Astronomy Answers
Julian Day Number

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\( \DeclareMathOperator{\trunc}{trunc} \DeclareMathOperator{\Div}{div} \DeclareMathOperator{\gcd}{gcd} \DeclareMathOperator{\lcm}{lcm} \DeclareMathOperator{\bdom}{dom} \def\dfloor#1{\left\lfloor #1 \right\rfloor} \def\dceil#1{\left\lceil #1 \right\rceil} \def\dparen#1{\left( #1 \right)} \def\dmod#1#2{\left\lfloor #1 \right\rceil_{#2}} \def\ddom#1#2{\left\lceil #1 \right\rfloor_{#2}} \def\dabs#1{\left| #1 \right|} \def\dfloorratio#1#2{\dfloor{\dfrac{#1}{#2}}} \def\dceilratio#1#2{\dceil{\dfrac{#1}{#2}}} \def\mod1ratio#1#2{\left\lfloor \dfrac{#1}{#2} \right\rceil_1} \def\dom1ratio#1#2{\left\lceil \dfrac{#1}{#2} \right\rfloor_1} \def\eqvide#1{\qquad\text{(vide \eqref{#1})}} \def\eqavide#1{\|\text{(vide \eqref{#1})}} \)

The first part of this page provides algorithms for some modern and historical calendars. The second part (from Section 12) explains how I deduced those algorithms.

There are many different algorithms in use for converting between calendar dates and day numbers. The only thing that counts is that they provide the correct answers for all day numbers and calendar dates. If there is only a limited number of input values and possible outcomes (for example for calculating the month number and day-within-the-month number from the day-within-the-year number), then you can usually find many different algorithms that all give just the right answers for those particular values (but perhaps very different answers for all other input values), even if there is no logical connection between that algorithm and the calendar. In such a case there is often also little or no logical connection between the algorithm that converts from date to day number and the algorithm that converts from day number to date, but the algorithms may be shorter then.

For every algorithm, you should know for which input values it was designed to work. Some calendar algorithms only work well with positive numbers and give wrong results for negative years or negative day numbers. The algorithms I give below are designed to work for all years (including negative ones), all months in each year, all days in each month, and for all Julian Day Numbers (including negative ones).

Some of the fomulas express a choice: If a certain condition is met, then a certain action should be taken (for example, apply a correction to something), and if that condition is not met, then a different action or no action at all should be taken. Such alternative paths are easy to use in manual calculations, in applications that process one date at a time, and in applications written in a programming language that requires compilation before the program can be executed (for example, written in C or Fortran) but are not very convenient for use in applications that can handle many dates at once (in an array) and that are written in a programming language that can be executed immediately (such as Basic or Perl or IDL), for which the execution of an if-then statement per date is much slower than the execution of a single fixed calculation for each date in an array.

If the calendrical calculations below here involve such choices, then where possible I give a formula that circumvents if-then constructions and is therefore convenient for application to arrays of dates in direct-execution programming languagues. The accompanying text explains which choice is being worked around.

1. Different Kinds of Day Numbers

For astronomical calculations that depend on time it is useful to have a continuous time scale that uses only a single unit, and not three like most calendars (days, months, years), and that in addition uses a fixed time zone so that there is no confusion about which precise instant of time is meant.

The IAU has adopted the Julian Date (JD) for this purpose. This should not be confused with "a date in the Julian Calendar". There are various other timescales that look like JD. The next table describes a few of them.

"Whole" means that that timescale uses whole numbers only and counts whole days only. "Fractional" means that that timescale uses fractional numbers (with a part after the decimal point) and indicates instants of time. The "Begins" column shows when each next calendar day begins. 12:00 UTC means that a new calendar day begins at 12:00 (noon) universal time, which is not 12:00 noon local time ― except if you happen to be in a time zone equal to UTC, such as in Great Britain in winter. 00:00 LT means that a new calendar day begins at midnight local time. The "From JD"-column shows how the other numbers can be calculated from JD. In that column, "TZ" stands for an adjustment for your local time zone; that adjustment is 0 for UTC.

In practice, "fractional" JDs and CJDs are usually written with at least one digit after the decimal marker (even if that digit is 0), and "whole" JDNs and CJDNs without digits after the decimal marker.

The zero point of JD (i.e., JD 0.0) corresponds to 12:00 UTC on 1 January −4712 in the Julian calendar. The zero point of CJD corresponds to 00:00 (midnight) local time on 1 January −4712. JDN 0 corresponds to the period from 12:00 UTC on 1 January −4712 to 12:00 UTC on 2 January −4712. CJDN 0 corresponds to 1 January −4712 (the whole day, in local time).

To avoid round-off errors it is best to do calendrical calculations with whole numbers only. In what follows we use CJDN for that.

2. Notation

2.1. Rounding

In the formulas given below, many numbers are rounded to a nearby whole number. It is important that those numbers get rounded in the right direction. Whole numbers are already rounded, so those don't change.

The types of rounding that we use here are downward rounding to the nearest whole number (for \( x \) this is indicated by \( ⌊x⌋ \)) or upward rounding to the nearest whole number (\( ⌈x⌉ \)). The standard rounding function on calculators usually rounds to the nearest whole number (our notation for that is \( [x] \)) and the standard conversion from a fractional number to a whole number in a computer language is usually by rounding in the direction of zero (our notation for that is \( \trunc(x) \)). We do not want those kinds of rounding here! The below table shows these three different kinds of rounding for some values.

2.2. Modular Arithmetic

If you divide the whole number \( y \) by the whole number \( x \) then you get a quotient \( q \) and a remainder \( r \). Sometimes we're interested in the quotient, and sometimes in the remainder. We choose to have the remainder never be negative. Then \( 0 ≤ r \lt |x| \) and

\begin{align} q \| = \dfloorratio{y}{x} \\ r \| = y \bmod x = y − x\dfloorratio{y}{x} \\ y \| = qx + r \end{align}

The notation \( y \bmod x \) means: the non-negative remainder from the division of \( y \) by \( x \).

We find

\begin{equation} \frac{y}{x} = \dfloorratio{y}{x} + \left( \frac{y}{x} \bmod 1 \right) = \dfloorratio{y}{x} + \frac{y \bmod x}{x} \end{equation}

and if \( x = 1 \) then

\begin{equation} y = ⌊y⌋ + (y \bmod 1) \end{equation}

The notation \( x ≡ y \pmod{n} \) means that \( x \) and \( y \) differ by some multiple of \( n \). One says that \( x \) and \( y \) are congruent modulo \( n \), and \( n \) is called the modulus. The equation \( x ≡ y \pmod{n} \) is called a congruence.

Modular arithmetic is arithmetic in which all calculations are modulo a fixed \( n \). That means that you can subtract arbitrary multiples of \( n \) from all terms and factors without changing the congruence. If \( x ≡ y \pmod{n} \) and \( p \) is a whole number, then also \( x + pn ≡ y \pmod{n} \), and \( px ≡ py \pmod{n} \), and also \( px ≡ py \pmod{pn} \).

Note the difference in notation: \( z = x \bmod y \) means that \( z \) is equal to the remainder of dividing \( x \) by \( y \), and \( z ≡ x \pmod{y} \) means that \( z \) is equal to \( x \) except for an arbitrary multiple of \( y \). If \( z = x \bmod y \) then also \( z ≡ x \pmod{y} \), but the opposite need not be true,

3. The Gregorian Calendar

See section 13.2.1 for the derivation of this algorithm.

3.1. From Gregorian Date to CJDN

One algorithm to calculate a CJDN \( J \) from a Gregorian date (calendar year \( j \), calendar month \( m \), calendar day \( d \)) is:

\begin{align} c_0 \| = \dfloorratio{m − 3}{12} \\ x_4 \| = j + c_0 \\ \{x_3, x_2\} \| = \Div(x_4, 100) \\ x_1 \| = m − 12c_0 − 3 \\ J \| = \dfloorratio{146097x_3}{4} + \dfloorratio{36525x_2}{100} + \dfloorratio{153x_1 + 2}{5} + d + 1721119 \end{align}

Here \( J_1 \) is the number of days from the beginning of calculation year 0 until the beginning of the current 100-calculation-year-period, \( J_2 \) the number of days since the beginning of the current 100-calculation-year-period, and \( J_3 \) the number of days minus one from the beginning of the current calculation year until the beginning of the current month. Calculation years run from 1 March to 1 March.

3.2. From CJDN to Gregorian Date

The algorithm to calculate a Gregorian date (calendar year \( j \), calendar month \( m \), calendar day \( d \)) from a CJDN \( J \) is:

\begin{align} \{x_3, r_3\} \| = \Div(4×J − 6884477, 146097) \\ \{x_2, r_2\} \| = \Div\left( 100\dfloorratio{r_3}{4} + 99, 36525 \right) \\ \{x_1, r_1\} \| = \Div\left( 5\dfloorratio{r_2}{100} + 2, 153 \right) \\ d \| = \dfloorratio{r_1}{5} + 1 \\ c_0 \| = \dfloorratio{x_1 + 2}{12} \\ j \| = 100x_3 + x_2 + c_0 \\ m \| = x_1 − 12c_0 + 3 \end{align}

Here \( x_3 \) is the number of calculation centuries since that time, \( x_2 \) the number of calculation years since that time, and \( x_1 \) the number of months since the beginning of the current calculation year.

4. The Milanković Calendar

See section 13.3 for the derivation of this algorithm.

4.1. From Milanković Date to CJDN

Some Eastern Orthodox churches use a calendar invented by Milutin Milanković, that differs from the Gregorian calendar only in the rules for which century years are leap years. One algorithm to calculate a CJDN \( J \) from a Milanković date (calendar year \( j \), calendar month \( m \), calendar day \( d \)) is:

\begin{align} c_0 \| = \dfloorratio{m − 3}{12} \\ x_4 \| = j + c_0 \\ x_3 \| = \dfloorratio{x_4}{100} \\ x_2 \| = x_4 \bmod 100 \\ x_1 \| = m − 12c_0 − 3 \\ J \| = \dfloorratio{328718x_3 + 6}{9} + \dfloorratio{36525x_2}{100} + \dfloorratio{153x_1 + 2}{5} + d + 1721119 \end{align}

4.2. From CJDN to Milanković Date

An algorithm to convert CJDN \( J \) to year \( j \), month \( m \), day \( d \) in the Milanković calendar is:

\begin{align} k_3 \| = 9×(J − 1721120) + 2 \\ x_3 \| = \dfloorratio{k_3}{328718} \\ k_2 \| = 100\dfloorratio{k_3 \bmod 328718}{9} + 99 \\ x_2 \| = \dfloorratio{k_2}{36525} \\ x_1 \| = \dfloorratio{5\dfloorratio{k_2 \bmod 36525}{100} + 2}{153} \\ c_0 \| = \dfloorratio{x_1 + 2}{12} \\ j \| = 100x_3 + x_2 + c_0 \\ m \| = x_1 − 12c_0 + 3 \\ d \| = \dfloorratio{k_1 \bmod 153}{5} + 1 \end{align}

5. The Julian Calendar

See section 13.1.1 for the derivation of this algorithm.

5.1. From Julian Date to CJDN

The algorithm to calculate a CJDN \( J \) from a Julian date (calendar year \( j \), calendar month \( m \), calendar day \( d \)) is:

\begin{align} J_0 \| = 1721117 \\ c_0 \| = \dfloorratio{m − 3}{12} \\ J_1 \| = \dfloorratio{1461×(j + c_0)}{4} \\ J_2 \| = \dfloorratio{153m − 1836c_0 − 457}{5} \\ J \| = J_1 + J_2 + d + J_0 \end{align}

Here \( J_1 \) is the number of days between the beginning of calculation year 0 and the beginning of the current calculation year, and \( J_2 \) the number of days between the beginning of the current calculation year and the beginning of the current month.

5.2. From CJDN to Julian Date

The algorithm to calculate a Julian date (calendar year \( j \), calendar month \( m \), calendar day \( d \)) from a CJDN \( J \) is:

\begin{align} y_2 \| = J − 1721118 \\ k_2 \| = 4y_2 + 3 \\ k_1 \| = 5\dfloorratio{k_2 \bmod 1461}{4} + 2 \\ x_1 \| = \dfloorratio{k_1}{153} \\ c_0 \| = \dfloorratio{x_1 + 2}{12} \\ j \| = \dfloorratio{k_2}{1461} + c_0 \\ m \| = x_1 − 12c_0 + 3 \\ d \| = \dfloorratio{k_1 \bmod 153}{5} + 1 \end{align}

6. The Islamic Calendar

See section 13.8 for the derivation of this algorithm.

6.1. From Islamic Date to CJDN

The religious Islamic calendar depends on observations and so cannot be caught in formulas. The administrative calendar has fixed rules so it can be caught in formulas. The difference between the religious and administrative calendars should usually be no more than 1 day.

The CJDN can be derived from the year number \( j \), month number \( m \), and day number \( d \) in the most commonly used administrative Islamic calendar (used by al-Fazārī, al-Khwārizmī, al-Battānī, and in the Toledan and Alfonsine Tables) as follows:

\begin{equation} J = \dfloorratio{10631j − 10617}{30} + \dfloorratio{325m − 320}{11} + d + 1948439 \end{equation}

6.2. From CJDN to Islamic Date

The following formulas calculate the year number \( j \), month number \( m \), and day number \( d \) in the most commonly used administrative Islamic calendar from the CJDN \( J \):

\begin{align} k_2 \| = 30×(J − 1948440) + 15 \\ k_1 \| = 11\dfloorratio{k_2 \bmod 10631}{30} + 5 \\ j \| = \dfloorratio{k_2}{10631} + 1 \\ m \| = \dfloorratio{k_1}{325} + 1 \\ d \| = \dfloorratio{k_1 \bmod 325}{11} + 1 \end{align}

7. The Babylonian Calendar

See section 13.5 for the derivation of this algorithm.

7.1. From Babylonian Date to CJDN

The Babylonians had a lunisolar calendar in which the beginning of each month was determined from observations of the Moon, but the number of months in each year followed a fixed 19-year pattern. The below formulas for an administrative Babylonian calendar uses that same 19-year pattern, and should usually deviate by at most one day from the calendar that the Babylonians used that depended on observations.

If \( j \) is the year number in the era of Seleukos, and \( m \) is the month number (beginning at 1) in the current year, and \( d \) is the day number (beginning at 1) in the current month, then the CJDN \( J \) can be found as follows:

\begin{align} y_1 \| = \dfloorratio{235j − 241}{19} + m \\ J \| = \dfloorratio{6940y_1}{235} + d + 1607557 \end{align}

7.2. From CJDN to Babylonian Date

From CJDN \( J \), we find the Babylonian year number \( j \), month number \( m \), and day number \( d \) as follows:

\begin{align} k_2 \| = 235×(J − 1607558) + 234 \\ k_1 \| = 19\dfloorratio{k_2}{6940} + 5 \\ j \| = \dfloorratio{k_1}{235} + 1 \\ m \| = \dfloorratio{k_1 \bmod 235}{19} + 1 \\ d \| = \dfloorratio{k_2 \bmod 6940}{235} + 1 \end{align}

8. The Jewish Calendar

See section 13.6 for the derivation of this algorithm.

