Calculate the day when Date is given as input using the Doomsday algorithm

Question

Was too bored in holidays...here's a code which finds the day of the week using the algorithm in this video. Please give a logical date in the correct format cause I didn't account for improper inputs cause I'm too lazy. My code is short and sweet.

day=["Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"] #some database numbers and strings
mdoom=[3,28,14,4,9,6,11,8,5,10,7,12]
while True:
 entry=input('\nEnter a date in DD/MM/YYYY format:') #Input date in a specified format
 date,month,year = map(int,entry.split('/')) #Get the date,month and year
 centdoom=[3,2,0,5]
 a=centdoom[((year//100)%4)-3]
 b=(year%100)//12
 c=(year%100)%12
 d=c//4 #Random calculations for doomsday :)
 e=a+b+c+d
 if (year%4==0) and ((year%100!=0) or (year%400==0)): #Checks for leap year
 mdoom[0],mdoom[1]=4,29
 ddoom=e%7 #finds a date in the 1st week which has the same day as the doomsday 
 r=mdoom[month-1] #Gives the doomsdate in the input month 
 r=r%7
 r=ddoom-r 
 date=date%7 #finds a date in the 1st week which has the same day as the input date
 print('\n',day[(r+date)%7]) #prints the output
 k=input('\nTry again?(y/n)')
 if k=='y':
 continue
 else:
 break

And any suggestions or improvements are welcome.

• dfhwze: I would slightly improve on leap year (en.wikipedia.org/wiki/Leap_year#Algorithm).
• random_npc: can you please send the edited part of the leap year condition...cause I can't plug in the 2 new conditions according to Wiki in my code.....I'm a complete noob :( ...thanks in advance

Answer 1

Your code is hard to understand, even after watching the video.

1. Make some functions. years_doomsday would help move some of the hard to understand information to be self-contained.
2. If you need to floor divide and get the remainder use divmod.
3. You can check the leap year using calendar.isleap.
4. You have a bug, if you ever enter a leap year then the non-leap years will return incorrect values for January and February.
5. You should make a function that calls years_doomsday, and returns the weekday.
6. You should make centdoom a global constant.
7. By rotating centdoom once you can remove the need for the -3.

import calendar
WEEKDAYS = ["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"]
MDOOMSDAY = [3, 28, 14, 4, 9, 6, 11, 8, 5, 10, 7, 12]
MDOOMSDAY_LEAP = list(MDOOMSDAY)
MDOOMSDAY_LEAP[:2] = [4, 29]
CDOOMSDAY = [2, 0, 5, 3]
def years_doomsday(year):
 a, b = divmod(year, 100)
 c, d = divmod(b, 12)
 e = d // 4
 return (CDOOMSDAY[a % 4] + c + d + e) % 7
def doomsday(year, month, day):
 y_doomsday = years_doomsday(year)
 mdoomsday = MDOOMSDAY_LEAP if calendar.isleap(year) else MDOOMSDAY
 return WEEKDAYS[(day - mdoomsday[month-1] + y_doomsday) % 7]
if __name__ == '__main__':
 tests = [
 (2305, 7, 13, 'Thursday'),
 (1776, 7, 4, 'Thursday'),
 (1969, 7, 20, 'Sunday'),
 (1984, 1, 6, 'Friday'),
 (1902, 10, 19, 'Sunday'),
 ]
 for test in tests:
 if doomsday(*test[:3]) != test[-1]:
 print('Broken for', test)

If you don't want any fun, you can replace your code with datetime.date.isoweekday.

import datetime
for test in tests:
 if WEEKDAYS[datetime.date(*test[:3]).isoweekday()%7] != test[-1]:
 print('Broken for', test)

• Justin: Nice answer! +1
• Maarten Fabré: why not do tests = [(2705, 7, 13), "Thursday",...] or even making it a dict, and then for date, answer in tests:if doormsday(*date) != answer:print(...); and I find your way of assembling MDOOMSDAY_LEAP rather strange
• Peilonrayz: @MaartenFabré Because I find ((1, 2, 3), 'd') is more annoying to type then (1, 2, 3, 'd').
• Maarten Fabré: it is 2 more brackets to type, yes, but a lot more clearer when using the test
• Peilonrayz: @MaartenFabré And I was testing this in IDLE where I kept having to write the same stuff over and over.