On 5/24/2022 8:57 AM, Dennis Bush wrote:
> On Monday, May 23, 2022 at 11:47:56 PM UTC-4, olcott wrote:
>> On 5/23/2022 10:16 PM, Dennis Bush wrote:
>>> On Monday, May 23, 2022 at 11:08:54 PM UTC-4, olcott wrote:
>>>> On 5/23/2022 10:05 PM, Dennis Bush wrote:
>>>>> On Monday, May 23, 2022 at 11:01:56 PM UTC-4, olcott wrote:
>>>>>> On 5/23/2022 9:57 PM, Dennis Bush wrote:
>>>>>>> On Monday, May 23, 2022 at 10:48:39 PM UTC-4, olcott wrote:
>>>>>>>> On 5/23/2022 9:40 PM, Dennis Bush wrote:
>>>>>>>>> On Monday, May 23, 2022 at 10:32:55 PM UTC-4, olcott wrote:
>>>>>>>>>> On 5/23/2022 9:23 PM, Dennis Bush wrote:
>>>>>>>>>>> On Monday, May 23, 2022 at 10:18:37 PM UTC-4, olcott wrote:
>>>>>>>>>>>> On 5/23/2022 9:09 PM, Dennis Bush wrote:
>>>>>>>>>>>>> On Monday, May 23, 2022 at 9:50:28 PM UTC-4, olcott wrote:
>>>>>>>>>>>>>> On 5/23/2022 8:40 PM, Richard Damon wrote:
>>>>>>>>>>>>>>> On 5/23/22 9:34 PM, olcott wrote:
>>>>>>>>>>>>>>>> On 5/23/2022 8:29 PM, Dennis Bush wrote:
>>>>>>>>>>>>>>>>> On Monday, May 23, 2022 at 9:24:47 PM UTC-4, olcott wrote:
>>>>>>>>>>>>>>>>>> On 5/23/2022 8:15 PM, Dennis Bush wrote:
>>>>>>>>>>>>>>>>>>> On Monday, May 23, 2022 at 9:08:54 PM UTC-4, olcott wrote:
>>>>>>>>>>>>>>>>>>>> On 5/23/2022 8:05 PM, Dennis Bush wrote:
>>>>>>>>>>>>>>>>>>>>> On Monday, May 23, 2022 at 8:57:46 PM UTC-4, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>> On 5/23/2022 7:44 PM, Dennis Bush wrote:
>>>>>>>>>>>>>>>>>>>>>>> On Monday, May 23, 2022 at 7:50:36 PM UTC-4, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>>>> On 5/23/2022 6:33 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>> On 5/23/22 3:05 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>>>>>> On 5/23/2022 1:52 PM, Mr Flibble wrote:
>>>>>>>>>>>>>>>>>>>>>>>>>>> A simple multiple choice question for Olcott:
>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>> All things being equal which is more likely:
>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>> (a) Olcott is correct and everybody else is incorrect
>>>>>>>>>>>>>>>>>>>>>>>>>>> (b) Olcott is incorrect and everybody else is correct
>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>> ?
>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>> /Flibble
>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>> Believability has the word [lie] embedded directly within
>>>>>>>>>>>>>>>>>>>>>>>>>> itself.
>>>>>>>>>>>>>>>>>>>>>>>>>> Instead of the fake measure of credibility one must employ
>>>>>>>>>>>>>>>>>>>>>>>>>> actual
>>>>>>>>>>>>>>>>>>>>>>>>>> validation.
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>> Actual validation conclusively proves that H(P,P)==0
>>>>>>>>>>>>>>>>>>>>>>>>>> is correct. This means that everyone that disagrees is either
>>>>>>>>>>>>>>>>>>>>>>>>>> insufficiently technically competent or a liar.
>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>> You consider that H(P,P) == 0 is correct when P(P) halts,
>>>>>>>>>>>>>>>>>>>>>>>>> when that is
>>>>>>>>>>>>>>>>>>>>>>>>> the DEFINITION of what H(P,P) is supposed to be answering?
>>>>>>>>>>>>>>>>>>>>>>>> The C function H correctly determines that there are no number
>>>>>>>>>>>>>>>>>>>>>>>> of steps
>>>>>>>>>>>>>>>>>>>>>>>> (0 to infinity) of its correct simulation of its input: a pair of
>>>>>>>>>>>>>>>>>>>>>>>> pointers to finite strings of x86 machine language that would
>>>>>>>>>>>>>>>>>>>>>>>> ever reach
>>>>>>>>>>>>>>>>>>>>>>>> the last instruction of this input.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> But since H has a fixed algorithm, it can't simulate for an
>>>>>>>>>>>>>>>>>>>>>>> infinite number of steps. It can only simulate P for some n
>>>>>>>>>>>>>>>>>>>>>>> number of steps.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> None-the-less on the basis of matching known behavior patterns H
>>>>>>>>>>>>>>>>>>>>>> can
>>>>>>>>>>>>>>>>>>>>>> determine what the behavior of the input would be if it did
>>>>>>>>>>>>>>>>>>>>>> simulate an
>>>>>>>>>>>>>>>>>>>>>> infinite number of steps.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> So because Pn(Pn) does not halt then Ha(Pa,Pa)==0 is correct?
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> No jackass, infinite loop does not halt because infinite loop is an
>>>>>>>>>>>>>>>>>>>> infinite loop.
>>>>>>>>>>>>>>>>>>>> _Infinite_Loop()
>>>>>>>>>>>>>>>>>>>> [000012c2](01) 55 push ebp
>>>>>>>>>>>>>>>>>>>> [000012c3](02) 8bec mov ebp,esp
>>>>>>>>>>>>>>>>>>>> [000012c5](02) ebfe jmp 000012c5
>>>>>>>>>>>>>>>>>>>> [000012c7](01) 5d pop ebp
>>>>>>>>>>>>>>>>>>>> [000012c8](01) c3 ret
>>>>>>>>>>>>>>>>>>>> Size in bytes:(0007) [000012
