Annotation of src/usr.bin/cal/README, Revision 1.1.1.1
1.1 deraadt 1: The cal(1) date routines were written from scratch, basically from first
2: principles. The algorithm for calculating the day of week from any
3: Gregorian date was "reverse engineered". This was necessary as most of
4: the documented algorithms have to do with date calculations for other
5: calendars (e.g. julian) and are only accurate when converted to gregorian
6: within a narrow range of dates.
7:
8: 1 Jan 1 is a Saturday because that's what cal says and I couldn't change
9: that even if I was dumb enough to try. From this we can easily calculate
10: the day of week for any date. The algorithm for a zero based day of week:
11:
12: calculate the number of days in all prior years (year-1)*365
13: add the number of leap years (days?) since year 1
14: (not including this year as that is covered later)
15: add the day number within the year
16: this compensates for the non-inclusive leap year
17: calculation
18: if the day in question occurs before the gregorian reformation
19: (3 sep 1752 for our purposes), then simply return
20: (value so far - 1 + SATURDAY's value of 6) modulo 7.
21: if the day in question occurs during the reformation (3 sep 1752
22: to 13 sep 1752 inclusive) return THURSDAY. This is my
23: idea of what happened then. It does not matter much as
24: this program never tries to find day of week for any day
25: that is not the first of a month.
26: otherwise, after the reformation, use the same formula as the
27: days before with the additional step of subtracting the
28: number of days (11) that were adjusted out of the calendar
29: just before taking the modulo.
30:
31: It must be noted that the number of leap years calculation is sensitive
32: to the date for which the leap year is being calculated. A year that occurs
33: before the reformation is determined to be a leap year if its modulo of
34: 4 equals zero. But after the reformation, a year is only a leap year if
35: its modulo of 4 equals zero and its modulo of 100 does not. Of course,
36: there is an exception for these century years. If the modulo of 400 equals
37: zero, then the year is a leap year anyway. This is, in fact, what the
38: gregorian reformation was all about (a bit of error in the old algorithm
39: that caused the calendar to be inaccurate.)
40:
41: Once we have the day in year for the first of the month in question, the
42: rest is trivial.