dow.py

'''
Determination of the day of the week

Jan 1st 1 AD is a Monday in Gregorian calendar.
So Jan 0th 1 AD is a Sunday [It does not exist technically].

Every 4 years we have a leap year. But xy00 cannot be a leap unless xy divides 4 with reminder 0.
y/4 - y/100 + y/400 : this gives the number of leap years from 1AD to the given year. As each year has 365 days (divdes 7 with reminder 1), unless it is a leap year or the date is in Jan or Feb, the day of a given date changes by 1 each year. In other case it increases by 2.
y -= m<3 : If the month is not Jan or Feb, we do not count the 29th Feb (if it exists) of the given year. 
So y + y/4 - y/100 + y/400  gives the day of Jan 0th (Dec 31st of prev year) of the year. (This gives the reminder with 7 of  the number of days passed before the given year began.)

Array t:  Number of days passed before the month 'm+1' begins.

So t[m-1]+d is the number of days passed in year 'y' upto the given date. 
(y + y/4 - y/100 + y/400 + t[m-1] + d) % 7 is reminder of the number of days from Jan 0  1AD to the given date which will be the day (0=Sunday,6=Saturday).

Description credits: Sai Teja Pratap (quora.com/How-does-Tomohiko-Sakamotos-Algorithm-work).
Python translation: Robert L Carr
'''
days = ["Sunday", "Monday", "Tuesday", "Wednesday","Thursday", "Friday", "Saturday"]

def dow(y, m, d):
  t = [0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4]
  y -= m < 3;
  return (y + y/4 - y/100 + y/400 + t[m-1] + d) % 7

days[dow(2013,5,13)]
days[dow(2016,6,14)]