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2024年06月4日 12:22

《七夕夜咏》 4/4

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All in one table

Julian calendar	700 1400	100 800 1500	200 900 1600	300 1000 1700	400 1100 1800	500 1200 1900	600 1300 2000	Paschal full moon date Year mod 19
Gregorian calendar	1700 2100		1800 2200		1900 2300	1600 2000		Gregorian (1900–2199)	Julian calendar
00 28 56 84	DC	ED	FE	GF	AG	BA	CB	14 Apr	5 Apr
01 29 57 85	B	C	D	E	F	G	A	3 Apr	25 Mar
02 30 58 86	A	B	C	D	E	F	G	23 Mar	13 Apr
03 31 59 87	G	A	B	C	D	E	F	11 Apr	2 Apr
04 32 60 88	FE	GF	AG	BA	CB	DC	ED	31 Mar	22 Mar
05 33 61 89	D	E	F	G	A	B	C	18 Apr	10 Apr
06 34 62 90	C	D	E	F	G	A	B	8 Apr	30 Mar
07 35 63 91	B	C	D	E	F	G	A	28 Mar	18 Apr
08 36 64 92	AG	BA	CB	DC	ED	FE	GF	16 Apr	7 Apr
09 37 65 93	F	G	A	B	C	D	E	5 Apr	27 Mar
10 38 66 94	E	F	G	A	B	C	D	25 Mar	15 Apr
11 39 67 95	D	E	F	G	A	B	C	13 Apr	4 Apr
12 40 68 96	CB	DC	ED	FE	GF	AG	BA	2 Apr	24 Mar
13 41 69 97	A	B	C	D	E	F	G	22 Mar	12 Apr
14 42 70 98	G	A	B	C	D	E	F	10 Apr	1 Apr
15 43 71 99	F	G	A	B	C	D	E	30 Mar	21 Mar
16 44 72	ED	FE	GF	AG	BA	CB	DC	17 Apr	9 Apr
17 45 73	C	D	E	F	G	A	B	7 Apr	29 Mar
18 46 74	B	C	D	E	F	G	A	27 Mar	17 Apr
19 47 75	A	B	C	D	E	F	G	Month
20 48 76	GF	AG	BA	CB	DC	ED	FE
21 49 77	E	F	G	A	B	C	D
22 50 78	D	E	F	G	A	B	C	Feb Mar Nov
23 51 79	C	D	E	F	G	A	B	Aug
24 52 80	BA	CB	DC	ED	FE	GF	AG	Jan May Oct
25 53 81	G	A	B	C	D	E	F	Apr Jul
26 54 82	F	G	A	B	C	D	E	Sep Dec
27 55 83	E	F	G	A	B	C	D	Jun
Day	1	2	3	4	5	6	7	Table of letters for the days of the year
	8	9	10	11	12	13	14
	15	16	17	18	19	20	21
	22	23	24	25	26	27	28
	29	30	31

Gauss's algorithm

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. He never published it. It was finally included in his collected works in 1927.

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

$R(1+5R(A-1,4)+4R(A-1,100)+6R(A-1,400),7)$

or

$R(1+5R(Y-1,4)+3(Y-1)+5R(C,4),7)$

from which a method for the Julian calendar can be derived

$R(6+5R(A-1,4)+3(A-1),7)$

or

$R(6+5R(Y-1,4)+3(Y-1)+6C,7)$

where $R(y,m)$ is the remainder after division of y by m, or y modulo m, Y = R(A, 100) and C = (A - Y)/100.

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

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

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

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

$R(5R(A-1,4)+4R(A-1,100)+6R(A-1,400),7)$

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

$R(5R(A,4)+4R(A,100)+6R(A,400)-1,7)$

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	$m=R(\left\lceil 2.6M\right\rceil ,7)$

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 mumber A is

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

or

$R(D+\left\lceil 2.6M\right\rceil +5R(A,4)+4R(A,100)+6R(A,400)-1,7)$

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

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

and

${\begin{aligned}&=R(23+\left\lceil 2.6\times 12\right\rceil +5R(1854,4)+4R(1854,100)+6R(1854,400)-1,7)\\&=R(2+4+3+6+5-1,7)\\&=5=Friday.\end{aligned}}$

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

Disparate variation

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

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

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)

${\begin{aligned}w&=\left(1+\lfloor 2.6\cdot 11-0.2\rfloor +(0-1)+\left\lfloor {\frac {0-1}{4}}\right\rfloor +\left\lfloor {\frac {20}{4}}\right\rfloor -2\cdot 20\right){\bmod {7}}\\&=(1+28-1-1+5-40){\bmod {7}}\\&=6={\text{Saturday}}\end{aligned}}$

