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On 3/10/24 9:52 AM, olcott wrote:If no source can be cited then the Olcott thesisOn 3/10/2024 10:50 AM, Richard Damon wrote:NOPE.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:Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ haltsOn 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.
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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.
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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.
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.
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.
*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
>
*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?
(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
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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.
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.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
H ⟨Ĥ⟩ ⟨Ĥ⟩ aborts right after Ĥ.Hq0 before it simulates ⟨Ĥ⟩ ⟨Ĥ⟩.
And if it does, as I said below, so will H^.H when it is run.
Yes.
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.
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.
ZFC simply tossed out the Russell's Paradox question as unsound.>So, ADMIT YOU ARE ANSWERING A DIFFERENT QUESTION
Thus in the same way that ZFC conquered Russell's Paradox
(by reframing the problem so that sets cannot refer to themselves)
the Halting Problem can be conquered by reframing it.
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ZFC simply tossed out the Russell's Paradox question as unsoundBoth H ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decide that:But the "Pathological Relationship" is ALLOWED.
(a) Their input halts H.qy
(b) Their input fails to halt or has a pathological
relationship to itself H.qn.
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So, you are just admitting that the Halting Problem, as originally stated, and as normally wanted answered, is just impossible.It is only impossible in the same way that this is impossible:
so, you are just claiming it is ok to lie if you don't know the answer.ZFC simply tossed out the Russell's Paradox question as unsound
That it is ok to turn an OBJECTIVE criteria into a useless subjective requirement.I am only claiming that both H and Ĥ.H correctly say YES
Yep, sound like you, the pathetic ignorant hypocritical pathological lying idiot.
And, you do not prove that you can actually solve your new definition, just that the simple proof based on that property doesn't work.
There are still an uncountable number of non-pathological machines that you won't be able to decide halting on, so you are still wrong.
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