olcott <NoOne@NoWhere.com> writes:
*My criterion measure with Ben's notational conventions*
H.q0 wM w ⊢* H.qy iff UTM(wM, w) halts
H.q0 wM w ⊢* H.qn iff UTM(wM, w) does not halt
It's not your criterion, it's mine. It's also Linz's. If yours is not
the same as his (and mine), it should be different in some way. (And we
can safely ignore it, since no one cares about your alternative
criteria.)
On 2021-12-29 21:51, olcott wrote:
On 12/29/2021 10:44 PM, Richard Damon wrote:
On 12/29/21 11:02 PM, olcott wrote:
On 12/29/2021 1:16 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
*My criterion measure with Ben's notational conventions*
H.q0 wM w ⊢* H.qy iff UTM(wM, w) halts
H.q0 wM w ⊢* H.qn iff UTM(wM, w) does not halt
It's not your criterion, it's mine. It's also Linz's. If yours is >>>>> not
the same as his (and mine), it should be different in some way.
(And we
can safely ignore it, since no one cares about your alternative
criteria.)
It has always been my criterion measure that halt deciders base
their halt status decision on the pure simulation of their inputs.
That your notational conventions very cleanly express this idea as
applied to the Linz H is a good breakthrough.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
As soon as people understand that this same criterion measure
applies recursively to every embedded_H at Ĥ.qx, then they will see
how this criterion measure provides the means for embedded_H to
correctly compute the mapping from ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn.
Yes each level can use the test to determine if the halt decider at
that level got the right answer or not.
The issue is that when we are doing this check, you check JUST that
level, and not lower levels.
No you are wrong.
No, he isn't.
The specification you give above (where YET AGAIN you omit the
conditions)
is for a case where the input to Ĥ is a description of
itself, but the actual specification of the machine can't make that assumption. So really we should be using:
Ĥ.q0 ⟨wM⟩ ⊢* Ĥ.qx wM wM⟩ ⊢* Ĥ.qy ∞ iff UTM(wM, wM) halts
Ĥ.q0 ⟨wM⟩ ⊢* Ĥ.qx wM wM⟩ ⊢* Ĥ.qn iff UTM(wM, wM) does not halt
Ĥ has absolutely no knowledge of what it will be given as an input. In
the case we considered above wM was ⟨Ĥ⟩, but wM could have been a description of any machine. Ĥ has no way of knowing if wM is a
description of itself, or a description of some TM which simply prints
its input string to the tape in reverse order (which wouldn't involve simulating its input at all).
So the conditions given above cannot
possibly apply to any 'lower level' since Ĥ knows nothing about any of
those 'lower levels'.
André
On 12/30/21 10:20 AM, olcott wrote:
On 12/30/2021 8:19 AM, Richard Damon wrote:
On 12/29/21 11:51 PM, olcott wrote:
On 12/29/2021 10:44 PM, Richard Damon wrote:
On 12/29/21 11:02 PM, olcott wrote:
On 12/29/2021 1:16 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
*My criterion measure with Ben's notational conventions*
H.q0 wM w ⊢* H.qy iff UTM(wM, w) halts
H.q0 wM w ⊢* H.qn iff UTM(wM, w) does not halt
It's not your criterion, it's mine. It's also Linz's. If yours >>>>>>> is not
the same as his (and mine), it should be different in some way.
(And we
can safely ignore it, since no one cares about your alternative
criteria.)
It has always been my criterion measure that halt deciders base
their halt status decision on the pure simulation of their inputs. >>>>>>
That your notational conventions very cleanly express this idea as >>>>>> applied to the Linz H is a good breakthrough.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
As soon as people understand that this same criterion measure
applies recursively to every embedded_H at Ĥ.qx, then they will
see how this criterion measure provides the means for embedded_H
to correctly compute the mapping from ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn.
Yes each level can use the test to determine if the halt decider at
that level got the right answer or not.
The issue is that when we are doing this check, you check JUST that
level, and not lower levels.
No you are wrong.
Source? (Some REAL reference)
H.q0 wM w ⊢* H.qy iff UTM(wM, w) halts
H.q0 wM w ⊢* H.qn iff UTM(wM, w) does not halt
The source is the correct reasoning of how the above criterion measure
would be applied when H is copied to Ĥ.qx.
We copy the ALGORITHM of H, the detailed step by step instructions
You keep thinking that a subsequent level UTM magically turns into
embedded_H and aborts its simulation.
There IS no UTM in the machine. PERIOD. (unless H IS a UTM)
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