Zeller's Congruence and Five Fr/Sa/Su in October

I saw a comment from one of my Facebook friends that October 2010 had five Fridays, Saturdays, and Sundays and that this occurred only once every 823 years. Now this didn't sound exactly right to me and it didn't take much time to figure out why. Take a look at the following calendar

   Su  Mo  Tu  We  Th  Fr  Sa
                        1   2
    3   4   5   6   7   8   9
   10  11  12  13  14  15  16
   17  18  19  20  21  22  23
   24  25  26  27  28  29  30
   31

You can see pretty easily from this, that a) we'll have the five Fr/Sa/Su anytime we start October on a Friday and b) no other configuration will have this. Now just thinking about this off the top of my head, I figured that this should happen about ever 7 years (disregarding leap years).

Now as it happens, I'd also just finished a short ruby method for Zeller's Congruence that I'd written from an article on Programming Praxis. So, I combined the two and came up with the following ...

# Zeller’s Congruence is a simple mathematical method for determining the day
# of the week for a given date.
# 
# In the 1880s, Christian Zeller, a German mathematician, noticed that, if you
# run a year from March through February, the cumulative number of days in each
# month forms a nearly straight line; that works because February, which would
# normally perturb the straight line, is moved to the end. He worked out the
# formula ⌊(13m−1)/5⌋ to give the number of weekdays that the start of the
# month moves each month, where m is the month number.
# 
# Then it is easy to calculate the day of the week for any given day: add the
# day of the month, the offset for the number of months since March, an offset
# for each year, and additional offsets for leap years and leap centuries
# (remembering to subtract one year for dates in January and February), taking
# the whole thing mod 7. It’s fun to work out the arithmetic yourself, but if
# you don’t want to take the time, the whole formula is shown in the solution.
# 
# Your task is to write a function that uses Zeller’s Congruence to calculate
# the day of the week for any given date. When you are finished, you are
# welcome to read or run a suggested solution, or to post your own solution or
# discuss the exercise in the comments below.#
#
# f = k + floor(13m - 1) / 5) + d + floor(d / 4) + floor(c / 4) - 2c mod 7
#
#   Su  Mo  Tu  We  Th  Fr  Sa
#                        1   2
#    3   4   5   6   7   8   9
#   10  11  12  13  14  15  16
#   17  18  19  20  21  22  23
#   24  25  26  27  28  29  30
#   31
#
#
#
def zeller(year, month, day)
    months = %w[march april may june july august september october november december january february]
    weekdays = %w[Sunday Monday Tuesday Wednesday Thursday Friday Saturday]
    k = day
    m = months.index(month.downcase) + 1
    y = (m <= 10) ? year : year-1
    d = y % 100
    c = y / 100
    f = (k + (((13*m) - 1) / 5).floor + d + (d/4).floor + (c/4).floor - (2*c)) % 7
    weekdays[f]
end

1900.upto(2300) do |y|
    puts "Five Friday, Saturday, and Sundays for October, #{y}" if zeller(y, "october", 1) == "Friday"
end

Here, we'll just go through the years from 1900 to 2300 or about 400 years and see how many times the five Fr/Sa/Su pattern occurs. If we run this and pipe it through wc -l, we end up with 56. Interestingly enough, 400 / 7 is 57 which is what we originally calculated above ignoring the leap years.

Let me know if you have any questions or comments.