Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05 --partial agreement--

On 3/7/24 10:30 PM, olcott wrote:

On 3/8/2024 12:09 AM, Richard Damon wrote:

On 3/7/24 9:38 PM, olcott wrote:

On 3/7/2024 9:53 PM, Richard Damon wrote:

On 3/7/24 7:29 PM, olcott wrote:

On 3/7/2024 8:34 PM, Richard Damon wrote:

On 3/7/24 6:02 PM, olcott wrote:

On 3/7/2024 7:35 PM, immibis wrote:

On 7/03/24 18:05, olcott wrote:

On 3/7/2024 8:47 AM, immibis wrote:

On 7/03/24 03:40, olcott wrote:

On 3/6/2024 8:22 PM, immibis wrote:

On 7/03/24 01:12, olcott wrote:

On 3/6/2024 5:59 PM, immibis wrote:

On 7/03/24 00:55, olcott wrote:

Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn
Correctly reports that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort its simulation.
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H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
Correctly reports that H ⟨Ĥ⟩ ⟨Ĥ⟩ need not abort its simulation.

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What are the exact steps which the exact same program with the exact same input uses to get two different results?
I saw x86utm. In x86utm there is a mistake because Ĥ.H is not defined to do exactly the same steps as H, which means you failed to do the Linz procedure.

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Both H(D,D) and H1(D,D) answer the exact same question:
Can I continue to simulate my input without ever aborting it?
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Both H(D,D) and H1(D,D) are computer programs (or Turing machines). They execute instructions (or transitions) in sequence, determined by their programming and their input.

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Yet because they both know their own machine address
they can both correctly determine whether or not they
themselves are called in recursive simulation.

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They cannot do anything except for exactly what they are programmed to do.

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H1(D,D) and H(D,D) are programmed to do this.
Because H1(D,D) simulates D(D) that calls H(D,D) that
aborts its simulation of D(D). H1 can see that its
own simulated D(D) returns from its call to H(D,D).
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An Olcott machine can perform an equivalent operation.
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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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To make the Linz proof work properly with Olcott machines, Ĥ should search for 4 spaces, delete its own machine description, and then insert the description of the original H. Then the Linz proof works for Olcott machines.

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That someone can intentionally break an otherwise correct
halt decider

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It always gives exactly the same answer as the working one, so how is it possibly broken?
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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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When this is executed in an Olcott machine then
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> is a different computation than H ⟨Ĥ⟩ ⟨Ĥ⟩ <H>

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WHY?
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The Master UTM will not change the usage at H^.H, because that is just an internal state of the Machine H^
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The ONLY thing that the master UTM does differently is append
the TMD to the TMD's own tape.

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Right, so it gives H and H^ that input.
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H^ can erase that input and replace it.
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At that point we have the IDENTICAL set of transitions (with just an equivalence mapping of state numbers) as H will have, and the EXACT same input as H

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it is stipulated by the definition of Olcott machines
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> // last element is <Ĥ> (not H)

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Nope.
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H^.H isn't a "Olcott Machine" it is a sub-machine of H^

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You already know that there is no such thing as sub-machines of
Turing machines. Turing machines have a set of states and a tape.
They have no sub-machines.

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A "Sub-Machine" of a Turing Machine is where you put a copy of another Turing Machine into the state space of the overall/parent machine,
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When the machine H^ reaches the state H^.H, then it encounters the EXACT sequence of states (after the eqivalence mapping generated when inserting them H's q0 -> H^.Hqo and so on)
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It is called a sub-machine because it acts largely as if that Turing Machihe "called" the equivalent Turing Machine as a subroutine, only the machine code was expanded "inline".

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There is no way that any conventional Turing machine can tell
that the states of another Turing machine are embedded withing it.

 It doesn't need to "KNOW", but it HAS THEM there.
 It was DESIGNED that way, so the programmer knows what he did.
 

I think your brain popped a gasket somewhere and is frantically trying to make sense of the world that is crashind down around you.
 

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Your Master UTM can't touch the tape here or it totally breaks the system.
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The master UTM is merely an ordinary UTM that always appends
the slaves TMD to the slaves tape. That is all that there is
to the master UTM.

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Right, so it can't change the tape when H^ enters state H^.H and then runs the exact sequence of steps if the machine H with the exact same

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Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> immediately detects that is about to simulate a
copy of itself with a copy of its own input thus immediately
detects recursive simulation just like H(D,D).

 But that isn't what happens.
 You seem stuck in your lie.
 H^ gets to H^.Hq0 (H^) (H^) <H> not <H^> because that is what H^ needs to do to meet its mission.
 Your are just stuck with your Strawman, maybe it is more of a tar-baby to you.
 

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H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> immediately detects that is NOT about to simulate a
copy of itself with a copy of its own input thus immediately
rejects recursive simulation just like H1(D,D).

 ?? H (H^) (H^) <H> isn't going to even TRY to see what H^ (H^) <H^> would do?
 I guess you admit if just FAILS as a any decider.
 

