your H should accept it.
>>>>
>>>> In other words x86 code is beyond your technical competence.
>>>> Nothing wrong with that except hiding it behind denigration.
>>>
>>> The reasons why you are wrong are clearly laid out.
>>
>> This is the key element and although it is self-evidently true it really
>> could use a much better proof.
>>
>> PREMISE ONE
>> Simulating Halt Decider Theorem (Olcott 2020):
>> A simulating halt decider correctly decides that any input that never
>> halts unless the simulating halt decider aborts its simulation of this
>> input is an input that never halts.
>>
> 
> Error: Definition of the Correct answer is does the machine that the
> input represent halt in a finite number of steps, or not. That answer is
> irrespective of WHY.
I am not providing this response to you, because it is beyond your 
capacity to understand. I am only providing this to those that can 
understand.

[Pathological Input] to a halt decider is stipulated to mean any input 
that was defined to do the opposite of whatever its corresponding halt 
decider decides as Sipser describes:

    Now we construct a new Turing machine D with H as a subroutine.
    This new TM calls H to determine what M does when the input to
    M is its own description ⟨M⟩. Once D has determined this
    information, it does the opposite.  (Sipser:1997:165)

When the input to the halt decider is pathological we must adapt the 
halt deciding criteria so that it can still correctly decide this 
Pathological input. The conventional halt decider criteria cannot 
correctly handle [Pathological Input].

-- 
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre 
minds." Einstein

[toc] | [prev] | [next] | [standalone]

#3423 — Re: That P(P) of main() halts does not contradict H(P,P)==0 [ ignorance ]

On 9/2/2021 6:28 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
> 
>> On 9/1/2021 8:42 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 9/1/2021 7:58 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 9/1/2021 9:44 AM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> Ĥ.qx applied to ⟨Ĥ⟩ ⟨Ĥ⟩ never halts [UNLESS] it aborts its simulation,
>>>>>>>> not very hard at all for people that care about truth as opposed to
>>>>>>>> and contrast with winning an argument.
>>>>>>>
>>>>>>> (correction added from your own follow-up)
>>>>>>> Ĥ.qx applied to ⟨Ĥ⟩ ⟨Ĥ⟩ halts.  It halts in rejecting state Ĥ.qn.  There
>>>>>>> is no dispute about this fact from you or anyone else.  The /reason/ it
>>>>>>> halts is interesting to you, but /not/ to anyone else.
>>>>>>>
>>>>>>> The facts remain: ⟨Ĥ⟩ ⟨Ĥ⟩ encodes a halting computation and you were
>>>>>>> flat-out wrong to say that is does not.  And H (the machine embedded in
>>>>>>> Ĥ at Ĥ.qx) is wrong to reject the string for that reason.  You will
>>>>>>> never admit either mistake.
>>>>>>> That you are wrong is so blinding obvious that any paper you write about
>>>>>>> the theorem will go in the editor's bin in seconds.  (Unless he or she
>>>>>>> decides it's worth pinning on the staff room notice board for fun.)
>>>>>>
>>>>>> The reason that I created the x86utm operating system was to enable
>>>>>> every single detail of the halting problem to be specified at the high
>>>>>> level of abstraction of C/x86 so that people don't merely imagine
>>>>>> details that are not true.
>>>>>
>>>>> You created it to distract from the massive lies you told in Dec 2018:
>>>>>
>>>>>      "I now have an actual H that decides actual halting for an actual (Ĥ,
>>>>>      Ĥ) input pair.  I have to write the UTM to execute this code, that
>>>>>      should not take very long.  The key thing is the H and Ĥ are 100%
>>>>>      fully encoded as actual Turing machines."
>>>>>      "Everyone has claimed that H on input pair (Ĥ, Ĥ) meeting the Linz
>>>>>      specs does not exist. I now have a fully encoded pair of Turing
>>>>>      Machines H / Ĥ proving them wrong."
>>>>>
>>>>> You need to concentrate the steaming pile of x86 code you are hiding
>>>>> rather than on the TM you lied about having.  And try to avoid saying
>>>>> anything clearly, because every time you do you get burned.  Your ⟨Ĥ⟩
>>>>> ⟨Ĥ⟩ encodes a halting computation and your H should accept it.
>>>>
>>>> In other words x86 code is beyond your technical competence.
>>>> Nothing wrong with that except hiding it behind denigration.
>>>
>>> The reasons why you are wrong are clearly laid out.
>>
>> This is the key element and although it is self-evidently true it
>> really could use a much better proof.
>>
>> PREMISE ONE
>> Simulating Halt Decider Theorem (Olcott 2020):
>> A simulating halt decider correctly decides that any input that never
>> halts unless the simulating halt decider aborts its simulation of this
>> input is an input that never halts.
> 
> Does this support the fact that your ⟨Ĥ⟩ ⟨Ĥ⟩ encodes a halting
> computation, and that your H should accept it?  If so, just say "sorry,
> I was wrong".  If it does /not/ support that fact, then you need to find
> out what is wrong with it.
> 

This might actually be legitimately beyond your capacity to understand:

// Simplified Linz Ĥ (Linz:1990:319)
// Strachey(1965) CPL translated to C
void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
}

int main()
{
   P((u32)P);
}

Ĥ.q0 ⟨Ĥ1⟩ ⊢* Ĥ.qx ⟨Ĥ1⟩ ⟨Ĥ2⟩ ⊢* Ĥ.qy ∞
if the simulated ⟨Ĥ1⟩ applied to ⟨Ĥ2⟩ halts, and

Ĥ.q0 ⟨Ĥ1⟩ ⊢* Ĥ.qx ⟨Ĥ1⟩ ⟨Ĥ2⟩ ⊢* Ĥ.qn
if the simulated ⟨Ĥ1⟩ applied to ⟨Ĥ2⟩ does not halt

This criteria merely relies on the fact that the UTM simulation of a 
machine description of a machine is computationally equivalent to the 
direct execution of this same machine:

Simulating Halt Decider Theorem (Olcott 2020):
A simulating halt decider correctly decides that any input that never 
halts unless the simulating halt decider aborts its simulation of this 
input is an input that never halts.

(A) The first line of main() halts.
This is analogous to the first element of Ĥ halts.

(B) The input to the first line of P never halts.
This is analogous to the input to Ĥ.qx ⟨Ĥ1⟩ ⟨Ĥ2⟩ never halts.

That you insist that we only pay attention to (A) and can safely ignore 
(B) might legitimately be simply ignorance on your part.




>> We have been around and around about this and none of the dishonest
>> dodges cracked the logical necessity at all.
> 
> ...but you keep dodging none the less.  You are wrong for very simple
> reasons.  So simple that you must post waffle like the above rather than
> address them.  You hope the waffle will draw the fire away from the fact
> that your string pair ⟨Ĥ⟩ ⟨Ĥ⟩ encodes a halting computation and your H
> should accept it.
> 
> The string pair <M> w is /defined/ as the encoding of the computation
> M(w) (that's M with w on the tape) and you know that Ĥ(⟨Ĥ⟩) halts and
> that H should not reject string pairs that encode halting computations.
> Can I put it any more simply than this?  Will you address this error?
> No.  You will post more stuff in the hope of getting people to talk
> about anything but the core mistake.
> 


-- 
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre 
minds." Einstein

[toc] | [prev] | [next] | [standalone]

#3426 — Re: That P(P) of main() halts does not contradict H(P,P)==0 [ ignorance ]

