Re: We finally know exactly how H1(D,D) derives a different result than H(D,D)

On 3/7/2024 9:28 PM, Richard Damon wrote:

On 3/7/24 4:00 PM, olcott wrote:

On 3/7/2024 5:18 PM, Richard Damon wrote:

On 3/7/24 3:02 PM, olcott wrote:

On 3/7/2024 4:32 PM, Richard Damon wrote:

On 3/7/24 1:05 PM, olcott wrote:

H1(D,D) maps its input + its own machine address 00001422 to its output.
  H(D,D) maps its input + its own machine address 00001522 to its output.
Thus both H1 and H are computable functions of their input.

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And thus you are admitting that Neither H or H1 are actually correct Halt Deciders, as Halt Deciders must be only a function of the description of the Compuation to be decided.
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It turns out that both H(D,D) and H1(D,D) do correctly determine
whether or not they must abort the simulation of their input.

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Which isn't the halting question, so you are LYING.

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As I completely explained yet you persistently ignore the
halting question can only be correctly answered indirectly
otherwise inputs that contradict the decider that is embedded
within these inputs have no answer at all.

 In other words you argue by lying.
 The QUESTION is, and always will be, does the computation described by the input Halt when run.
 The Computation so described is FIXED and UNCHANGING reguards of what the decider that is deciding does, as is the behavior of the H that it is built on.
 That was FIXED and made UNCHANGING when it was defined.
 Thus, the question does THIS H^(H^) halt? HAS a definite and fixed answer. SO you LIE when you said it doesn't.
 Your problem seems to be that you think "Get the Right Answer?" is a valid program instruction, or that H can somehow "change" itself after H^ gets defined. IT CAN'T.
 YOU persistently ignore this fact, likely because you are too stupid and ignorant to understand that fundamental nature of programs, that they will do what their programming says they will do, and that programming doesn't change, EVER, with out the creation of some NEW program that is different from its predicesor.
 YOU *NEVER* have the right to change the question for a problem.
 You can try to point out the the problem is inconsistant, and propose a NEW PROBLEM, but that doesn't change the old.
 You can talk about your new problem that you think is more useful than the actual Halting Problem, after all, someone might be more interested in the incorrect opinion of an admittedly faulty "Olcott-Halt Decider" than the actual behavior of the Computation they are interested in.
 NOT.
 What you can't to is say you are working on one problem, while trying to change it to mean something different. That is just call LYING, and you seem to know that you doing it (you might feel you have justified reasons to talk about a different problem) so the lie is DELIBERATE.
 

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That you or others consider this somehow improper does not change
the verified fact that they both correctly determine whether or
not they must abort their simulation.

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Which isn't the Halting Question, which you claim you are working on, so you are just LYING.
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Already fully explained many many times (including above)
yet your ignorance is very persistent.

 So, you think you can change the question and still be talking about the same question.
 You ARE the LIAR PARADOX.
 

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It is also the case that both H1(D,D) and H(D,D) are a pure function
of their inputs when we construe their own machine address to be an
element of these inputs.

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Which means they are not computing the Halting Function, which isn't a function of the decider, so again, you are LYING.
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Already fully explained many many times (including above)
yet your ignorance is very persistent.

 Yes, you have ADMITTED that you are LYING about working on the Halting Problem.
 

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Turing machines don't even have the idea of their own machine
address so this exact same thing cannot be Turing computable.

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And it isn't a Halt Decider even in Olcott machines as the algorithm is shown to vary by a parameter that it isn't allowed to vary to be a Halt Decider.
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Olcott machines entirely anchored in Turing machine notions
can compute the equivalent of H1(D,D) and H(D,D).
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Because Olcott machines are essentially nothing more than
conventional UTM's combined with Conventional Turing machine
descriptions their essence is already fully understood.
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The input to Olcott machines can simply be the conventional
space delimited Turing Machine input followed by four spaces.
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This is followed by the machine description of the machine
that the UTM is simulating followed by four more spaces.
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When this input is ignored Olcott machines compute the
exact same set as Turing machines.
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Unlike Turing machines, Olcott machines have the basis to
determine that they have been called with copies of their
own TMD.
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Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
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With Olcott machines Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> and H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> do
not have the same inputs thus can compute different outputs
when they do not ignore their own TMD.
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THen you build H^ wrong. Of course with your change in mechanics, the H^ that needs to be generated will be a bit different.
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That Olcott machines always know their own TMD is unconventional.

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And breaks much of the background of Turing Machines,

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Not at all. Not in the least little bit.
Olcott machines are 100% fully specified
in terms of Turing machines.

