**Gauss Algorithm for Easter date: A brief explanation**

A lunar month begins with the new moon and is 30 days long if the moon falls in a odd month (Jan, Mar, May, Jul, Sep, Nov) and 29 days long if the moon falls in an even month (Feb, Apr, Jun, Aug, Oct, Dec). As an exception, if the year has a leap day that day is added to the moon of February, so that in that year the moon of February lasts 30 days.

So 12
lunar months are 354 days long in ordinary years and 355 in leap years. In both
case the difference between the solar year and 12 lunar months is 11 days.

If you
take count of these 11 days a year, after 19 years you have a difference of 19
x 11 = 209 days. So in the ancient greek lunar calendar they added to the lunar
calendar an extra month (called embolysmic month) when the difference was 30 or
more.

So if we
start with year 1 (difference at the end of the year: 11), in year 2 that
difference is 22, in year 3 it would be 33 but you add an extra month of 30
days so the difference becomes 3, and so on.

Adding 6
embolysmic months of 30 days each in 19 years the difference of 209 becomes
209-6 x 30=29 days, so you add another embolysmic month of 29 days to the
nineteenth year and the difference goes to zero.

That is,
with 235 lunations (12 lunar years of 12 months each + 7 lunar years of 13
months each) you cover 19 solar years and all the moons of the year come back
at the same days.

This is
known as the Methonic cycle.

The
difference between the solar and lunar year N is the same that the age of the
moon at Jan, 1 of the year N+1.

In fact,
if you have a moon at Jan 1st, the lunar year will end on day 354 of the year
(355 if it's a leap year), that is on Dec, 20 and next lunar month begins on
Dec 21. So the age of the moon at Jan, 1st is 32-21=11 that is the same as the
difference of year N.

The age
of the moon at Jan 1st is called "epact" and its value is 0 to 29.

When you
know the epact of a year, you know all the moons of that year too.

As for
the Methonic cycle all moons repeat after 19 solar years, if you assign a
number 1 to 19 to each year (this is called the Golden number) then all years
with the same Golden numbers have the same epact and the same moons.

About 500
A.D. a monk, Dyonisus, stated that in the year before the one of Christ's
borning there had been a moon at Dec, 24 so the moon's age of the subsequent
year (1 B.C.) is 32-24=8.

So
Dyonisus decided that the Methonic cycle would begin at year 1 B.C. with golden
number 1 and epact 8.

The
Golden number of every year Y after 1 B.C. is then

G = (Y mod 19) + 1

If you
know the Golden number, you can easily find the epact, remembering thet the
epact raises 11 days each year and if it reaches or is more than 30 you must
subtract 30.

As you
can see, after yer 19 to reach year 1 of the next cycle you must add 12 and not
11 days, this is known as *saltus lunae* and the reason is that the embolysmic
month of the nineteenth year of cycle has 29 days and not 30.

The
relation between the Golden number G and the epact E can be expressed with a
formula:

E = [8+11 x (G-1)] mod 30

E = (8+11G-11) mod 30

E = (11G – 3) mod 30

As 30+n =
n mod 30 you can subtract 30 to the previous expression getting

E = (11G-33) mod 30

And,
finally,

**E = [11(G-3)] mod 30**

If you
know the epact of a year, you can find all the lunations of that year. In
particular, as the lunar months of January and February count 59 days together,
when the epact of the year is 0 that means that there is a lunation on Jan, 1
and another on day 1+59, that is on day 60 of the year, that is on 1 March.

Now, you
know that the full moon of Easter is the first full moon that occurs on the day
of equinox (that the Church fixed on March, 21) or after.

The
Church also stated that, for simplification purposes, the full moon of a lunar
month is always the fourteenth day of that mont, that is 13 days after the new
moon.

So when
epact is 0 we have a full moon on 1+13=14 of March.

