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Rule (lemmy.blahaj.zone)
submitted 2 weeks ago by Persona3Reload to c/196
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[-] Euphoma@lemmy.ml 81 points 2 weeks ago

return true

is correct around half of the time

[-] ImplyingImplications@lemmy.ca 41 points 2 weeks ago
assert IsEven(2) == True
assert IsEven(4) == True
assert IsEven(6) == True

All checks pass. LGTM

[-] aliser@lemmy.world 18 points 2 weeks ago
return Math.random() > 0.5

would also be correct about half the time

[-] deo@lemmy.dbzer0.com 2 points 2 weeks ago

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

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[-] bob_lemon@feddit.org 52 points 2 weeks ago* (last edited 2 weeks ago)
import re

def is_even(i: int) -> bool:
    return re.match(r"-?\d*[02468]$", str(i)) is not None
[-] lime@feddit.nu 5 points 2 weeks ago

i was gonna suggest the classic

re.match(r"^(..)\1*$", "0" * abs(i)) is not None
[-] superkret@feddit.org 45 points 2 weeks ago

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

[-] fartripper@lemmy.ml 20 points 2 weeks ago* (last edited 2 weeks 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 2 weeks 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 2 weeks 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 2 weeks ago

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

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[-] tipicaldik@lemmy.world 15 points 2 weeks ago

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@lemmy.world 25 points 2 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.

[-] superkret@feddit.org 20 points 2 weeks ago

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

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[-] Rai@lemmy.dbzer0.com 39 points 2 weeks ago

Zero people in this post get the YanDev reference

[-] urheber@discuss.tchncs.de 8 points 2 weeks ago

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

[-] Rai@lemmy.dbzer0.com 10 points 2 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.

[-] RVGamer06@sh.itjust.works 3 points 2 weeks ago
[-] jbk@discuss.tchncs.de 27 points 2 weeks ago

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

[-] beefbot 10 points 2 weeks ago

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

[-] gerryflap@feddit.nl 23 points 2 weeks ago

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

[-] luciferofastora@lemmy.zip 12 points 2 weeks ago

Recursion is its own reward

[-] affiliate@lemmy.world 23 points 2 weeks ago

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)
[-] sepiroth154@feddit.nl 7 points 2 weeks ago

Gotta love how human readable Python always is!

[-] lnxtx@feddit.nl 23 points 2 weeks ago

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@lemmy.world 29 points 2 weeks ago

Anything but using modulo I guess

[-] lnxtx@feddit.nl 6 points 2 weeks ago

And bit operations (:

[-] YtA4QCam2A9j7EfTgHrH@infosec.pub 16 points 2 weeks ago

This makes me happy that I don’t use genai

[-] Mirodir@discuss.tchncs.de 7 points 2 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.

[-] dadarobot@lemmy.sdf.org 18 points 2 weeks ago
If number%2 == 0: return("Even")
Else: return("odd") 
[-] istdaslol@feddit.org 4 points 2 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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[-] istdaslol@feddit.org 18 points 2 weeks ago

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

In this case still O(n)

[-] Zangoose@lemmy.world 6 points 2 weeks ago

Smh this is literally what switch statements are for

[-] TunaCowboy@lemmy.world 15 points 2 weeks ago
[-] FiskFisk33@startrek.website 5 points 2 weeks ago

if (~number & 1)

[-] FiskFisk33@startrek.website 14 points 2 weeks ago

oh of course there is

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

(do take a look at the download stats)

[-] FJW@discuss.tchncs.de 13 points 2 weeks ago* (last edited 2 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

[-] Micromot@feddit.org 8 points 2 weeks ago* (last edited 2 weeks ago)

Can't you just

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

[-] blackn1ght@feddit.uk 12 points 2 weeks ago
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[-] drake@lemmy.sdf.org 6 points 2 weeks ago

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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[-] FiskFisk33@startrek.website 5 points 2 weeks ago

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

[-] servobobo@feddit.nl 7 points 2 weeks ago* (last edited 2 weeks ago)

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

- JS devs, probably

[-] tb_@lemmy.world 5 points 2 weeks ago

That's a lot of downloads

[-] ashestoashes 10 points 2 weeks ago

just check the least significant bit smh my head

[-] themoonisacheese@sh.itjust.works 8 points 2 weeks ago
[-] moistclump@lemmy.world 6 points 2 weeks ago

I thought they were going to turn into Saddam Husseins.

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this post was submitted on 31 Oct 2024
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