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

On 5/11/2025 9:05 PM, olcott wrote:

On 5/11/2025 7:58 PM, dbush wrote:

On 5/11/2025 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.

>
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).

>
Done that.  It still stops.
>

>
If the computer cannot correctly decide whether or not DD halts,

>
The decider says it doesn't stop..
>

we have an undecidable computation,

>
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".
>
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.
>
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.
>
Decidability for just one single input is trivial and not intersting.
>

and therefore some computations are undecidable, so Turing's conclusion was right. Who knew? (Apart from practically everybody else, I mean.)

>
>
Mike.

>
DDD emulated by HHH according to the rules of
the computational language that DD is encoded
within

>
Doesn't happen, as you have admitted on the record:
>
>

 I admitted on record that DDD is not infinitely
emulated by HHH. Because you are only a damned
Troll you try to twist this into incorrect emulation.
 

False.  HHH aborts which goes against the rules of the x86 language, as you have admitted on the record:
On 5/5/2025 8:24 AM, dbush wrote:
 > On 5/4/2025 11:03 PM, dbush wrote:
 >> On 5/4/2025 10:05 PM, olcott wrote:
 >>> On 5/4/2025 7:23 PM, Richard Damon wrote:
 >>>> But HHH doesn't correct emulated DD by those rules, as those rules
 >>>> do not allow HHH to stop its emulation,
 >>>
 >>> Sure they do you freaking moron...
 >>
 >> Then show where in the Intel instruction manual that the execution of
 >> any instruction other than a HLT is allowed to stop instead of
 >> executing the next instruction.
 >>
 >> Failure to do so in your next reply, or within one hour of your next
 >> post on this newsgroup, will be taken as you official on-the-record
 >> admission that there is no such allowance and that HHH does NOT
 >> correctly simulate DD.
 >
 > Let the record show that Peter Olcott made the following post in this
 > newsgroup after the above message:
 >
 > On 5/4/2025 11:04 PM, olcott wrote:
 >  > D *WOULD NEVER STOP RUNNING UNLESS*
 >  > indicates that professor Sipser was agreeing
 >  > to hypotheticals AS *NOT CHANGING THE INPUT*
 >  >
 >  > You are taking
 >  > *WOULD NEVER STOP RUNNING UNLESS*
 >  > to mean *NEVER STOPS RUNNING* that is incorrect.
 >
 > And has made no attempt after over 9 hours to show where in the Intel
 > instruction manual that execution is allowed to stop after any
 > instruction other than HLT.
 >
 > Therefore, as per the above criteria:
 >
 > LET THE RECORD SHOW
 >
 > That Peter Olcott
 >
 > Has *officially* admitted
 >
 > That DD is NOT correctly simulated by HHH