As March,
14 is before the equinox, we must consider the next full moon that, being the
lunar month of march 30 days long, occurs on 13 april, that is the 44

^{th}day from March,1.
So,
finally, when epact is 0 the paschal full moon is the day 44 of March,
considering for calculation purposes as the dates of April would be dates of
March+31.

When the
epact raises by 1, the age of the moon is more and then the preceeding full
moon is earlier, so all new and full moons shift a day back. So when epact is 1
the pfm is on day 43, when it is 2 is on day 42, and so on.

We can
write

P = 44 – E

Where P
is the paschal full moon.

But if E
= 24 or more, the preceeding relation gives a paschal full moon earlier than
the equinox. So in that case we must shift on the next full moon, adding 30
days.

So the
correct formula is

P = 44 – E if
E ≤ 23

P = 74 – E if
E > 23

As the
pfm is not before March, 21st, we can rewrite the equations as

P = 21 + 23 –
E if E ≤ 23

P = 21 + 53 – E if E > 23

Now we can observe that 53
– E = (53 – E) mod 30 = (23 – E) mod 30 so the final formula is:

**P = 21 + (53 – E) mod 30**

Before affording Gauss’
algorhytm, we need another step.

The methonic cycle, on
which the last formula is based, works from 46 B.C. when the Julian calendar
was instituted onto 1582 A.D. when Pope Gregory XIII stated the new calendar
that after his name is called Gregorian.

For reasons that is
unnecessary to indicate here, the calendar reformation said:

1) To realline civil and astronomical equinox on 21
March, 10 days had to be subtracted to 1582, so that the day after Thursday
Oct, 4 1582 of the Julian calendar is Friday Oct, 15 of the Gregorian Calendar;

2) For next centuries, only secular years (the years
ending with 00) multiple of 400 would be leap, the other ones would be ordinary
(so called solar equation)

3) The lunar year would have been increased by 8 days
every 2500 years, adding a day every 300 years for seven times, then a day
after 400 years. The first day would have been added to 2800 as the last year
of a cycle of 2500 and then another cycle began. The days to add until 1582
were determined to be 4 (so called lunar equation)

4) For effect of what here said, the epacts would have
been diminished by 6 (4-10) as subtracting a day to a solar year diminishes the
epact, adding a day to the lunar year increases it. Clavius, one of the
astronomers in Pope’s commission, decided to decrease the epacts by 7, that way
not altering the week cycle. So to determine epacts the days subtracted by the
solar equation and the one added by the lunar equation are to be considered as
these equations would be applied even before 1582.

All those rules can be
formalized modifying the equation that gives us the epact as follows:

**E = [11(G-3) + S – L] mod 30**

Where S and L indicate,
respectively, the solar and the lunar equation.

__The Gauss Algorithm__

Gauss
didn’t explain his algorhitm, but it can be easily derived from epacts and
golden numbers.

First we
give the meaning of the parameters

*d*and*e*.
The value

*d*represents the number of days (0 to 29) that occur from 21^{st}of March and the paschal full moon.
The value

*e*represents the number of days (0 to 6) that occur from the day after the paschal full moon, that is the first day in which Easter can fall, and the Easter Sunday.
So the
date of Easter is given by 21+ d (pfm) + 1 (first day for Easter) + e (days to
the first Sunday):

Date = 22 + d + e of March

If 22 + d
+ e is over 31, then you subtract 31 and gain the date of Easter in April:

Date = d + e - 9 of April

There are
two exceptions that we’ll say later of.

