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On 3/8/2025 8:24 PM, olcott wrote:If the halting problem actually requires that the "decider"On 3/8/2025 6:56 PM, dbush wrote:In other words, you have no rebuttal to the fact that HHH doesn't meet the requirements to be a solution to the halting problem.On 3/8/2025 7:29 PM, olcott wrote:>On 3/8/2025 5:31 PM, dbush wrote:>On 3/8/2025 6:23 PM, olcott wrote:>On 3/8/2025 4:58 PM, dbush wrote:>On 3/8/2025 5:42 PM, olcott wrote:>On 3/8/2025 9:00 AM, dbush wrote:>On 3/8/2025 9:03 AM, olcott wrote:>>>
Apparently you don't understand that inputs to a
simulating termination analyzer specifying infinite
recursion or recursive emulation cannot possibly
reach their own final state and terminate normally.
Apparently you don't understand that inputs to a termination analyzer, simulating or otherwise, are specified by the specification that is the halting function:
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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
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And HHH(DD)==0 fails to meet the above specification
*THIS IS A SEMANTIC TAUTOLOGY THUS IMPOSSIBLY FALSE*
Replacing the code of HHH with an unconditional simulator and subsequently running HHH(DD) cannot possibly reach
its own "ret" instruction and terminate normally
because DD calls HHH(DD) in recursive emulation.
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It is ridiculously stupid to believe that HHH must
report on behavior other than the above behavior.
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It must if it is to be classified as a halt decider or termination analyzer as per the definition.
In other words you believe that HHH
Is required to map the halting function to meet the requirements to be a halt decider / termination analyzer.
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HHH must map from the input finite string DD
to the behavior that this finite string specifies
And what it specifies, to be considered a solution to the halting problem, is given by the specification:
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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
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In the same way that Sum(5,3) == 9
That is misconception is very widely held
does not make it not a misconception.
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