Gauss Formula for the Julian Date of Passover
Zvi Har’El
Department of Mathematics
Technion − Israel Institute of Technology
Haifa 32000, Israel
E ‐ Mail: rl@math.technion.ac.il
1. Definitions.
Nisan origin 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
molad, of Tishri and the beginning of 1 Tishri of the following Hebrew year.
Nisan origin of the year 1 AM1 is 29 Adar, 14 hours, because molad Tishri
of the year 2 AM was on Friday, 14 hours, and 1 Tishri was a Saturday.
March origin 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 dif‐
ference between that instant and March origin.
Note that Nisan origin of the year 1 AM is 33 March, 14 hours, because
the Julian date of 1 Nisan 1 AM was 3 April, i.e., 34 March. Here we assume
the Julian day begins in the previous evening (at 6 PM), as the Hebrew day
does, to simplify the discussion.
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 AM, in March days:
T=H(1)−J(1)=33:14:0=403/12. (1)
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:
K=29:12:793/19=765433/492480.
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:
L=365:6:0−235K=0:1:485/19=313/98496.
Note: Using decimals,
───────────
1 anno mundi.
‐2‐
K≈1.554241797,
L≈0.003177794022.
2. Nisan origin.
The length of a leap lunar year is 13⋅19K=247K, while the length of a
common lunar year is 12⋅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.
From these observations we get
H(A)−H(1)=365:6:0(A−1)−Δ(A),
where Δ(A) is the cumulative solar excess. It is given in the following table,
with leap years marked by an asterisk:
┌────┬─────────┐
│ A │ Δ(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 │
└────┴─────────┘
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(mod19), we get that the running value is −8+(7(A−8))|19, where
‐3‐
x|k is x−k[x/k]. Therefore, the coefficient of K is
−8+(7(A−8))|19=−8+(7A+1)|19
=−8+(18−(18−(7A+1))|19)
=10−(17−7A)|19
=10−(12A+17)|19
(using the identity x|k=(k−1)−((k−1)−x)|k). Denoting
a=(12A+17)|19,
we get
Δ(A)=(10−a)K+(A−1)L,
or finally
H(A)−H(1)=(A−1)365:6:0+(a−10)K−(A−1)L. (2)
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:
┌─────────┬─────────┬────────┬────────┬────────┐
│ 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 │
└─────────┴─────────┴────────┴────────┴────────┘
3. March origin.
Set
J(A)−J(1)=(A−1)365+δ(A),
where δ(A) is the number of Julian intercalary days (29 February) between 1
March 1 AM and 1 March of the year A. Since the Hebrew year 1 AM corresponds2
to the Julian year 3760 BCE, or −3759 CE 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(mod4). Thus δ(A)=[A/4], or, denoting
b=A|4,
───────────
2 At least, its major part containing 1 March.
‐4‐
we get
δ(A)=A/4−b/4.
Therefore
J(A)−J(1)=(A−1)365+A/4−b/4,
or, finally
J(A)−J(1)=(A−1)365:6:0−b/4+0:6:0. (3)
4. March date of Passover.
Subtracting (3) from (2) and using (1), we get
H(A)−J(A)=T+(a−10)K−(A−1)L+b/4−0:6:0,
or
H(A)−J(A)=(T−10K+L)+aK−AL+b/4−0:6:0.
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 later3, 1 Tishri is
postponed to the following day. Finally we add 14 days to get the March date
of 15 Nisan.
Setting
m0=T−10K+L+14=3156215/98496,
we get
M+m=m0+aK−AL+b/4,
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.
Note: Using decimals,
m0≈32.04409316.
5. Week day of Passover.
Calculating modulo 7, we obtain:
J(A)−J(1)≡(A−1)365:6:0−b/4+0:6:0
───────────
3 Molad Zaken.
‐5‐
≡5A/4−b/4−1
≡A−1+(A−b)/4
≡A−1+8(A−b)/4
≡A−1+2(A−b)
≡3A−2b−1
≡3A+5b−1(mod7)
Since March origin 1 AM was on Friday, we get for M March of the Hebrew year
A,
c=(M+3A+5b+5)|7.
c is the day in the week of M March, with c=0 for Saturday.
