.\" $Id: gauss.ms,v 1.24 2005/10/17 07:54:52 rl Exp $ .if !c\(*d .char \(*d delta .if !c\(*D .char \(*D DELTA .if n \{\ .nr PO 0 .nr LL 78n .\} .de LM .FS Last modified: \\$2 \\$3 .FE .. .EQ delim $$ tdefine / 'over' define mod 'roman "mod"' .EN .TL Gauss Formula for the Julian Date of Passover .AU Zvi Har'El .AI Department of Mathematics Technion \- Israel Institute of Technology Haifa 32000, Israel E - Mail: rl@math.technion.ac.il .NH Definitions. .PP \fINisan origin\fP is the instant occurring before the beginning of 1 Nisan, in a time difference which is equal to the difference between the new moon, or \fImolad\fP, of Tishri and the beginning of 1 Tishri of the following Hebrew year. .PP Nisan origin of the year 1 \s-2AM\s0\** .FS \fIanno mundi\fP. .FE is 29 Adar, 14 hours, because molad Tishri of the year 2 \s-2AM\s0 was on Friday, 14 hours, and 1 Tishri was a Saturday. .PP \fIMarch origin\fP is the instant occurring one day before the beginning of 1 March. Thus, the March date of a given instant will be in fact the time difference between that instant and March origin. .PP Note that Nisan origin of the year 1 \s-2AM\s0 is 33 March, 14 hours, because the Julian date of 1 Nisan 1 \s-2AM\s0 was 3 April, i.e., 34 March. Here we assume the Julian day begins in the previous evening (at 6 \s-2PM\s0), as the Hebrew day does, to simplify the discussion. .PP Let $H ( A ) $ denote the Nisan origin and $J ( A ) $ the March origin of the Hebrew year $A$. Let $T$ be Nisan origin of the year 1 \s-2AM\s0, in March days: .EQ (1) T = H ( 1 ) - J ( 1 ) = 33:14:0 = 403 / 12 . .EN Time intervals are given in the Talmudic units: An hour is divided into 1080 parts, so that $d$ days, $h$ hours and $p$ parts are written as $d:h:p = ( p /1080 + h ) /24 + d$. Let $K$ be 1/19 of the length of the lunar month: .EQ K = {29:12:793} / 19 = 765433 / 492480 . .EN A cycle of 19 consecutive lunar years contains 235 lunar months, arranged in 12 common years containing 12 months each, and 7 leap years, containing 13 month each. Conventionally, leap years are the 3rd, 6th, 8th, 11th, 14th, 17th, 19th in each cycle. Let $L$ be the average solar excess, i.e., difference in length between a solar year and an average lunar year: .EQ L = 365:6:0 - 235K = {0:1:485} / 19 = 313 / 98496 . .EN Note: Using decimals, .EQ T approx 33.58333333 , .EN .EQ K approx 1.554241797 , .EN .EQ L approx 0.003177794022 . .EN .NH Nisan origin. .PP The length of a leap lunar year is $13 cdot 19K = 247K$, while the length of a common lunar year is $12 cdot 19K = 228K$. Subtracting these quantities from $L + 235K$, the length of the solar year, we get the common solar excess, $L + 7K$, and the (negative) leap solar excess, $L - 12K$. .PP From these observations we get .EQ H ( A ) - H ( 1 ) = 365:6:0 ( A - 1 ) - Delta ( A ) , .EN where $Delta ( A ) $ is the cumulative solar excess. It is given in the following table, with leap years marked by an asterisk: .TS center box; n|c . $A$ $Delta ( A ) $ _ 1 $0$ 2 $L + 7K$ *3 $2L - 5K$ 4 $3L + 2K$ 5 $4L + 9K$ *6 $5L - 3K$ 7 $6L + 4K$ *8 $7L - 8K$ 9 $8L - K$ 10 $9L + 6K$ *11 $10L - 6K$ 12 $11L + K$ 13 $12L + 8K$ *14 $13L - 4K$ 15 $14L + 3K$ 16 $15L + 10K$ *17 $16L - 2K$ 18 $17L + 5K$ *19 $18L - 7K$ 20 $19L$ .TE .PP Each