8.1. From Jewish Date to CJDN

The Jewish calendar is a lunisolar calendar with complicated rules, and repeats itself only after 251,827,457 days = 35,975,351 weeks = 8,527,680 months = 689,472 years. The algorithm to transform a Jewish date into CJDN is long, and it features very large intermediate results. The algorithm that I provide here uses only whole numbers (to avoid round-off error), and assumes that all input values, intermediate results, and output values cannot be greater than 2³¹ = 2,147,483,648.

In the Jewish calendar, a year has 12 or 13 months, and a month has 29 or 30 days. New Year is the first day of the 7th month, Tishri. The months in a year of 12 months are: 1 = Nisan (נִיסָן‎), 2 = Iyar (אִיָּר / אייר‎), 3 = Sivan (סִיוָן / סיוון‎), 4 = Tammuz (תַּמּוּז‎), 5 = Av (אָב‎), 6 = Elul (אֱלוּל‎), 7 = Tishri (תִּשׁרִי‎), 8 = Ḥeshvan (מַרְחֶשְׁוָן / מרחשוון‎), 9 = Kislev (כִּסְלֵו / כסליו‎), 10 = Tevet (טֵבֵת‎), 11 = Shevat (שְׁבָט‎), 12 = Adar (אֲדָר‎). In a year of 13 months, embolismic month Adar Ⅰ (‎אֲדָר א׳) is inserted between Shevat and regular Adar, and that regular Adar is then temporarily renamed to Adar Ⅱ (אֲדָר ב׳‎). In a year of 13 months, we count Adar Ⅰ as month number 12, and Adar Ⅱ as month number 13.

One algorithm to transform calendar year \( j \), calendar month \( m \) (from 1 through 12 or 13), and calendar day \( d \) (from 1 through 29 or 30) into CJDN is then:

\begin{align} c_0 \| = \dfloorratio{13 − m}{7} \\ x_1 \| = j − 1 + c_0 \\ x_3 \| = m − 1 \\ z_4 \| = d − 1 \\ c_1(x_1) \| = \dfloorratio{235x_1 + 1}{19} \\ q(x_1) \| = \dfloorratio{c_1(x_1)}{1095} \\ r(x_1) \| = c_1(x_1) \bmod 1095 \\ υ_1(x_1) \| = 32336q(x_1) + \dfloorratio{15q(x_1) + 765433r(x_1) + 12084}{25920} \\ υ_2(x_1) \| = υ_1(x_1) + \left(\dfloorratio{6×(υ_1(x_1) \bmod 7)}{7} \bmod 2\right) \end{align}

Calculate in the same way also \( υ_2(x_1 + 1) \). Then

\begin{align} L_2(x_1) \| = υ_2(x_1 + 1) − υ_2(x_1) \\ L_2(x_1 − 1) \| = υ_2(x_1) − υ_2(x_1 − 1) \\ v_3 \| = 2\left(\dfloorratio{L_2(x_1) + 19}{15} \bmod 2\right) \\ v_4 \| = \dfloorratio{L_2(x_1 − 1) + 7}{15} \bmod 2 \\ c_2(x_1) \| = υ_2 + v_3 + v_4 \end{align}

Calculate in the same way also \( c_2(x_1 + 1) \) (which means you need to calculate \( υ_2(x_1 + 2) \), too). Then

\begin{align} L \| = c_2(x_1 + 1) − c_2(x_1) \\ c_8 \| = \dfloorratio{L + 7}{2} \bmod 15 \\ c_9 \| = −\left( \dfloorratio{385 − L}{2} \bmod 15 \right) \\ c_3 \| = \dfloorratio{384x_3 + 7}{13} + c_8\dfloorratio{x_3 + 4}{12} + c_9\dfloorratio{x_3 + 3}{12} \\ c_4(x_1,x_3) \| = c_2(x_1) + c_3(x_3) \\ J \| = J_0 − 177 + c_4(x_1,x_3) + z_4 = 347821 + c_2(x_1) + c_3 + z_4 \end{align}

8.2. From CJDN to Jewish Calendar

From CJDN \( J \) we go to Jewish year \( j \), month \( m \), and day \( d \):

\begin{align} y_4 \| = J − 347821 \\ q \| = \dfloorratio{y_4}{1447} \\ r \| = y_4 \bmod 1447 \\ y_1' \| = 49q + \dfloorratio{23q + 25920r + 13835}{765433} \\ γ_1 \| = y_1' + 1 \\ ξ_1 \| = \dfloorratio{19γ_1 + 17}{235} \\ μ_1 \| = γ_1 − \dfloorratio{235ξ_1 + 1}{19} \end{align}

Calculate \( c_{41}' = c_4(ξ_1, μ_1) \) in the way described in the previous section. Then

\begin{align} ζ_1 \| = y_4 − c_{41}' \\ γ_2 \| = γ_1 + \dfloorratio{ζ_1}{33} \\ ξ_2 \| = \dfloorratio{19γ_2 + 17}{235} \\ μ_2 \| = γ_2 − \dfloorratio{235ξ_2 + 1}{19} \end{align}

Calculate \( c_{42}' = c_4(ξ_2, μ_2) \) in the way described in the previous section. Then

\begin{align} ζ_2 \| = y_4 − c_{42}' \\ γ_3 \| = γ_2 + \dfloorratio{ζ_1}{33} \\ x_1 \| = ξ_3 = \dfloorratio{19γ_3 + 17}{235} \\ x_3 \| = μ_3 = γ_3 − \dfloorratio{235ξ_3 + 1}{19} \end{align}

Calculate \( c_{43}' = c_4(ξ_3, μ_3) \) in the way described in the previous section. Then

\begin{align} z_4 \| = ζ_3 = y_1 − c_{43}' \\ c \| = \dfloorratio{12 − x_3}{7} \\ j \| = x_1 + 1 − c \\ m \| = x_3 + 1 \\ d \| = z_4 + 1 \end{align}

9. The Egyptian Calendar

The ancient Egyptians had a very simple calendar indeed, without leap years and with 30 days to every month, except that the last month had 5 days. In the calendar according to the Era of Nabonassar the first day (1 Thoth of year 1) was equivalent to 26 February −746 in the Julian calendar.

9.1. From Egyptian Date to CJDN

The algorithm to translate an Egyptian date (calendar year \( j \), calendar month \( m \), calendar day \( d \)) to CJDN \( J \) is as follows:

\begin{equation} J = 365 j + 30 m + d + 1448242 \end{equation}

9.2. From CJDN to Egyptian Date

In the opposite direction, the calculations are simple as well.

\begin{align} y_2 \| = J − 1448638 \\ x_2 \| = \dfloorratio{y_2}{365} \\ y_1 \| = y_2 \bmod 365 \\ j \| = x_2 + 1 \\ m \| = \dfloorratio{y_1}{30} + 1 \\ d \| = y_1 − 30m + 31 = (y_1 \bmod 30) + 1 \end{align}

10. The Maya Calendar

10.1. From Maya Calendar to CJDN

The Maya from Anahuac (Central America) used three different calendars, of which two (the Tzolkin and Haab) were periodic with fairly short periods, and the third one (the Long Count) might have been intended to be periodic but has some periods that are so long that it can be considered to be continuous rather than repeating.

Different areas in Central America had slightly different versions of these calendars, with different names for days and months, and different ways to indicate years, and sometimes days were counted from 0 rather than from 1. Below we describe the calendars from the city of Tikal.

10.1.1. From Haab to CJDN

The Haab has a day number and a month, but no year number. There are 18 months of 20 days, plus a 19th month of 5 days, which makes 365 days in total, which we call a Haab year. There are no leap years. The months have a name, and the days have a number beginning at 0.

We write a date in the Haab as \( \{h_d,h_m\} \) where \( h_d \) is the day number (from 0 through 19) and \( h_m \) is the month number (from 1 through 19).

There is no year number in the Haab, so a particular Haab date \( \{h_d,h_m\} \) returns every 365 days. We need to apply additional information to pick the right one. We can find the last CJDN \( J \) on or before a particular CJDN \( J_0 \) that corresponds to this Haab date, as follows:

\begin{align} H \| = h_d + 20×(h_m − 1) \\ J \| = J_0 − ((J_0 − H + 65) \bmod 365) \end{align}

10.1.2. From Tzolkin to CJDN

The Tzolkin has a period of 20 days (the venteina) with a name for each day, and a period of 13 days (the trecena) with a number for each day (beginning at 1).

We write a date in the Tzolkin as \( \{t_t,t_v\} \), where \( t_t \) is the day number in the trecena (from 1 through 13) and \( t_v \) is the day number in the venteina (from 1 through 20).

The trecena and venteina increase simultaneously, so after day \( \{4,7\} \) comes day \( \{5,8\} \), and then \( \{6,9\} \), and so on. Because 13 and 20 are relatively prime, all possible combinations of venteina and trecena occur in this calendar, so after 13×20 = 260 days the dates repeat again.

There is no year number in the Tzolkin so a particular Tzolkin date returns every 260 days. We can find the last CJDN \( J \) that corresponds to Tzolkin date \( \{t_t,t_v\} \) on or before a particular CJDN \( J_0 \) as follows:

\begin{equation} J = J_0 − ((J_0 − 40 t_t + 39 t_v + 97) \bmod 260) \end{equation}

For the number of days \( T \) since the last \( \{1,1\} \) we have

\begin{equation} T = (40 t_t + 221 t_v − 1) \bmod 260 \end{equation}

10.1.3. From Tzolkin and Haab to CJDN

Sometimes a date is indicated in both Tzolkin and Haab. The CJDN \( J \) of the last Tzolkin/Haab date \( \{ h_d, h_m, t_t, t_v \} \) on or before CJDN \( J_0 \) is:

\begin{align} H \| = (h_d + 20 × (h_m − 1)) \bmod{365} \\ T \| = (40t_t + 221 t_v − 1) \bmod{260} \\ J \| = J_0 − ((J_0 − 365 T + 364 H − 7600) \bmod{18980}) \end{align}

The Tzolkin and Haab together have a period of 18980 days, which is approximately 52 years.

Beware! Not all possible combinations of \( \{ h_d, h_m, t_t, t_v \} \) occur in the calendar. You can fill in any combination you like in the preceding formulas, but you'll get useful results only for combinations that really occur in the calendar. Those are combinations for which \( H − T ≡ 65 − 96 ≡ 4 \pmod{5} \).

10.1.4. From the Long Count to CJDN

The Long Count counts days and has a series of increasingly longer periods. The smallest period is 20 days, the next one is 18 times longer (i.e., 360 days), and after that each next period is 20 times longer than the previous one. The number for each period begins at 0. Often dates in the Long Count are given with five numbers, but even longer periods are known, up to nine numbers. The longest known period corresponds to about 63 million years. Here we assume that the Long Count has five numbers, and that the fifth number can get arbitrarily long.

The epoch (0.0.0.0.0) of the Long Count probably corresponds to CJDN \( J_0 = 584283 \) (6 September −3113 in the Julian calendar).

Translating a Long Count to CJDN goes as follows:

\begin{equation} \begin{split} J \| = l_1 + 20×(l_2 + 18×(l_3 + 20×(l_4 + 20×l_5))) + J_0 \\ \| = l_1 + 20×l_2 + 360×l_3 + 7200×l_4 + 144000×l_5 \end{split} \end{equation}

10.2. From CJDN to the Maya Calendar

10.2.1. From CJDN to Haab

We calculate Haab date \( \{h_d,h_m\} \) from CJDN \( J \) as follows:

\begin{align} H \| = (J + 65) \bmod 365 \\ h_m \| = \dfloorratio{H}{20} + 1 \\ h_d \| = H \bmod 20 \end{align}

\( H \) is the number of days since the beginning of the current Haab year (with \( H = 0 \) for the first day of the Haab year).

10.2.2. From CJDN to Tzolkin

We translate CJDN \( J \) to a Tzolkin date as follows:

\begin{align} t_t \| = ((J + 5) \bmod{13}) + 1 \label{eq:trecena} \\ t_v \| = ((J + 16) \bmod{20}) + 1 \label{eq:venteina} \end{align}

10.2.3. From CJDN to the Long Count

Translating CJDN \( J \) to Long Count \( L ≡ l_5.l_4.l_3.l_2.l_1 \) goes as follows:

\begin{align} x_5 \| = J − J_0 \\ l_5 \| = \dfloorratio{x_5}{144000} \\ x_4 \| = x_5 \bmod 144000 \\ l_4 \| = \dfloorratio{x_4}{7200} \\ x_3 \| = x_4 \bmod 7200 \\ l_3 \| = \dfloorratio{x_3}{360} \\ x_2 \| = x_3 \bmod 360 \\ l_2 \| = \dfloorratio{x_2}{20} \\ l_1 \| = x_2 \bmod 20 \end{align}

10.3. A Lunar Calendar With Many Fixed Month Lengths

I've designed a lunisolar calendar in which nearly all months have the same number of days in every calendar year. The calendar follows the seasons by means of the Cycle of Meton, which lasts 19 years. Each year contains 12 or 13 months. The "shortest" years of 12 months contain 354 days. The "short" years of 12 months contain 355 days. The "long" years of 13 months contain 384 days. The year lengths are:

All together there are 235 months (125 long, 110 short) and 6940 days in the cycle of 19 years.

All odd months (except for the 13th) contain 30 days, and all even months contain 29 days, except that in short and long years (of 355 and 384 days) the 12th month contains 30 days instead of 29.

10.3.1. From Lunar Calendar to Running Day Number

Given year \( j \), month \( m \), and day \( d \) in this lunar calendar, the running day number \( y_2 \) can be calculated as follows:

\begin{equation} y_2 = 354j + 29m + 30\dfloorratio{7j − 6}{19} + \dfloorratio{4j + 8}{19} + \dfloorratio{7m}{13} + d − 384 \end{equation}

10.3.2. From Running Day Number to Lunar Calendar

Given running day number \( y_2 \), we calculate year \( j \), month \( m \), and day \( d \) in this lunar calendar as follows:

\begin{align} x' \| = \dfloorratio{19y_2 + 546}{6940} \\ x_2 \| = x' + \dfloorratio{y_2 − 354x' − 30\dfloorratio{7x' + 1}{19} − \dfloorratio{4x' + 12}{19}}{385} \\ k_1 \| = 13×\left( y_2 − 354x_2 − 30\dfloorratio{7x_2 + 1}{19} − \dfloorratio{4x_2 + 12}{19} \right) + 5 \\ j \| = x_2 + 1 \\ m \| = \dfloorratio{k_1}{384} + 1 \\ d \| = \dfloorratio{k_1 \bmod 384}{13} + 1 \end{align}

11. The Chinese Calendar

The Chinese calendar is a lunisolar calendar. The date of Chinese New Year depends on the position of the Sun and the Moon, but a precise method for calculating those positions has not been fixed. Here I discuss a few characteristics of the Chinese calendar.