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> Irrelevant, because this isn't part of any P.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> It not irrelevant jackass it proves that H can detect that an infinite
>>>>>>>>>>>>>>>>>> simulation would never halt without performing an infinite simulation.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> An infinite loop and the infinite simulation in Pn(Pn) are different,
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> It sure is _Infinite_Loop() is on topic and H(P,P) is on topic
>>>>>>>>>>>>>>>> and Pn(Pn) is a strawman error intentionally designed to deceive.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> If _Infinite_Loop, which isn't at all related to P is on topic, then Pn,
>>>>>>>>>>>>>>> which is one of the P's you talk about must be.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>> I get to decide what is on topic and what is off topic, I own the topic.
>>>>>>>>>>>>>>> Remember, you confusingly talk about a CLASS of H's since the H you keep
>>>>>>>>>>>>>>> on mentioning doesn't have a distinct rule (since if changes how much is
>>>>>>>>>>>>>>> simulates to be every possible length of simulation), thus giving them
>>>>>>>>>>>>>>> distinct names is a reasonable thing to do.
>>>>>>>>>>>>>> I am referring to one machine language immutable literal string named H
>>>>>>>>>>>>>> and another immutable machine language literal string named P.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Then we'll refer to H as Ha and refer to P as Pa.
>>>>>>>>>>>> No we will not.
>>>>>>>>>>>
>>>>>>>>>>> We all know exactly why not. Because by being clear about which H and which P we're talking about, it exposes the holes in your argument and makes it clear exactly where the problem is. So as Ben has said, clarity is your enemy.
>>>>>>>>>>>
>>>>>>>>>>> So explain exactly what is wrong with the below statement. Failure to explain in detail why it is wrong in your next post will be taken as not being able to explain why it is wrong and an acceptance that it is correct. Stating "strawman" without an explanation will be taken as a failure to explain.
>>>>>>>>>>>
>>>>>>>>>>> Simulating the input to Ha(Pa,Pa) up to an infinite number of steps is done by UTM(Pa,Pa) which halts, so Ha(Pa,Pa)==0 is wrong. And Hb(Pa,Pa)==1 which also shows that Ha(Pa,Pa)==0 is wrong. And yes the input to Ha(Pa,Pa) is the same as the input to Hb(Pa,Pa), and you have no basis to claim otherwise.
>>>>>>>>>>>
>>>>>>>>>> It is just like you are saying that because the dog named Spot is black
>>>>>>>>>> and the cat named Fluffy is white therefore the dog named Rover cannot
>>>>>>>>>> be brown.
>>>>>>>>>>
>>>>>>>>>> Ha(Pa,Pa)
>>>>>>>>>> Hb(Pa,Pa)
>>>
>>>>>>>>>> Simulate(Pa,Pa)
>>>>>>>>>>
>>>>>>>>>> are computationally distinct at the x86 machine language level
>>>>>>>>>> and you have always known this therefore you are a damned liar.
>>>>>>>>>
>>>>>>>>> I know that you *claim* they are distinct, but you have nothing to back that up. Both Ha and Hb are halt deciders and both are given the same exact input.
>>>>>>>>>
>>>>>>>> P does not call anything besides H
>>>>>>>
>>>>>>> And because the fixed algorithm of H aborts, then H is the same as Ha and P is therefore the same as Pa.
>>>>>> How dishonest can you get?
>>>>>> It is the same as if you claimed that 5 == 6
>>>>>>
>>>>>> Because they have entirely different execution traces
>>>>>
>>>>> They're identical up to the point that Ha aborts,
>>>> They are not identical therefore it is either ridiculously stupid to
>>>> claim that they should have the same behavior or in your case (because
>>>> we know that you are not stupid) it would be dishonest.
>>>
>>> If Ha(Pa,Pa) and Hb(Pa,Pa) are not identical because as you claim the traces differ, then that would also mean that
>> Since you always knew this: that you are a liar when
>> you claimed that they are equivalent.
>>>>>>>>> I know that you *claim* they are distinct, but you have nothing
>> to back that up.
>> I don't think that we are getting anywhere.
>
> Translation: You can't explain why I'm wrong and you're backed into a corner.
>
>>
>> All of the recent discussions are simply disagreement with an easily
>> verifiable fact. Any smart software engineer with a sufficient technical
>> background can easily confirm that H(P,P)==0 is correct:
>
> It is an easily verified fact that Ha(Pa,Pa)==0 is NOT correct > as Hb(Pa,Pa)==1 demonstrates that Ha aborted too soon. By the
definition of the problem any H is required to map the halting function,
which means the correct answer for Ha(Pa,Pa) and Hb(Pa,Pa) is 1 because
Pa(Pa) halts. Any claim that Ha(Pa,Pa) and Hb(Pa,Pa) are not deciding
the same thing is baseless.

