Re: Halting Problem: What Constitutes Pathological Input

On 5/7/2025 1:27 PM, dbush wrote:

On 5/7/2025 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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All simulating halt deciders must
PREDICT WHAT THE BEHAVIOR WOULD BE

 If the machine described by its input was executed directly, as per the requirements of a halt decider:
  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
 

I have proved that everyone has been wrong about this
for ninety years. Ignoring my proof is not any rebuttal.
My proof is probably totally over-your-head.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer