Determination of the day of the week

Every seventh day in a month has the same name as the previous:

"Corresponding months" are those months within the calendar year that start on the same day of the week. For example, September and December correspond, because September 1 falls on the same day as December 1 (as there are precisely thirteen 7-day weeks between the two dates). Months can only correspond if the number of days between their first days is divisible by 7, or in other words, if their first days are a whole number of weeks apart. For example, February of a common year corresponds to March because February has 28 days, a number divisible by 7, 28 days being exactly four weeks. In a leap year, January and February correspond to different months than in a common year, since adding February 29 means each subsequent month starts a day later.

January corresponds to October in common years and April and July in leap years. February corresponds to March and November in common years and August in leap years. March always corresponds to November, April always corresponds to July, and September always corresponds to December. August does not correspond to any other month in a common year. October doesn't correspond to any other month in a leap year. May and June never correspond to any other month.

In the months table below, corresponding months have the same number, a fact which follows directly from the definition.

There are seven possible days that a year can start on, and leap years will alter the day of the week after February 29. This means that there are 14 configurations that a year can have. All the configurations can be referenced by a dominical letter, but as February 29 has no letter allocated to it a leap year has two dominical letters, one for January and February and the other (one step back in the alphabetical sequence) for March to December.

For example, 2019 was a common year starting on Tuesday, indicating that the year as a whole corresponded to the 2013 calendar year. On the other hand, 2020 was a leap year starting on Wednesday which, on the whole, corresponds to the 1992 calendar year; specifically, its first 2 months, except for February 29, correspond to those of the 2014 calendar year, while, due to the 2020 leap day, its subsequent 10 months correspond to the 2015 calendar year.

Moreover:

For details see the table below.

Notes:

• Black means all the months of Common Year
• Red means the first 2 months of Leap Year
• Blue means the last 10 months of Leap Year

"Year 000" is, in normal chronology, the year 1 BC (which precedes AD 1). In astronomical year numbering the year 0 comes between 1 BC and AD 1. In the proleptic Julian calendar, (that is, the Julian calendar as it would have been if it had been operated correctly from the start), 1 BC starts on Thursday. In the proleptic Gregorian calendar, (so called because it wasn't devised until 1582), 1 BC starts on Saturday.

For Julian dates before 1300 and after 1999 the year in the table which differs by an exact multiple of 700 years should be used. For Gregorian dates after 2299, the year in the table which differs by an exact multiple of 400 years should be used. The values "r0" through "r6" indicate the remainder when the Hundreds value is divided by 7 and 4 respectively, indicating how the series extend in either direction. Both Julian and Gregorian values are shown 1500-1999 for convenience. Bold figures (e.g., 04) denote leap year. If a year ends in 00 and its hundreds are in bold it is a leap year. Thus 19 indicates that 1900 is not a Gregorian leap year, (but 19 in the Julian column indicates that it is a Julian leap year, as are all Julian x00 years). 20 indicates that 2000 is a leap year. Use Jan and Feb only in leap years.

For determination of the day of the week (1 January 2000, Saturday)

• the day of the month: 1 ~ 31 (1)
• the month: (6)
• the year: (0)
• the century mod 4 for the Gregorian calendar and mod 7 for the Julian calendar (0).
• adding 1+6+0+0=7. Dividing by 7 leaves a remainder of 0, so the day of the week is Saturday.

The formula is w = (d + m + y + c) mod 7.

Note that the date (and hence the day of the week) in the Revised Julian and Gregorian calendars is the same from 14 October 1923 to 28 February AD 2800 inclusive and that for large years it may be possible to subtract 6300 or a multiple thereof before starting so as to reach a year which is within or closer to the table.

To look up the weekday of any date for any year using the table, subtract 100 from the year, divide the difference by 100, multiply the resulting quotient (omitting fractions) by seven and divide the product by nine. Note the quotient (omitting fractions). Enter the table with the Julian year, and just before the final division add 50 and subtract the quotient noted above.

