Algorithm that returns the day of the week for a given date. Works for the Gregorian calendar which began on September 14, 1752 up till December 31, 2299.Based on the description found at http://www.kungfoomanchu.com/z-creations/mentalcalculationofdate.pdf

day_of_week.py

import datetime
import calendar

months = [6, 2, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4]
weekdays = ["Sunday", "Monday", "Tuesday", "Wednesday",
            "Thursday", "Friday", "Saturday", "Sunday"]
centuries = {17: 5,
             18: 3,
             19: 1,
             20: 0,
             21: -2,
             22: -4}

def day_of_week(day, month, year):
    """This algorithm is only valid for the Gregorian calendar,
    which began on September 14, 1752, and up till December 31, 2299"""
    if year <= 1752 and month <= 9 and day < 14:
        raise RuntimeError("This algorithm is only valid for the Gregorian " + \
                           "calendar which began on September 14, 1752.")
    if year >= 2300:
        raise RuntimeError("This algorithm is only valid for the Gregorian " + \
                           "calendar up till December 31, 2299.")


    # Take multiples of 28 from the the last 2 digits of the year
    y = divmod(year, 100)[1] % 28

    # Add a quarter of the nearest multiple of 4 below the number,
    y += divmod(y, 4)[0]

    # Take away 7 or multiples of 7. This leaves us the year code
    y = y % 7

    # The code for the month from the table above
    m = months[month - 1]

    # If it is a leap year AND the month is January or February, subtract 1
    if is_leap_year(year) and month in [1,2]:
        m -= 1

    # Take away 7 or multiples of 7 from the day
    d = day % 7

    # Add the codes for the year, the month and the day
    result = y + m + d

    # Add 1 if the date is in the 1900s
    result += centuries[divmod(year, 100)[0]]

    # Take away 7 or multiples of 7
    result = result % 7

    # The final number indicates day of the week
    return weekdays[result]

def is_leap_year(year):
    # Leap years are the years evenly divisible by 4
    # unless it ends in 00 and is a multiple of 400
    if not year % 400:
        return True
    elif not year % 100:
        return False
    elif not year % 4:
        return True
    return False


# alternate solution using builtin functions
def day_of_week_builtin(day, month, year):
    """ See http://stackoverflow.com/a/14941764/50065 by mgilson"""
    c = calendar.weekday(year, month, day)
    return calendar.day_name[c]

def date_generator(day, month, year):
    """Convenience function that yields (day, month, year) tuples"""
    d = datetime.date(year, month, day)
    while True:
        d += datetime.timedelta(days=1)
        yield d.day, d.month, d.year


if __name__ == '__main__':
    # checking all days from the beginning of the Gregorian
    # calendar till 2300
    d = dategenerator(14, 9, 1752)
    for day, month, year in d:
        if year == 2300:
            break
        assert day_of_week(day, month, year) == day_of_week_builtin(day, month, year)
               
    # checking my date of birth
    print day_of_week(19, 5, 1978)