Comment by hairtuq

The article leaves out optimizing is_leap_year. Here is an interesting version that is correct (in the Proleptic Gregorian calendar) for 0 <= year <= 102499 and takes only 3 instructions:

    ((year * 1073750999) & 3221352463) <= 126976

How this works is surprisingly complex, I am thinking of writing a whole blog post about it...