On 9/2/2021 8:29 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
> 
>> On 9/2/2021 8:05 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 9/2/2021 7:27 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>>>> Anyone that understands what I am saying can tell that their have been
>>>>>> zero actual rebuttals.
>>>>>
>>>>> Is that an empty set?  Does anyone in the world understand and agree
>>>>> with you?  Someone at the grocery store maybe?
>>> Well?  Does anyone in the world understand and agree with you?  Will you
>>> avoid every question put to you?
>>>
>>>> None of the rebuttals ever directly addressed any of the points that I
>>>> made. They always changed the subject to another different unrelated
>>>> point.
>>> Flat-out lie.  You made this point crucial point on 12th Aug:
>>>     "⟨Ĥ⟩ ⟨Ĥ⟩ is not a string that encodes a halting computation."
>>
>> Ĥ.q0 ⟨Ĥ1⟩     // You pay attention to this
>> Ĥ.qx ⟨Ĥ1⟩ ⟨Ĥ2⟩ // You utterly ignore this
> 
> Liar.  I have addressed this on many occasions.  Here, yet again, is
> what I have to say about it.  You show us that Ĥ.qx ⟨Ĥ1⟩ ⟨Ĥ2⟩ ⊦ Ĥ.qn and
> since ⟨Ĥ1⟩ = ⟨Ĥ2⟩ = ⟨Ĥ⟩ we see that H (embedded as it is at Ĥ.qx)
> rejects the string ⟨Ĥ⟩ ⟨Ĥ⟩.  How you think I knew that?  I don't make
> this stuff up, you do.
> 

The simple fact that you ignore is that the input to Ĥ.qx ⟨Ĥ1⟩ ⟨Ĥ2⟩
never halts unless Ĥ.qx aborts its simulation of this input.
You have dodged this 2017-03-11  (4.5 years ago)

On 3/11/2017 3:13 PM, peteolcott wrote:
 > http://LiarParadox.org/HP_Infinite_Recursion.pdf
 >
 > As this page 319 of An Introduction to Formal Languages and Automata
 > by Peter Linz 1990 indicates
 >
 >  From H' we construct another Turing machine H-Hat. This new machine 
takes as input Wm, copies it then behaves exactly like H'.
 >
 > q0 Wm |-* H-Hat q0 Wm Wm...
 >
 > Page 320 indicates that we apply H-Hat to itself as input.
 >
 > The problem is that every H-Hat needs a pair of inputs.
 >
 > H-Hat takes an H-Hat as input and copies it so that it
 > can analyze how its input H-hat would analyze the copy
 > of H-Hat that it just made.
 >
 > The input H-Hat would have to copy its own input H-Hat
 > so that it can analyze what its own input H-Hat would
 > do on its own input, on and on forever...
 >
 > Copyright 2016 and 2017 Pete Olcott.
 >


-- 
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre 
minds." Einstein

[toc] | [prev] | [next] | [standalone]

#3427 — Re: That P(P) of main() halts does not contradict H(P,P)==0 [ ignorance ]

On 9/2/2021 11:59 PM, André G. Isaak wrote:
> On 2021-09-02 22:09, olcott wrote:
>> On 9/2/2021 10:52 PM, Richard Damon wrote:
>>> On 9/2/21 11:18 PM, olcott wrote:
> 
>>>> Any input that never halts unless its simulation is aborted is an input
>>>> that never halts.
>>>>
>>>
>>> Bad Terminology. Inputs themselves don't halt. They only represent
>>> computations that might be halting or not.
>>>
>>
>>     In computability theory, the halting problem is the
>>     problem of determining, from a description of an arbitrary
>>     computer program and an input, whether the program will
>>     finish running, or continue to run forever.
>>     https://en.wikipedia.org/wiki/Halting_problem
>>
>> To bypass the pathology of pathological inputs:
>>     In computability theory, the halting problem is the
>>     problem of determining, from a description of an arbitrary
>>     computer program and an input,
>>
>>     whether the simulation of this input must be aborted to
>>     prevent it from running forever.
>>
>> The above criteria is valid on the basis of the known equivalence
>> between the direct execution of a computation and its simulation by a 
>> UTM. The same criteria universally works on all inputs.
> 
> But the direct execution of P(P) *does* halt. So if there is an 
> equivalence between direct execution and simulation by a UTM, then the 
> simulation by a UTM must also halt.
> 
> André
> 

The direct execution of P has a dependency on the H(P,P) that it calls. 
int main(){ P(P); } only halts because its call to H(P,P) returns 0. No 
instance of H(P,P) has any dependency on the return value of another 
function.