A software engineer with sufficient technical competence would disagree.

>
>>
>> Where H is a C function that correctly emulates its input pair of finite
>> strings of the x86 machine code of function P and criterion for
>> returning 0 is that the simulated P would never reach its "ret"
>> instruction.
>
> By the definition of the problem, the criteria for returning 0 is if Pa(Pa) does not halt, but it does.

A computer scientist with sufficient technical competence would
disagree. That you believe that a non-input must be mapped violates
the definition of a computable function.

Computable functions are the basic objects of study in
computability theory.
Computable functions are the formalized analogue of the intuitive
notion of
algorithms, in the sense that a function is computable if there
exists an algorithm
that can do the job of the function, i.e. given an input of the
function domain it
can return the corresponding output.
https://en.wikipedia.org/wiki/Computable_function

>
> That Ha(Pa,Pa) is unable to simulate to a final state doesn't mean that returning 0 is correct, otherwise that would mean Ha3(N,5)==0 is correct.
>
>>
>> The only other remaining objection is whether or not a halt decider must
>> compute the mapping of its input finite strings to an accept or reject
>> state on the basis of the actual behavior actually specified by its
>> input OR SOME OTHER MEASURE.
>
> It doesn't matter whether Ha or Hb fits the definition of what you think a halt decider should be. What matters is that Ha and Hb are required to map the halting function:
>
> D(M,w) == 1 if and only if M(w) halts, and
> D(M,w) == 0 if and only if M(w) does not halt
>
> So by definition the behavior of the actual behavior of the actual input to Ha(Pa,Pa) is the behavior of Pa(Pa).
>
> Ha (nor any other H) are unable to compute that mapping, therefore the halting function is not computable as the proofs show.

--
Copyright 2022 Pete Olcott

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