Saturday, December 30

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Daily News Stuff 30 December 2023

Bambeezled Edition

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Tech News

  • Nvidia is going all-out to compete with AMD's Radeon 7800 XT.  (Tom's Hardware)

    Except on price, which is probably going to render their efforts irrelevant in the gaming market.


  • Remember that program that could determine whether any unsigned 32-bit integer was even or odd?  Now there's a version that works for 64-bit integers.  (GitHub)

    This one cheats slightly.  Rather than having a separate check for each possible value, the developers pre-computed the result for every possible input and saved the results to a file.

    Then they compressed that file.

    Recursively.

    Reducing it from around two million terabytes to 13 kilobytes.


  • Don't ask why, just accept it.


  • In a similar vein, here's a fully-automated Christmas checker.  (XKCD)

    99.73% accurate, which is even better than the prime number checker.


  • Should we upgrade to M3 MacBooks?  (Incident)

    I don't know, let's throw a bunch of poorly-understood numbers into ChatGPT and see what it says.

    ChatGPT says yes!

    Also that it's an Amazon Affiliate but that in no way coloured its decision.



Disclaimer: Dad!  There's another dead clown on the roof!

Posted by: Pixy Misa at 06:26 PM | Comments (6) | Add Comment | Trackbacks (Suck)
Post contains 252 words, total size 3 kb.

1 Ok, I'm confused. Can't you just check the last digit? Convert to string, pull the last character, then at most 5 checks?

Posted by: Davd at Sunday, December 31 2023 04:56 AM (R7Z4D)

2 And I would assume you could do a binary check via some ror/rol calls, but would take me a bit to figure that out.

Posted by: Davd at Sunday, December 31 2023 04:57 AM (R7Z4D)

3 Ok, finish the walk to the refrigerator, and it's between 2 and 10 operations. You could probably order the checks to put 3 and 7 first to optimize if you're picking human-selected numbers first.
bool odd = false; if (a & 1 == 1)  || (a & 3 == 3) || (a & 5 == 5) || (a & 7 == 7) || (a & 9 == 9) {
odd = true; }

Posted by: Davd at Sunday, December 31 2023 05:07 AM (R7Z4D)

4 Couldn't you just, you know, check the low bit?

Posted by: Mauser at Sunday, December 31 2023 09:08 AM (sZ6tC)

5 Yeah, I mean, I think that's the joke here.

Posted by: normal at Sunday, December 31 2023 01:09 PM (bg2DR)

6 The canonical answer is the modulo operator, which basically amounts to, yes, checking the low bit. If you track back from the git repository Pixy linked, the original concept was "figure out even/odd without doing that." But then they went for the stupidest possible answer, and tried to optimize from there.

Posted by: Davd at Sunday, December 31 2023 07:03 PM (R7Z4D)

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Apple pies are delicious. But never mind apple pies. What colour is a green orange?




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