This makes int main(){ P(P); } a computationally distinct different 
computation than H(P,P) that has no such dependency.

You can dance all around this yet there exists no correct rebuttal in 
the universe that shows that these computations are equivalent or that 
computations that are not equivalent must have equivalent behavior.



    In computability theory, the halting problem is the
    problem of determining, from a description of an arbitrary
    computer program and an input, whether the program will
    finish running, or continue to run forever.
    https://en.wikipedia.org/wiki/Halting_problem

To bypass the pathology of pathological inputs:
    In computability theory, the halting problem is the
    problem of determining, from a description of an arbitrary
    computer program and an input,

    whether the simulation of this program must be aborted to
    prevent it from running forever.

The above criteria is valid on the basis of the known equivalence
between the direct execution of a computation and its simulation
by a UTM. The same criteria universally works on all inputs allowing
their halting status to be correctly decided.



-- 
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre 
minds." Einstein

[toc] | [prev] | [next] | [standalone]

#3428 — Re: That P(P) of main() halts does not contradict H(P,P)==0 [ ignorance ]

On 9/3/2021 8:41 AM, Ben Bacarisse wrote:
> Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes:
> 
>>> Programs will Halt, not inputs. Inputs are just the desciption used to
>>> specify the program.
>>>
>> There seems to be some sort of claim that  whilest H_Hat<H_Hat> is
>> halting, the string pair <H_Hat><H_Hat> (angle brackets represent tape
>> contents) is non-halting. I really can't get to the bottom of what exactly is
>> being claimed here.
> 
> I think that's deliberate, now. Several people have tried to get him to
> use terms correctly (or, as a last resort, invent new well-defined
> ones), but he won't do that because the obfuscation is all he has left.
> 

Ĥ.q0 ⟨Ĥ1⟩     // is equivalent to int main() { P(P); }
Ĥ.qx ⟨Ĥ1⟩ ⟨Ĥ2⟩ // is equivalent to H(P,P)

The H(P,P) C/x86 code makes a perfect and complete analogy to the
Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ Turing machine code. If you don't understand the x86 
language well enough you won't be able to ever understand what I am saying.

All of the actual Turing Machine base proofs must leave out almost all 
of the details because their programs would hundreds of thousands of 
pages. A TM with random access memory would be far less of a nightmare 
to deal with.

It is obvious that every simulation of P(P) would never halt unless H 
aborts its simulation of P(P) to everyone one sufficiently understanding 
the material.

It is obvious to everyone one sufficiently understanding the material 
that all "rebuttals" are based on either deception or failing to 
understand the material.

I show that X is true because of Y and the "rebuttal" essentially takes 
the form: "I just don't believe that".

> He has said
> 
>    "⟨Ĥ⟩ ⟨Ĥ⟩ is not a string that encodes a halting computation" (Aug
>    2021)
> 
> This is as wrong as saying "2 plus 2 is not 4" but when it's pointed out
> that ⟨Ĥ⟩ ⟨Ĥ⟩ /does/ encode a halting computation he says
> 
>    "the input to Ĥ.qx ⟨Ĥ1⟩ ⟨Ĥ2⟩ never halts" (Aug 2021)
> 
> which is akin to responding to the fact that 2+2=4 with "the input to 2a
> plus 2b goes not four".
> 
> The bad wording is seems suspiciously deliberate.  What else has he got?
> He hopes, I think, that people will argue about the bad wording and not
> challenge the basic error.
> 
>> ... unless you set yourself the insane goal of "solving the halting
>> problem".
> 
> PO has claimed, in a court of law, to be God.  I think we can safely say
> that his relationship with reality is... er... fragile.
> 