6. Exceptions.
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 excep‐
tion by adding 6 hours to Nisan origin. However, there are three more excep‐
tions.
The second exception is the rule that 1 Tishri is excluded from being a
Sunday, Wednesday or Friday4, 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.
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⋅19K=354:8:876 days, and the length of the leap
lunar year is 13⋅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
───────────
4 Adu.
5 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.
‐6‐
days5, while a leap Hebrew year has 383, 384 or 385 days6.
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 later7, 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.
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 later8, 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.
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.
Thus, setting
m1=(13⋅19K)|1=0:21:589=23269/25920,
m2=1−(12⋅19K)|1=0:15:204=1367/2160,
we find that the Julian date the first day of Passover is:
· M+1 March, if c=0, a≥12 and m≥m1,
· M+2 March, if c=1, a≥7 and m≥m2,
· M+1 March, if c=2, 4 or 6,
· M March, otherwise.
Note: Using decimals,
m1≈0.897723765,
m2≈0.63287037.
───────────
6 The intercalary month, Adar Rishon, has 30 days.
7 The molad being on Tuesday, 9 hours and 204 parts or later
(Gatrad).
8 The molad being on Monday, 15 hours and 589 parts or later
(Betu Takpat).
‐7‐
7. References.
1. Adler, Cyrus, Calendar, History of, in: Singer, Isidore (ed.), The Jewish
Encyclopedia, Vol. 3, pp. 498‐500. Ktav Publishing House, Inc., New York,
1901.
2. Dershowitz, N. and Reingold, E. M., Calendrical Calculations, Software −
Practice and Experience, 20 (1990), 899‐928.
3. Friedländer, Michael, Calendar, in: Singer, Isidore (ed.), The Jewish
Encyclopedia, Vol. 3, pp. 501‐508. Ktav Publishing House, Inc., New York,
1901.
4. Gauss, Karl Friedrich, Berechnung Des Jüdishen Osterfestes, Montaliche
Correspondenz zur Beförderung der Erd‐ und Himmels‐Kunde, herausgegeben
vom Freiherrn von Zach. Mai 1802. Werke, Vol 6, pp. 80‐81.
5. Gauss, Karl Friedrich, Berechnung Des Neumonds Tisri Für Jedes Jüdische
Jahr A, Handschriftlische Eintragung in Christian Wolf, Elementa mathe‐
seos universae, tomus IV. ‐ Von Gauss 1800 erworben. Werke, Vol. 11, pp.
215‐218.
6. Resnikoff, Louis A., Jewish Calendar Calculations, Scripta Math., 9
(1943), 191‐195.
8. Appendix: A Computer Implementation.
/*
* 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.
*/
/* Fundamental constants */
#define T (33. + 14. / 24.)
#define L ((1. + 485. / 1080.) / 24. / 19.)
#define K ((29. + (12. + 793. / 1080.) / 24. )/ 19.)
/* Derived constants */
#define m0 (T − 10. * K + L + 14.)
#define m1 ((21. + 589. / 1080.) / 24.) /* 13*19*K mod 1 */
#define m2 ((15. + 204. / 1080.) / 24.) /* 1 − (12*19*K mod 1) */
int
Gauss(int year, int g, int *day) Gauss
{ int a, b, c, M;
‐8‐
double m;
a = (12 * year + 17) % 19;
b = year % 4;
m = m0 + K * a + b / 4. − L * year;
if (m < 0) m−−;
M = m;
if (m < 0) m++;
m −= M;
switch (c = (M + 3 * year + 5 * b + 5) % 7) {
case 0:
if (a >= 12 && m >= m1) {
c = 1; M++;
}
break;
case 1:
if (a >= 7 && m >= m2) {
c = 3; M += 2;
}
break;
case 2:
c = 3; M++;
break;
case 4:
c = 5; M++;
break;
case 6:
c = 0; M++;
break;
}
if (day) *day = c;
if (g) /* Gregorian Calendar */
M += (year − 3760) / 100 − (year − 3760) / 400 − 2;
return M;
}
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Last modified: 2005/10/17 07:54:52