return true
is correct around half of the time
this post was submitted on 31 Oct 2024
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assert IsEven(2) == True
assert IsEven(4) == True
assert IsEven(6) == True
All checks pass. LGTM
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 NoneJust 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?
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(1 replies)
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.
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;
}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.
When you sacrifice memory for an O(1) algorithm.
In this case still O(n)
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
Can't you just
If (number % 2 == 0){return true}
return number % 2 === 0
Yeah, that's even simpler
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.
That would already cause an exception when calling the function because it has int number in the parameters
Javascript doesn’t have strongly-typed variables
yup, which is why I find the download stats truly horrifying
no
ok
"If it's not an npm package it's impossible"
- JS devs, probably
just check the least significant bit smh my head
=if((number/2)-round(number/2,0)=0,true,false)