Liste des Groupes | Revenir à theory |
On 5/5/2025 5:40 PM, Richard Heathfield wrote:There never has been any input that couldOn 05/05/2025 22:31, dbush wrote:Oh, he's agreed to it many times. Here's a partial list: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.
>
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
Les messages affichés proviennent d'usenet.