Sujet : Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior
De : polcott2 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory sci.logicDate : 10. Mar 2024, 07:14:16
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <usjfj8$2q613$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
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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:
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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.
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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.
>
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
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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.
>
Nope, your just being stuupid, perhaps intentionally.
(c) just moves around to its simulation of a
(a) H^.q0 (H^)
H^ then makes a copy of its input
(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^)
(c) = (2)
Which begins at the simulation of H^.q0 (H^)
(d = sim a) = (sim a)
Ths Simulated H^.q0 (H^) makes a copy of its input
(e = sim b) = (sim b)
The Simulated H^.H (H^) (H^) has is H begin the simulation of its input ...
and so on.
Both machine see EXACTLY the same level of details.
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.
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.
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.
-- 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