Sujet : Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --attribution--
De : polcott2 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory sci.logicDate : 11. Mar 2024, 03:00:36
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <usll3l$38f21$1@dont-email.me>
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On 3/10/2024 2:23 PM, Richard Damon wrote:
On 3/10/24 11:23 AM, olcott wrote:
On 3/10/2024 12:55 PM, Richard Damon wrote:
On 3/10/24 10:17 AM, olcott wrote:
On 3/10/2024 12:08 PM, Richard Damon wrote:
On 3/10/24 9:52 AM, olcott wrote:
On 3/10/2024 10:50 AM, Richard Damon wrote:
On 3/10/24 7:28 AM, olcott wrote:
On 3/10/2024 12:16 AM, Richard Damon wrote:
On 3/9/24 9:49 PM, olcott wrote:
On 3/9/2024 11:36 PM, Richard Damon wrote:
On 3/9/24 9:14 PM, olcott wrote:
On 3/9/2024 10:55 PM, Richard Damon wrote:
On 3/9/24 8:30 PM, olcott wrote:
On 3/9/2024 7:40 PM, immibis wrote:
On 10/03/24 02:37, olcott wrote:
On 3/9/2024 7:32 PM, immibis wrote:
On 10/03/24 02:29, olcott wrote:
On 3/9/2024 7:24 PM, immibis wrote:
On 10/03/24 01:30, olcott wrote:
On 3/9/2024 6:24 PM, immibis wrote:
On 10/03/24 01:22, olcott wrote:
On 3/9/2024 5:57 PM, immibis wrote:
On 10/03/24 00:26, olcott wrote:
On 3/9/2024 5:10 PM, immibis wrote:
On 9/03/24 23:22, olcott wrote:
On 3/9/2024 3:50 PM, immibis wrote:
On 9/03/24 22:34, olcott wrote:
>
What criteria would you use so that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ knows what
wrong answer to provide?
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Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective criteria that H ⟨Ĥ⟩ uses.
>
Simulating halt deciders must make sure that they themselves
do not get stuck in infinite execution. This means that they
must abort every simulation that cannot possibly otherwise halt.
>
This requires Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ to abort its simulation and does not
require H ⟨Ĥ⟩ ⟨Ĥ⟩ to abort its simulation when Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ aborts
its simulation.
>
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ does simulate itself in recursive simulation H ⟨Ĥ⟩ ⟨Ĥ⟩
does not simulate itself in recursive simulation.
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Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective criteria that H ⟨Ĥ⟩ uses.
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*Only because Ĥ.H is embedded within Ĥ and H is not*
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly get stuck in recursive simulation and
H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly get stuck in recursive simulation.
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You dishonestly ignored that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same OBJECTIVE criteria that H ⟨Ĥ⟩ uses.
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The above is true no matter what criteria that is used
as long as H is a simulating halt decider.
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Objective criteria cannot vary based on who the subject is. They are objective. The answer to different people is the same answer if the criteria are objective.
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It is objectively true that Ĥ.H can get stuck in recursive
simulation because Ĥ copies its input thus never runs
out of params.
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It is objectively true that Ĥ cannot possibly get stuck
in recursive because H does not copy its input thus runs
out of params.
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Wrong. Dead wrong. Stupidly wrong. So wrong that a dead monkey could do better. Write the Olcott machine (not x86utm) code for Ĥ and I would show you.
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*In other words you are denying these verified facts*
*In other words you are denying these verified facts*
*In other words you are denying these verified facts*
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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
>
That's not a verified fact, that's just something you want to be true.
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∞ means infinite loop. Infinite loop doesn't halt. You see how stupid it is, to say that an infinite loop halts?
>
Execution trace of Ĥ applied to ⟨Ĥ⟩
(a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
>
Execution trace of H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ BECAUSE IT IS PRECISELY IDENTICAL TO STEPS B AND C:
> (b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
> (c) which begins at Ĥ's own simulated ⟨Ĥ.q0⟩ to repeat the process
>
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*Yes and the key step of copying its input is left out so*
*H ⟨Ĥ⟩ ⟨Ĥ⟩ runs out of params and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ never runs out of params*
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that isn't how any of this works. Do you even know what words mean?
