Rule

return true

is correct around half of the time

assert IsEven(2) == True assert IsEven(4) == True assert IsEven(6) == True

All checks pass. LGTM
return Math.random() > 0.5

would also be correct about half the time
- Wouldn't that only be correct about 25% of the time?

import re

def is_even(i: int) -> bool:
    return re.match(r"-?\d*[02468]$", str(i)) is not None

Cursed
i was gonna suggest the classic

re.match(r"^(..)\1*$", "0" * abs(i)) is not None

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

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
- What if the remainder is greater than the first, but not less than the latter?
  
  Like, for example, 1?
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.
- 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.
- I believe that's the proper way to do it.

Zero people in this post get the YanDev reference

so nobody actually really got the joke. very sad Moment.
- 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.
I do :D

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

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

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

Recursion is its own reward

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;
}

Anything but using modulo I guess
- And bit operations (:
This makes me happy that I don’t use genai
- 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.

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)

Gotta love how human readable Python always is!

When you sacrifice memory for an O(1) algorithm.

In this case still O(n)

Smh this is literally what switch statements are for

If number%2 == 0: return("Even")
Else: return("odd")

Not all ARM CPUs support mod operations. It’s better to use bit operations. Check if the last bit is set. If set it’s odd else it’s even.
Deleted

if (!(number & 1))

if (~number & 1)

oh of course there is

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

(do take a look at the download stats)

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…
Can't you just

If (number % 2 == 0){return true}
- return number % 2 === 0
- but what if number isn’t an integer, or even a number at all? This code, and the improved code shared by the other user, could cause major problems under those conditions. Really, what you would want, is to validate that number is actually an integer before performing the modulo, and if it isn’t, you want to throw an exception, because something has gone wrong.
  
  That’s exactly what that NPM module does. And this is why it’s not a bad thing to use packages/modules for even very simple tasks, because they help to prevent us from making silly mistakes.
- yup, which is why I find the download stats truly horrifying
- no
"If it's not an npm package it's impossible"

- JS devs, probably
That's a lot of downloads

just check the least significant bit smh my head

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

I thought they were going to turn into Saddam Husseins.

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