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Rule (lemmy.blahaj.zone)

submitted 1 week ago by Persona3Reload@lemmy.blahaj.zone to c/196@lemmy.blahaj.zone

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[-] superkret@feddit.org 45 points 1 week ago

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

[-] fartripper@lemmy.ml 20 points 1 week ago* (last edited 1 week ago)

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

[-] superkret@feddit.org 8 points 1 week ago

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

Like, for example, 1?

[-] prime_number_314159@lemmy.world 3 points 1 week ago

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.

[-] fartripper@lemmy.ml 5 points 1 week ago

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

[-] prime_number_314159@lemmy.world 1 points 1 week 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.