-- 
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre 
minds." Einstein

[toc] | [prev] | [next] | [standalone]

#3430 — Re: That P(P) of main() halts does not contradict H(P,P)==0 [ Linz H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ ]

On 9/3/2021 7:07 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
> 
>> On 9/3/2021 4:40 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 9/3/2021 10:50 AM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 9/3/2021 10:16 AM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 9/3/2021 9:58 AM, Ben Bacarisse wrote:
> 
>>>>>>>>> What string encodes the halting computation of Ĥ applied ⟨Ĥ⟩?
>>>>>>>>
>>>>>>>> The input to Ĥ.qx ⟨Ĥ1⟩ ⟨Ĥ2⟩ never halts.
>>>>>>>
>>>>>>> It turns out you have ducked my question at least six times previously
>>>>>>> /without/ counting any other people who has asked!  This makes 7:
>>>>>>>
>>>>>>
>>>>>> I will quit dodging your dishonest dodge as soon as you acknowledge that
>>>>>> The input to Ĥ.qx ⟨Ĥ1⟩ ⟨Ĥ2⟩ never halts.
>>>>>
>>>>> Classic crank!  You'll answer when I accept a garbled falsehood, an
>>>>> obligation you are guaranteed to never have to meet!
>>>>>
>>>>> Anyway, you've dodged the question 9 times now (8 was in another
>>>>> thread).  Here's you chance to go for ten times:
>>>>>
>>>>> ------------------------------------------------------------
>>>>> What string encodes the halting computation of Ĥ applied ⟨Ĥ⟩?
>>>>> ------------------------------------------------------------
>>>
>>> Yes, we're up to ten dodges.  Do you want to try for 11 or will just
>>> ignore the post?
>>>
>>> What string encodes the halting computation of Ĥ applied ⟨Ĥ⟩?
>>
>> This string encodes Ĥ that halts and ⟨Ĥ1⟩ ⟨Ĥ2⟩ that never halt.
> 
> What string?
> 
>> Ĥ.q0 ⟨Ĥ1⟩ ⊢* Ĥ.qx ⟨Ĥ1⟩ ⟨Ĥ2⟩ ⊢* Ĥ.qn
>> if the simulated ⟨Ĥ1⟩ applied to ⟨Ĥ2⟩ does not halt
> 
> There are at least three strings mentioned on those two lines and even
> then you don't claim that any of them encodes the halting computation of
> Ĥ applied ⟨Ĥ⟩.  So no answer here.  That's 11 times you've avoided
> answering.
> 
> ------------------------------------------------------------
> What string encodes the halting computation of Ĥ applied ⟨Ĥ⟩?
> ------------------------------------------------------------
> 
> It's not a hard question.  In fact I am sure you know the answer.  You
> won't answer because you either have to say something silly, or admit
> you are wrong.
> 
> Would you like me to explain again why this question matters so much?
> 

A whole new way of making my point:
When the original Linz H defined to be a simulating halt decider 
according to this criteria:

    In computability theory, the halting problem is the
    problem of determining, from a description of an arbitrary
    computer program and an input,

    whether the simulation of this program must be aborted to
    prevent it from running forever.

The above criteria is valid on the basis of the known equivalence
between the direct execution of a computation and its simulation
by a UTM. The same criteria universally works on all inputs allowing
their halting status to be correctly decided.