Example: What is the day of the week of 27 January 8315?

8315-6300=2015, 2015-100=1915, 1915/100=19 remainder 15, 19x7=133, 133/9=14 remainder 7. 2015 is 700 years ahead of 1315, so 1315 is used. From table: for hundreds (13): 6. For remaining digits (15): 4. For month (January): 0. For date (27): 27. 6+4+0+27+50-14=73. 73/7=10 remainder 3. Day of week = Tuesday.

To find the Dominical Letter, calculate the day of the week for either 1 January or 1 October. If it is Sunday, the Dominical Letter is A, if Saturday B, and similarly backwards through the week and forwards through the alphabet to Monday, which is G.

Leap years have two Sunday Letters, so for January and February calculate the day of the week for 1 January and for March to December calculate the day of the week for 1 October.

Leap years are all years which divide exactly by four with the following exceptions:

In the Gregorian calendar - all years which divide exactly by 100 (other than those which divide exactly by 400).

In the Revised Julian calendar - all years which divide exactly by 100 (other than those which give remainder 200 or 600 when divided by 900).

This is an artefact of recreational mathematics. See doomsday rule for an explanation.

Use this table for finding the day of the week without any calculations.

Examples:

• For common method

December 26, 1893 (GD)

December is in row F and 26 is in column E, so the letter for the date is C located in row F and column E. 93 (year mod 100) is in row D (year row) and the letter C in the year row is located in column G. 18 ([year/100] in the Gregorian century column) is in row C (century row) and the letter in the century row and column G is B, so the day of the week is Tuesday.

October 13, 1307 (JD)

October 13 is a F day. The letter F in the year row (07) is located in column G. The letter in the century row (13) and column G is E, so the day of the week is Friday.

January 1, 2000 (GD)

January 1 corresponds to G, G in the year row (00) corresponds to F in the century row (20), and F corresponds to Saturday.

A pithy formula for the method: "Date letter (G), letter (G) is in year row (00) for the letter (F) in century row (20), and for the day, the letter (F) become weekday (Saturday)".

The Sunday Letter method

Each day of the year (other than 29 February) has a letter allocated to it in the recurring sequence ABCDEFG. The series begins with A on 1 January and continues to A again on 31 December. The Sunday letter is the one which stands against all the Sundays in the year. Since 29 February has no letter, this means that the Sunday Letter for March to December is one step back in the sequence compared to that for January and February. The letter for any date will be found where the row containing the month (in black) at the left of the "Latin square" meets the column containing the date above the "Latin square". The Sunday letter will be found where the column containing the century (below the "Latin square") meets the row containing the year's last two digits to the right of the "Latin square". For a leap year, the Sunday letter thus found is the one which applies to March to December.

So, for example, to find the weekday of 16 June 2020:

Column "20" meets row "20" at "D". Row "June" meets column "16" at "F". As F is two letters on from D, so the weekday is two days on from Sunday, i.e. Tuesday.

In a handwritten note in a collection of astronomical tables, Carl Friedrich Gauss described a method for calculating the day of the week for 1 January in any given year.[8] He never published it. It was finally included in his collected works in 1927.[9]

Gauss' method was applicable to the Gregorian calendar. He numbered the weekdays from 0 to 6 starting with Sunday. He defined the following operation: The weekday of 1 January in year number A is[8]

${\displaystyle R(1+5R(A-1,4)+4R(A-1,100)+6R(A-1,400),7)}$

or

${\displaystyle R(1+5R(Y-1,4)+3(Y-1)+5R(C,4),7)}$

from which a method for the Julian calendar can be derived

${\displaystyle R(6+5R(A-1,4)+3(A-1),7)}$

or

${\displaystyle R(6+5R(Y-1,4)+3(Y-1)+6C,7)}$

where ${\displaystyle R(y,m)}$ is the remainder after division of y by m,[9] or y modulo m, and Y + 100C = A.

For year number 2000, A - 1 = 1999, Y - 1 = 99 and C = 19, the weekday of 1 January is

${\displaystyle {\begin{aligned}&=R(1+5R(1999,4)+4R(1999,100)+6R(1999,400),7)\\&=R(1+1+4+0,7)\\&=6\end{aligned}}}$

${\displaystyle {\begin{aligned}&=R(1+5R(99,4)+3\times 99+5R(19,4),7)\\&=R(1+1+3+1,7)\\&=6=Saturday.\end{aligned}}}$

The weekday of the last day in year number A - 1 or 0 January in year number A is

${\displaystyle R(5R(A-1,4)+4R(A-1,100)+6R(A-1,400),7)}$

The weekday of 0 (a common year) or 1 (a leap year) January in year number A is

${\displaystyle R(6+5R(A,4)+4R(A,100)+6R(A,400),7)}$

In order to determine the week day of an arbitrary date, we will use the following lookup table.

Months	11 Jan	12 Feb	1 Mar	2 Apr	3 May	4 Jun	5 Jul	6 Aug	7 Sep	8 Oct	9 Nov	10 Dec	M
Common years	0	3	3	6	1	4	6	2	5	0	3	5	m
Leap years			4	0	2	5	0	3	6	1	4	6
Algorithm	${\displaystyle m=R(\left\lceil 2.6M\right\rceil ,7)}$

Note: minus 1 if M is 11 or 12 and plus 1 if M less than 11 in a leap year.

The day of the week for any day in year number A is

${\displaystyle R(D+m+5R(A-1,4)+4R(A-1,100)+6R(A-1,400),7)}$

or

${\displaystyle R(6+D+\left\lceil 2.6M\right\rceil +5R(A,4)+4R(A,100)+6R(A,400),7)}$

where D is the day of the month and A - 1 for Jan or Feb.

The weekdays for 30 April 1777 and 23 February 1855 are

${\displaystyle {\begin{aligned}&=R(30+6+5R(1776,4)+4R(1776,100)+6R(1776,400),7)\\&=R(2+6+0+3+6,7)\\&=3=Wednesday\end{aligned}}}$

and

${\displaystyle {\begin{aligned}&=R(6+23+\left\lceil 2.6\times 12\right\rceil +5R(1854,4)+4R(1854,100)+6R(1854,400),7)\\&=R(6+2+4+3+6+5,7)\\&=5=Friday.\end{aligned}}}$

This formula was also converted into graphical and tabular methods for calculating any day of the week by Kraitchik and Schwerdtfeger.[9][10]

Another variation of the above algorithm likewise works with no lookup tables. A slight disadvantage is the unusual month and year counting convention. The formula is

${\displaystyle w=\left(d+\lfloor 2.6m-0.2\rfloor +y+\left\lfloor {\frac {y}{4}}\right\rfloor +\left\lfloor {\frac {c}{4}}\right\rfloor -2c\right){\bmod {7}},}$

where

• Y is the year minus 1 for January or February, and the year for any other month
• y is the last 2 digits of Y
• c is the first 2 digits of Y
• d is the day of the month (1 to 31)
• m is the shifted month (March=1,...,February=12)
• w is the day of week (0=Sunday,...,6=Saturday). If w is negative you have to add 7 to it.

For example, January 1, 2000. (year - 1 for January)

Note: The first is only for a 00 leap year and the second is for any 00 years.

The term ${\displaystyle \lfloor 2.6m-0.2\rfloor \mod 7}$ gives the values of months: m

The term ${\displaystyle y+\left\lfloor {\frac {y}{4}}\right\rfloor \mod 7}$ gives the values of years: y

The term ${\displaystyle \left(\left\lfloor {\frac {c}{4}}\right\rfloor -2c\right)\mod 7}$gives the values of centuries: c

Now from the general formula: ${\displaystyle w=d+m+y+c{\bmod {7}}}$; January 1, 2000 can be recalculated as follows:

In Zeller's algorithm, the months are numbered from 3 for March to 14 for February. The year is assumed to begin in March; this means, for example, that January 1995 is to be treated as month 13 of 1994.[11] The formula for the Gregorian calendar is

${\displaystyle w\equiv \left(\left\lfloor {\frac {13(m+1)}{5}}\right\rfloor +\left\lfloor {\frac {y}{4}}\right\rfloor +\left\lfloor {\frac {c}{4}}\right\rfloor +d+y-2c\right){\bmod {7}}}$

where

• Y is the year minus 1 for January or February, and the year for any other month
• y is the last 2 digits of Y
• c is the first 2 digits of Y
• d is the day of the month (1 to 31)
• m is the shifted month (March=3,...January = 13, February=14)
• w is the day of week (1=Sunday,..0=Saturday)

The only difference is one between Zeller's algorithm (Z) and the Gaussian algorithm (G), that is Z - G = 1 = Sunday.

${\displaystyle (d+\lfloor (m+1)2.6\rfloor +y+\lfloor y/4\rfloor +\lfloor c/4\rfloor -2c){\bmod {7}}-(d+\lfloor 2.6m-0.2\rfloor +y+\lfloor y/4\rfloor +\lfloor c/4\rfloor -2c){\bmod {7}}}$

${\displaystyle =(\lfloor (m+2+1)2.6-(2.6m-0.2)\rfloor ){\bmod {7}}}$ (March = 3 in Z but March = 1 in G)

${\displaystyle =(\lfloor 2.6m+7.8-2.6m+0.2\rfloor ){\bmod {7}}}$

${\displaystyle =8{\bmod {7}}=1}$

So we can get the values of months from those for the Gaussian algorithm by adding one:

Wang's algorithm [12] for the Gregorian calendar is (the formula should be subtracted by 1 if m is 1 or 2 and the year is a leap year)

${\displaystyle w=\left(d-d_{0}(m)+y_{0}-y_{1}+\left\lfloor y_{0}/4-y_{1}/2\right\rfloor -2\left(c{\bmod {4}}\right)\right){\bmod {7}},}$

where

• ${\displaystyle y_{0}}$ is the last digit of the year
• ${\displaystyle y_{1}}$ is the last second digit of the year
• ${\displaystyle c}$ is the first 2 digits of the year
• ${\displaystyle d}$ is the day of the month (1 to 31)
• ${\displaystyle m}$ is the month (January=1,...December=12)
• ${\displaystyle w}$ is the day of the week (0=Sunday,..6=Saturday)
• ${\displaystyle d_{0}(m)}$ is the null-days function with values listed in the following table

m	${\displaystyle d_{0}(m)}$
1	1	A day
3	5	m + 2
5	7
7	9
9	3	m + 1
11	12
2	12	m + 3
4	2	m - 2
6	4
8	6
10	8
12	10

An algorithm for the Julian calendar can be derived from the algorithm above

${\displaystyle w=\left(d-d_{0}(m)+y_{0}-y_{1}+\left\lfloor y_{0}/4-y_{1}/2\right\rfloor -c\right){\bmod {7}},}$

where ${\displaystyle d_{0}(m)}$ is a doomsday.

m	${\displaystyle d_{0}(m)}$
1	3	C day
3	7	m + 4
5	9
7	11
9	5	m - 4
11	7
2	0	m - 2
4	4	m
6	6
8	8
10	10
12	12

In a partly tabular method by Schwerdtfeger, the year is split into the century and the two digit year within the century. The approach depends on the month. For m ≥ 3,

${\displaystyle c=\left\lfloor {\frac {y}{100}}\right\rfloor \quad {\text{and}}\quad g=y-100c,}$

so g is between 0 and 99. For m = 1,2,

${\displaystyle c=\left\lfloor {\frac {y-1}{100}}\right\rfloor \quad {\text{and}}\quad g=y-1-100c.}$

The formula for the day of the week is[9]

${\displaystyle w=\left(d+e+f+g+\left\lfloor {\frac {g}{4}}\right\rfloor \right){\bmod {7}},}$

where the positive modulus is chosen.[9]

The value of e is obtained from the following table:

The value of f is obtained from the following table, which depends on the calendar. For the Gregorian calendar,[9]

For the Julian calendar,[9]

Charles Lutwidge Dodgson (Lewis Carroll) devised a method resembling a puzzle, yet partly tabular in using the same index numbers for the months as in the "Complete table: Julian and Gregorian calendars" above. He lists the same three adjustments for the first three months of non-leap years, one 7 higher for the last, and gives cryptic instructions for finding the rest; his adjustments for centuries are to be determined using formulas similar to those for the centuries table. Although explicit in asserting that his method also works for Old Style dates, his example reproduced below to determine that "1676, February 23" is a Wednesday only works on a Julian calendar which starts the year on January 1, instead of March 25 as on the "Old Style" Julian calendar.

Algorithm:[13]

Take the given date in 4 portions, viz. the number of centuries, the number of years over, the month, the day of the month.

Compute the following 4 items, adding each, when found, to the total of the previous items. When an item or total exceeds 7, divide by 7, and keep the remainder only.

Century-item: For 'Old Style' (which ended 2 September 1752) subtract from 18. For 'New Style' (which began 14 September 1752) divide by 4, take overplus from 3, multiply remainder by 2.

Year-item: Add together the number of dozens, the overplus, and the number of 4s in the overplus.

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, "3"; for December, "12".

Day-item: The total, thus reached, must be corrected, by deducting "1" (first adding 7, if the total be "0"), if the date be January or February in a leap year, remembering that every year, divisible by 4, is a Leap Year, excepting only the century-years, in `New Style', when the number of centuries is not so divisible (e.g. 1800).

The final result gives the day of the week, "0" meaning Sunday, "1" Monday, and so on.

Examples:[13]

1783, September 18

17, divided by 4, leaves "1" over; 1 from 3 gives "2"; twice 2 is "4". 83 is 6 dozen and 11, giving 17; plus 2 gives 19, i.e. (dividing by 7) "5". Total 9, i.e. "2" The item for August is "8 from 10", i.e. "2"; so, for September, it is "2 plus 31", i.e. "5" Total 7, i.e. "0", which goes out. 18 gives "4". Answer, "Thursday".

1676, February 23

16 from 18 gives "2" 76 is 6 dozen and 4, giving 10; plus 1 gives 11, i.e. "4". Total "6" The item for February is "3". Total 9, i.e. "2" 23 gives "2". Total "4" Correction for Leap Year gives "3". Answer, "Wednesday".

Since 23 February 1676 (counting February as the second month) is, for Carroll, the same day as Gregorian 4 March 1676, he fails to arrive at the correct answer, namely "Friday," for an Old Style date that on the Gregorian calendar is the same day as 5 March 1677. Had he correctly assumed the year to begin on the 25th of March, his method would have accounted for differing year numbers - just like George Washington's birthday differs - between the two calendars.

It is noteworthy that those who have republished Carroll's method have failed to point out his error, most notably Martin Gardner.[14]

In 1752, the British Empire abandoned its use of the Old Style Julian calendar upon adopting the Gregorian calendar, which has become today's standard in most countries of the world. For more background, see Old Style and New Style dates.

In the C language expressions below, y, m and d are, respectively, integer variables representing the year (e.g., 1988), month (1-12) and day of the month (1-31).

        (d+=m<3?y--:y-2,23*m/9+d+4+y/4-y/100+y/400)%7  

In 1990, Michael Keith and Tom Craver published the foregoing expression that seeks to minimise the number of keystrokes needed to enter a self-contained function for converting a Gregorian date into a numerical day of the week.[15] It preserves neither y nor d, and returns 0 = Sunday, 1 = Monday, etc.

