Re: Halting Problem: What Constitutes Pathological Input

Op 06.mei.2025 om 01:30 schreef olcott:

On 5/5/2025 5:35 PM, dbush wrote:

On 5/5/2025 6:29 PM, olcott wrote:

On 5/5/2025 5:00 PM, dbush wrote:

On 5/5/2025 5:40 PM, Richard Heathfield wrote:

On 05/05/2025 22:31, dbush wrote:

It's just that no algorithm exists that can compute that mapping, as proven by Linz and other and as you have *explicitly* agreed is correct.

>
He's coming round to the idea, albeit slowly. He can't bring himself to describe the mapping as 'incomputable' or 'undecidable', but he's started to claim that such a mapping is 'incorrect', which is a tacit acknowledgement that it exists.
>

>
Oh, he's agreed to it many times.  Here's a partial list:
>
>
On 3/24/2025 10:07 PM, olcott wrote:
 > A halt decider cannot exist
>
On 4/28/2025 2:47 PM, olcott wrote:
 > On 4/28/2025 11:54 AM, dbush wrote:
 >> And the halting function below is not a computable function:
 >>
 >
 > It is NEVER a computable function
 >
 >> 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
>
On 3/14/2025 1:19 PM, olcott wrote:
 > When we define the HP as having H return a value
 > corresponding to the halting behavior of input D
 > and input D can actually does the opposite of whatever
 > value that H returns, then we have boxed ourselves
 > in to a problem having no solution.
>
On 6/21/2024 1:22 PM, olcott wrote:
 > the logical impossibility of specifying a halt decider H
 > that correctly reports the halt status of input D that is
 > defined to do the opposite of whatever value that H reports.
 > Of course this is impossible.
>
On 7/4/2023 12:57 AM, olcott wrote:
 > If you frame the problem in that a halt decider must divide up finite
 > strings pairs into those that halt when directly executed and those that
 > do not, then no single program can do this.
>
On 5/5/2025 5:39 PM, olcott wrote:
 > On 5/5/2025 4:31 PM, dbush wrote:
 >> Strawman.  The square root of a dead rabbit does not exist, but the
 >> question of whether any arbitrary algorithm X with input Y halts when
 >> executed directly has a correct answer in all cases.
 >>
 >
 > It has a correct answer that cannot ever be computed
>
>

>
There never has been any input that could
ever actually do the opposite of whatever
value that its termination analyzer returns.
That part of its code was ALWAYS unreachable.
>

>
None-the-less you have agreed on many occasions that the theorem which the halting problem proofs prove is correct.

 There are different ways of framing the problem.
The only one that matters is that HHH(DD) does
correctly determine that DD never halts.
 

It does not, because the programmer introduces a bug in HHH, which makes that it processes only part of the input. It misses relevant parts of the input, such as the code for Halt7.c, where a conditional abort is specified.
If the programmer of Sum(2,3) only processes 2 and not 3, that is an error of the programmer.
Similarly, the incomplete analysis of HHH makes its result incorrect.