${\begin{aligned}w&=\left(1+\lfloor 2.6\cdot 11-0.2\rfloor +(100-1)+\left\lfloor {\frac {100-1}{4}}\right\rfloor +\left\lfloor {\frac {20-1}{4}}\right\rfloor -2\cdot (20-1)\right){\bmod {7}}\\&=(1+28+99+24+4-38){\bmod {7}}\\&=6={\text{Saturday}}\end{aligned}}$

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

The term ⌊2.6m − 0.2⌋ mod 7 gives the values of months: m

Months	m
January	0
February	3
March	2
April	5
May	0
June	3
July	5
August	1
September	4
October	6
November	2
December	4

The term y + ⌊y/4⌋ mod 7 gives the values of years: y

y mod 28	y
01 07 12 18 --	1
02–13 19 24	2
03 08 14–25	3
-- 09 15 20 26	4
04 10–21 27	5
05 11 16 22 --	6
06–17 23 00	0

The term ⌊c/4⌋ − 2c mod 7 gives the values of centuries: c

c mod 4	c
1	5
2	3
3	1
0	0

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

${\begin{aligned}w&=1+0+5+0{\bmod {7}}=6={\text{Saturday}}\\d&=1,m=0\\y&=5(0-1{\bmod {2}}8=27)\\c&=0(20{\bmod {4}}=0)\end{aligned}}$

${\begin{aligned}w&=1+0+4+1{\bmod {7}}=6={\text{Saturday}}\\d&=1,m=0\\y&=4(99{\bmod {2}}8=15)\\c&=1(20-1{\bmod {4}}=3)\end{aligned}}$

Lewis Carroll's method

Algorithm:

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.

${\begin{aligned}{\text{Century-item NS}}&=2(3-c{\bmod {4}})\\&=6-2(c{\bmod {4}})\\&=\left(\left\lfloor {\frac {c}{4}}\right\rfloor -2c-1\right){\bmod {7}}\\&={\text{6 4 2 0}}\\{\text{OS}}&=(18-c){\bmod {7}}\\&=(5-c-1){\bmod {7}}\\&={\text{4 3 2 1 0 6 5}}.\end{aligned}}$

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

${\begin{aligned}{\text{Year-item}}&=\left(n+y-12n+\left\lfloor {\frac {y-12n}{4}}\right\rfloor \right){\bmod {7}}\\&=\left(n+y-5n+\left\lfloor {\frac {y}{4}}\right\rfloor -3n\right){\bmod {7}}\\&=\left(y+\left\lfloor {\frac {y}{4}}\right\rfloor \right){\bmod {7}}\\\end{aligned}}$

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".

${\begin{aligned}{\text{Month-item}}&={\text{Jan 0(6) 3(2) 3 6 1 4 6 2 5 0 3 5 Dec}}\\&{\text{January is the first month of the year.}}\\{\text{Month-item}}&=\left\lceil {\frac {13m}{5}}\right\rceil {\bmod {7}}{\text{ (modified)}}\\&={\text{(1 4) Mar 3 6 1 4 6 2 5 0 3 5 1 4 Feb}}\\&{\text{March is the first month of the year.}}\end{aligned}}$

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).

${\begin{aligned}{\text{Day-item}}=d{\bmod {7}}.\end{aligned}}$

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

${\begin{aligned}{\text{The total}}&=({\text{Century-item + Year-item + Month-item + Day-item}}){\bmod {7}}\\&={\text{0 1 2 3 4 5 6}}\\&={\text{Sun Mon Tue Wed Thu Fri Sat.}}\end{aligned}}$

Examples:

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".

${\begin{aligned}c&=17=16+1\\C&=(3-1)\times 2=4\end{aligned}}$

${\begin{aligned}y&=6\times 12+2\times 4+3\\Y&=(6+11+2){\bmod {7}}=5\end{aligned}}$

${\begin{aligned}m&=7\\M&=\left\lceil 2.6\times 7\right\rceil {\bmod {7}}=5\end{aligned}}$

${\begin{aligned}d&=18=14+4\\D&=4\end{aligned}}$

${\begin{aligned}T&=(4+5+5+4){\bmod {7}}=4\\W&=Thu\end{aligned}}$

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".

