On 3/23/2022 9:31 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 3/23/2022 8:47 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 3/23/2022 4:41 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 3/23/2022 10:19 AM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> 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.
>>>>>>> If H and Ĥ are Turing machines the copy of H at Ĥ.qx cannot make any
>>>>>>> different transitions than H does, since they both have identical
>>>>>>> inputs. If one compares some strings, the other will too and they will
>>>>>>> both arrive as the same result and they will both behave in the same
>>>>>>> way. You H and Ĥ are obviously not Turing machines.
>>>>>>> I regret dropping down into simplistic metaphors. I'd rather keep this
>>>>>>> about formal Turing machines, but you don't really know what they are,
>>>>>>> and you certainly can't reason about them, so that hits a brick wall
>>>>>>> pretty quickly.
>>>>>>>
>>>>>>>>> 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.
>>>>>>>
>>>>>>> I continue to believe that I know what a Turing machine is and that you
>>>>>>> don't. A Turing machine and an exact copy of it can't entail different
>>>>>>> sequences of transitions. Your H and Ĥ don't have this property. They
>>>>>>> are not Turing machines.
>>>>>>
>>>>>> I agree that a TM and an exact copy of it must entail the same
>>>>>> sequence of transitions when applied to identical input.
>>>>> OK, but what about your magic PO-machines? They don't behave like this
>>>>> (apparently). It's easy to agree to facts about Turing machines, if
>>>>> your plan is to keep talking about machines where
>>>>> "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"
>>>>
>>>> Not if H and Ĥ are identical and are not allowed to use their machine
>>>> address to tell them apart.
>>> H and Ĥ are never identical. You know what the hat means don't you? So
>>> your magic PO-machines have machine addresses, do they? If so, that's
>>> another difference with Turing machines. You won't be able to say
>>> anything about Linz's proof unless you start to talk about Turing
>>> machines.
>>>
>>>> In my prior claims either H and Ĥ were not identical finite stings
>>> You've lost the plot mate. Nether is even a string. If you made a
>>> prior claim about H or Ĥ being strings (whether identical or not), that
>>> was wrong too.
>>>
>>> As far as I can tell, when you write H and Ĥ we must assume that these
>>> refer to some kind of as yet unspecified machine that has "machine
>>> addresses" (so definitely not what Linz is taking about). And that
>>> these H and Ĥ have the property that H applied to some string can
>>> transition to a different state to an exact copy of H if it's embedded
>>> in some other machine. I suppose they don't actually have to be
>>> literally magic to do this, but they are certainly not Turing machines.
>>> Do you have anything to say about Linz's proof?
>>
>> All halt deciders must map their input to an accept or reject state
>> based on the actual behavior actually specified by this input as
>> measured by a finite number of N steps of the correct UTM simulation
>> of this input.
>
> No. You'll find what a halt decider must do in any good text book.
>
> It appears that after 17 years of work in this you are:
>
> (a) trying to redefine what that halting problem is;
>

Someone that understands these things deeper than learned by rote would
agree with me.

(a) You agreed that every decider computes the mapping from its inputs
to its own accept or reject state

(b) You agreed that a halt decider must form its halt status decision on
the actual behavior that is actually specified by its input.

(c) You disagreed that a correct measure of the actual behavior that is
actually specified by its input is a correct UTM simulation of N steps
of this input.

This would seem to indicate that you do not believe the direct execution
of a Turing machine is computationally equivalent to the UTM simulation
of the the machine description of this same machine.

We must have the constraint of a finite N number of steps of simulation
to require that the halt decider itself halts.

--
Copyright 2021 Pete Olcott

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