Does?

Does IQfy understand big O notation?
Hint: everything the redditor said is wrong

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  1. 3 weeks ago
    Anonymous

    [...]

    This thread again. Hopefully IQfy will be smart and not fall for the bait this time.

  2. 3 weeks ago
    Anonymous

    [math] f(x)=O(g(x))[/math] if and only if there exists some constant [math]c[/math] such that [math]f(x)leq xg(x)[/math] for all [math]x[/math]

    • 3 weeks ago
      Anonymous

      Wrong.

      • 3 weeks ago
        Anonymous

        Double wrong.

        • 3 weeks ago
          Anonymous

          [math] f(x)=O(g(x))[/math] if and only if there exists some constant [math]c[/math] such that [math]f(x)leq xg(x)[/math] for all [math]x[/math]

          x^2 <= x * (2x) for all x but x^2 is not in O(2x).

          • 3 weeks ago
            Anonymous

            Wrong.

          • 3 weeks ago
            Anonymous

            Wrong.

          • 3 weeks ago
            Anonymous

            x is not a constant you dumb Black person

          • 3 weeks ago
            Anonymous

            Wrong.

      • 3 weeks ago
        Anonymous

        Wrong. Downvoted.

      • 3 weeks ago
        Anonymous

        Double wrong.

        Wrong.

        Wrong. Downvoted.

        Wrong.

        [...]
        [...]
        [...]
        Wrong.

        Correct.

    • 3 weeks ago
      Anonymous

      >f(x)leq xg(x)
      you mean [math]f(x)leq cg(x)[math]

      • 3 weeks ago
        Anonymous

        Wrong.

    • 3 weeks ago
      Anonymous

      >math notation

  3. 3 weeks ago
    Anonymous

    That's not a hint and you're a Black person

  4. 3 weeks ago
    Anonymous

    Hard mode: Explain it without using the letter e

    • 3 weeks ago
      Anonymous

      Double wrong.

      [...]
      x^2 <= x * (2x) for all x but x^2 is not in O(2x).

      >Does IQfy understand big O notation?
      No, but I do understand
      >everything the redditor said is wrong
      The vast majority of posts on Reddit contain inaccurate or false information.
      In fact, the majority of the internet is inaccurate or false information.
      Which is why LLMs more often than not produce inaccurate or false information.

      Wrong.

    • 3 weeks ago
      Anonymous

      bump

  5. 3 weeks ago
    Anonymous

    its a big meme made out to be something when it really is nothing and computational complexity if you don't just conceptualize it in your head immediately then lol

    big o notation? more like big cope notation

  6. 3 weeks ago
    Anonymous

    >Does IQfy understand big O notation?
    No, but I do understand
    >everything the redditor said is wrong
    The vast majority of posts on Reddit contain inaccurate or false information.
    In fact, the majority of the internet is inaccurate or false information.
    Which is why LLMs more often than not produce inaccurate or false information.

  7. 3 weeks ago
    Anonymous

    Wrong.

    • 3 weeks ago
      Anonymous

      Wrong. Downvoted.

      Wrong.

      [...]
      [...]
      [...]
      Wrong.

      Wrong.

      [...]
      [...]
      [...]
      [...]
      [...]
      Correct.

      • 3 weeks ago
        Anonymous

        Is this OC?

  8. 3 weeks ago
    Anonymous

    >Hint: everything the redditor said is wrong
    Wrong.

  9. 3 weeks ago
    Anonymous

    >this thread
    kys op

  10. 3 weeks ago
    Anonymous

    Wrong.

  11. 3 weeks ago
    Anonymous

    big O is a good guideline when it comes to performance but its value is overrated since code runs on real hardware with cache-lines and not fairy magic

  12. 3 weeks ago
    Anonymous

    That previous thread made me realize IQfy is full of dunning-kruger moronic autistic homosexuals that wanted to sound smart

    • 3 weeks ago
      Anonymous

      Wrong.

  13. 3 weeks ago
    Anonymous

    Wrong

  14. 3 weeks ago
    Anonymous

    [...]

    Op is wrong, and my roll will be glorious

    • 3 weeks ago
      Anonymous

      shut up cia spook

  15. 3 weeks ago
    Anonymous

    big O has no use case

  16. 3 weeks ago
    Anonymous

    Literally everybody in this thread is wrong

    • 3 weeks ago
      Anonymous

      Wrong.

  17. 3 weeks ago
    Anonymous

    >reddit screencap thread
    >OP says "everything in it is LE WRONG!" (it's not)
    >30+ posts of useless vomit
    what an absolutely embarrassing thread. the fricking state of this board, i don't even recognize it anymore

    • 3 weeks ago
      Anonymous

      >>OP says "everything in it is LE WRONG!" (it's not)
      Name one correct thing that the redditor said.

      • 3 weeks ago
        Anonymous

        All of them. Go ahead and disprove them. You won't.

        • 3 weeks ago
          Anonymous

          > O(1) is constant time, which means it doesn't take longer as the input size increases.
          Wrong. It only means that it's bounded by a constant. It could always take longer with respect to the input size and still be O(1). For example, f(n) = 1 - 1/n always gets bigger with respect to n but is O(1).
          All other statements he made are similarly false.

          • 3 weeks ago
            Anonymous

            Wrong.

          • 3 weeks ago
            Anonymous

            >but is O(1).
            No it's not, it's O(1 - 1/n)

          • 3 weeks ago
            Anonymous

            It's wrong, simple as.

            How so?

            it's wrong in that what's measured is not necessarily time
            it can also be memory usage or some other factor

            That's not what's wrong with it.

            Utterly wrong. Read a fricking book.

  18. 3 weeks ago
    Anonymous

    Big O is about how much the time taken by an algo can grow as input increases, not about how much time it'll take as input increases. the time will obv increase as input increases even in O(1)

  19. 3 weeks ago
    Anonymous

    >print(timetocalculate(thing))
    What does big O even tell you? How cum onto a paper exam?

  20. 3 weeks ago
    Anonymous

    [...]

    Big rO(ll)

    • 3 weeks ago
      Anonymous

      NO WAY YOU GOT HER. NO WAY.
      FRICK OFF. AAAAAA LUCKY LUCKY LUCKY.

    • 3 weeks ago
      Anonymous

      NO WAY YOU GOT HER. NO WAY.
      FRICK OFF. AAAAAA LUCKY LUCKY LUCKY.

      >tfw now married to a girl I've never even seen outside of memes
      Time to finally start the series, it's a sign

  21. 3 weeks ago
    Anonymous

    all of this is masturbatory crap for academics anyway, in the real world there are too many factors to consider to be able to simplify the performance of an algorithm to big O. for small problems sizes, there are often algorithms with a worse complexity that run in less time, simply because they are able to use the hardware more efficiently (mainly the cache). there is substitute for testing your algorithms with real world data, N is not infinity in the real world.

    • 3 weeks ago
      Anonymous

      in the real world you can google or (in case Putler drops a few nuclear bombs on your ISP's buildings) look for the proper algorithm in a book and get copy the most efficient one from there. You never need to do this masturbatory academicuck wankery because you can just copy paste what the academigolem npcs came up with. Just steal it and use it in your code lmao.

      • 3 weeks ago
        Anonymous

        spoken like a true webshit dev soon to be replaced by AI

        • 3 weeks ago
          Anonymous

          sounds like you're poor

  22. 3 weeks ago
    Anonymous

    sure, the definition might be a little different, but I'd say it's close enough for anyone who isn't doing theoretical computer science

    • 3 weeks ago
      Anonymous

      It's wrong, simple as.

      • 3 weeks ago
        Anonymous

        How so?

        • 3 weeks ago
          Anonymous

          Gave an example here

          > O(1) is constant time, which means it doesn't take longer as the input size increases.
          Wrong. It only means that it's bounded by a constant. It could always take longer with respect to the input size and still be O(1). For example, f(n) = 1 - 1/n always gets bigger with respect to n but is O(1).
          All other statements he made are similarly false.

  23. 3 weeks ago
    Anonymous

    it's wrong in that what's measured is not necessarily time
    it can also be memory usage or some other factor

    • 3 weeks ago
      Anonymous

      That's not what's wrong with it.

  24. 3 weeks ago
    Anonymous

    I don't see what's wrong with what the redditor said. Technically when talking about O it means the wrost case and technically it doesn't have to mean that there exists that worst case but the claims he makes are correct. Furthermore proofs usually give an O bound and don't bother proving anything further (like that there is a worst case) even though an algorithm may actually be in Theta

    • 3 weeks ago
      Anonymous

      Oh I see now after reading this

      > O(1) is constant time, which means it doesn't take longer as the input size increases.
      Wrong. It only means that it's bounded by a constant. It could always take longer with respect to the input size and still be O(1). For example, f(n) = 1 - 1/n always gets bigger with respect to n but is O(1).
      All other statements he made are similarly false.

      . His mistake is claiming something about how it scales with input without previously expressing that he is talking about large input. For sufficiently large n, his claims are correct

      >but is O(1).
      No it's not, it's O(1 - 1/n)

      1-1/n is bound by a constant. That constant is 1

      • 3 weeks ago
        Anonymous

        >For sufficiently large n, his claims are correct
        Wrong.

        >but is O(1).
        No it's not, it's O(1 - 1/n)

        >No it's not
        Wrong
        >it's O(1 - 1/n)
        Correct.

        • 3 weeks ago
          Anonymous

          They are correct. For intsance in f(n) = Theta(g(n)) for any epsilon > 0 there is a sufficiently large m s.t. c*g(n) - epsilon < f(n) < c*g(n) + epsilon for all n > m. Additionally the number of steps an algorithm makes is an integer and therefore by picking an epsilon < 1/2 you get an m s.t. f(n) = c*g(n) for all n > m

          • 3 weeks ago
            Anonymous

            Wrong.

            That's not what's wrong with it.

            Wrong.

          • 3 weeks ago
            Anonymous

            Revise calculus

          • 3 weeks ago
            Anonymous

            Wrong.

          • 3 weeks ago
            Anonymous

            If anything was wrong and you knew calculus you would be able to point it out

        • 3 weeks ago
          Anonymous

          Not correct.

  25. 3 weeks ago
    Anonymous

    [...]

    What is the asymptotic time complexity of waifu acquisition?

    • 3 weeks ago
      Anonymous

      O(wrong).

  26. 3 weeks ago
    Anonymous

    [...]

    [...]
    What is the asymptotic time complexity of waifu acquisition?

    Rerolling because bees

  27. 3 weeks ago
    Anonymous

    Yes, its describes the order of efficiency of an algorithm which basically equates to the number of steps an algorithm needs to perform to solve a problem. O(n) would be linear i.e. roughly as many steps as the number of items in the array, O(n^2) would be quadratic i.e. n x n number of steps to complete the task for an array of n items.

    [...]

    also rolling for my waifu.

    • 3 weeks ago
      Anonymous

      >Yes, its describes the order of efficiency of an algorithm which basically equates to the number of steps an algorithm needs to perform to solve a problem. O(n) would be linear i.e. roughly as many steps as the number of items in the array, O(n^2) would be quadratic i.e. n x n number of steps to complete the task for an array of n items
      Wrong.

    • 3 weeks ago
      Anonymous

      Wrong.

      >Yes, its describes the order of efficiency of an algorithm which basically equates to the number of steps an algorithm needs to perform to solve a problem. O(n) would be linear i.e. roughly as many steps as the number of items in the array, O(n^2) would be quadratic i.e. n x n number of steps to complete the task for an array of n items
      Wrong.

      Wrong.

      [...]
      [...]
      Rerolling because bees

      Wrong.

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