11.1. The Year of the Monkey

In the traditional Chinese calendar, each year has a name from a cycle of 10 celestial stems and a name from a cycle of 12 terrestrial branches. For each next year both the celestial stem and terrestrial branch advance by one. The (untranslatable) celestial stems are 1. jiǎ, 2. yǐ, 3. bǐng, 4. dīng, 5. wù, 6. jǐ, 7. gēng, 8. xīn, 9. rén, 10. guǐ. The terrestrial branches are 1. zǐ (rat), 2. chǒu (ox), 3. yín (tiger), 4. mǎo (rabbit), 5. chén (dragon), 6. sì (snake), 7. wǔ (horse), 8. wèi (goat), 9. shēn (monkey), 10. yǒu (rooster), 11. xū (dog), 12. hài (pig).

You can calculate the celestial stem \( s \) (1 through 10) and terrestrial branch \( b \) (1 through 12) from the (astronomical) Gregorian year number \( j \) as follows:

\begin{align} s \| = ((j + 6) \bmod 10) + 1 \label{eq:jnaarchinees} \\ b \| = ((j + 8) \bmod 12) + 1 \end{align}

The number \( n \) in the combined cycle of 60 years is

\begin{equation} n = ((j + 56) \bmod 60) + 1 \end{equation}

The years that have a particular celestial stem \( s \) or terrestrial branch \( b \) or combination \( n \) are

\begin{align} j \| ≡ 1983 + s \pmod{10} \\ j \| ≡ 1983 + b \pmod{12} \\ j \| ≡ 1983 + n \pmod{60} \end{align}

The years that have celestial stem \( s \) as well as terrestrial branch \( b \) are

\begin{equation} j ≡ 1983 + 25b − 24s \pmod{60} \label{eq:chineesnaarj} \end{equation}

Beware! Not every combination of \( s \) and \( b \) is valid! Only combinations for which \( s ≡ b \pmod{2} \) are valid, so the difference between \( s \) and \( b \) must be even.

11.2. HYSN

The HYSN system is a special way to assign a number to a year. The conversion of a Gregorian (or Julian proleptic) year number to a HYSN year number goes as follows:

\begin{align} \{ H − 1, j_2 \} \| = \Div(j + 67016, 10800) \\ \{ Y − 1, j_3 \} \| = \Div(j_2, 360) \\ \{ S − 1, N − 1 \} \| = \Div(j_3, 30) \end{align}

In the other direction things are simpler. Given \( H \), \( Y \), \( S \), and \( N \), the Gregorian year is

\begin{equation} j = H×10800 + Y×360 + S×30 + N − 78207 \end{equation}

12. Derivation of the General Algorithms

We can derive many calendar formulas by finding straight lines that, with rounding, yield the correct results.

12.1. Notation

We use special notation for some things.

• \( C(x) \) has value 1 if test \( x \) is satisfied, and 0 otherwise. For example, \( C(3 \gt 2) = 1, C(3 \lt 2) = 0 \).

• \( x∥y \) asserts that \( x \) is a multiple of \( y \). For example, \( 8 ∥ 4 \) because 8 is a multiple of 4.

• \( x⊥y \) asserts that \( x \) is not a multiple of \( y \). For example, \( 7 ⊥ 4 \) because 7 is not a multiple of 4.

• \( \dfloor{x} \) is the floor function, which returns the nearest whole number not exceeding \( x \) (i.e., in the direction of \( −∞ \)). \( \dfloor{5.3} = 5, \dfloor{−2.8} = −3, \dfloor{7} = 7 \).

• \( \dmod{x}{p} \):

  \begin{equation} \dmod{x}{p} ≡ x \bmod p ≡ x − p\dfloorratio{x}{p} \label{eq:mod} \qquad (p \gt 0)\end{equation}

  is the modulus function that yields the non-negative difference between \( x \) and the nearest multiple of positive \( p \) not exceeding \( x \). Always (if \( p \gt 0 \))

  \begin{align} x \| = p\dfloorratio{x}{p} + \dmod{x}{p} \\ x \| = \dfloor{x} + \dmod{x}{1} \\ \dmod{x}{p} \| = p\left\lfloor \frac{x}{p} \right\rceil_1 \label{eq:mod1top} \\ 0 \| \le \dmod{x}{p} \lt p \\ \dmod{x + np}{p} \| = (x + np) − p\dfloorratio{x + np}{p} \notag \\ \| = x + np − p\dfloor{\dfrac{x}{p} + n} \notag \\ \| = x + np − p\dparen{\dfloorratio{x}{p} + n} \| \quad (n∥1)\notag \\ \| = x − p\dfloorratio{x}{p} = \dmod{x}{p} \label{eq:modn} \end{align}

  For example

  \({x}\)	\({4\dfloorratio{x}{4}}\)	\({\dmod{x}{4}}\)
  −2	−4	2
  −1	−4	3
  0	0	0
  1	0	1
  2	0	2
  3	0	3
  4	4	0
  5	4	1

• \( \dceil{x} \) is the ceiling function that yields the nearest whole number not less than \( x \) (i.e., in the direction of \( +∞ \)). For example, \( \dceil{5.3} = 6, \dceil{−2.8} = −2, \dceil{7} = 7 \).
• \( \ddom{x}{p}\):

  \begin{equation} \ddom{x}{p} ≡ x \bdom p ≡ p\dceilratio{x}{p} − x \qquad (p \gt 0) \label{eq:dom} \end{equation}

  is the domulus function that yields the non-negative difference between \( x \) and the nearest multiple of positive \( p \) not less than \( x \). I made up the notation \( \bdom \), which is \( \bmod \) spelled backwards. And I invented the name "domulus", like "modulus" but with "dom" instead of "mod". Always (if \( p \gt 0 \))

  \begin{align} x \| = p\dceilratio{x}{p} − \ddom{x}{p} \\ x \| = \dceil{x} − \ddom{x}{1} \\ \ddom{x}{p} \| = p\left\lceil \frac{x}{p} \right\rfloor_1 \label{eq:dom1top} \\ 0 \| \le \ddom{x}{p} \lt p \\ \ddom{x + np}{p} \| = p\dceilratio{x + np}{p} − (x + np) \notag \\ \| = p\dceil{\dfrac{x}{p} + n} − x − np \notag \\ \| = p\dparen{\dceil{\dfrac{x}{p}} + n} − x − np \| \quad (n∥1) \notag \\ \| = p\dceil{\dfrac{x}{p}} − x = \ddom{x}{p} \label{eq:domn} \end{align}

  For example

  \({x}\)	\({4\dceilratio{x}{4}}\)	\({\ddom{x}{4}}\)
  −2	0	2
  −1	0	1
  0	0	0
  1	4	3
  2	4	2
  3	4	1
  4	4	0
  5	8	3

• \( \{ m, n \} = \Div(x,y) \) is a function that yields two values \( m, n \) such that \( x = my + n \) with \( m \) a whole number, \( 0 \le n \lt y \gt 0 \), and

  \begin{align*} m \| = \dfloorratio{x}{y} \\ n \| = \dmod{x}{y} = x \bmod y \end{align*}

A few useful relationships (for \( p \gt 0 \)):

1. \begin{equation} \dceil{\dceil{x}} = \dfloor{\dceil{x}} = \dceil{x}\end{equation}
2. \begin{equation} \dfloor{\dfloor{x}} = \dceil{\dfloor{x}} = \dfloor{x} \end{equation}
3. \begin{equation} \dmod{\dmod{x}{p}}{p} = \ddom{\dmod{x}{p}}{p} = \dmod{x}{p} \label{eq:modmod} \end{equation}
4. \begin{equation} \ddom{\ddom{x}{p}}{p} = \dmod{\ddom{x}{p}}{p} = \ddom{x}{p} \label{eq:domdom} \end{equation}
5. \begin{equation} \dmod{x ± y}{p} = \dmod{\dmod{x}{p} ± \dmod{y}{p}}{p} \label{eq:dmod2} \end{equation}
6. \begin{equation} \ddom{x ± y}{p} = \ddom{\ddom{x}{p} ± \ddom{y}{p}}{p} \label{eq:ddom2} \end{equation}
7. \begin{equation} \dceil{x} − \dfloor{x} = C(x⊥1) \label{eq:ceilfloor} \end{equation}
8. \begin{equation} \dmod{x}{p} + \ddom{x}{p} = pC(x⊥p) \end{equation}
9. \begin{equation} \dfloor{−x} = −\dceil{x} = −\dfloor{x} − C(x⊥1) \label{eq:minusfloor} \end{equation}
10. \begin{equation} \dceil{−x} = −\dfloor{x} = −\dceil{x} + C(x⊥1) \label{eq:minusceil} \end{equation}
11. \begin{equation} \dmod{−x}{p} = \ddom{x}{p} = −\dmod{x}{p} + pC(x⊥p) \label{eq:minusmod} \end{equation}
12. \begin{equation} \ddom{−x}{p} = \dmod{x}{p} = −\ddom{x}{p} + pC(x⊥p) \label{eq:minusdom} \end{equation}
13. \begin{equation} \dfloor{x + y} = \dfloor{x} + \dfloor{y} + C\dparen{\dmod{x}{1} + \dmod{y}{1} ≥ 1} \label{eq:splitfloor} \end{equation}
14. \begin{equation} \dfloor{x − y} = \dfloor{x} − \dfloor{y} − C\dparen{\dmod{x}{1} − \dmod{y}{1} \lt 0} \label{eq:floorm} \end{equation}
15. \begin{equation} \dmod{x + y}{p} = \dmod{x}{p} + \dmod{y}{p} − pC\dparen{\dmod{x}{p} + \dmod{y}{p} ≥ p} \label{eq:modp} \end{equation}
16. \begin{equation} \dmod{x − y}{p} = \dmod{x}{p} − \dmod{y}{p} + pC\dparen{\dmod{x}{p} − \dmod{y}{p} \lt 0} \end{equation}
17. \begin{equation} \dceil{x + y} = \dceil{x} + \dceil{y} − C\dparen{\dmod{x}{1} + \dmod{y}{1} ≥ 1} \end{equation}
18. \begin{equation} \dceil{x − y} = \dceil{x} − \dceil{y} + C\dparen{\dmod{x}{1} − \dmod{y}{1} \lt 0} \end{equation}
19. \begin{equation} \ddom{x + y}{p} = \ddom{x}{p} + \ddom{y}{p} − pC\dparen{\ddom{x}{p} + \ddom{y}{p} ≥ p} \label{eq:domp} \end{equation}
20. \begin{equation} \ddom{x − y}{p} = \ddom{x}{p} − \ddom{y}{p} + pC\dparen{\ddom{x}{p} − \ddom{y}{p} \lt 0} \label{eq:domm} \end{equation}
21. \begin{equation} n\dmod{x}{1} − \dmod{nx}{1} = \dfloor{n\dmod{x}{1}} \qquad{(n ∥ 1)} \label{eq:dmoddiff1} \end{equation}

22. \begin{equation} x − y − 1 \lt \dfloor{x} − \dfloor{y} \lt x − y + 1 \label{eq:floordiffrange} \end{equation}
23. \begin{equation} \dceil{x − y} − 1 \le \dfloor{x} − \dfloor{y} \le \dfloor{x − y} + 1 \label{eq:floordiffrange2} \end{equation}
24. \begin{equation} x − y − 1 \lt \dceil{x} − \dceil{y} \lt x − y + 1 \label{eq:ceildiffrange} \end{equation}
25. \begin{equation} \dceil{x − y} − 1 \le \dceil{x} − \dceil{y} \le \dfloor{x − y} + 1 \label{eq:ceildiffrange2} \end{equation}
26. \begin{equation} \dceilratio{x}{y} = \dfloorratio{x − 1}{y} + 1 \qquad (x, y ∥ 1) \label{eq:ceil2floor} \end{equation}
27. \begin{equation} \ddom{x}{y} = y − 1 − \dmod{x − 1}{y} \qquad (x, y ∥ 1) \label{eq:ddom2dmod} \end{equation}
28. \begin{equation} \dfloor{\dmod{x}{y}} = \dmod{\dfloor{x}}{y} \qquad (y ∥ 1) \label{eq:modfloor} \end{equation}

And their proofs:

1. \( \dceil{\dceil{x}} = \dfloor{\dceil{x}} = \dceil{x}\) because \( \dceil{x} \) is a whole number.
2. \( \dfloor{\dfloor{x}} = \dceil{\dfloor{x}} = \dfloor{x} \) because \( \dfloor{x} \) is a whole number.
3. \( \dmod{\dmod{x}{p}}{p} = \ddom{\dmod{x}{p}}{p} = \dmod{x}{p} \) because \( 0 \le \dmod{x}{p} \lt p \).
4. \( \ddom{\ddom{x}{p}}{p} = \dmod{\ddom{x}{p}}{p} = \ddom{x}{p} \) because \( 0 \le \ddom{x}{p} \lt p \).
5. \( \dmod{x ± y}{p} = \dmod{\dmod{x}{p} ± \dmod{y}{p}}{p} \) because

   \begin{align} \dmod{x ± y}{p} \| = p\dmod{\dfrac{x ± y}{p}}{1} \notag \eqavide{eq:mod1top} \\ \| = p\dmod{\dfrac{p\dfloorratio{x}{p} + \dmod{x}{p} ± \dparen{p\dfloorratio{y}{p} + \dmod{y}{p}}}{p}}{1} \eqavide{eq:mod} \notag \\ \| = p\dmod{\dfloorratio{x}{p} ± \dfloorratio{y}{p} + \dfrac{\dmod{x}{p} ± \dmod{y}{p}}{p}}{1} \notag \\ \| = p\dmod{\dfrac{\dmod{x}{p} ± \dmod{y}{p}}{p}}{1} \eqavide{eq:modn} \notag \\ \| = \dmod{\dmod{x}{p} ± \dmod{y}{p}}{p} \eqavide{eq:mod1top} \end{align}

6. \( \ddom{x ± y}{p} = \ddom{\ddom{x}{p} ± \ddom{y}{p}}{p} \) because

   \begin{align} \ddom{x ± y}{p} \| = p\ddom{\dfrac{x ± y}{p}}{1} \notag \eqavide{eq:dom1top} \\ \| = p\ddom{\dfrac{p\dceilratio{x}{p} − \ddom{x}{p} ± \dparen{p\dceilratio{y}{p} − \ddom{y}{p}}}{p}}{1} \eqavide{eq:dom} \notag \\ \| = p\ddom{\dceilratio{x}{p} ± \dceilratio{y}{p} − \dparen{\dfrac{\ddom{x}{p} ± \ddom{y}{p}}{p}}}{1} \notag \\ \| = p\ddom{−\dfrac{\ddom{x}{p} ± \ddom{y}{p}}{p}}{1} \eqavide{eq:domn} \label{eq:xpyp} \end{align}

   so also

   \begin{align} \ddom{\ddom{x}{p} ± \ddom{y}{p}}{p} \| = p\ddom{−\dfrac{\ddom{\ddom{x}{p}}{p} ± \ddom{\ddom{y}{p}}{p}}{p}}{1} \eqavide{eq:xpyp} \notag \\ \| = p\ddom{−\dfrac{\ddom{x}{p} ± \ddom{y}{p}}{p}}{1} \eqavide{eq:domdom} \notag \\ \| = \ddom{x ± y}{p} \eqavide{eq:xpyp} \end{align}

7. \( \dceil{x} − \dfloor{x} = C(x⊥1) \) because when \( x \) is a whole number (so \( x∥1 \)) then \( \dceil{x} \) and \( \dfloor{x} \) are equal to \( x \), and otherwise the next higher whole nuber (\( \dceil{x} \)) is one greater than the next lower whole number (\( \dfloor{x} \)).