The Linz H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ correctly transitions to H.qy

This is analogous to int main() { H1(P,P; } in section V3

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation


-- 
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre 
minds." Einstein

[toc] | [prev] | [next] | [standalone]

#3422

On 9/2/2021 4:40 AM, Malcolm McLean wrote:
> On Thursday, 2 September 2021 at 00:06:40 UTC+1, richar...@gmail.com wrote:
>> On 9/1/21 9:27 AM, olcott wrote:
>>> On 9/1/2021 4:19 AM, Malcolm McLean wrote:
>>>> On Wednesday, 1 September 2021 at 05:33:09 UTC+1, olcott wrote:
>>>>>
>>>>> For the moment let's just hypothesize that my "theorem" is true and that
>>>>> it has been agreed that int main() { P(P); } never halts unless H(P,P)
>>>>> aborts its input. Then we can conclude that the input to H(P,P) never
>>>>> halts.
>>>>>
>>>> I suggested this a long time ago. The halt issued by the copy of H
>>>> embedded
>>>> in P is considered special. That idea was rejected.
>>>>
>>>
>>> There is a single master instance of H that allocates all of its tape to
>>> its slave instances.
>>>
>> So?
>>
>> If it doesn't create proper 'virtual' copies of the machines and
>> simulate them accurately it doesn't prove anything.
>>
> It makes it easier to get confused.
> With the Linz set up, there is a near copy of H on the tape, but it is separate
> code. H is a physical machine (or more likely an electronic emulation).
> With PO's set up, we have C routines. And the same physical copy of H
> is called (not emulated, which is surprising and we still haven't got to
> the bottom of that).

All of the inputs to any halt decider are emulated by an x86 emulator. 
Functions that are called by this emulated code are also emulated.

> So when you say "the program under test was halted by H" it's not clear
> exactly what you are talking about.

int main { P(P); } only halts because the program under test of H(P,P) 
called by P(P) had its simulation aborted.

>>
>> The fact that simulated copies of H(H^,^) are apparently non-halting
>> means that something is broken, either H is inaccurate in its simulating
>> or H isn't a Computation, and thus can't be a decider.
>>
> It's common ground that H_Hat<H_Hat> halts, and H<H_Hat><H_Hat>
> returns false (non-halting). But the claim is that H is nevertheless correct.

We have to adapt the halt deciding criteria so that it can correctly 
handle pathological inputs.

Pathological Input to a halt decider is stipulated to mean any input 
that was defined to do the opposite of whatever its corresponding halt 
decider decides as Sipser describes:

      Now we construct a new Turing machine D with H as a subroutine.
      This new TM calls H to determine what M does when the input to M
      is its own description ⟨M⟩. Once D has determined this information,
      it does the opposite.  (Sipser:1997:165)

This criteria merely relies on the fact that the UTM simulation of a 
machine description of a machine is computationally equivalent to the 
direct execution of this same machine:

Simulating Halt Decider Theorem (Olcott 2020):
A simulating halt decider correctly decides that any input that never 
halts unless the simulating halt decider aborts its simulation of this 
input is an input that never halts.

> I've suggested to PO that maybe that's because the halts issued by the copy
> of H in H_Hat are "special". But he hasn't agreed with that. He writes
> [PO}
>>> When we examine the x86 execution trace of the simulation of the input
>>> to H(P,P) we can determine that unless H aborts its simulation that its
>>> input never halts.
>>>
> So this begs the question.
> 


-- 
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre 
minds." Einstein

[toc] | [prev] | [next] | [standalone]

#3424 — Re: That P(P) of main() halts does not contradict H(P,P)==0 [ pathological inputs ]

On 9/2/2021 10:19 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
> 
>> On 9/2/2021 4:40 AM, Malcolm McLean wrote:
> 
>>> It's common ground that H_Hat<H_Hat> halts, and H<H_Hat><H_Hat>
>>> returns false (non-halting). But the claim is that H is nevertheless correct.
>>
>> We have to adapt the halt deciding criteria so that it can correctly
>> handle pathological inputs.
> 
> Explicit admission (again) that you are not deciding halting.  Does
> anyone care about the criterion you are applying and confusing calling
> halting?  I don't think so.
> 
>> Pathological Input to a halt decider is stipulated to mean any input
>> that was defined to do the opposite of whatever its corresponding halt
>> decider decides as Sipser describes:
>>
>>       Now we construct a new Turing machine D with H as a subroutine.
>>       This new TM calls H to determine what M does when the input to M
>>       is its own description ⟨M⟩. Once D has determined this information,
>>       it does the opposite.  (Sipser:1997:165)
> 
> There is no algorithm that can detect what you call "Pathological
> Input".  You keep quoting me saying that, so presumably you agree with
> me.  Why are you wasting time on an undecidable set ("pathological
> inputs") that is not even interesting?
> 

