Thanks to goodness, finally. A (giggle & snort) solid algorithm. There ya’s go set yer clocks & go get a haircut.

This is why this code is good. Opens MS paint. When I worked at Blizzard-

Good if you are rated by an AI that pays for LOCs.

this is like the making chess one

no unit tests huh.

/s

private?

This code would run a lot faster as a hash table look up.

def is_even(n: int) -> bool:
    if n < 0:
        return is_even(-n)
    r = True
    for _ in range(n):
        r = not r
    return r

Y'all laugh but this man has amazing code coverage numbers.

Can you imagine being a TA and having to grade somebody's hw and you get this first thing? lmao

Plot twist: they used a script to generate that code.

This is YandereDev levels of bad.

pro hacker tip: you can optimize this by using "num" for the variable name instead of "number"

def even(n: int) -> bool:
    code = ""
    for i in range(0, n+1, 2):
        code += f"if {n} == {i}:\n out = True\n"
        j = i+1
        code += f"if {n} == {j}:\n out = False\n"
    local_vars = {}
    exec(code, {}, local_vars)
    return local_vars["out"]

scalable version

This joke was not written by the dude pictured. The author wrote a book of funny code jokes.

You don't get it, it runs on a smart fridge so there's no reason to change it

That code is so wrong. We're talking about Jason "Thor" Hall here—that function should be returning 1 and 0, not booleans.

To be fair, the question is "Write a function that simultaneously determines if the number is even and works as a timer"

I hope that the language's ints are at most 32 bits. For 8 bits it could even be written by hand & the source code for a 32 bit version would only take up avg_line_len * 4GiB space for the source code of the function. But it might take a bit of time to compile a version that supports the full range of 64 or 128 bit ints.

I'll join in

const isEven = (n) 
  => !["1","3","5","7","9"]
    .includes(Math.round(n).toString().slice(-1)) 

ftfy

bool IsEven(int number) {
  return !IsOdd(number);
}

bool IsOdd(int number) {
  return !IsEven(number);
}