Re: Halting Problem: What Constitutes Pathological Input

On 5/7/25 11:04 PM, olcott wrote:

On 5/7/2025 9:14 PM, Richard Damon wrote:

On 5/7/25 2:24 PM, olcott wrote:

On 5/7/2025 5:58 AM, Richard Damon wrote:

On 5/6/25 10:00 PM, olcott wrote:

On 5/6/2025 5:49 PM, Richard Damon wrote:

On 5/6/25 2:05 PM, olcott wrote:

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:

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Which starts with the assumption that an algorithm exists that performs the following mapping:
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Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
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A solution to the halting problem is an algorithm H that computes the following mapping:
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(<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
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DO COMPUTE THAT THE INPUT IS NON-HALTING
IFF (if and only if) the mapping FROM INPUTS
IS COMPUTED.

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

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

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Not at all. The COMPUTER SCIENCE of your requirements IS WRONG!
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No, YOU don't understand what Computer Science actually is talking about.
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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.

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Right, "Computed by a model of computation", that
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HHH(DD) must emulate DD according to the rules
of the x86 language.

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Right, which is doesn't do.
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Remember, your HHH stop processing at a CALL HHH instruction.
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<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
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*input D* // the actual input

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Which calls the original H
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*would never stop running unless aborted*
// A hypothetical HHH/DD pair where HHH and DD are
// exactly the same except that this HHH does not abort.
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No, your hypothetical HHH (like  your HHH1) paired with the originl DD which uses the original HHH.
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That is NOT what this means:
*simulated D would never stop running unless aborted*

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Right, CORRECTLY SIMULATED D would never stop running unless aborted, as that is the only type of simulation that shows behavior.
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And D is, and only is, the program that we started with as the input, which calls the original H, and only that H.
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Where do you ee
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All simulating halt deciders must
PREDICT WHAT THE BEHAVIOR WOULD BE
and thus cannot simply wait and see
what the behavior is.

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Right, to be correct, they need to do that, but they can't.
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 HHH does correctly predict that its input DD
*would never stop running unless aborted*

Nope becuase the fact is that the correct emulation of the DD PROGRAM given as its input will Halt, since the program HHH DOES abort its emulation of DD and return 0 to all callers
The fact that you also have admitted that your decider and input aren't even the needed programs says that NOTHING you have claimed has any factual basis as your whole argument is based on category errors.

 

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If they don't analyze what
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the behavior of the input WOULD BE if
they did not abort their simulation

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But that is what they must do to be correct.
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they would get stuck and never halt.
Simulating halt decider must PREDICT BEHAVIOR.
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Right, which is why they have problems, and are not correct.
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 HHH does correctly predict that its input DD
*would never stop running unless aborted*
 <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
      H can abort its simulation of D and correctly report that D
     specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>

Right, if the (CORRECT) simulation of D would never stop running unless aborted.
SInce PROGRAM D calls the original PROGRAM H that does abort and return 0, this D can be simulate to its end without needing to abort THIS simulation (it does simulate an simulation that aborts, but that isn't what is being refered to).

 

Your logic is just demonstrating that you are admitting that you logic insists that it is ok to lie about the correct answer and be wrong, but still claim to be right.
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 HHH does correctly predict that its input DD
*would never stop running unless aborted*
THUS MEETING THE REQUIRED SPEC.

But it does stop running when correctly emulated.
You are just too stupid to understand that.

 

That concept is what has made you into the pathological liar you are. You have erased the definition of truth from your mind, by convincing yourself that it must be ok to lie so you can claim to do what is actually impossible.
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This is what makes you utterly stupid.

 You are very intelligent you pretend that you
do not understand very simple things so that
you can play sadistic head games.
 void DDD()
{
   HHH(DDD);
   return;
}
 DDD simulated (according to the rules of the C
programming language) by any HHH that can possibly
exist cannot possibly reach its own "return" instruction.
 

But your HHH doesn't meet those requirement, so that logic doesn't apply.
If your HHH aborts its simulation, it never did a correct simulation, and thus the fact that a DIFFERENT DDD based on that DIFFERENT HHH wouldn't halt is irrelevernt.
All you are proving is that you just don't understand what a program is, and are just lying by creating an equivocation trying to make it seem that you are talking about the same programs.
Sorry, you are just proving how much of a stupid liar you are.