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On 7/5/2025 8:14 AM, Richard Damon wrote:Nope.On 7/4/25 6:11 PM, olcott wrote:In order to have an honest dialogue you must payOn 7/4/2025 3:53 PM, Richard Damon wrote:>On 7/4/25 4:43 PM, olcott wrote:>On 6/3/2025 10:02 PM, dbush wrote:>On 6/3/2025 10:58 PM, olcott wrote:>On 6/3/2025 9:46 PM, dbush wrote:>On 6/3/2025 10:34 PM, olcott wrote:>On 6/3/2025 9:12 PM, dbush wrote:>>>
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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Yes there is no algorithm that does that
Excellent!
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Let The Record Show
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That Peter Olcott
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Has *EXPLICITLY* admitted
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That no algorithm H exists that meets the above requirements, which is precisely the theorem that the halting problem proofs prove.
In the exact same way that there is no set of all set
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.
Which arose because the axioms of naive set theory created a contradiction.
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Likewise with halt deciders that are required to report
on the behavior of directly executed Turing machines.
And what is the CONTRADICTION?
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The result is just some things are not computable.
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The result is that there cannot possibly be
an *ACTUAL INPUT* that does the opposite of
whatever its partial halt decider decides
thus the HP proof fails before it begins.
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Sure there is.
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100% complete attention to every single word.
You can't just erase one of the words that I said
and then form a rebuttal on that basis.
Directly executed Turing machines have always been
outside of the domain of every Turing machine based
decider.
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