year we add $L + 7K$, unless the year is leap, when we add $L - 12K$ (since we compute in effect the next molad Tishri). In this way, the coefficient of $L$ is incremented continuously, while the coefficient of $K$ is increased by 7 each time, until a moment when it becomes 11 or higher, when it is decreased by 19. Since the lowest possible value of this coefficient is - 8, and this value is obtained at $A = 8 ( mod~19 ) $, we get that the running value is $ - 8 + ( 7 ( A - 8 ) ) |19$, where $x|k$ is $x - k [x/k]$. Therefore, the coefficient of $K$ is .EQ - 8 + ( 7 ( A - 8 ) ) |19 mark = - 8 + ( 7A + 1 ) |19 .EN .EQ lineup = - 8 + ( 18 - ( 18 - ( 7A + 1 ) ) |19 ) .EN .EQ lineup = 10 - ( 17 - 7A ) |19 .EN .EQ lineup = 10 - ( 12A + 17 ) |19 .EN (using the identity $x|k = ( k - 1 ) - ( ( k - 1 ) - x ) |k$). Denoting .EQ a = ( 12A + 17 ) |19 , .EN we get .EQ Delta ( A ) = ( 10 - a ) K + ( A - 1 ) L , .EN or finally .EQ (2) H ( A ) - H ( 1 ) = ( A - 1 ) 365:6:0 + ( a - 10 ) K - ( A - 1 ) L . .EN .PP As an added bonus, we can divide the cycle years into 4 categories, according to the value of the coefficient of $K$ in the cumulative solar excess: .TS center box; c|c|c|c|c. $10 - a$ $a$ $A - 1$ $A$ $A + 1$ _ \-8...\-2 18...12 common leap common \-1...3 11...7 leap common common 4...5 6...5 leap common leap 6...10 4...0 common common leap .TE .NH March origin. .PP Set .EQ J ( A ) - J ( 1 ) = ( A - 1 ) 365 + delta ( A ) , .EN where $delta ( A ) $ is the number of Julian intercalary days (29 February) between 1 March 1 \s-2AM\s0 and 1 March of the year $A$. Since the Hebrew year 1 \s-2AM\s0 corresponds\** .FS At least, its major part containing 1 March. .FE to the Julian year 3760 \s-2BCE\s0, or \-3759 \s-2CE\s0 which gives a remainder of 1 when divided by 4, we obtain that the year $A$ will contain a intercalary day if and only if $A == 0 ~ ( mod~4 )$. Thus $delta ( A ) = [A/4]$, or, denoting .EQ b = A|4 , .EN we get .EQ delta ( A ) = A /4 - b /4 . .EN Therefore .EQ J ( A ) - J ( 1 ) = ( A - 1 ) 365 + A /4 - b /4 , .EN or, finally .EQ (3) J ( A ) - J ( 1 ) = ( A - 1 ) 365:6:0 - b /4 + 0:6:0 . .EN .NH March date of Passover. .PP Subtracting (3) from (2) and using (1), we get .EQ H ( A ) - J ( A ) = T + ( a - 10 ) K - ( A - 1 ) L + b /4 - 0:6:0 , .EN or .EQ H ( A ) - J ( A ) = ( T - 10K + L ) + aK - AL + b /4 - 0:6:0 . .EN This is the March date of Nisan origin of the Hebrew year $A$. We add 6 hours to implement the rule that if molad Tishri is at noon or later\**, .FS \fIMolad Zaken\fP. .FE 1 Tishri is postponed to the following day. Finally we add 14 days to get the March date of 15 Nisan. .PP Setting .EQ m sub 0 = T - 10K + L + 14 = 3156215 / 98496 , .EN we get .EQ M + m = {m sub 0} + aK - AL + b /4 , .EN where M is the integral part and m the fractional part of the right hand side. Unless further exceptions apply (see below), $M$ is the Julian March date of the first day of Passover of the Hebrew year $A$. .PP Note: Using decimals, .EQ m sub 0 approx 32.04409316 . .EN .NH Week day of Passover. .PP Calculating modulo 7, we obtain: .EQ J ( A ) - J ( 1 ) mark == ( A - 1 ) 365:6:0 - b /4 + 0:6:0 .EN .EQ lineup == ( A - 1 ) 1:6:0 - b /4 + 0:6:0 .EN .EQ lineup == 5A /4 - b /4 - 1 .EN .EQ lineup == A - 1 + ( A - b ) /4 .EN .EQ lineup == A - 1 + 8 ( A - b ) /4 .EN .EQ lineup == A - 1 + 2 ( A - b ) .EN .EQ lineup == 3A - 2b - 1 .EN .EQ lineup == 3A + 5b - 1 ~ ( mod~7 ) .EN Since March origin 1 \s-2AM\s0 was on Friday, we get for $M$ March of the Hebrew year $A$, .EQ c = ( M + 3A + 5b + 5 ) |7 . .EN $c$ is the day in the week of $M$ March, with $c = 0$ for Saturday. .NH Exceptions. .PP In the discussion above, we assumed that 1 Tishri is the day on which molad Tishri has taken place, and established that the Julian date of 15 Nisan is $M$ March. We already mentioned one exception. If molad Tishri is at noon or later, 1 Tishri is postponed to the following day. We implemented this exception by adding 6 hours to Nisan origin. However, there are three more exceptions. .PP The second exception is the rule that 1 Tishri is excluded from being a Sunday, Wednesday or Friday\**, .FS \fIAdu\fP. .FE and is postponed to the following day. To implement this rule, we notice that 15 Nisan and the following 1 Tishri are 152 days apart, i. e., 22 weeks minus 2 days. Thus, 15 Nisan is excluded from being a Friday, Monday of Wednesday, respectively. .PP The last two exceptions are derived from the previous one, and from a restriction on the length of the Hebrew year. As we have seen, the length of the common lunar year is $12 cdot 19K = 354:8:876$ days, and the length of the leap lunar year is $13 cdot 19K = 383:21:589$ days. Of course, a calendar year must have an integral number of days. Thus, a common Hebrew year has 353, 354 or 355 days\**, .FS 12 months, alternating between 30 and 29 days each, give a total of 354 days. This number may increase by adding one to the 29 days of Heshvan, or decrease by subtracting one from the 30 days of Kislev. .FE while a leap Hebrew year has 383, 384 or 385 days\**. .FS The intercalary month, Adar Rishon, has 30 days. .FE .PP The third exception follows from restricting the common year to have at most 355 days. Molad Tishri of a common year $A+1$ and its successor are 354:8:876 days apart, i. e., 51 full weeks minus 2:15:204 days. Thus, if molad Tishri of $A+1$, after being moved 6 hours ahead, is on Tuesday, 15 hours and 204 parts or later\**, .FS The molad being on Tuesday, 9 hours and 204 parts or later (\fIGatrad\fP). .FE its successor is on Sunday. Then, 1 Tishri $A+2$ is a Monday, and if 1 Tishri $A+1$ is not postponed from Tuesday (to Thursday, as Wednesday is excluded), the year $A+1$ will have 356 days. .PP Similarly, the fourth exception follows from restricting the leap year to have at least 383 days. Molad Tishri of a leap year $A$ and its successor are 383:21:589 days apart, i. e., 54 full weeks plus 5:21:589 days. Thus, if molad Tishri of $A+1$, after being moved 6 hours ahead, is on Monday, 21 hours and 589 parts or later\**, .FS The molad being on Monday, 15 hours and 589 parts or later (\fIBetu Takpat\fP). .FE its predecessor is on Wednesday. Then, 1 Tishri $A$ is a Thursday, and if 1 Tishri $A+1$ is not postponed from Monday (to Tuesday), the year $A$ will have 382 days. .PP To implement the last two exceptions, we notice that that 1 Tishri $A+1$ being a Monday or Tuesday implies that 15 Nisan $A$ is a Saturday or Sunday, respectively. Also, if we consider the table