Re: Halting Problem: What Constitutes Pathological Input

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
 *input D* // the actual input

Which calls the original H

 *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.
 

No, your hypothetical HHH (like  your HHH1) paired with the originl DD which uses the original HHH.
You are just proving your stupidity.
Your logic is just based on lies and equivocation, proving you are just a pathological liar.
The fact you can just repeat your same claims and not actually deal with the errors pointed out shows that we have reached the point where you errors have been exposed. You can't go more basic, as that will make clear even to you that you are just stupidily wrong.