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It can only add the description at the beginning, as it doesn't know what state in that machine are "sub-machines" and what are just parts of the machine running.
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An ordinary UTM writes the input to the slaves tape.
The master UTM of Olcott machines simply adds one more step to this.

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Right.
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The master UTM is merely an ordinary UTM that always appends
the slaves TMD to the slaves tape. That is all that there is
to the master UTM.

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Right. so it can't interfere with H^ "lying" to the copy of H that it has at H^.H to make it think it is the top level machine running.
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You don't seem to understand much about Turing machines.
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I seem to remember you bailed out after like the first part of the first lesson.
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You can't point out an errors because there are no errors.
You keep saying things at least slightly incorrectly like
sub-machines of Turing machines, there is no such thing.

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Sure there is.
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You are just too dumb to understand.
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In fact, if you actually try to do work with Turing Machines, the concept of sub-machines is important to manage the complexity.
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Yes, the actual encoding of the Turing Machine doesn't indicate that a submachine exist at a location, except maybe the state names (which get lost when you "compile" them into numbers) but often at the design level they show up in the names.
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The machine itself has no idea of this thus cannot use it in
its computation in any way. It cannot copy the states of an
external H because it does not even know that an external H exists.

 Machines do what they are programmed to do.
 Thus, if programmed to do that, they will.
 You are just stuck in your "thinking machine" model aren't you, which just shows how ignorant you are of how computers actually work.
 That is why you are just a pathetic ignorant pathological lying idiot.
 

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Why do you think the state is called H^.Hq0, that is the q0 state of the H sub-machine of the Turing Machine H^.

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That is only for human consumption machines only have natural numbers.

 Nope. Turing Machines have "states" that can be represented how ever we want.
 When we represent them, we might use a numeric encoding for convience, but that is only an ENCODING of a number, not a number itself.
 Turing Machine can't handle actual numbers, just encodings of numbers.
 Don't you remeber you lessons on how to write Turing Machines and how to encode inputs?
 

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   H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> // last element is <H> (not Ĥ)
That the last element of the input to Ĥ.H and H <is> different.
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Until the code in H^ between q0 and H^.H changes it.

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I have no idea what you are saying yet the only category that
it can be is that Ĥ tries to thwart its own internal termination
analyzer.

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Right, it sets up its copy of H so that it doesn't know it is part of a larger computation so it answer just like the actual top level H.
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And of course you have no idea, because you have no idea how Turing machines actually work.
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No matter how Ĥ ⟨Ĥ⟩ screws itself up this can have no
effect on H ⟨Ĥ⟩ ⟨Ĥ⟩.
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The Olcott machine H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> can see that it does not
call itself in recursive simulation.

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As will the machine at H^.H, since the description it thinks it is doesn't match the machine that gets called.

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A simply string comparison of the finite strings
⟨Ĥ⟩ and <Ĥ> proves that they are the same.
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The Olcott machine Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> can see that it calls
itself in recursive simulation unless it discards this
ability. In either case it cannot fool H, it either halts
or fails to halt.
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But that isn't what H^.H gets called with.

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Olcott machines always append the TMD to the end of the tape
of this TMD.

So H^ STARTS with <H^> at the end. The Tape, BY DEFINITION, is writable.
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That is sort of one problem with your x86UTM, its "tape" equivalent was Read Only since it was executable code space.
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This means that they are an actual input to every TMD that
does not ignore them.

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Right, and H^ can change it.
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The code between H^.q0 and H^.H removes the <H^> and replaces it with <H>, so the copy of H at H^.H can't tell the difference.
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Ĥ cannot possibly do this because it has no access to nor
even knows the existence of any external ⟨H⟩.

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Of course it know about the ONE H that it is on a mission to make wrong.

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Most of the halting problem proofs fail to understand
that an actual Turing machine cannot actually call
another external Turing machine. This means that it
cannot actually confound an external Turing machine.

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But it CAN copy the algorithm of one.
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One Turing machine cannot copy another one.

 The person creating it can.
 

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You don't seem to understand how that works.
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It is sort of like an inline function in C, the function that "calls" the inline function actually has no call instruction, but the code acts like it did as far as executing all the instructions of the subroutine at the right part of the code.
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I guess it is because you are thinking of Turing machines in terms of the goal they are designed to do, and not the actual algorithm they implement to acheive that goal.
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Conventional Turing Machines.

 Right,
 Which is also the fundamental engine of Olcott-Machines since you didn't want to work out anything else.
 Turing Machines are just a collection of state transition rules (incluing the tape ops) connected to a tape with the data it is to process..
 The machine itself know NOTHING about its goal, it just follows its instructions.
 

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Maybe that shows that you do your programming wrong too.
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So that H, and only that H gets confonded.
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You sure seem dumb about that.
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You don't seem to understand what a Turing Machine CAN do.
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