Re: ZFC solution to incorrect questions: reject them

On 3/12/24 7:41 PM, olcott wrote:

On 3/12/2024 9:17 PM, immibis wrote:

On 13/03/24 02:47, olcott wrote:

On 3/12/2024 8:05 PM, immibis wrote:

On 13/03/24 01:18, olcott wrote:

On 3/12/2024 7:10 PM, immibis wrote:

So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?

Exactly one element of Q differs by writing a 1 instead of a 0.

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That's part of δ but this mistake doesn't matter.
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It wasn't clear whether you were talking about a Turing machine that was somehow identical but gave a different return value, or one that was not identical. Now you have explained it is not identical.
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They are identical except for their return value that is specified
in a single state that is different.
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*This means that they implement the exact same algorithm*

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OK. Well, one of them gets the right answer and one of them gets the wrong answer. What is the confusion?

 The Linz Ĥ.H machine gets the wrong answer on its own
machine description no matter how its Linz H is defined.
 This means that it gets the wrong answer on YES and the
wrong answer on NO.
 

Not quite. It always gets the wrong answer, but only one of them for each quesiton.
For EACH SEPARATE definition of H, and thus H^, we have a different question.
Note, the machine H^ isn't DEFINED to just get H^ as an input.
H^ is defined to get as an input, the description of ANY Turing Machine, and to ask H what that machine applied to its description will do, and then it does the opposite.
Thus, for every different H we go to test, we get a DIFFERENT H^ machine. and when we look at the question to H (or H^.H) about the description (H^) (H^),
If H (H^) (H^) goes to qn, then H^ (H^) goes to qn too and halts, so the correct answer would have been to go to qy.
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy too, and loops, so the correct answer would have been to go to qn.
So, each case HAS a correct answer, just not the one that H (or H^.H) goes to,
So yes, which ever one it goes to (and a given machine will only go to one with this input) will be wrong, but the other one would have been right, and an H* machine that answer the opposite of H would have been correct.