Re: How the requirements that Professor Sipser agreed to are exactly met

On 5/13/2025 1:22 AM, olcott wrote:

On 5/13/2025 12:11 AM, Richard Heathfield wrote:

On 13/05/2025 03:01, olcott wrote:

If the Goldbach conjecture is true (and there is
no short-cut)

>
We don't know that. Fermat's Last Theorem had a short cut, but it took 358 years to find it. At that rate, we won't find Goldbach's short cut (if it has one) for another 75 years.
>

this requires testing against every
element of the set of natural numbers an infinite
computation.

>
Only if it's true. So if it's true, the testing program will never halt. But if it's false, the tester will eventually find the counter- example, print it, and stop.
>
So a program that can tell whether another program will halt can tell us whether Goldbach's conjecture is true.
>
It would be the short cut that you say doesn't exist.
>

 I am refuting the key halting problem

You mean the proof when you admitted *multiple time* that the theorem is proves is correct?
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