this post was submitted on 31 Oct 2024
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[–] [email protected] 76 points 3 weeks ago (2 children)

return true

is correct around half of the time

[–] [email protected] 40 points 3 weeks ago
assert IsEven(2) == True
assert IsEven(4) == True
assert IsEven(6) == True

All checks pass. LGTM

[–] aliser 18 points 3 weeks ago (1 children)
return Math.random() > 0.5

would also be correct about half the time

[–] [email protected] 1 points 3 weeks ago (1 children)

Wouldn't that only be correct about 25% of the time?

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[–] [email protected] 49 points 3 weeks ago* (last edited 3 weeks ago) (2 children)
import re

def is_even(i: int) -> bool:
    return re.match(r"-?\d*[02468]$", str(i)) is not None
[–] [email protected] 21 points 3 weeks ago
[–] [email protected] 5 points 3 weeks ago

i was gonna suggest the classic

re.match(r"^(..)\1*$", "0" * abs(i)) is not None
[–] [email protected] 43 points 3 weeks ago (3 children)

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

[–] [email protected] 19 points 3 weeks ago* (last edited 3 weeks ago) (1 children)

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
[–] [email protected] 8 points 3 weeks ago (1 children)

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

Like, for example, 1?

[–] prime_number_314159 3 points 3 weeks ago (1 children)

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.

[–] [email protected] 5 points 3 weeks ago (1 children)

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

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[–] tipicaldik 13 points 3 weeks 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.

[–] Korne127 24 points 3 weeks 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.

[–] [email protected] 19 points 3 weeks ago

I believe that's the proper way to do it.

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[–] [email protected] 38 points 3 weeks ago (2 children)

Zero people in this post get the YanDev reference

[–] [email protected] 8 points 3 weeks ago (1 children)

so nobody actually really got the joke. very sad Moment.

[–] [email protected] 9 points 3 weeks ago

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.

[–] [email protected] 3 points 3 weeks ago
[–] [email protected] 25 points 3 weeks ago (1 children)

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

[–] [email protected] 9 points 3 weeks ago

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

[–] affiliate 22 points 3 weeks ago (1 children)

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)
[–] [email protected] 7 points 3 weeks ago

Gotta love how human readable Python always is!

[–] [email protected] 21 points 3 weeks ago (1 children)

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

[–] [email protected] 11 points 3 weeks ago

Recursion is its own reward

[–] [email protected] 21 points 3 weeks ago (2 children)

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;
}
[–] Sanctus 29 points 3 weeks ago (1 children)

Anything but using modulo I guess

[–] [email protected] 6 points 3 weeks ago

And bit operations (:

[–] [email protected] 13 points 3 weeks ago (1 children)

This makes me happy that I don’t use genai

[–] [email protected] 6 points 3 weeks ago

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.

[–] [email protected] 17 points 3 weeks ago (1 children)

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

In this case still O(n)

[–] Zangoose 6 points 3 weeks ago

Smh this is literally what switch statements are for

[–] [email protected] 15 points 3 weeks ago (2 children)
If number%2 == 0: return("Even")
Else: return("odd") 
[–] [email protected] 4 points 3 weeks ago

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.

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[–] TunaCowboy 14 points 3 weeks ago (1 children)
[–] [email protected] 4 points 3 weeks ago

if (~number & 1)

[–] [email protected] 13 points 3 weeks ago (4 children)

oh of course there is

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

(do take a look at the download stats)

[–] [email protected] 13 points 3 weeks ago* (last edited 3 weeks ago)

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

[–] [email protected] 7 points 3 weeks ago* (last edited 3 weeks ago) (4 children)

Can't you just

If (number % 2 == 0){return true}

[–] [email protected] 11 points 3 weeks ago (1 children)
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[–] [email protected] 6 points 3 weeks ago (3 children)

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.

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[–] [email protected] 5 points 3 weeks ago

yup, which is why I find the download stats truly horrifying

[–] [email protected] 6 points 3 weeks ago* (last edited 3 weeks ago)

"If it's not an npm package it's impossible"

- JS devs, probably

[–] tb_ 5 points 3 weeks ago

That's a lot of downloads

[–] [email protected] 10 points 3 weeks ago

just check the least significant bit smh my head

[–] [email protected] 8 points 3 weeks ago
[–] moistclump 6 points 3 weeks ago

I thought they were going to turn into Saddam Husseins.

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