**Int -> Bool**and indicate whether a year is a leap year (under Gregorian rules, ignoring when various locales made the switch to keep it simple).

Some submissions are from people not yet comfortable with assigning/returning Booleans other than the constants

**True**or

**False**, or maybe not comfortable with the lazy nature of Python's

**and**and

**or**. Their versions are full of

**if**statements. I bashed out

and it worked fine... but then I fell in. How aboutdef is_leap_year(year): return year % (400 if year % 100 == 0 else 4) == 0

Nah... no need to make the next guy, who could be me, do the prime factorization of 100 again and figure out how that ties to the leap year rules? OK, here's one way to look at it...return year % 16 == 0 if year % 25 == 0 else year % 4 == 0

return (year // 100 if year % 100 == 0 else year) % 4 == 0

*i.e.*on century boundaries, the century has to be divisible by four, otherwise the year has to be. Well, that's one way to look at it, but one that doesn't go with the definition as clearly as one would like...

...and then the path of premature optimization that Knuth warns against beckoned.

The idea: three fourths of numbers get ruled out by a check for divisibility by four, which can be done cheaply if the compiler or interpreter notice. (That works for any power of two, not just four, which with another nod to factorization means one could writereturn year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)

which means, if you're lucky, only doing one divisibility check with an actual divide.)return year % 4 == 0 and (year % 100 != 0 or year % 16 == 0)

Final decision? Leave it alone