in Section 2, we notice that $A$ is leap if $a >= 12$ and $A + 1$ is common if $a >= 7$. .PP Thus, setting .EQ m sub 1 = (13 cdot 19K)|1 = 0:21:589 = 23269 / 25920 , .EN .EQ m sub 2 = 1 - (12 cdot 19K)|1 = 0:15:204 = 1367 / 2160 , .EN we find that the Julian date the first day of Passover is: .IP \(bu $M + 1$ March, if $c = 0$, $a >= 12$ and $m >= m sub 1$, .IP \(bu $M + 2$ March, if $c = 1$, $a >= 7$ and $m >= m sub 2$, .IP \(bu $M + 1$ March, if $c = 2$, 4 or 6, .IP \(bu $M$ March, otherwise. .PP Note: Using decimals, .EQ m sub 1 approx 0.897723765 , .EN .EQ m sub 2 approx 0.63287037 . .EN .NH References. .PP .IP 1. Adler, Cyrus, \fICalendar, History of\fP, in: Singer, Isidore (ed.), The Jewish Encyclopedia, Vol. 3, pp. 498-500. Ktav Publishing House, Inc., New York, 1901. .IP 2. Dershowitz, N. and Reingold, E. M., \fICalendrical Calculations\fP, Software \- Practice and Experience, 20 (1990), 899-928. .IP 3. Friedl\(:ander, Michael, \fICalendar\fP, in: Singer, Isidore (ed.), The Jewish Encyclopedia, Vol. 3, pp. 501-508. Ktav Publishing House, Inc., New York, 1901. .IP 4. Gauss, Karl Friedrich, \fIBerechnung Des J\(:udishen Osterfestes\fP, Montaliche Correspondenz zur Bef\(:orderung der Erd- und Himmels-Kunde, herausgegeben vom Freiherrn von Zach. Mai 1802. Werke, Vol 6, pp. 80-81. .IP 5. Gauss, Karl Friedrich, \fIBerechnung Des Neumonds Tisri F\(:ur Jedes J\(:udische Jahr A\fP, Handschriftlische Eintragung in Christian Wolf, Elementa matheseos universae, tomus IV. - Von Gauss 1800 erworben. Werke, Vol. 11, pp. 215-218. .IP 6. Resnikoff, Louis A., \fIJewish Calendar Calculations\fP, Scripta Math., 9 (1943), 191-195. .NH Appendix: A Computer Implementation. .sp 2 .nf .ie \n(.g .fam C .el \{\ .fp 1 CW .fp 2 CI .fp 3 CB .fp 4 CX \} \s-2\f1/*\f2 * Gauss formula for Passover * * Arguments: * year \- Hebrew year (anno mundi) * g \- boolean flag, 0 for Julian dates, 1 for Gregorian. * day \- optional pointer to an integer, * to return the day\-of\-the week. * Return value: * March date of the first day of Passover. \f1*/ /*\f2 Fundamental constants \f1*/ #\f3define\f1 T (33. + 14. / 24.) #\f3define\f1 L ((1. + 485. / 1080.) / 24. / 19.) #\f3define\f1 K ((29. + (12. + 793. / 1080.) / 24. )/ 19.) /*\f2 Derived constants \f1*/ #\f3define\f1 m0 (T \- 10. * K + L + 14.) #\f3define\f1 m1 ((21. + 589. / 1080.) / 24.) /*\f2 13*19*K mod 1 \f1*/ #\f3define\f1 m2 ((15. + 204. / 1080.) / 24.) /*\f2 1 \- (12*19*K mod 1) \f1*/ \f3int\f1 Gauss(\f3int\f1 year, \f3int\f1 g, \f3int\f1 *day)\kx\s+2\f4\h'|\n(.lu-\w'Gauss'u'Gauss\f1\s0\h'|\nxu' .ds r Gauss { \f3int\f1 a, b, c, M; \f3double\f1 m; a = (12 * year + 17) % 19; b = year % 4; m = m0 + K * a + b / 4. \- L * year; \f3if\f1 (m < 0) m\-\-; M = m; \f3if\f1 (m < 0) m++; m \-= M; \f3switch\f1 (c = (M + 3 * year + 5 * b + 5) % 7) { \f3case\f1 0: \f3if\f1 (a >= 12 && m >= m1) { c = 1; M++; } \f3break\f1; \f3case\f1 1: \f3if\f1 (a >= 7 && m >= m2) { c = 3; M += 2; } \f3break\f1; \f3case\f1 2: c = 3; M++; \f3break\f1; \f3case\f1 4: c = 5; M++; \f3break\f1; \f3case\f1 6: c = 0; M++; \f3break\f1; } \f3if\f1 (day) *day = c; \f3if\f1 (g) /*\f2 Gregorian Calendar \f1*/ M += (year \- 3760) / 100 \- (year \- 3760) / 400 \- 2; \f3return\f1 M; }\s0 .LM $$Date: 2005/10/17 07:54:52 $$