8. \( \dmod{x}{p} + \ddom{x}{p} = pC(x⊥p) \) because

   \begin{align} \dmod{x}{p} + \ddom{x}{p} \| = x − p\dfloorratio{x}{p} + p\dceilratio{x}{p} − x \| \text{(vide \eqref{eq:mod}, \eqref{eq:dom})} \notag \\ \| = p \dparen{\dceilratio{x}{p} − \dfloorratio{x}{p}} \notag \\ \| = p C\dparen{\dfrac{x}{p} ⊥ 1} \notag \eqavide{eq:ceilfloor} \\ \| = p C\dparen{x ⊥ p} \end{align}

9. For \( \dfloor{−x} = −\dceil{x} = −\dfloor{x} − C(x⊥1) \) the second \( = \) follows from Equation \eqref{eq:ceilfloor}.

   We prove the first \( = \) as follows.

   \begin{align} \dfloor{−x} \| = \dfloor{−\dfloor{x} − \dmod{x}{1}} \eqavide{eq:mod} \notag \\ \| = −\dfloor{x} + \dfloor{−\dmod{x}{1}} \\ \dceil{x} \| = \dceil{\dfloor{x} + \ddom{x}{1}} \eqavide{eq:dom} \notag \\ \| = \dfloor{x} + \dceil{\ddom{x}{1}} \\ \dfloor{−x} + \dceil{x} \| = \dfloor{−\dmod{x}{1}} + \dceil{\ddom{x}{1}} \end{align}

   If \( x \) is a whole number then \( \dmod{x}{1} = \ddom{x}{1} = 0 \) so \( \dfloor{−x} + \dceil{x} = 0 \).

   And if \( x \) is not a whole number then

   \begin{equation} 0 \lt \dmod{x}{1} \lt 1 ⇒ \dceil{\dmod{x}{1}} = 1 \end{equation}

   and also

   \begin{equation} −1 \lt −\dmod{x}{1} \lt 0 ⇒ \dfloor{−\dmod{x}{1}} = −1 \end{equation}

   so then, too, \( \dfloor{−x} + \dceil{x} = \dceil{\dmod{x}{1}} + \dfloor{−\dmod{x}{1}} = 0 \) and so \( \dfloor{−x} = −\dceil{x} \).

10. For \( \dceil{−x} = −\dfloor{x} = −\dceil{x} + C(x⊥1) \), the second \( = \) follows from Equation \eqref{eq:ceilfloor}.

    The first \( = \) follows from Equation \eqref{eq:minusfloor} if you replace every \( x \) by \( −x \).

11. For \( \dmod{−x}{p} = \ddom{x}{p} = −\dmod{x}{p} + pC(x⊥p) \) the first \( = \) follows from

    \begin{align} \dmod{−x}{p} \| = −x − p\dfloorratio{−x}{p} \eqavide{eq:mod} \notag \\ \| = p\dceilratio{x}{p} − x \eqavide{eq:minusfloor} \notag \\ \| = \ddom{x}{p} \eqavide{eq:dom} \end{align}

    and the second \( = \) from

    \begin{align} \dmod{x}{p} \| = x − p\dfloorratio{x}{p} \notag \eqavide{eq:mod} \\ \| = x − p\dparen{\dceilratio{x}{p} − C\dparen{\dfrac{x}{p} ⊥ 1}} \notag \eqavide{eq:ceilfloor} \\ \| = x − p\dceilratio{x}{p} + pC\dparen{x ⊥ p} \notag \\ \| = −\ddom{x}{p} + pC\dparen{x ⊥ p} \eqavide{eq:dom} \end{align}

12. \( \ddom{−x}{p} = \dmod{x}{p} = −\ddom{x}{p} + pC(x⊥p) \) follows from the previous one if you replace every \( x \) with \( −x \).

13. \( \dfloor{x + y} = \dfloor{x} + \dfloor{y} + C\dparen{\dmod{x}{1} + \dmod{y}{1} ≥ 1} \) because

    \begin{align} \dfloor{x ± y} \| = x ± y − \dmod{x ± y}{1} \eqavide{eq:mod} \notag \\ \| = \dfloor{x} + \dmod{x}{1} ± \dparen{\dfloor{y} + \dmod{y}{1}} − \dmod{x ± y}{1} \eqavide{eq:mod} \notag \\ \| = \dfloor{x} ± \dfloor{y} + \dmod{x}{1} ± \dmod{y}{1} − \dmod{x ± y}{1} \notag \\ \| = \dfloor{x} ± \dfloor{y} + \dmod{x}{1} ± \dmod{y}{1} − \dmod{\dmod{x}{1} ± \dmod{y}{1}}{1} \eqavide{eq:dmod2} \label{eq:floorpm} \end{align}

    We have

    \begin{align} \dmod{\dmod{x}{1} + \dmod{y}{1}}{1} \| = \dmod{x}{1} + \dmod{y}{1} − C\dparen{\dmod{x}{1} + \dmod{y}{1} \ge 1} \\ \dmod{\dmod{x}{1} − \dmod{y}{1}}{1} \| = \dmod{x}{1} − \dmod{y}{1} + C\dparen{\dmod{x}{1} − \dmod{y}{1} \lt 0} \label{eq:mod1m} \end{align}

    because \( 0 \le \dmod{•}{1} \lt 1 \). So

    \begin{equation} \dfloor{x + y} = \dfloor{x} + \dfloor{y} + C\dparen{\dmod{x}{1} + \dmod{y}{1} \ge 1} \end{equation}

14. \( \dfloor{x − y} = \dfloor{x} − \dfloor{y} + C\dparen{\dmod{x}{1} − \dmod{y}{1} \lt 0} \) follows from Eqs. \eqref{eq:floorpm} and \eqref{eq:mod1m}.
15. \( \dmod{x + y}{p} = \dmod{x}{p} + \dmod{y}{p} − pC\dparen{\dmod{x}{p} + \dmod{y}{p} ≥ p} \) follows from

    \begin{align} \dmod{x ± y}{p} \| = x ± y − p\dfloorratio{x ± y}{p} \eqavide{eq:mod} \notag \\ \| = p\dfloorratio{x}{p} + \dmod{x}{p} ± \dparen{p\dfloorratio{y}{p} + \dmod{y}{p}} \notag \\ \| − p\dfloorratio{p\dfloorratio{x}{p} + \dmod{x}{p} ± \dparen{p\dfloorratio{y}{p} + \dmod{y}{p}}}{p} \notag \eqavide{eq:mod} \\ \| = p\dfloorratio{x}{p} + \dmod{x}{p} ± p\dfloorratio{y}{p} ± \dmod{y}{p} \notag \\ \| − p\dfloorratio{x}{p} ∓ p\dfloorratio{y}{p} − p\dfloorratio{\dmod{x}{p} ± \dmod{y}{p}}{p} \notag \\ \| = \dmod{x}{p} ± \dmod{y}{p} − p\dfloorratio{\dmod{x}{p} ± \dmod{y}{p}}{p} \label{eq:i15a} \end{align}

    We have

    \begin{align} \dfloorratio{\dmod{x}{p} + \dmod{y}{p}}{p} \| = C\dparen{\dmod{x}{p} + \dmod{y}{p} \ge p} \label{eq:i15c} \\ \dfloorratio{\dmod{x}{p} − \dmod{y}{p}}{p} \| = −C\dparen{\dmod{x}{p} − \dmod{y}{p} \lt 0} \label{eq:i15b} \end{align}

    because \( 0 \le \dmod{•}{p} \lt p \), so

    \begin{equation} \dmod{x}{p} + \dmod{y}{p} − p\dfloorratio{\dmod{x}{p} + \dmod{y}{p}}{p} = \dmod{x}{p} + \dmod{y}{p} − pC\dparen{\dmod{x}{p} + \dmod{y}{p} \ge p} \end{equation}

    which completes the proof.

16. \( \dmod{x − y}{p} = \dmod{x}{p} − \dmod{y}{p} + pC\dparen{\dmod{x}{p} − \dmod{y}{p} \lt 0} \) follows from Eqs. \eqref{eq:i15a} and \eqref{eq:i15b}.
17. \( \dceil{x + y} = \dceil{x} + \dceil{y} − C\dparen{\dmod{x}{1} + \dmod{y}{1} ≥ 1} \) because

    \begin{align} \dceil{x ± y} \| = x ± y + \ddom{x ± y}{p} \notag \eqavide{eq:dom} \\ \| = \dceil{x} − \ddom{x}{1} ± \dparen{\dceil{y} − \ddom{y}{1}} + \ddom{x ± y}{p} \notag \eqavide{eq:dom} \\ \| = \dceil{x} ± \dceil{y} − \ddom{x}{1} ∓ \ddom{y}{1} + \ddom{x ± y}{p} \label{eq:i17} \end{align}

    That leads to

    \begin{align} \dceil{x + y} \| = \dceil{x} + \dceil{y} − \ddom{x}{1} − \ddom{y}{1} + \ddom{x + y}{p} \notag \\ \| = \dceil{x} + \dceil{y} − \ddom{x}{1} − \ddom{y}{1} + \ddom{x}{1} + \ddom{y}{1} − C\dparen{\ddom{x}{1} + \ddom{y}{1} ≥ 1} \eqavide{eq:domp} \notag \\ \| = \dceil{x} + \dceil{y} − C\dparen{\ddom{x}{1} + \ddom{y}{1} ≥ 1} \end{align}

18. \( \dceil{x − y} = \dceil{x} − \dceil{y} + C\dparen{\dmod{x}{1} − \dmod{y}{1} \lt 0} \) because

    \begin{align} \dceil{x − y} \| = \dceil{x} − \dceil{y} − \ddom{x}{1} + \ddom{y}{1} + \ddom{x − y}{p} \notag \eqavide{eq:i17} \\ \| = \dceil{x} − \dceil{y} − \ddom{x}{1} + \ddom{y}{1} + \ddom{x}{1} − \ddom{y}{1} + C\dparen{\ddom{x}{1} − \ddom{y}{1} \lt 0} \eqavide{eq:modp} \notag \\ \| = \dceil{x} − \dceil{y} + C\dparen{\ddom{x}{1} − \ddom{y}{1} \lt 0} \end{align}

19. \( \ddom{x + y}{p} = \ddom{x}{p} + \ddom{y}{p} − pC\dparen{\ddom{x}{p} + \ddom{y}{p} ≥ p} \) because

    \begin{align} \ddom{x ± y}{p} = \| p\dceilratio{x ± y}{p} − \dparen{x ± y} \notag \eqavide{eq:dom} \\ = \| p\dceilratio{p\dceilratio{x}{p} − \ddom{x}{p} ± \dparen{p\dceilratio{y}{p} − \ddom{y}{p}}}{p} \notag \\ \| − \dparen{p\dceilratio{x}{p} − \ddom{x}{p} ± \dparen{p\dceilratio{y}{p} − \ddom{y}{p}}} \notag \eqavide{eq:dom} \\ = \| p\dceilratio{x}{p} ± p\dceilratio{y}{p} + p\dceilratio{−\ddom{x}{p} ∓ \ddom{y}{p}}{p} \notag \\ \| − \dparen{p\dceilratio{x}{p} ± p\dceilratio{y}{p} − \dparen{\ddom{x}{p} ± \ddom{y}{p}}} \notag \\ = \| \ddom{x}{p} ± \ddom{y}{p} + p\dceilratio{−\ddom{x}{p} ∓ \ddom{y}{p}}{p} \label{eq:i19a} \end{align}

    We have

    \begin{align} \dceilratio{−\ddom{x}{p} − \ddom{y}{p}}{p} \| = −C\dparen{\ddom{x}{p} + \ddom{y}{p} \ge p} \\ \dceilratio{−\ddom{x}{p} + \ddom{y}{p}}{p} \| = C\dparen{\ddom{x}{p} − \ddom{y}{p} \lt 0} \label{eq:i19b} \end{align}

    because \( 0 \le \ddom{•}{p} \lt 1 \), so

    \begin{equation} \ddom{x}{p} + \ddom{y}{p} + p\dceilratio{\ddom{x}{p} + \ddom{y}{p}}{p} = \ddom{x}{p} + \ddom{y}{p} − pC\dparen{\ddom{x}{p} + \ddom{y}{p} \ge p} \end{equation}

    which completes the proof.

20. \( \ddom{x − y}{p} = \ddom{x}{p} − \ddom{y}{p} + pC\dparen{\ddom{x}{p} − \ddom{y}{p} \lt 0} \) follows from Eqs. \eqref{eq:i19a} and \eqref{eq:i19b}.

21. \begin{eqnarray} n\dmod{x}{1} − \dmod{nx}{1} \| = \| n\dmod{x}{1} − \dmod{n\dfloor{x} + n\dmod{x}{1}}{1} \notag \\ \| = \| n\dmod{x}{1} − \dmod{n\dmod{x}{1}}{1} \qquad{(n ∥ 1)} \eqavide{eq:modp} \notag \\ \| = \| \dfloor{n\dmod{x}{1}} \eqavide{eq:mod} \end{eqnarray}
22. \begin{align} \dfloor{x} − \dfloor{y} \| = \dparen{x − \dmod{x}{1}} − \dparen{y − \dmod{y}{1}} \notag \\ \| = x − y + \dmod{y}{1} − \dmod{x}{1} \end{align}

    Then

    \begin{equation} x − y − 1 \lt \dfloor{x} − \dfloor{y} \lt x − y + 1 \end{equation}

    because

    \begin{equation} 0 \le \dmod{•}{1} \lt 1 \end{equation}

23. \begin{equation} \dceil{x − y} − 1 \le \dfloor{x} − \dfloor{y} \le \dfloor{x − y} + 1 \end{equation}

    because

    \begin{align} x − y − 1 \lt \dfloor{x} − \dfloor{y} \lt x − y + 1 \eqavide{eq:floordiffrange} \\ x − y \le \dceil{x − y} \| \quad (\dfloor{x} − \dfloor{y} ∥ 1) \end{align}

24. \begin{align} \dceil{x} − \dceil{y} \| = \dparen{x + \ddom{x}{1}} − \dparen{y + \ddom{y}{1}} \notag \\ \| = x − y + \ddom{x}{1} − \ddom{y}{1} \end{align}

    Then

    \begin{equation} x − y − 1 \lt \dceil{x} − \dceil{y} \lt x − y + 1 \end{equation}

    because

    \begin{equation} 0 \le \ddom{•}{1} \lt 1 \end{equation}

25. \begin{equation} \dceil{x − y} − 1 \le \dceil{x} − \dceil{y} \le \dfloor{x − y} + 1 \end{equation}

    because

    \begin{align} x − y − 1 \lt \dceil{x} − \dceil{y} \lt x − y + 1 \eqavide{eq:ceildiffrange} \\ x − y \le \dceil{x − y} \| \quad (\dfloor{x} − \dfloor{y} ∥ 1) \end{align}