This criteria merely relies on the fact that the UTM simulation of a 
machine description of a machine is computationally equivalent to the 
direct execution of this same machine:

Simulating Halt Decider Theorem (Olcott 2020):
A simulating halt decider correctly decides that any input that never 
halts unless the simulating halt decider aborts its simulation of this 
input is an input that never halts.

The above criteria circumvents the pathological aspect of pathological 
input. Because the halt decider acts as a pure simulator until after it 
makes its halt status decision the pathological feedback loop is 
eliminated.



-- 
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre 
minds." Einstein

[toc] | [prev] | [next] | [standalone]

#3425 — Re: That P(P) of main() halts does not contradict H(P,P)==0 [ pathological inputs ]

On 9/2/2021 11:10 AM, Malcolm McLean wrote:
> On Thursday, 2 September 2021 at 16:19:57 UTC+1, Ben Bacarisse wrote:
>> olcott <No...@NoWhere.com> writes:
>>
>>> On 9/2/2021 4:40 AM, Malcolm McLean wrote:
>>
>>>> It's common ground that H_Hat<H_Hat> halts, and H<H_Hat><H_Hat>
>>>> returns false (non-halting). But the claim is that H is nevertheless correct.
>>>
>>> We have to adapt the halt deciding criteria so that it can correctly
>>> handle pathological inputs.
>> Explicit admission (again) that you are not deciding halting. Does
>> anyone care about the criterion you are applying and confusing calling
>> halting? I don't think so.
>>> Pathological Input to a halt decider is stipulated to mean any input
>>> that was defined to do the opposite of whatever its corresponding halt
>>> decider decides as Sipser describes:
>>>
>>> Now we construct a new Turing machine D with H as a subroutine.
>>> This new TM calls H to determine what M does when the input to M
>>> is its own description ⟨M⟩. Once D has determined this information,
>>> it does the opposite. (Sipser:1997:165)
>> There is no algorithm that can detect what you call "Pathological
>> Input". You keep quoting me saying that, so presumably you agree with
>> me. Why are you wasting time on an undecidable set ("pathological
>> inputs") that is not even interesting?
>>
> You've nailed it exactly. PO has said, in effect, "Why not detect the self-contradictory
> case and handle it specially?". According to you, when you teach this material
> to computer science freshmen, there's always someone who raises that
> question.
> 

Pathological Input to a halt decider is stipulated to mean any input 
that was defined to do the opposite of whatever its corresponding halt 
decider decides as Sipser describes:

      Now we construct a new Turing machine D with H as a subroutine.
      This new TM calls H to determine what M does when the input to M
      is its own description ⟨M⟩. Once D has determined this information,
      it does the opposite.  (Sipser:1997:165)

This criteria merely relies on the fact that the UTM simulation of a 
machine description of a machine is computationally equivalent to the 
direct execution of this same machine:

Simulating Halt Decider Theorem (Olcott 2020):
A simulating halt decider correctly decides that any input that never 
halts unless the simulating halt decider aborts its simulation of this 
input is an input that never halts.

The above criteria handles all inputs correctly including pathological 
inputs.

Because the halt decider acts as a pure simulator until after its halt 
status decision has been made the pathological feedback loop has been 
eliminated from the halt status decision of pathological inputs.

-- 
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre 
minds." Einstein

[toc] | [prev] | [standalone]

Back to top | Article view | comp.software-eng

csiph-web