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From: alopez-o@neumann.uwaterloo.ca (Alex Lopez-Ortiz)
Newsgroups: sci.math,news.answers,sci.answers
Subject: sci.math FAQ: dayofweek
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Date: 17 Feb 2000 22:55:53 GMT
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Archive-name: sci-math-faq/dayofweek
Last-modified: February 20, 1998
Version: 7.5
How to determine the day of the week, given the month, day and year
First a brief explanation: In the Gregorian Calendar, over a period of
four hundred years, there are 97 leap years and 303 normal years. Each
normal year, the day of January 1 advances by one; for each leap year
it advances by two.
303 + 97 + 97 = 497 = 7 * 71
As a result, January 1 year N occurs on the same day of the week as
January 1 year N + 400. Because the leap year pattern also recurs with
a four hundred year cycle, a simple table of four hundred elements,
and single modulus, suffices to determine the day of the week (in the
Gregorian Calendar), and does it much faster than all the other
algorithms proposed. Also, each element takes (in principle) only
three bits; the entire table thus takes only 1200 bits; on many
computers this will be less than the instructions to do all the
complicated calculations proposed for the other algorithms.
Incidental note: Because 7 does not divide 400, January 1 occurs more
frequently on some days than others! Trick your friends! In a cycle of
400 years, January 1 and March 1 occur on the following days with the
following frequencies:
Sun Mon Tue Wed Thu Fri Sat
Jan 1: 58 56 58 57 57 58 56
Mar 1: 58 56 58 56 58 57 57
Of interest is that (contrary to most initial guesses) the occurrence
is not maximally flat.
In the Mathematical Gazette, vol. 53,, pp.127-129, it is shown that
the 13th of the month is more likely to be a Friday than any other
day.The author is a 13 year old S.R.Baxter.
The Gregorian calendar was introduced in 1582 in parts of Europe; it
was adopted in 1752 in Great Britain and its colonies, and on various
dates in other countries. It replaced the Julian Calendar which has a
four-year cycle of leap years; after four years January 1 has advanced
by five days. Since 5 is relatively prime to 7, a table of 4 * 7 = 28
elements is necessary for the Julian Calendar.
There is still a 3 day over 10,000 years error which the Gregorian
calendar does not take into account. At some time such a correction
will have to be done but your software will probably not last that
long!
Here is a standard method suitable for mental computation:
1. Take the last two digits of the year.
2. Divide by 4, discarding any fraction.
3. Add the day of the month.
4. Add the month's key value: JFM AMJ JAS OND 144 025 036 146
5. Subtract 1 for January or February of a leap year.
6. For a Gregorian date, add 0 for 1900's, 6 for 2000's, 4 for
1700's, 2 for 1800's; for other years, add or subtract multiples
of 400.
7. For a Julian date, add 1 for 1700's, and 1 for every additional
century you go back.
8. Add the last two digits of the year.
9. Divide by 7 and take the remainder.
Now 1 is Sunday, the first day of the week, 2 is Monday, and so on.
The following formula, which is for the Gregorian calendar only, may
be more convenient for computer programming. Note that in some
programming languages the remainder operation can yield a negative
result if given a negative operand, so mod 7 may not translate to a
simple remainder.
W = (k + floor(2.6m - 0.2) - 2C + Y + floor(Y/4) + floor(C/4)) mod 7
where floor() denotes the integer floor function,
k is day (1 to 31)
m is month (1 = March, ..., 10 = December, 11 = Jan, 12 = Feb) Treat
Jan & Feb as months of the preceding year
C is century (1987 has C = 19)
Y is year (1987 has Y = 87 except Y = 86 for Jan & Feb)
W is week day (0 = Sunday, ..., 6 = Saturday)
Here the century and 400 year corrections are built into the formula.
The floor(2.6m - 0.2) term relates to the repetitive pattern that the
30-day months show when March is taken as the first month.
The following short C program works for a restricted range, it returns
0 for Monday, 1 for Tuesday, etc.
dow(m,d,y){y-=m<3;return(y+y/4-y/100+y/400+"-bed=pen+mad."[m]+d)%7;}
The program appeared was posted by sakamoto@sm.sony.co.jp (Tomohiko
Sakamoto) on comp.lang.c on March 10th, 1993.
A good mnemonic rule to help on the computation of the day of the week
is as follows. In any given year the following days come on the same
day of the week:
4/4
6/6
8/8
10/10
12/12
to remember the next four, remember that I work from 9-5 at a 7-11 so
9/5
5/9
7/11
11/7
and the last day of Feb.
"In 1995 they come on Tuesday. Every year this advances one other than
leap-years which advance 2. Therefore for 1996 the day will be
Thursday, and for 1997 it will be Friday. Therefore ordinarily every 4
years it advances 5 days. There is a minor correction for the century
since the century is a leap year iff the century is divisible by 4.
Therefore 2000 is a leap year, but 1900, 1800, and 1700 were not."
Even ignoring the pattern over for a period of years this is still
useful since you can generally figure out what day of the week a given
date is on faster than someone else can look it up with a calender if
the calender is not right there. (A useful skill that.)
References
Winning Ways for your mathematical plays. Elwyn R. Berlekamp, John H.
Conway, and Richard K. Guy London ; Toronto : Academic Press, 1982.
Mathematical Carnival. Martin Gardner. New York : Knopf, c1975.
Elementary Number Theory and its applications. Kenneth Rosen. Reading,
Mass. ; Don Mills, Ont. : Addison-Wesley Pub. Co., c1993. p. 156.
Michael Keith and Tom Craver. The Ultimate Perpetual Calendar? Journal
of Recreational Mathematics, 22:4, pp. 280-282, 19
--
Alex Lopez-Ortiz alopez-o@unb.ca
http://www.cs.unb.ca/~alopez-o Assistant Professor
Faculty of Computer Science University of New Brunswick