Sujet : Re: Halting Problem: What Constitutes Pathological Input
De : mikko.levanto (at) *nospam* iki.fi (Mikko)
Groupes : comp.theoryDate : 08. May 2025, 10:35:52
Autres entêtes
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Message-ID : <vvhtto$1m1ok$1@dont-email.me>
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On 2025-05-07 20:47:29 +0000, olcott said:
On 5/7/2025 1:27 PM, dbush wrote:
On 5/7/2025 2:24 PM, olcott wrote:
On 5/7/2025 5:58 AM, Richard Damon wrote:
On 5/6/25 10:00 PM, olcott wrote:
On 5/6/2025 5:49 PM, Richard Damon wrote:
On 5/6/25 2:05 PM, olcott wrote:
On 5/6/2025 5:59 AM, Richard Damon wrote:
On 5/5/25 10:18 PM, olcott wrote:
On 5/5/2025 8:59 PM, dbush wrote:
On 5/5/2025 8:57 PM, olcott wrote:
On 5/5/2025 7:49 PM, dbush wrote:
Which starts with the assumption that an algorithm exists that performs the following mapping:
Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes the following mapping:
(<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
DO COMPUTE THAT THE INPUT IS NON-HALTING
IFF (if and only if) the mapping FROM INPUTS
IS COMPUTED.
i.e. it is found to map something other than the above function which is a contradiction.
The above function VIOLATES COMPUTER SCIENCE.
You make no attempt to show how my claim
THAT IT VIOLATES COMPUTER SCIENCE IS INCORRECT
you simply take that same quote from a computer
science textbook as the infallible word-of-God.
All you are doing is showing that you don't understand proof by contradiction,
Not at all. The COMPUTER SCIENCE of your requirements IS WRONG!
No, YOU don't understand what Computer Science actually is talking about.
Every function computed by a model of computation
must apply a specific sequence of steps that are
specified by the model to the actual finite string
input.
Right, "Computed by a model of computation", that
HHH(DD) must emulate DD according to the rules
of the x86 language.
Right, which is doesn't do.
Remember, your HHH stop processing at a CALL HHH instruction.
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its
*input D* until H correctly determines that its simulated D
*would never stop running unless aborted* then
*input D* // the actual input
Which calls the original H
*would never stop running unless aborted*
// A hypothetical HHH/DD pair where HHH and DD are
// exactly the same except that this HHH does not abort.
No, your hypothetical HHH (like your HHH1) paired with the originl DD which uses the original HHH.
That is NOT what this means:
*simulated D would never stop running unless aborted*
All simulating halt deciders must
PREDICT WHAT THE BEHAVIOR WOULD BE
If the machine described by its input was executed directly, as per the requirements of a halt decider:
Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes the following mapping:
(<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
I have proved that everyone has been wrong about this
for ninety years. Ignoring my proof is not any rebuttal.
My proof is probably totally over-your-head.
The halting problem defined in the definition 12.1 at
https://www.liarparadox.org/Linz_Proof.pdfWhat dbush said above means the same so is correct.
You have never presented any proof of anything.
-- Mikko
Re: Halting Problem: What Constitutes Pathological Input
Re: Halting Problem: What Constitutes Pathological Input