Lewis Carroll – computing the day of the week for any given date (1887)

JH Conway used a different technique[1] which I have swutch to when computing days in the current year. It's quicker and easier, but I find that it's harder to compute "The Doomsday" for other years (it's Thursday this year), so I revert to my standard method[2].

Example: Today is 2024/05/24

    Years since 2012 is 12
    Leap years is 12/4 = 3
    Magic month number = 2
    Date is 24
    Add mod 7 is (12+3+2+24) = (-2)+3+2+3 = 6 = Friday

  144 : Jan, Feb, Mar
  025 : Apr, May, Jun
  036 : Jul, Aug, Sep
  146 : Oct, Nov, Dec

[2] All values computed mod 7. Take years since 1900. Add number of leap years since 1900 (so (YYYY-1900)/4 rounded down). Add the magic number for the month. Add the day of the month. Then Sunday=1, Wednesday=4, etc.

Because the calendar repeats every 28 years, for more recent dates you can start with 2012 instead of 1900.

From [1]: By the 18th century, the English legal year – used for legal, financial and other civil purposes – had for centuries begun on 25 March, or Lady Day.[13][i] Thus, for example, 24 March 1707 was immediately followed by 25 March 1708, while the day following 31 December 1708 was 1 January 1708, with 1709 still nearly three months away.

Thank goodness that happened.

[1] https://en.m.wikipedia.org/wiki/Calendar_(New_Style)_Act_175...

From the article:

The Month-Item. — If it begins or ends with a vowel, subtract the number, denoting its place in the year, from 10. This, plus its number of days, gives the item for the following month. The item for January is ‘0’; for February or March (the 3rd month), ‘3’; for December (the 12th month), ’12.’ [So, for clarity, the required final numbers after division by 7 are January, 0; February, 3; March, 3; April, 6; May, 1; June, 4; July, 6; August 2; September, 5; October, 0; November, 3; and December, 5.]

My attempt to implement:

For January, the preceding month is December, which neither begins nor ends with a vowel, so I do not subtract 10. The days in December are 31, so 12+31 = 43 mod 7 = 1, not the 0 of the article.

For February, the preceding month is January, which ends in a vowel, so I subtract 1 from 10, giving 9, add the days in January 1+31 = 32 mod 7 = 4, not the 3 of the article.

For March, the preceding month is February, which ends in a vowel, so I subtract 2 from 10 giving 8 and add February's 28 days 8+28 = 36 mod 7 = 1, not the 3 of the article. Had I used a leap year 29 for the days in February at this step the result would be 2, which still does not match the article result. Since the article has a constant result for March, it is either always using 28 or always using 29 days in February.

For April, the preceding month is March, which neither begins nor ends in a vowel so I do not subtract it from 10. 31 days in March gives 3+31 = 34 mod 7 = 6, which does match the article result.

For May, the preceding month is April, which starts with a vowel, so I subtract its position from 10, 10-4 = 6, add April's 30 days 6+30 = 36 mod 7 = 1 which does match the article.

For June, the preceding month is May, ending in a vowel, so I subtract May's 5 from 10, and adding May's 31 days gives 5+31= 36 mod 7 = 1 which does not match the article.

This should suffice to illustrate my misunderstanding; what have I got wrong?

[edit: typos]

Very clever, and obvious once you know.

[0] https://www.ted.com/talks/arthur_benjamin_a_performance_of_m... ... starts at 7:47

The Century-Item:

The year is 2024. The century is 20.

20/4 = 5, with no remainder. Overplus is 0. 3-0 = 3. 3x2 = 6.

The Year-Item:

The year part is 24. Number of dozens: 2

24/12=2. Overplus: 24 mod 12 = 0. Number of 4's in the overplus: 0. 2+0+0 = 2.

The Month-Item: May: 1 (from the text).

The Day-Item: The day is 24.

Adding all items:

6+2+1+24=33. 33 mod 7 = 5, which is Friday.

I'd go as far as saying that in mathematics little other than notation matters and the same is true in computer science.

This is why lisp like languages are the obvious choice when variables are explicit.

He didn't consider himself a "rapid computer".

# Enter a date (yyyy-mm-dd): 2024-05-24

# The day of the week for 2024-05-24 is Friday.

def compute_day_of_week(date):

    day, month, year = date
    century = year // 100
    year_part = year % 100

    # Compute Century-Item
    if year < 1752 or (year == 1752 and (month < 9 or (month == 9 and day < 14))):
        century_item = (18 - century) % 7
    else:
        century_item = ((3 - (century % 4)) * 2) % 7
    
    # Compute Year-Item
    year_item = (year_part + (year_part // 4)) % 7
    
    # Compute Month-Item
    month_items = [0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5]
    month_item = month_items[month - 1]
    
    # Compute Day-Item
    day_item = day % 7
    
    # Total
    total = (century_item + year_item + month_item + day_item) % 7
    
    # Correction for Leap Year
    if month <= 2 and (year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)):
        total = (total - 1 + 7) % 7
    
    return total

    days = ["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"]
    return days[day_index]

date_input = input("Enter a date (yyyy-mm-dd): ")

year, month, day = map(int, date_input.split('-'))

date = (day, month, year)

# Compute and print the day of the week

day_index = compute_day_of_week(date) day_name = day_of_week_string(day_index) print(f"The day of the week for {date_input} is {day_name}.")

We really should just declared the year to be 360 days long, with 12 months, 30 days each, split into 6 day long weeks. The extra 5 days? Everybody take a vacation. Leap days go in there too somehow. Every month starts with a Monday and ends with a Sunday. Wednesday will be removed because it was no good anyway.

Let Y = 10 T + U, i.e., T is the first digit of the two digit year and U is the second digit.

If T is even, the year part is:

  2 T + U       if U = 0, 1, 2, or 3
  2 T + U + 1   if U = 4, 5, 6, or 7
  2 T + U + 2   if U = 8 or 9

  3 + 2 T + U       if U = 0 or 1
  3 + 2 T + U + 1   if U = 2, 3, 4, or 5
  3 + 2 T + U + 2   if U = 6, 7, 8 or 9

  Compute 2 T + U
  If T is even
      Add 1 if T >= 4 and add another 1 if T >= 8
  else
      Add 3
      Add 1 if T >= 2 and add another 1 if T >= 6

Example: for 2024, T=2 and U=4. Mentally my train of thought would be:

  2 doubles to 4, plus 4 = 8 = 1 mod 7, even decade so add 1 for 4 <= T < 8 giving 2.

Example: for 2099, T = 9 and U = 9. Mentally that goes something like this:

  9 = 2 mod 7 doubles to 4, add 9 = 2 mod 7 giving 6
  T was odd so add 3 giving 9 = 2 mod 7
  T was odd so the so there were leap years at 2 and 6, which each add 1
  That gives us 4

With my method you never have to deal with a number above 32, and that's only if you do no reductions mod 7 along the way. If you reduce mod 7 whenever you can you never have to deal with anything above 12.

Note: when reducing mod 7 along the way it is crucial to remember that whether you take the T is even case or the T is odd case depends on the actual value of T. E.g., when I am doing 2099 I reduce both T and U to 2 right away but first note that 9 is odd so will need to take the odd case.

https://en.m.wikipedia.org/wiki/International_Fixed_Calendar