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(b) and (c) are not the same as (1) and (2)
Execution trace of H applied to ⟨Ĥ⟩ ⟨Ĥ⟩
(1) H applied ⟨Ĥ⟩ ⟨Ĥ⟩ simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(2) which begins at simulated ⟨Ĥ.q0⟩
(a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
>
This means that Turing machine H ⟨Ĥ⟩ ⟨Ĥ⟩ can see one more execution
trace of Ĥ ⟨Ĥ⟩ than its simulated Turing machine Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can see.
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Nope, your just being stuupid, perhaps intentionally.
(c) just moves around to its simulation of a
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(a) H^.q0 (H^)
H^ then makes a copy of its inp
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(b) H^.H (H^) (H^) == (1) H (H^) (H^)
The algorithm of H begins a simulation of its input, watching the behaior of H^ (H^)
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(c) = (2)
Which begins at the simulation of H^.q0 (H^)
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(d = sim a) = (sim a)
Ths Simulated H^.q0 (H^) makes a copy of its input
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(e = sim b) = (sim b)
The Simulated H^.H (H^) (H^) has is H begin the simulation of its input ...
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and so on.
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Both machine see EXACTLY the same level of details.
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Yes, the top level H is farther along at any given time then its simulated machine, and that is H's problem, it has to act before it sees how its simulation will respond to its copy of its actions.
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Thus, if it stops, it needs to make its decision "blind" and not with an idea of how the machine it is simulating will perform.
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If it doesn't stop, the level of recursion just keeps growing and no answer ever comes out.
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The earliest point that H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly see to abort
its simulation is immediately before Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ would begin
its simulation. Right before its cycle repeats the first time.
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So?
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If it DOES abort there, then so will H^.H when it gets to that point in its simulation, which will be AFTER The point that H has stopped simulating it, so H doesn't know what H^ will do.
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Thus, if H DOES abort there, we presume from your previous answer it will think the input will not halt and answer qn.
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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
H ⟨Ĥ⟩ ⟨Ĥ⟩ aborts right after Ĥ.Hq0 before it simulates ⟨Ĥ⟩ ⟨Ĥ⟩.
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And if it does, as I said below, so will H^.H when it is run.
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Yes.
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And thus, H^.H will give the same answer as H,
so H^ will act contrary to what H says,
so H will give the wrong answer.
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Unlike anything else that anyone else has ever done both H ⟨Ĥ⟩ ⟨Ĥ⟩
and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly determine that they must abort their own
simulation to prevent their own infinite execution.
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NOPE.
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If no source can be cited then the Olcott thesis
"that no one did this before" remains unrefuted.
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Since, BY THE DEFINITIONS of what H MUST do to be correct, and what H^ WILL do by its design, as shown in the Linz Proof.
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If no source can be cited that shows a simulating halt decider can
correctly determine that it must abort its simulation of the Halting
Problem's pathological input to prevent its own non-termination, then
innovation remains attributable to me.
>
<snip>
Of course it can abort its simulation.
It just needs some way to get the right answer.
*I have always been using this long before I read about it*
blind variation and selective retention (BVSR)...
Two common phenomena characterize BVSR thinking: superfluity and backtracking. Superfluity means that the creator generates a variety of ideas, one or more of which turn out to be useless.
Backtracking signifies that the creator must often return to an earlier approach after blindly going off in the wrong direction.
https://www.scientificamerican.com/article/the-science-of-genius/*I am aware of no one else that had the idea to apply a simulating* *termination analyzer to the halting problem counter-example input*
Professor Hehner had a seed of this idea before I did.
From a programmer's point of view, if we apply an interpreter to a
program text that includes a call to that same interpreter with that
same text as argument, then we have an infinite loop. A halting program
has some of the same character as an interpreter: it applies to texts
through abstract interpretation. Unsurprisingly, if we apply a halting
program to a program text that includes a call to that same halting
program with that same text as argument, then we have an infinite loop.
https://www.cs.toronto.edu/~hehner/PHP.pdf*Turing Machine and Olcott machine implementations seem to be dead*
*This the (possibly augmented) RASP machine equivalent of x86*
Every machine must be able to get its own machine address.
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer
Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior
Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior