• SpeakinTelnet@programming.dev
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    2 years ago
    def is_even(n):
        match n:
            case 1:
                return False
            case 0:
                return True
            # fix No1
            case n < 0:
                return is_even(-1*n)
            case _:
                return is_even(n-2)
    
    • misophist@lemmy.world
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      2 years ago

      This is confusing. I’m already using the iSeven API to determine if a number is 7. I’m getting a namespace collision error when I try to load this new API. Bug report filed.

  • neidu@feddit.nl
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    2 years ago

    My solution in perl back in the day when I was a teenage hobbyist who didn’t know about the modulus operator: Divide by 2 and use regex to check for a decimal point.

    if ($num / 2 =~ /\./) { return “odd” }
    else { return “even” }

      • Chobbes@lemmy.world
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        2 years ago

        You know, I was going to let this slide under the notion that we’re just ignoring the limited precision of floating point numbers… But then I thought about it and it’s probably not right even if you were computing with real numbers! The decimal representation of real numbers isn’t unique, so this could tell me that “2 = 1.9999…” is odd. Maybe your string coercion is guaranteed to return the finite decimal representation, but I think that would be undecidable.

        • backgroundcow@lemmy.world
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          2 years ago

          Ackchyually-- IEEE 754 guarantees any integer with absolute value less than 2^24 to be exactly representable as a single precision float. So, the “divide by 2, check for decimals” should be safe as long as the origin of the number being checked is somewhat reasonable.

  • Anticorp@lemmy.world
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    2 years ago

    Back when I was learning programming a lot of lessons would make you do something like this, and then show you the real way to do it in the next lesson. My reaction was always “why didn’t you lead with this?”.

  • noddy@beehaw.org
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    2 years ago

    I know how to fix this!

    bool IsEven(int number) {
        bool even = true;
        for (int i = 0; i < number; ++i) {
            if (even == true) {
                even = false;
            }
            else if (even == false) {
                even = true;
            }
            else {
                throw RuntimeException("Could not determine whether even is true or false.");
            }
        }
    
        if (even == true) {
            return even ? true : false;
        }
        else if (even == false) {
            return (!even) ? false : true;
        }
        else {
            throw RuntimeException("Could not determine whether even is true or false.");
        }
    }
    
      • noddy@beehaw.org
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        2 years ago

        I know an even better way. We can make it run in O(1) by using a lookup table. We only need to store 2^64 booleans in an array first.

  • recursive_recursion [they/them]@programming.dev
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    2 years ago

    modulo

    pseudocode:

    if number % 2 == 0
      return "number is even" (is_num_even = 1 or true)
    else
      return "number is odd" (is_num_even = 0 or false)
    

    plus you’d want an input validation beforehand