On 3/23/2022 8:39 PM, Richard Damon wrote:
> On 3/23/22 9:29 PM, olcott wrote:
>> On 3/23/2022 7:15 PM, Richard Damon wrote:
>>> On 3/23/22 7:50 PM, olcott wrote:
>>>> On 3/23/2022 6:26 PM, Richard Damon wrote:
>>>>> On 3/23/22 7:20 PM, olcott wrote:
>>>>>> On 3/23/2022 6:04 PM, Richard Damon wrote:
>>>>>>> On 3/23/22 9:09 AM, olcott wrote:
>>>>>>>> On 3/23/2022 6:19 AM, Richard Damon wrote:
>>>>>>>>> On 3/23/22 12:00 AM, olcott wrote:
>>>>>>>>>> On 3/22/2022 10:37 PM, Ben Bacarisse wrote:
>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>
>>>>>>>>>>>> On 3/22/2022 9:32 AM, Ben Bacarisse wrote:
>>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> On 3/21/2022 10:22 PM, Ben Bacarisse wro0te:
>>>>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> A copy of Linz H is embedded at Ĥ.qx as a simulating
>>>>>>>>>>>>>>>> halt decider (SHD).
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>>>>>> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would
>>>>>>>>>>>>>>>> reach its final state.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>>>>> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would
>>>>>>>>>>>>>>>> never reach its
>>>>>>>>>>>>>>>> final state.
>>>>>>>>>>>>>>> But for your "PO-machines":
>>>>>>>>>>>>>>>      "Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn
>>>>>>>>>>>>>>>      corresponds to
>>>>>>>>>>>>>>>      H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qy"
>>>>>>>>>>>>>>> and
>>>>>>>>>>>>>>>      "The copy of H at Ĥ.qx correctly decides that its
>>>>>>>>>>>>>>> input never halts.
>>>>>>>>>>>>>>>      H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its
>>>>>>>>>>>>>>> input halts"
>>>>>>>>>>>>>>> so this has nothing to do with Linz.  He is talking about
>>>>>>>>>>>>>>> Turing
>>>>>>>>>>>>>>> machines.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> The Linz conclusion only pertains to the behavior the copy
>>>>>>>>>>>>>> of H
>>>>>>>>>>>>>> embedded within Ĥ applied to ⟨Ĥ⟩ ⟨Ĥ⟩.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Everything Linz says, everything, is predicated on what a
>>>>>>>>>>>>> Turing machine
>>>>>>>>>>>>> is.  Unlike Turing machines, your machines are magic --
>>>>>>>>>>>>> identical state
>>>>>>>>>>>>> transition functions can entail different configuration
>>>>>>>>>>>>> sequences for
>>>>>>>>>>>>> the same input.  Nothing you say has any relevance to
>>>>>>>>>>>>> Linz's Turing
>>>>>>>>>>>>> machines until you categorically repudiate this nonsense.
>>>>>>>>>>>>
>>>>>>>>>>>> That your only rebuttal to what I say now is dredging up
>>>>>>>>>>>> what I said
>>>>>>>>>>>> many months ago proves that you are being dishonest.
>>>>>>>>>>>
>>>>>>>>>>> You said this:
>>>>>>>>>>>
>>>>>>>>>>>    "The copy of H at Ĥ.qx correctly decides that its input
>>>>>>>>>>> never halts.
>>>>>>>>>>>    H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its input halts"
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> If ⟨H⟩ and ⟨Ĥ⟩ were identical finite strings then they must
>>>>>>>>>> derive the same result. They are not identical final strings.
>>>>>>>>>>
>>>>>>>>>>> four days ago and you haven't retracted it.  Until you do,
>>>>>>>>>>> when you
>>>>>>>>>>> write Ĥ your readers must assume that you are referring to
>>>>>>>>>>> something
>>>>>>>>>>> about which this quote applies.
>>>>>>>>>>>
>>>>>>>>>>> What's more, for your remarks to have any bearing on Linz's Ĥ
>>>>>>>>>>> you must
>>>>>>>>>>> not only repudiate what you said, you must accept the converse,
>>>>>>>>>>> i.e. that if
>>>>>>>>>>>
>>>>>>>>>>>    Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* Ĥ.qn
>>>>>>>>>>>
>>>>>>>>>>> then
>>>>>>>>>>>
>>>>>>>>>>>    H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qn
>>>>>>>>>>>
>>>>>>>>>>> So, do you retract what you said and accept this fact about
>>>>>>>>>>> Linz's H and
>>>>>>>>>>> Ĥ?
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> You you continue to say that you believe that a decider must
>>>>>>>>>> report on its own behavior when you already know damn well
>>>>>>>>>> that a decider only computes the mapping from its inputs to
>>>>>>>>>> its own final state.
>>>>>>>>>
>>>>>>>>> A Decider must report on its own behavior (or the behavior of a
>>>>>>>>> copy of it) if that is what the input asks for.
>>>>>>>>>
>>>>>>>> You know that a decider only computes the mapping from its input
>>>>>>>> finite strings to its own final state thus you know that you lie
>>>>>>>> what you say that a decider must compute the mapping from a
>>>>>>>> non-finite sting non-input.
>>>>>>>>
>>>>>>>> WHY LIE ?  WHY LIE ?  WHY LIE ?  WHY LIE ?
>>>>>>>>
>>>>>>>
>>>>>>> I don't, but you seem to like to.
>>>>>>>
>>>>>>> I never said that H needs to compute a mapping of anything but
>>>>>>> what has been given as an input.
>>>>>>>
>>>>>>> The only thing H needs to compute the mapping of is <H^> <H^>,
>>>>>>> which is EXACTLY the string on its input.
>>>>>>>
>>>>>>> The problem which you don't seem to understand is that the
>>>>>>> MAPPING it needs to try and compute (and which is not guaranteed
>>>>>>> to BE Computable), is the Behavior of the machine it represents,
>>>>>>> H^ applied to <H^>, as that is the mapping of the Halting Function.
>>>>>>
>>>>>> That is a non finite string non input, so you lied.
>>>>>>
>>>>>
>>>>> <H^> <H^> is a finite string, which is what needs to be mapped, so
>>>>> you are just lying.
>>>>>
>>>>
>>>> On 3/23/2022 6:04 PM, Richard Damon wrote:
>>>>  > the MAPPING it needs to try and compute (and which is
>>>>  > not guaranteed to BE Computable), is the Behavior of
>>>>  > the machine it represents, H^ applied to <H^>,
>>>>
>>>> So you are just bald faced liar then.
>>>> It does not map H^ applied to <H^> to anything.
>>>> It maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn
>>>>
>>>>
>>>
>>> I never said it did.
>>>
>>> It maps the INPUT: <H^> <H^> to (OUTPUT) Qy or Qn based on (THE
>>> FUNCTION) whether H^ applied to <H^> will Halt.
>>
>> It must map the input to an accept or reject state based on the actual
>> behavior actually specified by this input as measured by N steps of
>> the correct UTM simulation of this input.
>>
>>
>
> Almost, it must map the INPUT <H^> <H^> to an OUTPUT Qy or Qn, based on
> the FUNCTION it is computing.
>
> There is NO requirement that it be based on its on simulation, or only N
> steps of a UTM.
>

None-the-less if the behavior that is being measured is not exactly the
same behavior as the UTM simulation of the input then the behavior being
measured is measured incorrectly.

A finite number of N steps is a mandatory constraint otherwise it would
be OK to report that infinite execution never halts after an infinite
number of steps.

--
Copyright 2021 Pete Olcott

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