${\begin{aligned}c&=16\\C&=18-16=2\end{aligned}}$

${\begin{aligned}y&=76=6\times 12+1\times 4\\Y&=(6+4+1){\bmod {7}}=4\end{aligned}}$

${\begin{aligned}m&=12\\M&=\left\lceil 2.6\times 12\right\rceil {\bmod {7}}=4\end{aligned}}$

${\begin{aligned}d&=23=21+2\\D&=2\end{aligned}}$

${\begin{aligned}T&=(2+4+4+2){\bmod {7}}=5\\W&=Fri\end{aligned}}$

ISO week date

Julian centuries (mod 7)	Gregorian centuries (mod 4)	Days of the week							Dates
04 11 18	19 23 27	Sun	Mon	Tue	Wed	Thu	Fri	Sat	Jan	Apri	Jul			01	08	15	22	29
03 10 17		Mon	Tue	Wed	Thu	Fri	Sat	Sun				Sep	Dec	02	09	16	23	30
02 09 16	18 22 26	Tue	Wed	Thu	Fri	Sat	Sun	Mon			Jun			03	10	17	24	31
01 08 15		Wed	Thu	Fri	Sat	Sun	Mon	Tue	Feb	Mar			Nov	04	11	18	25
00 07 14	17 2 25	Thu	Fri	Sat	Sun	Mon	Tue	Wed	Feb			Aug		05	12	19	26
–1 06 13		Fri	Sat	Sun	Mon	Tue	Wed	Thu			May			06	13	20	27
–2 05 12	16 20 24	Sat	Sun	Mon	Tue	Wed	Thu	Fri	Jan				Oct	07	14	21	28
Years		00	01	02	03		04	05
		06	07		08	09	10	11
			12	13	14	15		16
		17	18	19		20	21	22
		23		24	25	26	27

Table

每年的一月和二月都有四天的週數是固定的。除外以星期四開始的閏年(三月份起週數加一)，每個月都有四到五天的週數是固定的(詳見下表)。

月份	日期					週數
01月	04	11	18	25		01–04
02月	01	08	15	22		05–08
03月	01	08	15	22	29	09–13
04月	05	12	19	26		14–17
05月	03	10	17	24	31	18–22
06月	07	14	21	28		23–26
07月	05	12	19	26		27–30
08月	02	09	16	23	30	31–35
09月	06	13	20	27		36–39
10月	04	11	18	25		40–43
11月	01	08	15	22	29	44–48
12月	06	13	20	27		49–52

Table 1

主日字母表不僅可以用來查找格里曆(CD)任意一年的主日字母(DL)和任意一天的日字母(dl)及星期(w)，而卻可以查找ISO周日曆(WD)的週數(n)及其相應的日期(D)，因此可以利用該表來實現這兩種日曆的相互轉換。

位於世紀(c)列和年(y)行交叉處(c, y)的字母就是該年的主日字母(本世紀的主日字母位於A列)，而位於日列(d)和月(m)行交叉點(d, m)的字母就是該日的字母，知道了主日字母就可以確定其它日字母的星期。由字母D就可以查到週數固定的日期，當遇到以週四開始的閏年(DC)從三月份起週數加一。

例一查找2032年10月1日的星期及週數:

c = 20，y = 32 mod 28 = 4，d = 1，m = 10；

DL = (20，04/04) = DC，dl = (1，10) = A，D = (4，10)(40 + 1)；

C = 週日，A = 週五，D = 週一(41)；

n = 41 - 1 = 40，w = 5；

WD = 2032–W40–5。

例二查找1980–W40–1的日期:

c = 19，y = 80 mod 28 = 24，n = 40，w = 1 = 週一；

DL = (19，24/24) = FE，D = (4，10)；

E = 週日，D = 週六(40)，F = (6，10)週一(41) = (-1，10)(29，9)週一(40)；

CD = 1980年9月29日週一。

週數			日期			01 08 15 22 29	02 09 16 23 30	03 10 17 24 31	04 11 18 25 --	05 12 19 26 --	06 13 20 27 --	07 14 21 28 --	主日字母星期表
01–04		40–43	01月		10月	A	B	C	D	E	F	G	00	06	12	17	23
14–17		27–30	04月		07月	G	A	B	C	D	E	F	01	07	12	18	24
36–39		49–52	09月		12月	F	G	A	B	C	D	E	02	08	13	19	24
	23–26			06月		E	F	G	A	B	C	D	03	08	14	20	25
05–08	09–13	44–48	02月	03月	11月	D	E	F	G	A	B	C	04	09	15	20	26
	31–35			08月		C	D	E	F	G	A	B	04	10	16	21	27
	18–22			05月		B	C	D	E	F	G	A	05	11	16	22	00
年的前2位數 mod 4						20 00 16		21 01 17		22 02 18		23 03 19	年的後2位 數 mod 28

This table can be used to look up dominical letters (DL), day letters (dl), weekdays (w), week numbers (n), and week dates (W). Letters both in a century column (A C E G) and year rows are dominical letters for years of the century (c, y). Letters both in day columns and a month row are day letters for days of the month (d, m).

