Re: Proof that H(D,D) meets its abort criteria --mistake--

On 3/15/2024 10:28 PM, Richard Damon wrote:

On 3/15/24 8:14 PM, olcott wrote:

On 3/15/2024 6:26 PM, Richard Damon wrote:

On 3/15/24 3:41 PM, olcott wrote:

On 3/15/2024 5:10 PM, Richard Damon wrote:

On 3/15/24 2:13 PM, olcott wrote:

On 3/15/2024 3:27 PM, Richard Damon wrote:

On 3/15/24 12:38 PM, olcott wrote:

On 3/15/2024 2:30 PM, Richard Damon wrote:

On 3/15/24 12:14 PM, olcott wrote:

On 3/15/2024 2:06 PM, Richard Damon wrote:

On 3/15/24 11:39 AM, olcott wrote:

On 3/15/2024 1:38 PM, immibis wrote:

On 15/03/24 18:52, olcott wrote:

On 3/15/2024 12:36 PM, Richard Damon wrote:

On 3/15/24 9:20 AM, olcott wrote:

Best selling author of Theory of Computation textbooks:
*Introduction To The Theory Of Computation 3RD, by sipser*
https://www.amazon.com/Introduction-Theory-Computation-Sipser/dp/8131525295/
>
Date 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)
(a) 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
(b) H can abort its simulation of D and correctly report that D specifies a non-halting sequence of configurations.
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*When we apply the abort criteria* (elaborated above)
Will you halt if you never abort your simulation?
*Then H(D,D) is proven to meet this criteria*
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*Proof that H(D,D) meets its abort criteria*
>
int D(int (*x)())
{
   int Halt_Status = H(x, x);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
int main()
{
   Output("Input_Halts = ", H(D,D));
}
>
  machine   stack     stack     machine    assembly
  address   address   data      code       language
  ========  ========  ========  =========  =============
[00001d22][00102fc9][00000000] 55         push ebp      ; begin main()
[00001d23][00102fc9][00000000] 8bec       mov ebp,esp
[00001d25][00102fc5][00001cf2] 68f21c0000 push 00001cf2 ; push DD
[00001d2a][00102fc1][00001cf2] 68f21c0000 push 00001cf2 ; push D
[00001d2f][00102fbd][00001d34] e8eef7ffff call 00001522 ; call H(D,D)
>
H: Begin Simulation   Execution Trace Stored at:113075
Address_of_H:1522
[00001cf2][00113061][00113065] 55         push ebp ; enter D(D)
[00001cf3][00113061][00113065] 8bec       mov ebp,esp
[00001cf5][0011305d][00103031] 51         push ecx
[00001cf6][0011305d][00103031] 8b4508     mov eax,[ebp+08]
[00001cf9][00113059][00001cf2] 50         push eax ; push D
[00001cfa][00113059][00001cf2] 8b4d08     mov ecx,[ebp+08]
[00001cfd][00113055][00001cf2] 51         push ecx ; push D
[00001cfe][00113051][00001d03] e81ff8ffff call 00001522 ; call H(D,D)
H: Recursive Simulation Detected Simulation Stopped
                           H(D,D) returns 0 to main()
>
*That was proof that H(D,D) meets its abort criteria*
H(D,D) correctly determines that itself is being called with its same inputs and there are no conditional branch instructions between the invocation of D(D) and its call to H(D,D).
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Except that D calling H(D,D) does NOT prove the required (a), since the simulated D WILL stop running because *ITS* H will abort *ITS* simulation and returm 0 so that simulated D will halt.

You keep saying that H(D,D) never really needs to abort the
simulation of its input because after H(D,D) has aborted the
simulation of this input it no longer needs to be aborted.
>

You keep thinking there is more than one H(D,D) and then when it's convenient for you you think there is only one H(D,D). Why is that?

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The first H(D,D) to see that the abort criteria has been met
(the outermost one) must abort the simulation of its input or
none of them ever abort.
>

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But since it does, which is your definition of H, the others

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never begin to be simulated.

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But D(D) started to be simulated, and we can know what D(D) actually does, which includes it using its version of H.
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We cannot reference the behavior of what D(D) does after H(D,D)
has already aborted the simulation of its input at the point
in time before H(D,D) aborts its input as any criterion measure
for this H(D,D).
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WHy not?
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That is what Correct Simulation refers to.
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I guess you are just admiting to being a LIAR (or stupid).

>
*I am not a liar or stupid and you admitted your mistake*
*I am not a liar or stupid and you admitted your mistake*
*I am not a liar or stupid and you admitted your mistake*
*I am not a liar or stupid and you admitted your mistake*

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So, do you admit that the definition of a "Correct Simulation" for the purposes of that criteria are the complete not-aborted simulation done by possibly some other simulator?
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(a) 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
>
Not at all the words don't say anything like that.
"H correctly simulates its input D until"
specifically means a partial simulation.
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Means H uses a partial simulation to make its decision.
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Finally you get this.
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The correctness of the decision is measured by the full simulation, even past where H simulated. Thus, is based on things H might not know.
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No. the correctness of the decision is essentially anchored
in something like mathematical induction that correctly
predicts that complete simulation would never end.

 But, since the Execution DOES end, so will a correct simulation.
 

*YOU KEEP FORGETTING THIS*
There cannot possibly exist any H(D,D) that is called by
D where H(D,D) simulates its input and D(D) stops running
and H never aborts its simulation.

You seem to think that it is possible to prove a false statement.
 

>
I would estimate that an actual proof fully anchored in actual
mathematical induction can be derived. I do remember from my
independent studies course on proof of program correctness that
this does typically anchor in actual mathematical induction.

 So, you REALLY beleive that you can prove that the correct simulation of a halting computation can go on forever.
 

There cannot possibly exist any H(D,D) that is called by
D where H(D,D) simulates its input and D(D) stops running
and H never aborts its simulation.

You think you can prove an INCORRECT program as correct.
 (Since D(D) Halts, H(D,D) returning 0 is just WRONG for a Halt Decider).
 

H(D,D) fails to make the required mistake of reporting on what it does not see.

>
I still have two textbooks on this.

 Really? do you read and understand them?
 "Something like mathematical induction" sounds dubious for you really understanding it.
 Try to define the induction (if you understand what to do).
 

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You don't seem to understand the meaning of CORRECT.
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I guess you are just showing you think it is ok to LIE to make it look like you have done something.
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--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer