Re: Incorrect requirements --- Computing the mapping from the input to HHH(DD)

On 5/11/2025 9:49 PM, dbush wrote:

On 5/11/2025 10:46 PM, olcott wrote:

On 5/11/2025 9:38 PM, dbush wrote:

On 5/11/2025 10:36 PM, olcott wrote:

On 5/11/2025 9:28 PM, dbush wrote:

On 5/11/2025 10:14 PM, olcott wrote:

On 5/11/2025 8:59 PM, dbush wrote:

On 5/11/2025 9:56 PM, olcott wrote:

On 5/11/2025 8:27 PM, Richard Damon wrote:

On 5/11/25 8:48 PM, olcott wrote:

On 5/11/2025 7:38 PM, Mike Terry wrote:

On 11/05/2025 18:11, Richard Heathfield wrote:

On 11/05/2025 17:44, olcott wrote:

Any yes/no question where both yes and no are the
wrong answer is an incorrect polar question.

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Either DD stops or it doesn't (once it's been hacked around to get it to compile and after we've leeched out all the dodgy programming).

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Done that.  It still stops.
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If the computer cannot correctly decide whether or not DD halts,

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The decider says it doesn't stop..
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we have an undecidable computation,

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No no, that doesn't make sense.  DD stops, and there are lots of partial halt deciders that will decide that particular input correctly.  PO's DD isn't "undecidable".
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No single computation can be undecidable, considered on its own! There are only two possibilities: it halts or it doesn't. In either case there is a decider which decides that /one specific input/ correctly. By extension, any finite number of computations is decidable - we just have a giant switch statement followed by returning halts/ neverhalts as appropriate.  If the input domain has just n inputs, there are 2^n trivial deciders that together cater for every combination of each input halting or never halting.  One of those deciders is a correct decider for that (finite domain) problem.
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The HP is asking for a TM (or equiv.) that correctly decides EVERY (P,I) in its one finite algorithm.  That is what is proven impossible.  The trick of having a big switch statement no longer works because there are infinitely many possible inputs.
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Decidability for just one single input is trivial and not intersting.
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and therefore some computations are undecidable, so Turing's conclusion was right. Who knew? (Apart from practically everybody else, I mean.)

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Mike.

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DDD emulated by HHH according to the rules of
the computational language that DD is encoded
within already proves that the HP "impossible"
input specifies a non-halting sequence of
configurations.

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No it doesn't.
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_DDD()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov  ebp,esp  ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404     add  esp,+04
[00002182] 5d         pop  ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
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Show all the steps of DDD emulated by simulating
termination analyzer HHH according to the rules
of the x86 language

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Which it doesn't do as you have admitted on the record:
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I am daring you to show what they should be.
You know you can't because you know you are a liar.
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Category error.  Algorithm HHH does one thing and one thing only. There is no "what it should be" because it *is* only one thing.
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Show the exact sequence of machine address steps of DDD
such that the DDD emulated by some HHH

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Changing the input is not allowed.

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There is no change of input you are a liar.
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If you change function HHH, you no longer have algorithm DDD, which means you're changing the input.
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Changing the input is not allowed.

 I am talking about every element of an infinite set
you nitwit.