The Doomsday algorithm, devised by mathematician J. H. Conway, computes the day of the week any given date fell on. The algorithm is designed to be simple enough to memorize and use for mental calculation.
Example. With the algorithm, we can compute that July 4, 1776 (the day the United States declared independence from Great Britain) was a Thursday.
The algorithm is based on the fact that for any year, several dates always fall on the same day of the week, called the doomsday for the year. These dates include 4/4, 6/6, 8/8, 10/10, and 12/12.
Example. The doomsday for 2016 is Monday, so in 2016 the dates above all fell on Mondays. The doomsday for 2017 is Tuesday, so in 2017 the dates above will all fall on Tuesdays.
The doomsday algorithm has three major steps:
Each step is explained in detail below.
The doomsday for the first year in a century is called the anchor day for that century. The anchor day is needed to compute the doomsday for any other year in that century. The anchor day for a century $c$ can be computed with the formula: $$ a = \bigl( 5 (c \bmod 4) + 2 \bigr) \bmod 7 $$ The result $a$ corresponds to a day of the week, starting with $0$ for Sunday and ending with $6$ for Saturday.
Note. The modulo operation $(x \bmod y)$ finds the remainder after dividing $x$ by $y$. For instance, $12 \bmod 3 = 0$ since the remainder after dividing $12$ by $3$ is $0$. Similarly, $11 \bmod 7 = 4$, since the remainder after dividing $11$ by $7$ is $4$.
Example. Suppose the target year is 1954, so the century is $c = 19$. Plugging this into the formula gives $$a = \bigl( 5 (19 \bmod 4) + 2 \bigr) \bmod 7 = \bigl( 5(3) + 2 \bigr) \bmod 7 = 3.$$ In other words, the anchor day for 1900-1999 is Wednesday, which is also the doomsday for 1900.
Exercise 1.1. Write a function that accepts a year as input and computes the anchor day for that year's century. The modulo operator % and functions in the math module may be useful. Document your function with a docstring and test your function for a few different years. Do this in a new cell below this one.
#Define the function
def anchor(year):
#get c by dividing year/100
c=year/100
#anchor formula
a=(5*(c%4)+2)%7
return a
anchor(2000)
Once the anchor day is known, let $y$ be the last two digits of the target year. Then the doomsday for the target year can be computed with the formula: $$d = \left(y + \left\lfloor\frac{y}{4}\right\rfloor + a\right) \bmod 7$$ The result $d$ corresponds to a day of the week.
Note. The floor operation $\lfloor x \rfloor$ rounds $x$ down to the nearest integer. For instance, $\lfloor 3.1 \rfloor = 3$ and $\lfloor 3.8 \rfloor = 3$.
Example. Again suppose the target year is 1954. Then the anchor day is $a = 3$, and $y = 54$, so the formula gives $$ d = \left(54 + \left\lfloor\frac{54}{4}\right\rfloor + 3\right) \bmod 7 = (54 + 13 + 3) \bmod 7 = 0. $$ Thus the doomsday for 1954 is Sunday.
Exercise 1.2. Write a function that accepts a year as input and computes the doomsday for that year. Your function may need to call the function you wrote in exercise 1.1. Make sure to document and test your function.
def doomsday(year):
y=year%100
d=(y+(y/4)+anchor(year))%7
return d
doomsday(1954)
The final step in the Doomsday algorithm is to count the number of days between the target date and a nearby doomsday, modulo 7. This gives the day of the week.
Every month has at least one doomsday:
Example. Suppose we want to find the day of the week for 7/21/1954. The doomsday for 1954 is Sunday, and a nearby doomsday is 7/11. There are 10 days in July between 7/11 and 7/21. Since $10 \bmod 7 = 3$, the date 7/21/1954 falls 3 days after a Sunday, on a Wednesday.
