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

On 5/13/25 1:14 AM, olcott wrote:

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.
>

>
In other words you have no such actual reasoning.

>
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.
 

Coming from the person who insists that he can change definitions he doesn;'t like.
The problem IS that "Termination Analyer" as a term DOES mean universal unless restricted by the context, and YOU can't change it.
Your problem is you have always misinterpreted the meaning of the words you read, because you just don't understsnd what you are reading.
If you mean you are going to try to refute the Halting Problem by redefing the problem to say the decider only needs to get one machine correctly, that is just an admission that you are a crooked liar.
And, to be correct, the criteria can't be changed, since DD() Halts when run, the only correct answer for HHH(DD) is to answer halting, but it answer non-halting, so it didn't even get that one correct.
Of course, the fact that your setup breaks so many of the rules, like you DD isn't even the required program, just shows how much of a liar you actually are.

>
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
does not halt EXECUTION TRACES DON'T LIE

Sure they do, as has been pointed out,
A *CORRECT* trace of the input DD needs to follow the call HHH into HHH itelf and then show the code of HHH running.
It will NEVER show DD being run agian, as it never it, The closest is that HHH emulates it, but the emulation needs to show the actions of HHH, not the interpreation of what it is doing.
Again, you logic is just based on LYING.

 

>
HHH(DD) does not compute the mapping from its input
to BEHAVIOR THAT THIS INPUT DOES NOT SPECIFY.
>
The whole issue is that everyone here has been
indoctrinated into baselessly believing that
HHH should in some cases compute the mapping
to BEHAVIOR THAT THE INPUT DOES NOT SPECIFY.

>
That's because if I want to know if any arbitrary algorithm X with input Y will halt when executed directly, that requires an H that meets these requirements, not your alternate requirements:
>
>
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
>

>
No one can possibly provide any correct reasoning
why HHH must do this because no such correct
reasoning exists. On this issue it is merely
THAT IS NOT THE WAY THAT I MEMORIZED IT.
>

>
A LIE, as you were given numerous reasons below why this mapping is useful and your alternate mapping is not useful, and your failure to respond to this constitutes your acceptance of this fact.
>
>
On 5/12/2025 10:22 PM, dbush wrote:
 > On 5/12/2025 10:01 PM, olcott wrote:
 >> If the Goldbach conjecture is true (and there is
 >> no short-cut) this requires testing against every
 >> element of the set of natural numbers an infinite
 >> computation.
 >
 > And if we take that computation and feed it into an H that meets these
 > requirements:
 >
 >
 > 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
 >
 >
 > Then if H reports non-halting the Goldbach conjecture is true, and if H
 > reports halting the Goldbach conjecture is false.
 >
 > If on the other hand we feed the computation to an H that tells us if it
 > "specifies non-halting behavior to H", we won't necessarily get that
 > answer.  We have to worry about how the computation is implemented. With
 > the above requirements we don't need to worry about that.
 >
 > The same applies to *any* process which may or may not be infinite to
 > tell us whether or not it is actually an infinite process.