Shortly afterwards, Hans Lachman streamlined their algorithm for ease of use on low-end devices. As designed originally for four-function calculators, his method needs fewer keypad entries by limiting its range either to A.D. 1905–2099, or to historical Julian dates. It was later modified to convert any Gregorian date, even on an abacus. On Motorola 68000-based devices, there is similarly less need of either processor registers or opcodes, depending on the intended design objective.[16]

The tabular forerunner to Tøndering's algorithm is embodied in the following K&R C function.[17] With minor changes, it was adapted for other high level programming languages such as APL2.[18] Posted by Tomohiko Sakamoto on the comp.lang.c Usenet newsgroup in 1992, it is accurate for any Gregorian date.[19][20]

    dayofweek(y, m, d)	/* 1 <= m <= 12,  y > 1752 (in the U.K.) */
    {
        static int t[] = {0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4};
        if( m < 3 )
        {
          y -= 1;
        }
        return (y + y/4 - y/100 + y/400 + t[m-1] + d) % 7;
    }

The function does not always preserve y, and returns 0 = Sunday, 1 = Monday, etc. In contrast, the following expression

    dow(m,d,y) { y-=m<3; return(y+y/4-y/100+y/400+"-bed=pen+mad."[m]+d)%7; }

posted simultaneously by Sakamoto is not easily adaptable to other languages, and may even fail if compiled on a computer that encodes characters using other than standard ASCII values (e.g. EBCDIC), or on C compilers that enforce ANSI C compliance (even on code that is still compliant with the original K&R C specification, where omitted type declarations are assumed to be integer). For the latter consideration alone, Sakamoto's more-verbose version might be considered non-portable, as might also that of Keith and Craver.

IBM's Rata Die method requires that one know the "key day" of the proleptic Gregorian calendar i.e. the day of the week of January 1, AD 1 (its first date). This has to be done to establish the remainder number based on which the day of the week is determined for the latter part of the analysis. By using a given day August 13, 2009 which was a Thursday as a reference, with Base and n being the number of days and weeks it has been since 01/01/0001 to the given day, respectively and k the day into the given week which must be less than 7, Base is expressed as

                      Base = 7n + k       (i)

Knowing that a year divisible by 4 or 400 is a leap year while a year divisible by 100 and not 400 is not a leap year, a software program can be written to find the number of days. The following is a translation into C of IBM's method for its REXX programming language.

int daystotal (int y, int m, int d)
{
	static char daytab[2][13] =
	{
		{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
		{0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}
	};
	int daystotal = d;
	for (int year = 1 ; year <= y ; year++)
	{
		int max_month = ( year < y ? 12 : m-1 );
		int leap = (year%4 == 0);
		if (year%100 == 0 && year%400 != 0)
			leap = 0;
		for (int month = 1 ; month <= max_month ; month++)
		{
			daystotal += daytab[leap][month];
		}
	}
	return daystotal;
}

It is found that daystotal is 733632 from the base date January 1, AD 1. This total number of days can be verified with a simple calculation: There are already 2008 full years since 01/01/0001. The total number of days in 2008 years not counting the leap days is 365 *2008 = 732920 days. Assume that all years divisible by 4 are leap years. Add 2008/4 = 502 to the total; then subtract the 15 leap days because the years which are exactly divisible by 100 but not 400 are not leap. Continue by adding to the new total the number of days in the first seven months of 2009 that have already passed which are 31 + 28 + 31 + 30 + 31 + 30 + 31 = 212 days and the 13 days of August to get Base = 732920 + 502 - 20 + 5 + 212 + 13 = 733632.

What this means is that it has been 733632 days since the base date. Substitute the value of Base into the above equation (i) to get 733632 = 7 *104804 + 4, n = 104804 and k = 4 which implies that August 13, 2009 is the fourth day into the 104805th week since 01/01/0001. 13 August 2009 is Thursday; therefore, the first day of the week must be Monday, and it is concluded that the first day 01/01/0001 of the calendar is Monday. Based on this, the remainder of the ratio Base/7, defined above as k, decides what day of the week it is. If k = 0, it's Monday, k = 1, it's Tuesday, etc.[21]