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d4a28ab0 CW |
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. |