Exercise 1.3. Write a function to determine the day of the week for a given day, month, and year. Be careful of leap years! Your function should return a string such as "Thursday" rather than a number. As usual, document and test your code.
days = ['Sunday', 'Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday']
def day(month, day, year):
if ((year%4 == 0 & year%100 != 0) | year%400 == 0): #account for leap year
leap = [11, 29, 21, 4, 9, 6, 11, 8, 5, 10, 7, 12] #the 'day' value
inbetween = ((day - leap[month - 1])%7) #number of days in between: input day minus leap doomsday value
else:
nonleap = [10, 28, 21, 4, 9, 6, 11, 8, 5, 10, 7, 12]
inbetween = ((day - nonleap[month - 1])%7)
#finding the day of the week for the date
the_doomsday = (doomsday(year) + inbetween)%7
return days[the_doomsday] #corresponding day value
day(1,23,2017)
Exercise 1.4. How many times did Friday the 13th occur in the years 1900-1999? Does this number seem to be similar to other centuries?
def fri13(cen1, cen2):
ans=sum([sum([1 for i in range(1, 13)
if day(i, 13, j)=="Friday"]) for j in range(cen1, cen2)])
return ans
print fri13(500,600)
print fri13(1700, 1800)
print fri13(1900, 2000)
Friday the 13th occured 173 times, which is fairly similar to other centuries.
Exercise 1.5. How many times did Friday the 13th occur between the year 2000 and today?
fri13_2017= sum([sum([1 for i in range(1,2) #for this year
if day(i, 13, j)=="Friday"]) for j in range(2017,2018)])
fri13_2017 #add this
print(fri13(2000,2017)+fri13_2017)
There have been 28 Friday the 13ths since 2000.
Exercise 2.1. The file birthdays.txt contains the number of births in the United States for each day in 1978. Inspect the file to determine the format. Note that columns are separated by the tab character, which can be entered in Python as \t. Write a function that uses iterators and list comprehensions with the string methods split() and strip() to convert each line of data to the list format
[month, day, year, count]
The elements of this list should be integers, not strings. The function read_birthdays provided below will help you load the file.
def read_birthdays(file_path):
"""Read the contents of the birthdays file into a string.
Arguments:
file_path (string): The path to the birthdays file.
Returns:
string: The contents of the birthdays file.
"""
with open(file_path) as file:
return file.read()
import re
b=read_birthdays("birthdays.txt")
birthdays=[x.replace("\t","/").split("/") for x in b.split("\n")[6:]][:365]
final=map( lambda k: map(int, k), birthdays)
final
Exercise 2.2. Which month had the most births in 1978? Which day of the week had the most births? Which day of the week had the fewest? What conclusions can you draw? You may find the Counter class in the collections module useful.
months = Counter()
for date in final:
months[date[0]] += date[3]
print months
August had 302,795 births, which makes it the highest birthing month in 1978.
weeks = {}
for i, date in enumerate(final):
if i % 7 not in weeks:
weeks[i%7] = date[3]
else:
weeks[i%7] += date[3]
print weeks
Tuesday was the most popular birthing day with 504,858 births, and Sunday was the least popular with 421,400. There are less births on weekends than weekdays. Friday is the second most popular day.
Exercise 2.3. What would be an effective way to present the information in exercise 2.2? You don't need to write any code for this exercise, just discuss what you would do.
We could graph this information in order to display the counts for each category, having a separate graph for months and days of the week. It could also be sorted and analyzed by way of a table, but a bar chart would probably be the best visual representation.
Notes: Collaborated with fellow 141 students: Sierra Tevlin, Hannah Kosinovsky, Chad Pickering, Ricky Safran, and a few others in passing. Used materials from stackoverflow and other sources. Here are a few: https://docs.python.org/2/library/collections.html http://stackoverflow.com/questions/7368789/convert-all-strings-in-a-list-to-int http://pyladiespdx.github.io/listcomps/ There were certainly more and can be retrieved if necessary from my browser history.