26. If \( y = 1 \):

    \begin{align} \dceilratio{x}{y} \| = \dceilratio{x}{1} = x \| \qquad (x ∥ 1) \\ \| = (x − 1) + 1 = \dfloorratio{x − 1}{1} + 1 = \dfloorratio{x − 1}{y} + 1 \end{align}

    If \( y \gt 1 \):

    \begin{align} \dceilratio{x}{y} \| = \dfloorratio{x}{y} + C(x⊥y) \eqavide{eq:ceilfloor} \notag \\ \| = \dfloorratio{x}{y} + C\dparen{\dmod{x}{y} \ge 1} \| \qquad (x, y ∥ 1) \notag \\ \| = \dfloorratio{x}{y} + C\dparen{\dmod{\dfrac{x}{y}}{1} \ge \dfrac{1}{y}} \notag \\ \| = \dfloorratio{x}{y} + C\dparen{\dmod{\dfrac{x}{y}}{1} \ge \dmod{\dfrac{1}{y}}{1}} \| (y \gt 1) \notag \\ \| = \dfloorratio{x}{y} − \dfloorratio{1}{y} + C\dparen{\dmod{\dfrac{x}{y}}{1} − \dmod{\dfrac{1}{y}}{1} \ge 0} \| (y \gt 1) \notag \\ \| = \dfloorratio{x}{y} − \dfloorratio{1}{y} + 1 − C\dparen{\dmod{\dfrac{x}{y}}{1} − \dmod{\dfrac{1}{y}}{1} \lt 0} \notag \\ \| = \dfloorratio{x − 1}{y} + 1 \eqavide{eq:floorm} \end{align}

27. If \( y = 1 \):

    \begin{equation} \ddom{x}{y} = \ddom{x}{1} = 0 = y − 1 − \ddom{x − 1}{y} \qquad (x ∥ 1) \end{equation}

    If \( y \gt 1 \):

    \begin{align} \ddom{x}{y} \| = y\ddom{\dfrac{x}{y}}{1} \notag \\ \| = y\dparen{\dceil{\dfrac{x}{y}} − \dfrac{x}{y}} \eqavide{eq:dom} \notag \\ \| = y\dparen{\dfloorratio{x − 1}{y} + 1 − \dfrac{x}{y}} \eqavide{eq:ceil2floor} \notag \\ \| = y\dparen{\dfrac{x − 1}{y} − \dmod{\dfrac{x − 1}{y}}{1} + 1 − \dfrac{x}{y}} \eqavide{eq:mod} \notag \\ \| = x − 1 − y\dmod{\dfrac{x − 1}{y}}{1} + y − x \notag \\ \| = y − 1 − \dmod{x − 1}{y} \end{align}

28. Let

    \begin{align} \{ k, α \} \| = \Div(x, y) \\ \{ m, β \} \| = \Div(α, 1) \end{align}

    so

    \begin{equation} x = ky + α = ky + m + β \end{equation}

    where

    \begin{align} k, m, y \| ∥ 1 \\ 0 \| ≤ m ≤ y − 1 \\ 0 \| ≤ β \lt 1 \end{align}

    then

    \begin{align} \dfloor{\dmod{x}{y}} \| = \dfloor{\dmod{ky + m + β}{y}} \notag \\ \| = \dfloor{\dmod{m + β}{y}} \notag \\ \| = \dfloor{m + β} \| (0 ≤ m + β \lt y) \notag \\ \| = m \end{align}

    and also

    \begin{align} \dmod{\dfloor{x}}{y} \| = \dmod{\dfloor{ky + m + β}}{y} \notag \\ \| = \dmod{ky + m}{y} = m \end{align}

    so

    \begin{equation} \dfloor{\dmod{x}{y}} = \dmod{\dfloor{x}}{y} \end{equation}

12.2. Large Intermediate Results

Many of the calendar calculations that are described below are of the type

\begin{align} y \| = \dfloorratio{fx + t}{d} \label{eq:template} \\ e \| = \dmod{fx + t}{d} = (fx + t) \bmod d \label{eq:template-e} \end{align}

i.e.,

\[ \{y, e\} = \Div(fx + t, d) \]

with

\begin{equation} fx + t = dy + e \end{equation}

where \( f \), \( t \), \( d \) are fixed whole numbers with \( f, d \gt 1 \) and \( 0 ≤ e \lt d \) and \( y \), \( x \) are variable but whole numbers. The intermediate result \( fx + t \) is then usually much greater in magnitude than \( x \) and than the end result. This can give trouble on calculators and in computer programs where whole numbers cannot exceed a certain size that we'll refer to as \( w \). (Floating-point numbers like 1.234 × 10²⁰ can be much greater, but have limited accuracy, and then you have to worry about round-off errors.)

Intermediate results greater than \( w \) give trouble, so to avoid trouble we'd need roughly \( |fx| ≤ w \), so \( |x| ≤ w/f \), which is much less than \( w \) itself.

If \( |t| ≥ d \), then we can rewrite equation \eqref{eq:template}ff to

\begin{align} fx + t \| = fx + \dfloorratio{t}{d} d + \dmod{t}{d} = dy + e \eqavide{eq:mod} \\ y \| = \dfloorratio{fx + t}{d} = \dfloorratio{t}{d} + \dfloorratio{fx + \dmod{t}{d}}{d} \\ e \| = \dmod{fx + t}{d} = \dmod{fx + \dmod{t}{d}}{d} \eqavide{eq:modmod} \end{align}

where the second term of the last two equations has the same form as equation \eqref{eq:template}, if you rename \( \dmod{t}{d} \) to a new \( t \). Below, we assume that that has been done, which means that \( 0 ≤ t \lt d \).

If \( f \gt d \), then we can rewrite equation \eqref{eq:template}ff to

\begin{align} fx + t \| = \dparen{\dfloorratio{f}{d} d + \dmod{f}{d}} x + t = dy + e \eqavide{eq:mod} \\ y \| = \dfloorratio{f}{d} x + \dfloorratio{\dmod{f}{d} x + t}{d} \label{eq:template2} \\ e \| = \dmod{\dmod{f}{d} x + t}{d} \end{align}

Then \( |x| \) may be roughly as great as \( w/\dmod{f}{d} \) before we run into trouble.

Equations \eqref{eq:template2}ff again have the form of equations \eqref{eq:template}ff, if you rename \( \dmod{f}{d} \) to a new \( f \). Below, we assume that this has been done, so that \( f \lt d \).

If we could first do the division by \( d \) and then the multiplication by \( f \), then the intermediate results would stay smaller than \( x \) itself. We'd like to calculate something like \( f\dfloorratio{x}{d} \) which is roughly equal to \( y \), and then apply a correction to get the real \( y \).

We can rewrite equation \eqref{eq:template}ff to

\begin{align} fx + t \| = f\dparen{ \dfloorratio{x}{d} d + \dmod{x}{d}} + t \eqavide{eq:mod} \\ y \| = f\dfloorratio{x}{d} + \dfloorratio{f\dmod{x}{d} + t}{d} \label{eq:detour1} \\ e \| = \dmod{f\dmod{x}{d} + t}{d} \end{align}

Now the greatest intermediate result is no longer \( fx + t \), which can get as large as \( fw \), but \( f\dmod{x}{d} + t \), which cannot exceed \( fd \lt d^2 \).

Equations \eqref{eq:detour1}ff again have the form of equations \eqref{eq:template}ff, if you rename \( \dmod{x}{d} \) to a new \( x \).

Sometimes an intermediate result of (nearly) \( fd \) is still too great, because the product of two numbers can be much greater than \( w \) even if the two numbers are each much less than \( w \). If the product must remain less than \( w \), then the two numbers cannot both be greater than \( \sqrt{w} \), which is a lot less than \( w \) itself. For example, for 32-bit numbers, \( w = 2^{31} − 1 = 2147483647 \) and \( \sqrt{w} = 46341 \), so even if \( d \) and \( f \) are both just a bit greater than 46431, which isn't really very large, then \( fd \) is already too great to fit into a 32-bit number.

Suppose we pick a whole number \( P \gt 0 \) and then calculate one time

\begin{align} \{Q, R\} \| = \Div(fP, d) \label{eq:PQR} \\ Q \| = \dfloorratio{fP}{d} \\ R \| = \dmod{fP}{d} \end{align}

so

\begin{equation} fP = dQ + R \end{equation}

Then, for each desired \( x \), calculate

\begin{align} \{q, r\} \| = \Div(x, P) \\ q \| = \dfloorratio{x}{P} \\ r \| = \dmod{x}{P} \end{align}

so

\begin{equation} x = qP + r \end{equation}

and then

\begin{align} fx + t \| = fqP + fr + t = qdQ + qR + fr + t \\ y \| = \dfloorratio{qdQ + qR + fr + t}{d} = qQ + \dfloorratio{qR + fr + t}{d} \label{eq:detour2} \\ e \| = \dmod{qdQ + qR + fr + t}{d} = \dmod{qR + fr + t}{d} \end{align}

Now the greatest intermediate result is \( m ≡ qR + fr + t \). To stay out of trouble we need \( \dabs{m} ≤ w \), so (because \( r \lt P \))

\begin{eqnarray} \dabs{x}\frac{R}{P} + fP + t ≤ w \\ \dabs{x} ≤ X ≡ \dparen{w − fP − t}\dfrac{P}{R} \end{eqnarray}

We have \( f \lt d \) so \( f \) is not a multiple of \( d \) so \( R \gt 0 \).

We must have \( 0 \lt P \lt \dfrac{w − t}{f} \), otherwise certainly \( X ≤ 0 \).

We want to make the range of \( x \) as large as possible, so we seek \( P \) and \( R \) such that \( X \) becomes as great as possible, given \( w \), \( f \), \( t \). This means that \( R \) should preferably be small.

If \( P \) increases slowly (with steps of 1) then \( R = \dmod{fP}{d} \) varies strongly between 0 and \( d − 1 \), so the value of \( R \) is independent of the global magnitude of \( P \), and many different values of \( P \) are possible for the same smallest value of \( R \).

If \( R \) has the smallest possible value (greater than zero), then roughly what should the value of \( P \) be that yields the greatest value of \( X \)?

Then \( X \) is a quadratic function of \( P \). If \( P \) could have any value then the greatest possible value of \( X \) would occur for

\begin{equation} P = P_0 = \dfrac{w − t}{2f} \end{equation}

and be equal to

\begin{equation} \max(X) = \dfrac{\dparen{w − t}^2}{4Rf} = \dfrac{fP_0^2}{R} \end{equation}

To get \( \max(X) ≥ w \) we'd certainly need

\begin{equation} R ≤ \dfrac{w}{4f}\dparen{1−\dfrac{t}{w}}^2 \label{eq:maxR} \end{equation}

The least possible value that \( R \) can have is equal to the greatest common divisor \( \gcd(f, d) \) of \( f \) and \( d \).

\begin{equation} R = R_1 ≡ \gcd(f, d) \end{equation}

The Extended Algorithm of Euclid produces that value, and also numbers \( x \) and \( y \) such that

\begin{equation} fx + dy = R \end{equation}

The \( x \) and \( y \) that you get are in some sense the smallest, but the following values also work (with \( k \) an arbitrary whole number):

\begin{equation} f\dparen{x + k\dfrac{d}{R}} + d\dparen{y − k\dfrac{f}{R}} = R \end{equation}

Because \( R \) is a divisor of \( d \) and also of \( f \), the divisions by \( R \) in this formula always produce whole numbers.

The corresponding possible values of \( P \) and \( Q \) are then:

\begin{align} P \| = x + k\dfrac{d}{R} \\ Q \| = −y + k\dfrac{f}{R} \end{align}

for arbitrary whole \( k \) for which \( P, Q \gt 0 \). The smallest usable positive \( P \) and the corresponding value of \( Q \) are

\begin{align} P_1 \| = \dmod{x}{d/R_1} \\ Q_1 \| = \dmod{−y}{f/R_1} \end{align}

These \( P_1, Q_1, R_1 \) comply with equations \eqref{eq:PQR}ff, because

\begin{equation} \dfloorratio{fP_1}{d} = \dfloorratio{dQ_1 + R_1}{d} = Q_1 + \dfloorratio{R_1}{d} = Q_1 \end{equation}

because \( \dfloor{R_1/d} = 0 \), because \( R_1 = \gcd(f,d) \lt d \), because\( f \lt d \). And if \( Q_1 \) fits then \( R_1 \) fits, too.

To get \( X \gt 0 \) we certainly need \( P ≤ w/f \), so the greatest usable \( P \) is

\begin{equation} P_\text{max} = \dfrac{w}{f} − \dmod{\dfrac{w}{f} − P_1}{d/R_1} \end{equation}

The step size between successive values of \( P \) that fit with this \( R \) is \( d/R \) so if \( d/R \gt w/f \) (so \( fd/R \gt w \)) then there may be no acceptable value of \( P \) for this \( R \) at all. In that case \( P_\text{max} \lt 0 \).

If you cannot find an acceptable value of \( P \) for \( R = R_1 \), then try ever greater multiples of \( R_1 \) until you do find an acceptable \( P \). That is possible, because we have

\begin{equation} fP_1 − dQ_1 = R_1 \end{equation}

so also

\begin{equation} fnP_1 − dnQ_1 = nR_1 \end{equation}

for arbitrary \( n \).

But \( nP_1 \) must not be greater than \( w/f \) so \( n \) must not be greater than \( w/(fP_1) \) which often isn't very large.

