On 3/17/22 5:30 PM, olcott wrote:
On 3/17/2022 3:36 PM, Richard Damon wrote:
On 3/17/22 4:00 PM, olcott wrote:
On 3/17/2022 2:30 PM, Richard Damon wrote:
On 3/17/22 10:06 AM, olcott wrote:
On 3/16/2022 10:02 PM, Richard Damon wrote:
Ĥ applied to ⟨Ĥ⟩ reaches its final state only when embedded_H >>>>>> applied to ⟨Ĥ⟩ ⟨Ĥ⟩ aborts its simulation.
Right, but it DOES reach its final state, so embedded_H, to be
correct, must take that into account. it doesn't, so it gives the
WRONG answer.
No BRAIN DEAD MORON the simulated ⟨Ĥ⟩ never reaches its own final >>>> state.
The CORRECT Simulation shows that it does.
H's simulation only doesn't reach that final state because H aborted
it simulation.
We are not talking about the simulation reaching the final state of
the simulation BRAIN DEAD MORON we are talking about the simulated
input reaching its own final state.
When Ĥ is applied to ⟨Ĥ⟩
Ĥ copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then embedded_H simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩
Then these steps would keep repeating:
Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩
⟨Ĥ5⟩...
The simulated input never reaches its own final state whether or not
embedded_H aborts this simulation.
No, because if embedded_H aborts its simulation, then the trace must be:
When H^ is applied <H^>
H^0 copies its input <H^1> to <H^2> then H0 simulates <H^1> <H^2>
H^1 copies its input <H^2> to <H^3> then H1 simulates <H^2> <H^3>
H^2 copies its input <H^3> to <H^4> then H2 simulates <H^3> <H^4>
... n levels (to when H0 aborts
H0 goes to H0.Qn and H^0 HALTS.
I.E. if embedded_H will only simulate a finite number of iterations, and
them abort its simulation, there is NO infinite simulation that happens.
olcott <NoOne@NoWhere.com> writes:
No BRAIN DEAD MORON this is not true, I keep calling you a BRAIN DEAD
MORON because after I have explain all the details you cannot remember
what I just said.
I see the petulant six-year-old is in residence today.
A decider maps its inputs to its own accept reject state.
A halt decider does not compute the halt status of itself.
A halt decider, let's call it H, maps the input <Ĥ> <Ĥ> to its accept
state if (and only if) Ĥ enter a final state on input <Ĥ>,
and it maps
<Ĥ> <Ĥ> to its reject state if (and only if) Ĥ does not enter a final
state on input <Ĥ>.
No such H exists.
You implicitly accept this fact because you propose an H which does not
meet this specification but I can't see why you think anyone would care
about it.
On 3/18/22 1:41 PM, olcott wrote:
On 3/18/2022 11:54 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
No BRAIN DEAD MORON this is not true, I keep calling you a BRAIN DEAD
MORON because after I have explain all the details you cannot remember >>>> what I just said.
I see the petulant six-year-old is in residence today.
A decider maps its inputs to its own accept reject state.
A halt decider does not compute the halt status of itself.
A halt decider, let's call it H, maps the input <Ĥ> <Ĥ> to its accept
state if (and only if) Ĥ enter a final state on input <Ĥ>,
No that is entirely incorrect.
It the DEFINITION!!!!
embedded_H maps its input <Ĥ> <Ĥ> to its Ĥ.qn final reject state if
the correctly simulated input <Ĥ> <Ĥ> would never reach its own final
state of <Ĥ>.qn in any finite number of steps of simulation.
WRONG CRITERIA.
That is just your POOP which no one cares about. We care
about the simulation by a REAL UTM, which matches the actual behavior of
the Turing Machine / Input combination represented by the input, not the behavior that H sees in its partial simulation.
and it maps
<Ĥ> <Ĥ> to its reject state if (and only if) Ĥ does not enter a final >>> state on input <Ĥ>.
No such H exists.
You implicitly accept this fact because you propose an H which does not
meet this specification but I can't see why you think anyone would care
about it.
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