[kseistrup]

I really like the “2053” pattern, it's easy to remember. But I doubt I will be able to remember the rest. I think I'd rather do like this:

Given a year CCXX, we know the doomsday of year CC00 from the “2053“ rule, and then:

    (a, b) = divmod(XX, 12)
    c = b // 4
    offset = a + b + c

Then add the offset to the day found by the “2053” rule, and you have the doomsday of the year.

E.g., for 2020 we have:

    start = 2  # from the 2053 rule
    (a, b) = divmod(20, 12)  # (1, 8)
    c = 8 // 4  # 2
    offset = 1 + 8 + 2 = 4 (mod 7)
    doomsday = 2 + 4 = 6  # saturday