To yet find more smaller values of \( P \) we can make use of the modulus function. If we use for \( P \)

\begin{equation} P = \dmod{nP_1}{(d/R_1)} \end{equation}

then

\begin{align} \dfrac{nP_1R_1}{d} \| = n\dfrac{R_1}{f}\dfrac{dQ_1 + R_1}{d} \\ P \| = \dfrac{d}{R_1}\dmod{\dfrac{nP_1R_1}{d}}{1} \\ \dfrac{fP}{d} \| = \dfrac{f}{R_1}\dmod{\dfrac{nP_1R_1}{d}}{1} = \dfrac{f}{R_1}\dmod{n\dfrac{R_1}{f}\dfrac{dQ_1 + R_1}{d}}{1} \notag \\ \| = \dmod{nQ_1 + \dfrac{nR_1}{d}}{(f/R_1)} \\ Q \| = \dfloorratio{fP}{d} = \dfloor{\dmod{nQ_1 + \dfrac{nR_1}{d}}{(f/R_1)}} \notag \\ \| = \dmod{\dfloor{nQ_1 + \dfrac{nR_1}{d}}}{(f/R_1)} = \dmod{nQ_1 + \dfloor{\dfrac{nR_1}{d}}}{(f/R_1)} \eqvide{eq:modfloor} \\ R \| = fP − dQ \notag \\ \| = d\dmod{nQ_1 + \dfrac{nR_1}{d}}{(f/R_1)} − d\dmod{nQ_1 + \dfloor{\dfrac{nR_1}{d}}}{(f/R_1)} \notag \\ \| = d\dmod{nQ_1 + \dfloor{\dfrac{nR_1}{d}} + \dmod{\dfrac{nR_1}{d}}{1}}{(f/R_1)} − d\dmod{nQ_1 + \dfloor{\dfrac{nR_1}{d}}}{(f/R_1)} \notag \\ \| = d\dmod{nQ_1 + \dfloor{\dfrac{nR_1}{d}}}{(f/R_1)} + d\dmod{\dmod{\dfrac{nR_1}{d}}{1}}{(f/R_1)} \notag \\ \| − \dfrac{fd}{R_1} C\dparen{\dmod{nQ_1 + \dfloor{\dfrac{nR_1}{d}}}{(f/R_1)} + \dmod{\dmod{\dfrac{nR_1}{d}}{1}}{(f/R_1)} ≥ \dfrac{f}{R_1}} \notag \\ \| − d\dmod{nQ_1 + \dfloor{\dfrac{nR_1}{d}}}{(f/R_1)} \notag \\ \| = d\dmod{\dmod{\dfrac{nR_1}{d}}{1}}{(f/R_1)} \notag \\ \| − \dfrac{fd}{R_1} C\dparen{\dmod{nQ_1 + \dfloor{\dfrac{nR_1}{d}}}{(f/R_1)} + \dmod{\dmod{\dfrac{nR_1}{d}}{1}}{(f/R_1)} ≥ \dfrac{f}{R_1}} \end{align}

We have \( f/R_1 ≥ 1 \) and \( 0 ≤ \dmod{nR_1/d}{1} \lt 1 \) so the \( \dmod{•}{(f/R_1)} \) surrounding that has no effect, so

\begin{equation} d\dmod{\dmod{\dfrac{nR_1}{d}}{1}}{(f/R_1)} = d\dmod{\dfrac{nR_1}{d}}{1} = \dmod{nR_1}{d} \end{equation}

Likewise, the formula within the \( C(•) \) can be rewritten as

\begin{equation} \dmod{nQ_1 + \dfloor{\dfrac{nR_1}{d}}}{(f/R_1)} + \dmod{\dfrac{nR_1}{d}}{1} ≥ \dfrac{f}{R_1} \end{equation}

We have

\begin{align} nQ_1 + \dfloorratio{nR_1}{d} \le \dfrac{f}{R_1} − 1 \\ \dmod{\dfrac{nR_1}{d}}{1} \lt 1 \end{align}

because \( nQ_1 \) and \( \dfloor{nR_1/d} \) and also \( f/R_1 \) are whole numbers. So the sum of those two is less than \( f/R_1 \), so the condition within the \( C(•) \) is never met, so that term drops away. All in all we find

\begin{align} P \| = \dmod{nP_1}{(d/R_1)} \\ Q \| = \dmod{nQ_1 + \dfloorratio{nR_1}{d}}{(f/R_1)} \\ R \| = \dmod{nR_1}{d} \end{align}

For calculating these numbers the following form is more convenient:

\begin{align} P \| = \dmod{nP_1}{(d/R_1)} \\ \{ ρ, R \} \| = \Div(nR_1, d) \\ Q \| = \dmod{nQ_1 + ρ}{(f/R_1)} \end{align}

As long as \( n \lt d/R_1 \), also \( \dfloor{nR_1/d} = 0 \) so \( Q = \dmod{nQ_1}{(f/R_1)} \). That will usually be the case in practice.

To find

\begin{equation} X = (w − fP − t)\dfrac{P}{R} \gt 0 \end{equation}

we need \( P \) to be close enough to \( P_0 = (w − t)/(2f) \).

With \( P = \dmod{nP_1}{(d/R_1)} \), the smallest positive value \( n_0 \) of \( n \) for which exactly \( P = P_0 \) is \( n_0 = P_0/P_1 \) and each next value of \( n \) for which \( P = P_0 \) is \( d/P_1R_1 \) greater than the previous one, so the values \( n_k \) of \( n \) for which \( P = P_0 \) are

\begin{equation} n_k = \dfrac{P_0}{P_1} + k\dfrac{d}{P_1R_1} \end{equation}

for whole values \( 0 ≤ k \lt d \).

The whole number \( [n_k] = \dfloor{n_k + 1/2} \) nearest to \( n_k \) is then the candidate value of \( n \). Calculate \( X \) for that value of \( n \) to see if it meets the condition that \( X \gt w \).

If we use equations \eqref{eq:detour1}ff then it is sufficient if \( X ≥ d \) to be able to handle all values \( |x| \lt w \), because then we can use equations \eqref{eq:detour1}ff to first make \( |x| \lt d \). If we do not use equations \eqref{eq:detour1}ff, then we must have \( X ≥ w \) to be able to handle all values \( |x| \lt w \).

There can be multiple combinations of \( P\), \( Q \), \( R \) that yield \( X ≥ w \) or \( X ≥ d \), and those are all good enough. Within that group, we prefer small \( P\), \( Q\), \( R \), because that makes for calculations that are easier to perform and read.

I've investigated for all \( w \) from 3 through 1024 what the greatest values of \( f \) and \( d \) are for which solutions can be found with \( X ≥ d \) or \( X ≥ w \), and what the greatest \( P \), \( Q \), \( R \), and \( X \) are among those solutions, and what the greatest number \( c \) of such solutions is, and the total count \( N \) of combinations of \( f \) and \( d \) for which there is at least one solution. I find

\begin{align} \max(f_w) \| ≈ \dfloorratio{w}{4} \\ \max(f_d) \| ≈ \dfloorratio{w}{3} \\ \max(d_w) \| ≈ w − 5 + ⌊w⌉_2 \\ \max(d_d) \| ≈ w − 1 \\ \max(c_w) \| ≈ \dfloorratio{w+8}{16} \\ \max(c_d) \| ≈ \dfloorratio{w−1}{2} \\ \max(P_w) \| ≈ \dfloorratio{w}{2} − 2 + ⌊w⌉_2 \\ \max(P_d) \| ≈ \dfloorratio{w}{2} − 1 \\ \max(Q) \| ≈ \dfloorratio{2\sqrt{w − 1} − 3}{2} \\ \max(R_w) \| ≈ \dfloorratio{\sqrt{w + 1} − 1}{2} \\ \max(R_d) \| ≈ \frac{\sqrt{w(w + 1100)}}{25} \\ \max(X) \| ≈ \dfloorratio{w^2}{8} \\ N_w \| ≈ \frac{1}{5}w^{5/3} \\ N_d \| ≈ \frac{1}{3}w^{5/3} \end{align}

where the variables with subscript \( w \) apply only to \( X \gt w \), those with subscript \( d \) apply only to \( X \gt d \), and those without a subscript apply to both. The formulas for \( \max(R_d) \), \( N_w \), and \( N_d \) are approximations; the other formulas are exact for \( 10 ≤ w ≤ 1024 \).

If we extrapolate the above formules to \( w = 2^{31} − 1 \) (for values with a width of 32 bits), then we find, approximately

\begin{align*} \max(f_w) \| ≈ 5.4×10^8 \\ \max(f_d) \| ≈ 7.2×10^8 \\ \max(d) \| ≈ 2.1×10^9 \\ \max(c_w) \| ≈ 1.3×10^8 \\ \max(c_d) \| ≈ 1.1×10^9 \\ \max(P) \| ≈ 1.1×10^9 \\ \max(Q) \| ≈ 46340 \\ \max(R_w) \| ≈ 23170 \\ \max(R_d) \| ≈ 8.6×10^7 \\ \max(X) \| ≈ 5.8×10^{17} \\ N_w \| ≈ 7.1×10^{14} \\ N_d \| ≈ 1.2×10^{15} \end{align*}

so if we seek a solution for \( X ≥ w \) (so that we do not need equations \eqref{eq:detour1}ff), then we need try only up to 23170 different values of \( R \). If such a solution does not exist, then we seek a solution for \( X ≥ d \) (for which we do need to use equations \eqref{eq:detour1}ff) and then we must try up to 86 million values of \( R \), which is still much less than the up to 1100 million values of \( P \).

For a given \( w \) there are many combinations of \( d \) and \( f \) for which there are no solutions with \( X ≥ w \) or \( X ≥ d \). For example, for \( w = 499 \) there are only 6554 combinations of \( d \) and \( f \) with at least one solution with \( X ≥ w \), and 10560 combinations with at least one solution with \( X ≥ d \), but there are \( \frac{1}{2} (w − 2)×(w − 3) = 123256 \) combinations with \( 2 ≤ f \lt d \lt w \). The greater \( f \) is, the smaller is the chance that there is a solution for the combination of that \( f \) and an arbitrary \( d \gt f\).

However, when \( f \) and \( d \) are considerably less than \( w \), then there is a good chance of a solution. For example, for \( w = 2^{31} − 1 \) and for \( f \) and \( d \) near \( f ≈ 25920 \) and \( d ≈ 765433 \) there is no solution with \( X ≥ d \) for only roughly 1 out of every 9100 combinations of \( f \) and \( d \), and no solution with \( X ≥ w \) for only roughly 1 out of about 54 combinations.

12.3. The Simple Calendar

For convenience, we'll work with a calendar that has only "days" and "months", where the months are longer than the days, but the formulas also work for other units of time. Later, we'll discuss calendars with more than two time periods.

We'll indicate with \( m \) the month number, with \( d \) the day number within the month, and with \( s \) the running day number that is the start or endpoint of the calculation.

12.3.1. From Month and Day Number Within the Month to Running Day Number

In the simplest case, the average length of the month is equal to \( p \) days, and all months are either \( ⌊p⌋ \) or \( ⌊p⌋ + 1 \) days long. We define \( q \) as the length of the short months, and \( ψ \) as the fraction of months that are long. Then

\begin{eqnarray} q \| = \| \dfloor{p} \\ p \| = \| q + ψ \end{eqnarray}

The running day number \( s \) depends on the month number \( m \) and the day number \( d \) within the current month as follows:

\begin{eqnarray} σ \| ≡ \| ⌊pm + b⌋ \label{eq:sm} \\ s \| ≡ \| σ + d \label{eq:s} \end{eqnarray}

where \( m, d, s, σ \) are whole numbers, and \( p ≥ 1 \) and \( b \) are fixed values that depend on the calendar. The first month has \( m = 0 \) and the first day of the month has \( d = 0 \), for easier calculation. \( σ \) is the running day number of the beginning of the month. If values of \( σ \) for multiple months are of interest then we write \( σ[m] \) for the value of \( σ \) that goes with month \( m \).

The first day (\( d = 0 \)) of the first month (\( m = 0 \)) has running day number \( s = 0 \), so

\begin{equation} ⌊b⌋ = 0 ⇔ 0 ≤ b \lt 1 \end{equation}

So \( b \) must be between 0 and 1 (0 is allowed, 1 is not).

12.3.2. Month Length

The length \( L(m) \) of month \( m \) is

\begin{eqnarray} L(m) \| = \| σ(m + 1) − σ(m) \notag \\ \| = \| \dfloor{q (m + 1) + ψ (m + 1) + b} − \dfloor{qm + ψm + b} \notag \\ \| = \| q + \dfloor{ψm + ψ + b} − \dfloor{ψm + b} \notag \\ \| = \| q + \dfloor{ψm + b} + C\dparen{\dmod{ψm + b}{1} + ψ ≥ 1} − \dfloor{ψm + b} \eqavide{eq:splitfloor} \notag \\ \| = \| q + C\dparen{\dmod{ψm + b}{1} + ψ ≥ 1} \end{eqnarray}

All months are \( q \) or \( q + 1 \) days long. Month \( m \) is a long month (with \( q + 1 \) days) if \( \dmod{ψx + b}{1} ≥ 1 − ψ \). On average there is a long month after every

\begin{equation} Q = \frac{1}{ψ} \label{eq:Q} \end{equation}

months, but that isn't always a whole number, so in practice the number of months between two successive long months is equal to \( ⌊Q⌋ \) or \( ⌈Q⌉ \).

Also

\begin{align} L(m) \| = q + \dfloor{ψm + ψ + b} − \dfloor{ψm + b} \notag \\ \| = q + ψm + ψ + b − \dmod{ψm + b + ψ}{1} − \dparen{ψm + b − \dmod{ψm + b}{1}} \notag \\ \| = q + ψ + \dmod{ψm + b}{1} − \dmod{ψm + b + ψ}{1} \notag \\ \| = p + \dmod{ψm + b}{1} − \dmod{ψm + b + ψ}{1} \end{align}

12.3.3. Shifting the Pattern of Months

With this we get a certain pattern of long and short months. Suppose that that pattern would be exactly right if only it were shifted by \( ∆m \) months. So if \( σ_* \) goes with the old calendar (with \( b_* \)) and \( σ \) goes with the new calendar (with \( b = b_* + ∆b \)) then we want that the following holds for all values of \( m \):

\begin{equation} σ(m + 1) − σ(m) = σ_*(m + ∆m + 1) − σ_*(m + ∆m) \end{equation}

We have

\begin{align} \| σ(m + 1) − σ(m) \notag \\ \| = \dfloor{p(m + 1) + b} − \dfloor{pm + b} \notag \\ \| = \dfloor{pm + b + p} − \dfloor{pm + b} \notag \\ \| = \dfloor{\dfloor{pm + b} + \dmod{pm + b}{1} + p} − \dfloor{pm + b} \notag \\ \| = \dfloor{pm + b} + \dfloor{\dmod{pm + b}{1} + p} − \dfloor{pm + b} \notag \\ \| = \dfloor{\dmod{pm + b}{1} + p} \end{align}

and also

\begin{align} \| σ_*(m + ∆m + 1) − σ_*(m + ∆m) \notag \\ \| = \dfloor{p(m + ∆m + 1) + b_*} − \dfloor{p(m + ∆m) + b_*} \notag \\ \| = \dfloor{pm + p∆m + b_* + p} − \dfloor{pm + p∆m + b_*} \notag \\ \| = \dfloor{\dfloor{pm + p∆m + b_*} + \dmod{pm + p∆m + b_*}{1} + p} − \dfloor{pm + p∆m + b_*} \notag \\ \| = \dfloor{pm + p∆m + b_*} + \dfloor{\dmod{pm + p∆m + b_*}{1} + p} − \dfloor{pm + p∆m + b_*} \notag \\ \| = \dfloor{\dmod{pm + b_* + p∆m}{1} + p} \end{align}

so we get what we want if

\begin{equation} b = \dmod{b_* + p∆m}{1} \end{equation}

12.3.4. From Running Day Number to Month and Day Number Within the Month

We go from running month number \( m \) and day number \( d \) within the month to running day number \( s \) via a formula like

\begin{equation} s = \dfloor{pm + v} + d = σ + d \label{eq:sv} \end{equation}

where \( v \) is an arbitrary number. Equation \eqref{eq:s} is like Eq. \eqref{eq:sv} if \( v = b \). For finding \( m \) from \( s \) we look for a formula like:

\begin{equation} m = \dceilratio{s + u − p}{p} = \dceilratio{s + u}{p} − 1 \end{equation}

What should \( u \) be to make this work?

\begin{align} m \| = \dceilratio{s + u − p}{p} \notag \\ \| = \dceilratio{\dfloor{pm + v} + d + u − p}{p} \eqavide{eq:s} \notag \\ \| = \dceilratio{d + u + pm + v − \dmod{pm + v}{1} − p}{p} \notag \\ \| = m + \dceilratio{d + u + v − \dmod{pm + v}{1} − p}{p} \end{align}

so then we need

\begin{eqnarray} \dceilratio{d + u + v − \dmod{pm + v}{1} − p}{p} = 0 \notag \\ −1 \lt \dfrac{d + u + v − \dmod{pm + v}{1} − p}{p} ≤ 0 \notag \\ −p \lt d + u + v − \dmod{pm + v}{1} − p ≤ 0 \notag \\ \dmod{pm + v}{1} − v − d \lt u ≤ \dmod{pm + v}{1} − v − d + p \end{eqnarray}

This must hold for all values of \( d \) and \( m \) that can occur in the calendar. For the left-hand inequality \( … \lt u \) this means that we should substitute the least value that \( d \) can have, because greater values of \( d \) meet the inequality more easily than lesser values do. So we can substitute \( d = 0 \).