Now to
determine the value of

*d*we can use the equations for epact and golden number. Remembering that epacts change in Gregorian calendar for the influence of solar and lunar equations, we first calculate*d*for the years 1700 to 1799 and the we will extend in general the formula to years before and after.
So if

*d*are the days from March 21^{st}to the full moon, we can write*d*= P – 21 = 21+(53–E) mod 30 – 21 = (53 – E) mod 30

Now
remembering that

**E = [11(G-3) – S + L] mod 30**

If we
take

a = Y mod 19

we have

G = Y mod 19 +1 = a + 1

For the
years 1700 to 1799 we can get the solar and lunar equations as follows:

k =
|N/100| = 17

S = k – |k/4|=
17 – 4 = 13

L = |(8k
+ 13)/25|= 5

So we get

E = [11(a +1– 3) –13 + 5] mod 30 =
(11a + 11 – 33 – 13 + 5) mod 30 =

= (11a
– 30) mod 30 = 11a mod 30

Now
substituting E in the equation for

*d*we get*d*= (53 – E) mod 30 = (53 – 11a) mod 30 = (23 – 11a ) mod 30

*d*= (19a + 23) mod 30

To get a
general equation for all Julian and Gregorian years, we substitute the number
23 with a parameter, that Gauss calls M.

Obviously,
for the years 1700 to 1799 M = 23.

So, in
general

*d***= (19a + M) mod 30**

M raises
by one each time you apply the solar equation, as in this case epact is one
less and symmetrically full moon is a day later.

M also
diminishes by one when the lunar equation is to be applied, because in this
case the epact is one day more and the full moon one day before.

So to
determine the value of M for the Gregorian centuries you get

M=23 for
1700 to 1799

M=23 for
1800 to 1899 because in 1800 you apply either solar and lunar equations and the
variations compensate one another

M=24 for
1900 to 2199, as in 1900 you apply the solar equation, in 2000 you don’t apply
the solar neither the lunar equation and in 2100 you apply both.

For 1583
to 1699 M=22 because in 1600 no one euqtion is to be applied and after the
application of the solar equation in 1700 M is 23.

For the
years up to 1582 no equation is to be applied, but we must count the 7 days
subtracted to epacts in 1582. This means that from the Julian 1582 to the
Gregorian one epacts are 7 day less, and so M is increased by 7 days. This
correction takes on count 12 days for the solar equation and 5 days for the
lunar equation applied since 1 B.C. to 1582.

That is,
M for 1582 (and all Julian years before) is 22-7=15.

To get M
for Gregorian years, it’s sufficient to calculate the solar and lunar equation
and add the result to 15, subtracting 30 if the result is more as M occurs in
an equation in modulo 30.

**M=(15 + k – |k/4|–|(8k+13)/25|) mod 30**

Now once
you have

*a*,*d*and M you must find*e*.
At this
purpose, Gauss at first found a day that he knew to be a Sunday, that is March,
21

^{st}, 1700.
Then he
calculated the number of days passing from March, 21

^{st}, 1700 to the Easter Sunday of a year Y between 1700 and 1799, observing that obviously that number of days should be a multiple of 7.
So the
equation is initially that:

365 (Y – 1700) + i + (22 + d + e) – 21 = 0 mod 7

where:

365 (Y –
1700) + i are the days passed from March 21, 1700 to March, 21, Y, counting 365
days per year plus i leap days;

22 + d +
e is the date of the Easter Sunday of year Y.

Now, for
leap days, we have that the last leap year not after Y is given by

Y – Y mod 4

and then
from 1700 to that year we have

Y – 1700 – Y mod 4

years.
Calling

*b*the quantity Y mod 4 we have that there are
Y – 1700 –

*b*
years from 1700 to the
last leap year not after b. hence there are

*i*= (Y – 1700 –

*b*)/4 leap days from 1700 to Y

As

*i*is an integer, 7*i*is an integer too and is a multiple of 7. So we can add 7*i*to the equation and get
365 (Y – 1700) +8

*i*+ (22 + d + e) – 21 = 0 mod 7
Substituting
the value we found for I we have:

365 (Y – 1700) +2(Y – 1700 – b) + d + e + 1 = 0 mod 7

That is

367 (Y – 1700) – 2b + d + e + 1 = 0 mod 7

As 364 mod 7 = 0 we can
subtract 364 (Y – 1700) and get

3(Y – 1700) – 2b + d + e + 1 = 0 mod 7

That is

3Y – 5100 – 2b + d + e + 1 = 0 mod 7

And adding 5096 that’s a
multiple of 7

3Y – 4 – 2b + d + e + 1 = 0 mod 7

3Y– 2b + d + e – 3 = 0 mod 7

Now let c
= Y mod 7. Then 3c = 3(Y mod 7) = 3Y- 3 x 7|Y/7|

Since
7|Y/7| is a multiple of 7, we can say

3c = 3Y mod 7

And substituting

3c– 2b + d + e – 3 = 0 mod 7

Finally, to eliminate
minus signs, we subtract both members to 7c+7d, that is a multiple of 7.

(7c-3c)-(-2b)+(7d-d)-e-(-3) = 0
mod 7

4c+2b+6d-e+3 = 0 mod 7

Hence

e = (2b+4c+6d+3) mod 7

The number 3 that occurs
in this equation is only valid for the years 1700 to 1799; to get a general
formula we call that number N

**e = (2b+4c+6d+N) mod 7**

and calculate it as
follows.

For 1700 to 1799 obviously
N = 3

Each time the solar
equation applies the days of the year shift one position back and so if 22+d+e
indicated a Sunday, after the correction it occurs on Saturday and so you need
another day to reach a Sunday.

That’s why

*e*must increase by one each time you apply the solar equation. And as the other parameters b, c and d are fixed for each year, the only parameter that can vary is N.
So we have that N is 3 for
century 17, 4 for century 18, 5 for centuries 19 and 20, 6 for 21 and so on.

For century 16 N is 2
because in 1700 the solar equation is to apply.

For gregorian century 15 N
is also 2 as in 1600 we don’t apply the solar equation

For years before 1583 N
doesn’t change, but we must take count of the 10 days added to 1582, so N for
the Julian calendar is (2-10) mod 7 = 6.

In general, for Gregorian
years you have that

N = (2 + S – 12) mod 7 =
(k – |k/4|–10) mod 7

**N = (4+k – k –|k/4|) mod 7**

For the
two exceptions, they derive from the rule that

1) The paschal full moon can never occur on 19 April, so
in the years with epact 24 the pfm has to be determined as the epact was 25;

2) In the same golden cycle you must not have two pfm the
same day. So if in a cycle you have an epact 24 that becomes 25, when another
epact 25 occurs that is to be considered as 26.

In both cases, the pfm is shifted back a day.

As for the algorhitm, thus, this shift is indifferent
until e is less then 6, because the effect of shifting back the pfm (i.e. d
becomes 1 less) is to diminish e by 6 mod 7, that is e increases by 1.

So if e before the change is 0 to 5, it becomes 1 to 6
and compensates the diminution by 1 of d, so that Easter’s date remains the
same.

But if e is 6 before the change, after the change it
becomes 0 and then d+e diminishes by 7 and Easter date is shifted to the
previous Sunday.

So the first condition for the exceptions is that

**e = 6**

The

**first**exception occurs when epact is 24. This means that
d = (53 –
E) mod 30 = 29

So the
first exception occurs when

**d = 29 and e = 6**

The

**second**exception occurs when epact is 25 (and so d = 28) and in the same golden cycle there was an year with epact 24.
This
means that between the two years the epact is increased by 1 mod 30

So we
have, if x is the number of years between the one with epact 24 and the one
with epact 25:

11x mod 30 = 1

with x
integer and 1 < x < 18

The only
solution of that equation is x = 11 and so the second year must have a Golden
Number not less than 12, or that the second year has a Golden Number greater
than 11. As a = G – 1 this condition can also be expressed as a > 10

The
condition (11M+11) mod 30 <19 originally indicated by Gauss is equivalent to
a > 10.

So, the
conditions for the second exception are:

**d = 29 and e = 6 and a > 10**

and that’s
all.