• For 1 October 2032

c = 20, y = 32 mod 28 = 4, d = 1, m = Oct;

DL = (20, 04/04) = DC, dl = (1, Oct) = A, D = 4 Oct;

C = Sun, A = Fri = 5, D = Mon;

n = 40 + 1 - 1 = 40, w = 5;

W = 2032–W40–5.

• For 1980–W40–1

c = 19, y = 80 mod 28 = 24, n = 40, w = 1 = Mon;

DL = (19, 24/24) = FE, m = Oct, D = 4;

E = Sun, D = Sat = 6;

d = 4 + 1 - 6 = -1 Oct = 29 September 1980.

Week day table for Julian calendar

Date	Month	Year	N	Day	Mutiple
01 08 15 22 29	Jun	01 07 12 18	1	Mon
02 09 16 23 30	Sep Dec	02 13 19 24	2	Tue
03 10 17 24 31	Apr Jul Jan	03 08 14 25	3	Wed
04 11 18 25	Jan Oct	09 15 20 26	4	Thu
05 12 19 26	May	04 10 21 27	5	Fri
06 13 20 27	Aug Feb	05 11 16 22	6	Sat
07 14 21 28	Feb Mar Nov	06 17 23 28	0	Sun	056 084 112 140 168 196 224 252

Table 2

To the day of	13 Jan	14 Feb	3 Mar	4 Apr	5 May	6 Jun	7 Jul	8 Aug	9 Sep	10 Oct	11 Nov	12 Dec	i
Add	000	031(4+3)	059(8+3)	090(12+6)	120(17+1)	151(21+4)	181(25+6)	212(30+2)	243(34+5)	273(39+0)	304(43+3)	334(47+5)	003
Leap year	000	031	060	091	121	152	182	213	244	274	305	335	002
Algorithm	od = 30 (m - 1) + floor (0.6 (m + 1)) - i + d
Year's last 2–digit mod 28 (y)							01	02	03	04		05	06
							07	08		09	10	11	12
								13	14	15	16		17
							18	19	20		21	22	23
							24		25	26	27	00
Correction (c) Year's first 2–digit mod 4 (C)						00	00	01	02	03	- 3	- 2	- 1
						01	- 2	- 1	00	01	02	03	- 3
						02	03	- 3	- 2	- 1	00	01	02
						03	01	02	03	- 3	- 2	- 1	00
Algorithm	c = (y + floor ((y - 1)/4) + 5 C - 1) mod 7 - 7 if the result > 3

${\begin{aligned}{\text{weeks}}({\text{year}})&=52+{\begin{cases}1{\text{ (long)}}&{\text{if }}p({\text{year}})=4\\&{\text{or }}p({\text{year}}-1)=3\\0{\text{ (short)}}&{\text{otherwise}}\end{cases}}\\p({\text{year}})&=\left({\text{year}}+\left\lfloor {\frac {\text{year}}{4}}\right\rfloor -\left\lfloor {\frac {\text{year}}{100}}\right\rfloor +\left\lfloor {\frac {\text{year}}{400}}\right\rfloor \right){\bmod {7}}\end{aligned}}$

${\begin{aligned}{\text{Week date}}{\begin{cases}{\text{week number}}&=\left\lceil {\frac {od+c}{7}}\right\rceil \\{\text{weekday number}}&=(od+c){\bmod {7}}\end{cases}}\end{aligned}}$

for od and c look up the table above or use the algorithm to calculate. There are 53 weeks in any year (c = 3) or in leap years (c = 2), otherwise there are 52 weeks in a year.

Ceiling the quotient equals the week number and the remainder is the weekday number (0 = Sunday = 7).

For 26 September 2008

ceil ((244 + 26 + 01)/7) = 39

(244 + 26 + 01) mod 7 = 5

the week date is 2008W395.

Summary

Year–starting on (G/W)	Commom–year 365 − 1 or + 6	Leap–year 366 − 2 or + 5
Mon/01 Jan	+ 0 − 1	+ 0 − 2
Tue/31 Dec	+ 1 − 2	+ 1 − 3
Wed/30 Dec	+ 2 − 3	+ 2 + 3
Thu/29 Dec	+ 3 + 3	+ 3 + 2
Fri/04 Jan	− 3 + 2	− 3 + 1
Sat/03 Jan	− 2 + 1	− 2 + 0
Sun/02 Jan	− 1 + 0	− 1 − 1