 Yes, BUT if you talk about an Olcott machine, you MUST include the added data as part of the description of that machine, or you are just LYING.
 

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so if you what to use ANY establish property of Turing Machine, you must include that now extra data EXPLICITLY.
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It is already explicitly included in the definition of an Olcott machine.

 Which means that you can't actually write an Olcott-Machine that matches the requirements for a Halt Decider.
 A Halt Decider MUST be able to defined as taking JUST the description of the Computation to be decided (the Allgorithm and the Data). In general this also holds, to be a Foo decider, the decider must be give JUST the information about the thing that we are deciding Foo on.
 Since the mapping function for no practial problem actually will depend on the decider that is deciding it, NO Olcott-Machine meets the basic definition to be any of those deciders.
 You are forced to have the machine take an extra parameter that you then need to constrain that the machine can not actually depend on it.
 

Only Olcott machines can correctly determine that they themselves
are called in recursive simulation, Turing machines cannot possibly
do that.

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That their own TMD is correctly construed as an additional input
to their computation (whenever they don't ignore it) does provide
the reason why Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> and H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> can compute different
results and still be computations.

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But not the RIGHT computation.
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Not the same computation that conventional wisdom expects.

 SO not the compution DEFINED for the problem.
 Remember, the problem STARTS with the mapping function, which defines what the decider is allowed to have as inputs, and what answer each value of those inputs should generate.
 

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Right answer to the wrong question is still a WRONG ANSWER.
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The right answer to a different question that can always
be construed as the right answer to the original question
by H ⟨Ĥ⟩ ⟨Ĥ⟩ is the right answer.

 NOPE, more LIES.
 If H(H^,H^) returns non-halting, then H^.H(H^,H^) will ALSO return non-halting, and H^ will halt, meaning H was wrong, and you are LYING.

Olcott machines can compute the difference between H1(D,D) and H(D,D)
because Olcott machines can correctly determine whether or not they
themselves were called in recursive simulation.
For H1(D,D) and H(D,D)it is the extra inputs of 00001422 and 00001522
For H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> the extra inputs are shown.
When I keep repeating the above a few hundred more times
you will quit immediately forgetting that I ever said it.

 If H(H^,H^) returns halting, then H^.H(H^,H^) will also return halting, and H^ will loop forever.
 When we transition to Olcott Machines, H^,H just needs to supply to that algorithm, the description of the H it is supposed to confound, so it will.
 

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So, you are just shown again to be a LIAR.
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H^ needs to have a copy of the algorithm of the decider at that point, and give it exactly the same input as the decider will get.
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Both Ĥ and H have a copy of their own non-identical TMD's.

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Yep, and H^ can give its H^.H the same TMD of the machine that will decide it to its copy of H's algorithm,

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You are not saying that Ĥ does not have its own TMD
you are only saying that it can screw this up too.

 No, I am saying that when H^ enters the state H^.H, it has prepared the tape by removing the TMD of itself from the tape, and replaced it with the TMD of the original H.
 Nothing prevents it from doing this.
 

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No matter what Ĥ does it will either halt or fail to
halt in a way that can be determined by the actual H.

 Nope. H must make its decision before it sees H^ make its.
 YOU have proven that.
 

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so it gets the same answer as that H wil produce, and thus that H will give the WRONG answer to the Halting Question,
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Since H will get on its tape the string (H^) (H^) (H), that is exactly what H^ (H^) must put on its tape,

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It will not get this, where the Hell did you get this from?
Ĥ ⟨Ĥ⟩ gets only two inputs and both of them are ⟨Ĥ⟩.
The latter one is appended to its tape by the master UTM.
Ĥ never gets any separate <H>.

 But it can GENERATE it. After all, the designer of H^ knows the H it is supposed to confound.
 Are you THAT stupid to not think that Turing Machines can generate data for the tape?
 

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Not at all. Olcott machines are exactly the same as Turing machines
yet always have a copy of their own machine description appended to
the end of their own tape. That you assert that H must have a copy of
some other TMD at the end of its tape is quite nutty.

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WHAT WAS WRONG WITH WHAT I SAID.
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Where did I say that a machine when started had anything else.
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The SUB-MACHINE H^.H can have on its tape whatever H^ wants to put there, if that puts in the "Your TMD" slot the description of some other verison of it, that is fine and you can't stop it.
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Ĥ ⟨Ĥ⟩ cannot possibly put a separate <H> anywhere on its tape
it has no access to any such separate <H>.

 Of course it can have access to that, i just needs a long series of states saying write the first symbol of <H>

Turing machine Ĥ does not even know that an external
Turing machine H exists.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer