Re: How the requirements that Professor Sipser agreed to are exactly met

Op 13.mei.2025 om 07:14 schreef olcott:

On 5/12/2025 10:31 PM, dbush wrote:

On 5/12/2025 11:22 PM, olcott wrote:

On 5/12/2025 9:58 PM, dbush wrote:

On 5/12/2025 10:46 PM, olcott wrote:

On 5/12/2025 8:00 PM, dbush wrote:

On 5/12/2025 8:56 PM, olcott wrote:

On 5/12/2025 7:36 PM, dbush wrote:

On 5/12/2025 8:34 PM, olcott wrote:

On 5/12/2025 7:27 PM, dbush wrote:

On 5/12/2025 8:25 PM, olcott wrote:

On 5/12/2025 7:12 PM, dbush wrote:

On 5/12/2025 7:53 PM, olcott wrote:

>
Simulating Termination analyzers cannot possibly report
on the actual behavior of non-terminating inputs
because this would cause themselves to never terminate.
>
They must always hypothesize what the behavior of the
input would be if they themselves never aborted.
>

>
False.  They must always hypothesize what the behavior of algorithm described by the input would be if it was executed directly, as per the requirements:
>

>
Show the actual reasoning of how it makes sense
that a simulating termination analyzer should
ignore the behavior (to its own peril) that the
input actually specifies.

>
There is no requirement that building a termination analyzer, simulating or otherwise, is possible.  In fact, it has proved to not be possible by Linz and others, which you have *explicitly* agreed with.
>

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In other words you have no such actual reasoning.

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The reasoning is that there is no requirement that building a termination analyzer is possible.

>
So you have no actual reasoning that addresses my
actual point.
>
 >>>> Show the actual reasoning of how it makes sense
 >>>> that a simulating termination analyzer should
 >>>> ignore the behavior (to its own peril) that the
 >>>> input actually specifies.
>

>
It makes sense because that's what's required to tell me if any arbitrary algorithm X with input Y will halt when executed directly.
>

>
A simulating termination analyzer(STA) reports on
the behavior of the direct execution of the
algorithm specified by its input except in the
case where the input calls this STA to try to fool it.
>
What you are proposing would cause HHH to get stuck
in infinite execution. How is getting stuck in
infinite execution better than not getting stuck?

>
In other words, if you assume that a termination analyzer exists,

>
Unlike a halt decider that is either ALL KNOWING
or wrong a termination analyzer is correct even
if it can only compute the mapping from a single
input (that has no inputs) to the behavior that
this input specifies and thus its halt status.

>
False, as a universal termination analyzer must still answer for all algorithms the following mapping:
>

 You don't have the authority to rename elements of my work.
I chose termination analyzers intentionally avoiding the term
deciders. I don't want them to be universal.
 

>
Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
>
A solution to the halting problem is an algorithm H that computes the following mapping:
>
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly
>
>

>
HHH(DD) does correctly compute the mapping from its
input to the behavior that this input specifies.

>
False, because it specifies the description of an algorithm that halts when executed directly.
>

 It may be very similar to the DD() that
halts but this different instance actually

fails to see the halting behaviour.
HHH has a bug to abort before it reaches the end.
This bug in HHH does not change anything in the specification of the input, only HHH does not see it.
The traces of HHH and HHH1 are identical up to the point where HHH suddenly stops the simulation.

EXECUTION TRACES DON'T LIE