Re: Halting Problem: What Constitutes Pathological Input

On 5/6/25 10:09 PM, olcott wrote:

On 5/6/2025 5:52 PM, Richard Damon wrote:

On 5/6/25 4:37 PM, olcott wrote:

On 5/6/2025 3:22 PM, joes wrote:

Am Tue, 06 May 2025 13:05:15 -0500 schrieb olcott:

On 5/6/2025 5:59 AM, Richard Damon wrote:

On 5/5/25 10:18 PM, olcott wrote:

On 5/5/2025 8:59 PM, dbush wrote:

On 5/5/2025 8:57 PM, olcott wrote:

On 5/5/2025 7:49 PM, dbush wrote:

>

DO COMPUTE THAT THE INPUT IS NON-HALTING IFF (if and only if) the
mapping FROM INPUTS IS COMPUTED.

i.e. it is found to map something other than the above function
which is a contradiction.

The above function VIOLATES COMPUTER SCIENCE. You make no attempt to
show how my claim THAT IT VIOLATES COMPUTER SCIENCE IS INCORRECT you
simply take that same quote from a computer science textbook as the
infallible word-of-God.

What does it violate?
>

All you are doing is showing that you don't understand proof by
contradiction,

Not at all. The COMPUTER SCIENCE of your requirements IS WRONG!

No, YOU don't understand what Computer Science actually is talking
about.

Every function computed by a model of computation must apply a specific
sequence of steps that are specified by the model to the actual finite
string input.

>

You are very confused. An algorithm or program computes a function.
>

>
Nothing computes a function unless it applies a specific
set of rules to its actual input to derive its output.
Anything that ignores its input is not computing a function.
>

>
Right, so HHH needs to apply the rules that it was designed with.
>
And that means it breaks the criteria that you say it needs to do to get the right answer,
>
And thus it gets the wrong answer.
>

 It needs to emulate DD according to the rules of
the x86 language. This includes emulating itself
emulating DD until it recognizes that if it kept
doing this that DD would never halt.

No, to be a correct emulator it needs to continue until it reaches the end,
It can get the right answer if it emulates the input to the point that it can show that a

 <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its
     input D until H correctly determines that its simulated D
     *would never stop running unless aborted* then

Right, that UTM(D) would never halt.

 *would never stop running unless aborted*
Is the hypothetical HHH/DD pair where HHH does not abort.

Nope, can't change DD, it is your hypothetical HHH, which has become UTM, when given the ORIGINAL DD, which calls the ORIGINAL HHH, as that code was part of the definition of DD.

 

Note, the rules of how to compute say NOTHING about what is the correct answer to the question, the definition of the function to be computed does that, and that clearly says to look at the behavior of the program the input represents

 That has always been stupidly wrong.

No, you have always been stupidly wrong, because you decided not to learn the meaning of the words you were using.

 Look at the behavior that the finite string input specifies.
Not look at some behavior that cannot possibly be derived by
applying the rules of the x86 language to this finite string input.
 

WHich first means your finite string of just the code of DD is incorrect, your finite string must include the code of the HHH that it calls.
This shows your first major lie.
Then, when we DO emulate this completely per the definition of the x86 language, which demands that call instructions be followed by the processing of the function called, we see that DD calls HHH which will emulated DD for a while and then abort (incorrectly) its emulation and return to DD which then returns to the systme.
Thus, you claim that it can not is shown to be a stupid and damned lie.
You are just proving how utterly stupid you are, claiming things that are false, even when they are explained to you.