Vote on which of Hacker News' challenges for AI have been met
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Unofficial Hacker News client; not affiliated with Y Combinator.
Vote on which of Hacker News' challenges for AI have been met
Unofficial Hacker News client; not affiliated with Y Combinator.
hatthew · · focus · HN ↗
Another thing to note is that the (presumably AI-generated) summary of my challenge does not accurately represent what I wrote, listing only half the things I said and saying "or" rather than "and".
an0malous · · focus · HN ↗
senordevnyc · · focus · HN ↗
I find it mind boggling that anyone thinks these agents only solved this problem because they maybe could possibly have seen the unfinished work of researchers who were working on a simpler version of the problem, (also with AI).
Looking forward to the cope when the next big problem falls.
dormento · · focus · HN ↗
I hope I'm not being too blunt, but the other alternative is to "just trust me bro" the hyperscalers, who are pretty much locked into a battle for profitability and have all the incentives to make up things to prop up their stock, no? I don't think this is the way.
pessimizer · · focus · HN ↗
senordevnyc · · focus · HN ↗
someonebaggy · · focus · HN ↗
eviks · · focus · HN ↗
_superposition_ · · focus · HN ↗
someonebaggy · · focus · HN ↗
off_with_their_ · · focus · HN ↗
[dead]
olmo23 · · focus · HN ↗
tsunamifury · · focus · HN ↗
stronglikedan · · focus · HN ↗
tsunamifury · · focus · HN ↗
ben_w · · focus · HN ↗
Simply put: For the same reason computers have not already solved all problems in mathematics.
More concretely:
Consider the Collatz conjecture. It's a very simple rule to write down:
Trivial to write a program to test numbers starting at 1 and going up. We know it holds up to at least 2.36e21 (according to Wikipedia), but to prove it is true with such a program requires testing all of the infinite set of positive integers.But you may notice some things about the rules, that they suggest a subset of numbers will trivially always converge to 1, so that you don't need to even test them: any integer 2^n where n is also a positive integer.
You may find other easy wins, or ways to simplify the test, e.g. once you know the numbers up to m will converge to 1, you can terminate your loop early if you test m+1 and it ever has an intermediate value less than or equal to m. You can combine that with applying one of the rules in reverse, and know that all even numbers between m and 2m will on their first move be halved, making them smaller than m, which means you know they'll eventually converge.
But actually proving this is fully general? Nobody knows. You can't just throw arithmetic at the problem directly, you have to figure out patterns that would let you prove that it always holds, no matter what.
Or, you may find many such patterns and directly calculate some number not in any of them, to find one which doesn't converge to 1.
Jtariiiii · · focus · HN ↗
The solution to NS was categorically not stolen, nobody is alleging that OpenAI stole a complete solution to NS. The alleged theft was about a different set of related equations.
someonebaggy · · focus · HN ↗
3uruiueijjj · · focus · HN ↗
Talk about bad luck!
irthomasthomas · · focus · HN ↗