389
Rule (lemmy.blahaj.zone)
submitted 2 years ago by to c/196@lemmy.blahaj.zone
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[–] 45 points 2 years ago (3 children)

Just divide the number into its prime factors and then check if one of them is 2.

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  • [–] 20 points 2 years ago* (1 child)

    or divide the number by two and if the remainder is greater than

    -(4^34)

    but less than

    70 - (((23*3*4)/2)/2)

    then

    true
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  • [–] 8 points 2 years ago (1 child)

    What if the remainder is greater than the first, but not less than the latter?

    Like, for example, 1?

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  • [–] 3 points 2 years ago (1 child)

    Then you should return false, unless the remainder is also greater than or equal to the twenty second root of 4194304. Note, that I've only checked up to 4194304 to make sure this works, so if you need bigger numbers, you'll have to validate on your own.

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  • [–] 5 points 2 years ago (1 child)

    i hate to bring this up, but we also need a separate function for negative numbers

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  • [–] 1 point 2 years ago

    You can just bitwise AND those with ...000000001 (for however many bits are in your number). If the result is 0, then the number is even, and if it's 1, then the number is odd. This works for negative numbers because it discards the negative signing bit.

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  • [–] 14 points 2 years ago (2 children)

    I remember coding actionscript in Flash and using modulo (%) to determine if a number was even or odd. It returns the remainder of the number divided by 2 and if it equals anything other than 0 then the number is odd.

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  • [–] 25 points 2 years ago

    Yeah. The joke is that this is the obvious solution always used in practise, but the programmer is that bad that they don't know it and use some ridiculous alternative solutions instead.

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