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Persona3Reload@lemmy.blahaj.zone to 196@lemmy.blahaj.zoneEnglish · 9 months ago

Rule

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Rule

lemmy.blahaj.zone

Persona3Reload@lemmy.blahaj.zone to 196@lemmy.blahaj.zoneEnglish · 9 months ago
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  • Euphoma@lemmy.ml
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    9 months ago

    return true

    is correct around half of the time

    • ImplyingImplications@lemmy.ca
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      9 months ago
      assert IsEven(2) == True
      assert IsEven(4) == True
      assert IsEven(6) == True
      

      All checks pass. LGTM

    • aliser@lemmy.world
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      9 months ago
      return Math.random() > 0.5
      

      would also be correct about half the time

      • deo@lemmy.dbzer0.com
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        9 months ago

        Wouldn’t that only be correct about 25% of the time?

        • Rain World: Slugcat Game@lemmy.world
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          9 months ago

          for even, 50% chance of correctness, same for odd.

  • Rai@lemmy.dbzer0.com
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    9 months ago

    Zero people in this post get the YanDev reference

    • psychOdelic@discuss.tchncs.de
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      9 months ago

      so nobody actually really got the joke. very sad Moment.

      • Rai@lemmy.dbzer0.com
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        9 months ago

        It’s really just us… I’ve seen the basic programming joke a bunch of times, but people really aren’t understanding the YanDev/font embellishment. Sad indeed.

    • RVGamer06@sh.itjust.works
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      9 months ago

      I do :D

  • jbk@discuss.tchncs.de
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    9 months ago

    so did someone draw this by hand or was it a filter

    • beefbot@lemmy.blahaj.zone
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      9 months ago

      tbh it looks like an AI broke this down slightly & reconstructed it

  • affiliate@lemmy.world
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    9 months ago

    a wise programmer knows to always ask the question “can i solve this problem in python using metaprogramming?” in this instance, the answer is yes:

    def is_even(n: int):
        s = "def is_even_helper(number: int):\n"
        b = True
        for i in range(0, abs(n)+2):
            s += f"\tif (abs(number) == {i}): return {b}\n"
            b = not b
        exec(s)
        return locals().get("is_even_helper")(n)
    
    • sepiroth154@feddit.nl
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      9 months ago

      Gotta love how human readable Python always is!

  • lnxtx (xe/xem/xyr)@feddit.nl
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    9 months ago

    Ask AI:

    public static boolean isEven(int number) {
        // Handle negative numbers
        if (number < 0) {
            number = -number; // Convert to positive
        }
        
        // Subtract 2 until we reach 0 or 1
        while (number > 1) {
            number -= 2;
        }
        
        // If we reach 0, it's even; if we reach 1, it's odd
        return number == 0;
    }
    
    • Sanctus@lemmy.world
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      9 months ago

      Anything but using modulo I guess

      • lnxtx (xe/xem/xyr)@feddit.nl
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        9 months ago

        And bit operations (:

    • YtA4QCam2A9j7EfTgHrH@infosec.pub
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      9 months ago

      This makes me happy that I don’t use genai

      • Mirodir@discuss.tchncs.de
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        9 months ago

        I’m not sure how fucked up their prompt is (or how unlucky they were). I just did 3 tries and every time it used modulo.

        I’m assuming they asked it specifically to either not use modulo or to do a suboptimal way to make this joke.

  • gerryflap@feddit.nl
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    9 months ago

    Using Haskell you can write it way more concise:

    iseven :: Int -> Bool
    iseven 0 = True
    iseven 1 = False
    iseven 2 = True
    iseven 3 = False
    iseven 4 = True
    iseven 5 = False
    iseven 6 = True
    iseven 7 = False
    iseven 8 = True
    ...
    

    However, we can be way smarter by only defining the 2 base cases and then a recursive definition for all other numbers:

    iseven :: Int -> Bool
    iseven 0 = True
    iseven 1 = False
    iseven n = iseven (n-2)
    

    It’s having a hard time with negative numbers, but honestly that’s quite a mood

    • luciferofastora@lemmy.zip
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      9 months ago

      Recursion is its own reward

  • TunaCowboy@lemmy.world
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    9 months ago

    if (!(number & 1))

    • FiskFisk33@startrek.website
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      9 months ago

      if (~number & 1)

  • dadarobot@lemmy.sdf.org
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    9 months ago
    If number%2 == 0: return("Even")
    Else: return("odd") 
    
    • lol_idk@lemmy.ml
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      9 months ago

      Deleted

  • FiskFisk33@startrek.website
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    9 months ago

    oh of course there is

    https://www.npmjs.com/package/is-even

    (do take a look at the download stats)

    • Fiona@discuss.tchncs.de
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      9 months ago

      And that isn’t even the worst thing about it…

      The implementation looks like this:

      function isEven(i) {
        return !isOdd(i);
      };
      

      And yes, is-odd is a dependency that in turn depends on is-number…

    • servobobo@feddit.nl
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      9 months ago

      “If it’s not an npm package it’s impossible”

      - JS devs, probably

    • tb_@lemmy.world
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      9 months ago

      That’s a lot of downloads

  • ashestoashes@lemmy.blahaj.zone
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    9 months ago

    just check the least significant bit smh my head

  • themoonisacheese@sh.itjust.works
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    9 months ago

    https://codegolf.stackexchange.com/q/275739/88192

  • moistclump@lemmy.world
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    9 months ago

    I thought they were going to turn into Saddam Husseins.

  • MuffinHeeler@aussie.zone
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    9 months ago

    =if((number/2)-round(number/2,0)=0,true,false)

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