FM / util-linux

/usr/share/doc/util-linux

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examples/	-	0755	rm
releases/	-	0755	rm
00-about-docs.txt	260	0644	edit dl rm
AUTHORS.gz	15113	0644	edit dl rm
blkid.txt	3258	0644	edit dl rm
cal.txt	2262	0644	edit dl rm
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Edit: /usr/share/doc/util-linux/cal.txt (2262B)

1 Jan 1 is a Saturday because that's what cal says and I couldn't change
that even if I was dumb enough to try.  From this we can easily calculate
the day of week for any date.  The algorithm for a zero based day of week:

calculate the number of days in all prior years (year-1)*365
	add the number of leap years (days?) since year 1
		(not including this year as that is covered later)
	add the day number within the year
		this compensates for the non-inclusive leap year
		calculation
	if the day in question occurs before the gregorian reformation
		(3 sep 1752 for our purposes), then simply return
		(value so far - 1 + SATURDAY's value of 6) modulo 7.
	if the day in question occurs during the reformation (3 sep 1752
		to 13 sep 1752 inclusive) return THURSDAY. This is my
		idea of what happened then. It does not matter much as
		this program never tries to find day of week for any day
		that is not the first of a month.
	otherwise, after the reformation, use the same formula as the
		days before with the additional step of subtracting the
		number of days (11) that were adjusted out of the calendar
		just before taking the modulo.

It must be noted that the number of leap years calculation is sensitive
to the date for which the leap year is being calculated.  A year that occurs
before the reformation is determined to be a leap year if its modulo of
4 equals zero.  But after the reformation, a year is only a leap year if
its modulo of 4 equals zero and its modulo of 100 does not.  Of course,
there is an exception for these century years.  If the modulo of 400 equals
zero, then the year is a leap year anyway.  This is, in fact, what the
gregorian reformation was all about (a bit of error in the old algorithm
that caused the calendar to be inaccurate.)

Once we have the day in year for the first of the month in question, the
rest is trivial.

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