this post was submitted on 31 Oct 2024
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[–] [email protected] 74 points 4 days ago (2 children)

return true

is correct around half of the time

[–] [email protected] 39 points 4 days ago
assert IsEven(2) == True
assert IsEven(4) == True
assert IsEven(6) == True

All checks pass. LGTM

[–] aliser 18 points 4 days ago (2 children)
return Math.random() > 0.5

would also be correct about half the time

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[–] [email protected] 37 points 4 days ago (2 children)

Zero people in this post get the YanDev reference

[–] [email protected] 2 points 2 days ago
[–] [email protected] 7 points 4 days ago (1 children)

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

[–] [email protected] 8 points 4 days 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] 48 points 4 days ago* (last edited 4 days ago) (2 children)
import re

def is_even(i: int) -> bool:
    return re.match(r"-?\d*[02468]$", str(i)) is not None
[–] [email protected] 20 points 4 days ago
[–] [email protected] 5 points 4 days ago

i was gonna suggest the classic

re.match(r"^(..)\1*$", "0" * abs(i)) is not None
[–] [email protected] 25 points 4 days ago (1 children)

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

[–] [email protected] 9 points 4 days ago

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

[–] [email protected] 21 points 4 days 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 4 days ago

Recursion is its own reward

[–] [email protected] 42 points 4 days ago (3 children)

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

[–] [email protected] 19 points 4 days ago* (last edited 4 days 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 4 days 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 4 days 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 4 days 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 4 days 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 23 points 4 days 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 4 days ago

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

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[–] affiliate 22 points 4 days 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 days ago

Gotta love how human readable Python always is!

[–] [email protected] 21 points 4 days 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 4 days ago (1 children)

Anything but using modulo I guess

[–] [email protected] 6 points 4 days ago

And bit operations (:

[–] [email protected] 13 points 4 days ago (1 children)

This makes me happy that I don’t use genai

[–] [email protected] 6 points 4 days 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] 13 points 4 days 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 4 days ago* (last edited 4 days 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 4 days ago* (last edited 4 days ago) (4 children)

Can't you just

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

[–] [email protected] 11 points 4 days ago (1 children)
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[–] [email protected] 6 points 4 days ago (2 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 4 days ago

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

[–] [email protected] 6 points 4 days ago* (last edited 4 days ago)

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

- JS devs, probably

[–] tb_ 5 points 4 days ago

That's a lot of downloads

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

if (~number & 1)

[–] [email protected] 17 points 4 days ago (1 children)

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

In this case still O(n)

[–] Zangoose 6 points 4 days ago

Smh this is literally what switch statements are for

[–] [email protected] 15 points 4 days ago (2 children)
If number%2 == 0: return("Even")
Else: return("odd") 
[–] [email protected] 4 points 4 days 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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[–] [email protected] 10 points 4 days ago

just check the least significant bit smh my head

[–] [email protected] 8 points 4 days ago
[–] moistclump 6 points 4 days ago

I thought they were going to turn into Saddam Husseins.

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