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On 5/7/2025 5:58 AM, Richard Damon wrote:Right, CORRECTLY SIMULATED D would never stop running unless aborted, as that is the only type of simulation that shows behavior.On 5/6/25 10:00 PM, olcott wrote:That is NOT what this means: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:>>>
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.
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.
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!
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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.
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.
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
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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*simulated D would never stop running unless aborted*
All simulating halt deciders mustRight, to be correct, they need to do that, but they can't.
PREDICT WHAT THE BEHAVIOR WOULD BE
and thus cannot simply wait and see
what the behavior is.
If they don't analyze whatBut that is what they must do to be correct.
the behavior of the input WOULD BE if
they did not abort their simulation
they would get stuck and never halt.Right, which is why they have problems, and are not correct.
Simulating halt decider must PREDICT BEHAVIOR.
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