For the right-hand inequality \( u ≤ … \) we should substitute the greatest value of \( d \) that can occur, because lesser values of \( d \) meet the inequality more easily than greater values do. So we can substitute \( d = L(m) − 1 \), the day number (since the beginning of the month) of the last day in month \( m \). We rewrite it into a more form that is more convenient here.

\begin{align} L(m) \| = σ(m + 1) − σ(m) \notag \\ \| = \dfloor{p(m + 1) + v} − \dfloor{pm + v} \notag \\ \| = \dparen{pm + v + p − \dmod{pm + v + p}{1}} − \dparen{pm + v − \dmod{pm + v}{1}} \notag \\ \| = p + \dmod{pm + v}{1} − \dmod{pm + v + p}{1} \end{align}

Then we find

\begin{equation*} \dmod{pm + v}{1} − v \lt u ≤ \dmod{pm + v}{1} − v − \dparen{p + \dmod{pm + v}{1} − \dmod{pm + v + p}{1} − 1} + p \end{equation*}

\begin{equation} \dmod{pm + v}{1} − v \lt u ≤ \dmod{pm + v + p}{1} − v + 1 \end{equation}

For the left-hand inequality \( … \lt u \) we should substitute the greatest possible value of \( \dmod{pm + v}{1} \). That value is just less than 1, so if we substitute 1 then we can change the \( \lt \) to \( ≤ \). For the right-hand inequality \( u ≤ … \) we should substitute the least possible value of \( \dmod{pm + v + p}{1} \), which is 0. Then we find

\begin{equation} 1 − v ≤ u ≤ 1 − v \end{equation}

Those inequalities have exactly one solution:

\begin{equation} u = 1 − v \end{equation}

so

\begin{equation} m = \dceilratio{s + 1 − v − p}{p} = \dceilratio{s + 1 − v}{p} − 1 \label{eq:svtom} \end{equation}

In the above derivation it was necessary to use the inequality \( d ≤ L(m) − 1 \). If we had instead used \( d \lt L(m) \) then we would not have found a solution. That means that you may get the wrong answer if \( s \) is a fractional number that belongs to a moment of time during the last day of the month.

The transition from one value of \( m \) to the next happens when

\[ \dfrac{s + 1 − v − p}{p} \]

is exactly a whole number (which we'll call \( μ \)), so when

\[ s = pμ + p + v − 1 = p(μ + 1) + v − 1 \]

\( p \) and \( v \) need not be whole numbers, so the value of \( s \) at which Equation \eqref{eq:svtom} switches to a new month need not be a whole number, either.

So Equation \eqref{eq:svtom} requires \( s \) to be a whole number, otherwise you may get the wrong month number \( m \).

And then we can also find \( d \).

\begin{align} pm \| = p\dceilratio{s + 1 − v}{p} − p \notag \\ \| = p\dparen{\dfrac{s + 1 − v}{p} + \ddom{\dfrac{s + 1 − v}{p}}{1}} − p \eqavide{eq:dom} \notag \\ \| = s + 1 − v − p + \ddom{s + 1 − v}{p} \\ d \| = s − \dfloor{pm + v} \notag \\ \| = s − \dfloor{s + 1 − p + \ddom{s + 1 − v}{p}} \notag \\ \| = −\dfloor{1 − p + \ddom{s + 1 − v}{p}} \eqavide{eq:minusfloor} (s ∥ 1) \notag \\ \| = \dceil{p − 1 − \ddom{s + 1 − v}{p}} \eqavide{eq:minusceil} \end{align}

So if we split \( s + 1 − v \) into

\begin{equation} s + 1 − v = pμ − δ = p\dceilratio{s + 1 − v}{p} − \ddom{s + 1 − v}{p} \end{equation}

then

\begin{align} m \| = μ − 1 \label{eq:mviaμ} \\ d \| = \dceil{p − 1 − δ} \label{eq:dviaδ} \end{align}

We can rewrite this so we can use the \( \Div \) function, through

\begin{equation}\{−μ,δ\} = \Div(−(s + 1 − v), p)\end{equation}

For the simple calendar \( v = b \) so then

\begin{align} \{−μ,δ\} \| = \Div(−(s + 1 − b), p) \\ m \| = μ − 1 = \dceilratio{s + 1 − b}{p} − 1 \label{eq:ynaarx} \\ d \| = \dceil{p − 1 − δ} = \dceil{p − 1 − \ddom{s + 1 − b}{p}} \end{align}

12.4. With Whole Numbers Only

Calculating machines often work with a limited number of decimals behind the decimal mark, so you can get into trouble with round-off errors. If \( (s + 1 − b)/p = 6.99999999998 \) but because of a very small round-off error your calculating machine thinks that \( (s + 1 − b)/p = 7.00000000001 \), then your calculating machine thinks that \( \dceil{(s + 1 − b)/p} = 7 \) rather than 6, and then you find the wrong month.

Converting equations with \( ⌈•⌉ \) or \( ⌈•⌋_1 \) into equations with \( ⌊•⌋ \) or \( ⌊•⌉_1 \) or \( \Div \) is easier when only whole numbers are used, because for arbitrary whole numbers \( x \) and \( y \) (\( y \gt 0 \))

\begin{align} \dceilratio{x}{y} \| = \dfloorratio{x − 1}{y} + 1 \eqavide{eq:ceil2floor} \\ \ddom{x}{y} \| = y − 1 − \dmod{x − 1}{y} \eqavide{eq:ddom2dmod} \end{align}

If the average length \( p \) is a ratio (of whole numbers), then we can avoid round-off errors by rewriting the formulas based on ratios. We define

\begin{align} p \| = \dfrac{f}{g} = q + \dfrac{h}{g} \\ q \| = \dfloorratio{f}{g} \\ h \| = \dmod{f}{g} \end{align}

with \( f \), \( g \) whole numbers greater than zero. Then we find, for an arbitrary value \( x \),

\begin{equation} \dmod{x}{f/g} = \dmod{x}{p} = p\dmod{\dfrac{x}{p}}{1} = \dfrac{f \dmod{\dfrac{gx}{f}}{1}}{g} = \dfrac{\dmod{gx}{f}}{g} \label{eq:gmodp} \end{equation}

and likewise

\begin{equation} \ddom{x}{f/g} = \dfrac{\ddom{gx}{f}}{g} \label{eq:gdomp} \end{equation}

Then we go from month \( m \) and day \( d \) to running day number \( s \) through

\begin{equation} t ≡ vg \end{equation}

\begin{align} s \| = \dfloor{pm + v} + d \notag \\ \| = \dfloor{\dfrac{fm + vg}{g}} + d \notag \\ \| = \dfloor{\dfrac{fm + t}{g}} + d \end{align}

where we demand that \( t \) is a whole number, too. If \( v \) is a ratio then you can always achieve that by a suitable choice of \( g \).

For the month length we find

\begin{equation} L(m) = q + C\dparen{\dmod{hm + t}{g} ≥ g − h}\end{equation}

For the calculation of \( m \) and \( d \) from \( s \) we find:

\begin{align} m \| = \dceilratio{s + 1 − v}{p} − 1 = \dceilratio{gs + g − t}{f} − 1 \notag \\ \| = \dfloorratio{gs + g − t − 1}{f} \eqavide{eq:ceil2floor} (t,g,s,f ∥ 1) \label{eq:ynaarx2} \\ d \| = \dceil{p − 1 − \ddom{s + 1 − v}{p}} \notag \\ \| = \dceilratio{f − g − \ddom{gs + g − t}{f}}{g} \eqavide{eq:gdomp} \notag \\ \| = \dfloorratio{f − g − \ddom{gs + g − t}{f} − 1}{g} + 1 \eqavide{eq:ceil2floor} (t,g,s,f ∥ 1) \notag \\ \| = \dfloorratio{f − g − (f − 1 − \dmod{gs + g − t − 1}{f}) − 1}{g} + 1 \eqavide{eq:ddom2dmod} \notag \\ \| = \dfloorratio{\dmod{gs + g − t − 1}{f}}{g} \eqavide{eq:ddom2dmod} \end{align}

Summarizing, from running month number \( m \) and day number \( d \) in the month to running day number \( s \):

\begin{equation} s = \dfloorratio{fm + t}{g} + d \end{equation}

and from \( s \) to \( m \) and \( d \):

\begin{align} w \| = gs + g − t − 1 \\ m \| = \dfloorratio{w}{f} \label{eq:ynaarxr} \\ d \| = \dfloorratio{\dmod{w}{f}}{g} \end{align}

Now you can do all calculations with ratios, which have no numbers after the decimal mark and hence no round-off errors.

12.5. Very Unequal Months

So far we have assumed that a long month was only one day longer than a short month, but that difference can be larger. Suppose that short months are \( q \) days long and that long months are \( q + r \) days long, with \( q\), \( r \) whole numbers greater than 0, and that a fraction \( ψ \) (between 0 and 1) of all months is long. Then the average length of a month is equal to

\begin{equation} p = q + rψ \end{equation}

days. The formula to calculate the running day number \( σ \) of the first day of month \( m \) is then

\begin{equation} σ = qm + r\dfloor{ψm + b} \label{eq:sm2} \end{equation}

If we substitute \( r = 1 \) then we find

\begin{eqnarray} σ \| = \| qm + \dfloor{ψm + b} \notag \\ \| = \| \dfloor{(q + ψ)m + b} \notag \\ \| = \| \dfloor{pm + b} \end{eqnarray}

which is the same result as Eq. \eqref{eq:sm}.

With Eq. \eqref{eq:sm2}, the length \( L(m) \) of month \( m \) is

\begin{eqnarray} L(m) \| = \| σ(m + 1) − σ(m) \notag \\ \| = \| q(m + 1) + r\dfloor{ψ(m + 1) + b} − (qm + r\dfloor{ψm + b}) \notag \\ \| = \| q + rC\dparen{\dfloor{ψm + ψ + b} − \dfloor{ψm + b}} \notag \\ \| = \| q + rC\dparen{\ddom{ψm + b}{1} + ψ ≥ 1} \eqavide{eq:splitfloor} \end{eqnarray}

so the long months are those for which \(\ddom{ψm + b}{1} + ψ ≥ 1\), just like we found earlier for the simple calendar.

The formula to calculate the running day number \( s \) from month number \( m \) and day number \( d \) in the current month is then

\begin{eqnarray} s \| = \| σ + d \notag \\ \| = \| qm + r⌊ψm + b⌋ + d \label{eq:xnaary3} \end{eqnarray}

How do we go in the opposite direction? To figure that out we compare the running day number that comes from equation \eqref{eq:xnaary3} for the first day (\( d = 0 \)) of month \( m \) with the running day number that follows for the same \( m \) from the most similar simplest calendar. We write the outcome for the desired calendar as \( σ \) and for the simpler comparison calendar as \( σ_* \).

\begin{align} σ \| = qm + r\dfloor{ψm + b} \\ σ_* \| = qm + \dfloor{r(ψm + b)} = \dfloor{pm + rb} \label{eq:σ*} \\ ∆σ ≡ σ_* − σ \| = \dfloor{r(ψm + b)} − r\dfloor{ψm + b} \notag \\ \| = r(ψm + b) − \dmod{r(ψm + b)}{1} − \dparen{r(ψm + b) − r\dmod{ψm + b}{1}} \notag \\ \| = r\dmod{ψm + b}{1} − \dmod{r(ψm + b)}{1} \notag \\ \| = r\dmod{ψm + b}{1} − \dmod{r\dmod{ψm + b}{1}}{1} \notag \\ \| = \dfloor{r\dmod{ψm + b}{1}} \end{align}

This difference satisfies

\begin{equation} 0 \le σ_* − σ \le r − 1 \end{equation}

so the beginning of month \( m \) in the target calendar is between \( 0 \) and \( r − 1 \) days (inclusive) before the beginning of month \( m \) in the simple calendar.

For that simpler calendar based on \( σ_* \) we have Equation \eqref{eq:svtom} to calculate the month number from the running day number:

\[ m_*(s) = \dceilratio{s + 1 − v}{p} − 1 \]

That means that

\begin{equation} m_↑ = m_*(s + \max(∆σ)) = \dceilratio{s + \max(∆σ) + 1 − v}{p} − 1 \end{equation}

is greater than or equal to the correct month number. And

\begin{equation} m_↓ = m_*(s + \min(∆σ)) = \dceilratio{s + \min(∆σ) + 1 − v}{p} − 1 \end{equation}

is less than or equal to the correct month number.

With \( m_↑ \) we find

\begin{equation} d_↑ = s − σ(m_↑) \end{equation}

If \( d_↑ \lt 0 \) then \( m_↑ \) is too great. If \( d_↑ \ge 0 \) then we have the correct month number but \( d_↑ \) is probably not the correct day number.

Can we find a recipe to calculate \( d \) with a fixed number of steps? Something like

\begin{align} m_↑ \| ≡ \dceilratio{s + \max(∆σ) + 1 − v}{p} − 1 \\ d_↑ \| = s − σ(m_↑) \\ m \| = m_↑ + ξ \\ d \| = s − σ(m) \end{align}

where \( ξ \) is the correction to apply to \( m_↑ \) to get \( m \). We cannot always find \( ξ \) in one go, but can sometimes find it if \( ξ = 0 \) or \( ξ = 1 \), i.e., when the difference between \( m_↑ \) and \( m \) is at most 1.

We then need \( ξ = 0 \) if \( d_↑ ≥ 0 \), and \( ξ = −1 \) if \( d_↑ \lt 0 \). That suggests something like

\begin{equation} ξ = \dfloorratio{d_↑}{Ξ} \end{equation}

because that yields \( ξ = 0 \) if \( 0 \le d_↑ \lt Ξ \) and yields \( ξ = −1 \) if \( −Ξ \le d_↑ \lt 0 \). What value should we use for \( Ξ \)? That depends on the least and greatest value that \( d_↑ \) can have, so let's deduce those first.

\begin{align} s \| = σ(m) + d \notag \\ \| = σ_*(m) − ∆σ(m) + d \notag \\ \| = \dfloor{pm + v} − ∆σ(m) + d \notag \\ \| = pm + v − \dmod{pm + v}{1} − ∆σ(m) + d \end{align}

so

\begin{align} m_↑ − m \| = \dceilratio{s + \max(∆σ) + 1 − v}{p} − 1 − m \notag \\ \| = \dceilratio{pm + v − \dmod{pm + v}{1} − ∆σ(m) + d + \max(∆σ) + 1 − v}{p} − 1 − m \notag \\ \| = \dceilratio{1 − \dmod{pm + v}{1} + \max(∆σ) − ∆σ(m) + d}{p} − 1 \end{align}

Always \( 0 \lt 1 − \dmod{•}{1} \le 1 \), and \( \max(∆σ) − ∆σ(m) \ge 0 \), and \( d ≥ 0 \), so the part between \( \dceil{} \) is always greater than 0, so \( m_↑ − m ≥ 0 \) as desired. When is certainly \( m_↑ − m ≤ 1 \)? Then we must have

\begin{align*} \dceilratio{1 − \dmod{pm + v}{1} + \max(∆σ) − ∆σ(m) + d}{p} ≤ 2 \\ \dfrac{1 − \dmod{pm + v}{1} + \max(∆σ) − ∆σ(m) + d}{p} ≤ 2 \\ 1 − \dmod{pm + v}{1} + \max(∆σ) − ∆σ(m) + d ≤ 2p \end{align*}

The greatest value that \( d \) can have is \( \max(L) − 1 \), so we certainly have \( m_↑ − m \le 1 \) if

\begin{equation} ρ ≡ \max(∆σ) − \min(∆σ) + \max(L) − 2p ≤ 0 \label{eq:singlepass} \end{equation}

If \( m_↑ = m \) then the least value of \( d_↑ \) is 0 and the greatest value is \( \max(L) − 1 \).

If \( m_↑ = m + 1 \) then

\begin{align} d_↑ \| = s − σ(m_↑) \notag \\ \| = s − σ(m + 1) \notag \\ \| = \dparen{pm + v − \dmod{pm + v}{1} − ∆σ(m) + d} \notag \\ \| − \dparen{pm + v + p − \dmod{pm + v + p}{1} − ∆σ(m + 1)} \notag \\ \| = d − p + \dparen{\dmod{pm + v + p}{1} − \dmod{pm + v}{1}} + ∆σ(m + 1) − ∆σ(m) \notag \\ \| \gt 0 − p + \min\dparen{\dmod{pm + v + p}{1} − \dmod{pm + v}{1}} + \min(∆σ(m + 1) − ∆σ(m)) \notag \\ \| = −p + (\dmod{p}{1} − 1) + \min(∆σ(m + 1) − ∆σ(m)) \eqavide{eq:domp} \notag \\ \| = −p + \dmod{p}{1} − 1 + \min(L_*(m) − L(m)) \notag \\ \| = −\dfloor{p} − 1 + \min(L_*(m) − L(m)) \eqavide{eq:mod} \notag \\ \| ≥ −\dfloor{p} − 1 + \min(L_*) − \max(L) \notag \\ \| = −\dfloor{p} − 1 + \dfloor{p} − \max(L) \notag \\ \| = −1 − \max(L) \notag \\ d_↑ \| \gt −1 − \max(L) \\ d_↑ \| ≥ −\max(L) \end{align}

so

\begin{equation} −\max(L) ≤ d_↑ ≤ \max(L) − 1 \end{equation}

For

\[ ξ = \dfloorratio{d_↑}{Ξ} \]

we then need both of the following conditions satisfied:

\begin{align*} \dfloorratio{−\max(L)}{Ξ} = −1 \\ \dfloorratio{\max(L) − 1}{Ξ} = 0 \end{align*}

Both those conditions are equivalent to

\begin{equation} Ξ ≥ \max(L) \end{equation}

So if we use

\begin{equation} Ξ = \max(L) \end{equation}

(or a greater value) then we meet both constraints.

Summarizing, if

\[ ρ = \max(∆σ) − \min(∆σ) + \max(L) − 2p ≤ 0 \]

then

\begin{align} m_↑ \| = \dceilratio{s + \max(∆σ) + 1 − v}{p} − 1 \\ d_↑ \| = s − σ_↑ = s − σ(m_↑) \\ ξ \| = \dfloorratio{d_↑}{\max(L)} \\ m \| = m_↑ + ξ \\ d \| = s − σ(m) \end{align}

With

\begin{align*} v \| = rb \\ \max(∆σ) \| = r − 1 \\ \min(∆σ) \| = 0 \\ \max(L) \|= q + r \\ p \| = q + rψ \end{align*}

we find

\begin{align} m_↑ \| = \dceilratio{s + r − rb}{p} − 1 \label{eq:ynaarx3} \\ d_↑ \| = s − σ_↑ = s − σ(m_↑) \\ ρ \| = 2r(1 − ψ) − 1 − q \\ ξ \| = \dfloorratio{d_↑}{q + r} \\ m \| = m_↑ + ξ \\ d \| = s − σ(m) \end{align}

so \( ρ ≤ 0 \) corresponds to

\begin{equation} r ≤ \dfrac{1}{2} \dfrac{q + 1}{1 − ψ} ≥ q + 1 \end{equation}

If \( ρ \gt 0 \) then the recipe doesn't always work so then you shouldn't use it. Then see Section 12.7.

If \( p = \dfrac{f}{g} \), \( ψ = \dfrac{h}{g}\) and \( b = \dfrac{t}{g} \) are ratios of whole numbers (\( f, g, h, t \) are whole numbers with \( f \gt g \gt 0 \), \( 0 \le h, t \lt g \)), then we find

\begin{align} s \| = σ + d = qm + r\dfloorratio{hm + t}{g} + d \label{eq:xnaary3r} \\ m_↑ \| = \dceilratio{s + \max(∆σ) + 1 − v}{p} − 1 \notag \\ \| = \dceilratio{gs + g\max(∆σ) + g − gv}{f} − 1 \notag \\ \| = \dceilratio{gs + g\max(∆σ) + g − t}{f} − 1 \notag \\ \| = \dfloorratio{gs + g\max(∆σ) + g − t − 1}{f} \eqavide{eq:ceil2floor} \\ ρ \| = 2r\dparen{1 − \dfrac{h}{g}} − 1 − q \\ d_↑ \| = s − qm_↑ − r\dfloorratio{hm_↑ + t}{g} \\ ξ \| = \dfloorratio{d_↑}{q + r} \\ m \| = m_↑ + ξ \\ d \| = s − qm − r\dfloorratio{hm + t}{g} \end{align}

12.6. Many Kinds of Unequal Months

We now allow there to be more than two kinds of months with different month lengths. In equation \eqref{eq:xnaary3}, \( \dfloor{ψm + b} \) gave the pattern of long months (with a shift as desired). We now allow an arbitrary number of such pattern, each with its own length difference \( r_i \) (that can be positive or negative), relative frequency \( 0 \lt ψ_i \lt 1 \), and pattern shift \( b_i \). Then we find

\begin{align} σ \| = qm + \sum_{i}r_i \dfloor{ψ_im + b_i} \\ s = σ + d \| = qm + \sum_{i}r_i \dfloor{ψ_im + b_i} + d \label{eq:xnaary4} \end{align}

The derivation of the formulas for the opposite direction goes analogous to that of equation \eqref{eq:ynaarx3}. We find

\begin{align} σ_* \| = qm + \dfloor{\sum_i r_i (ψ_i m + b_i)} \notag \\ \| = \dfloor{pm + \sum_i r_i b_i} \notag \\ \| = pm + \sum_i r_i b_i − \dmod{\sum_i r_i \dparen{ψ_i m + b_i}}{1} \label{eq:σ*∑} \eqavide{eq:mod} \\ ∆σ = σ_* − σ \| = \dfloor{\sum_{i}r_{i}(ψ_{i}m + b)} − \sum_{i} r_{i}\dfloor{ψ_{i}m + b_{i}} \notag \\ \| = \dfloor{\sum_{i} r_{i}\dmod{ψ_{i}m + b_{i}}{1}} \eqavide{eq:dmoddiff1} \notag \end{align}

so

\begin{align} v \| = \sum_i r_ib_i \\ \max(L) \| ≤ q + \sum_{r_i \gt 0} r_i \\ \max(∆σ) \| ≤ \dparen{\sum_{r_i \gt 0} r_i} − 1 \\ \min(∆σ) \| ≥ \sum_{r_i \lt 0} r_i \\ ρ \| = \max(∆σ) − \min(∆σ) + \max(L) − 2p \notag \\ \| ≤ 2\dparen{\sum_{r_i \gt 0} r_i} − \dparen{\sum_{r_i \lt 0} r_i} − q − 1 − 2\dparen{\sum_i r_i ψ_i} \end{align}

It is permissible to use a higher value for \( \max(L) \) in the below formulas, such as the value given above at the right hand side of \( \max(L) ≤ \), and similarly for the other inequalities. For convenience, we pretend that those inequalities are equalities.

And so

\begin{align} m_↑ \| = \dceilratio{s + \max(∆σ) + 1 − v}{p} − 1 \notag \\ \| = \dceilratio{s + \dparen{\sum_{r_i \gt 0} r_i} − \sum_i r_i b_i}{p} − 1 \label{eq:ynaarx4} \\ d_↑ \| = s − qm_↑ − \sum_i r_i\dfloor{ψ_i m_↑ + b_i} \\ ρ \| = 2\dparen{\sum_{r_i \gt 0} r_i} − \dparen{\sum_{r_i \lt 0} r_i} − q − 1 − 2\dparen{\sum_i r_i ψ_i} \end{align}

If \( ρ ≤ 0 \) then \( m \) and \( m_↑ \) differ by at most 1. That condition corresponds to

\begin{align} 2\dparen{\sum_{r_i \gt 0} r_i} \| − \dparen{\sum_{r_i \lt 0} r_i} − q − 1 − 2\dparen{\sum_i r_i ψ_i} ≤ 0 \notag \\ \dparen{\sum_{r_i \gt 0} r_i} \| + \dparen{\sum_{r_i \gt 0} r_i} − \dparen{\sum_{r_i \lt 0} r_i} − q − 1 − 2\dparen{\sum_i r_i ψ_i} ≤ 0 \notag \\ \dparen{\sum_{r_i \gt 0} r_i} \| + \dparen{\sum_{i} |r_i|} − q − 1 − 2\dparen{\sum_i r_i ψ_i} ≤ 0 \notag \\ \sum_{r_i \gt 0} r_i \| ≤ q + 1 + 2\dparen{\sum_i r_i ψ_i} − \dparen{\sum_{i} |r_i|} \eqavide{eq:singlepass} \end{align}

Then

\begin{align} ξ \| = \dfloorratio{d_↑}{q + \sum_{r_i \gt 0} r_i} \\ m \| = m_↑ + ξ \label{eq:xζ4} \\ d \| = s − qm − \sum_i r_i\dfloor{ψ_i m + b_i} \end{align}

If all \( ψ_i = h_i/g_i \) and \( b_i = t_i/g_i \) are ratios of whole numbers, and if condition \eqref{eq:singlepass} is met, then we find

\begin{align} p \| = q + \sum_i r_iψ_i = q + \sum_i \dfrac{r_i h_i}{g_i} \\ g \| = \lcm(g_i) \\ γ_i \| = \dfrac{g}{g_i} \\ f \| = pg = gq + \sum_i γ_i r_i h_i \\ s \| = qm + \sum_i r_i\dfloorratio{h_im + t_i}{g_i} + d \label{eq:xnaary4r} \\ m_↑ \| = \dceilratio{s + \dparen{\sum_{r_i \gt 0} r_i} − \dparen{\sum_i r_i b_i}}{f/g} − 1 \notag \\ \| = \dceilratio{gs + g\dparen{\sum_{r_i \gt 0} r_i} − \dparen{\sum_i γ_i r_i t_i}}{f} − 1 \notag \\ \| = \dfloorratio{gs + g\dparen{\sum_{r_i \gt 0} r_i} − \dparen{\sum_i γ_i r_i t_i} − 1}{f} \label{eq:ynaarx4r} \\ d_↑ \| = s − qm_↑ − \sum_i r_i\dfloorratio{h_im_↑ + t_i}{g_i} \\ ρ \| = 2\dparen{\sum_{r_i \gt 0} r_i} − \dparen{\sum_{r_i \lt 0} r_i} − q − 1 − 2\dparen{\sum_i \dfrac{r_i h_i}{g_i}} \\ ξ \| = \dfloorratio{d_↑}{q + \sum_{r_i \gt 0} r_i} \\ m \| = m_↑ + ξ \\ d \| = s − qm − \sum_i r_i\dfloorratio{h_im + t_i}{g_i} \end{align}

\( g \) is the least common multiple of the denominators of all \( ψ_i \) and \( b_i \).

12.7. Very Unequal Month Lengths

The preceding methods for calculating the calendar date from the running day number for calendars with very unequal month lengths assumed that the variation in month lengths is sufficiently small that the month number calculated with the corresponding simple calendar is at most one off: \( 2\dparen{\sum_{r_i \gt 0} r_i} − \dparen{\sum_{r_i \lt 0} r_i} − q − 1 − 2\dparen{\sum_i r_iψ_i} ≤ 0 \). What to do if that condition is not met?

In that case you'll have to find the correct month by searching for it. First try month number \( m_↑ \) and calculate the corresponding \( d_↑ \) and \( ξ \). Is that \( ξ \lt 0 \)? Then subtract 1 from \( m_↑ \) and try again, until \( ξ = 0 \): then you have found the correct \( m \).

Warning! In previous sections we added \( ξ \) to \( m_↑ \) to correct the month number, but here we are not certain that that could never produce a month number that is too small. A month number that is too large is easy to detect (then \( d_↑ \lt 0 \)) but a month number that is too small is not easy to detect. So we must not add \( ξ \). If \( ξ \lt 0 \), then subtract 1 from the month number and try again with that new month number, and repeat until \( ξ = 0 \).

If you want to, then you can use this procedure also if condition \eqref{eq:singlepass} is met.

The greatest number of months that you may have to try before you find the right one is equal to the greatest value that \( m_↑ − m_↓ \) can attain, plus 1.

\begin{align} m_↑ − m_↓ \| = \dceilratio{s + \max(∆σ) + 1 − v}{p} − \dceilratio{s + \min(∆σ) + 1 − v}{p} \\ \| ≤ \dceilratio{\max(∆σ) − \min(∆σ)}{p} + 1 \eqavide{eq:ceildiffrange} \end{align}

so the number of months that you have to try at most is

\begin{equation} N = \dceilratio{\max(∆σ) − \min(∆σ)}{p} + 2 \end{equation}

For a calendar with two month lengths this means

\begin{equation} N = \dceilratio{r − 1}{p} + 2 \end{equation}

and for a calendar with more month lengths

\begin{align} N \| ≤ \dceilratio{\dparen{\sum_{r_i \gt 0} r_i} − \dparen{\sum_{r_i \lt 0} r_i}}{p} + 2 \\ \| = \dceilratio{\sum_i |r_i|}{p} + 2 \end{align}

12.8. Month Lengths Without Internal Patterns

Previously we looked at calendars of which the month lengths can be expressed as the sum of various patterns, for example with an extra day every second month and two additional days every third month, so that each sixth month gets three extra days in total. But what if the months repeat after some time but show no other discernable pattern?

The calendar has \( f \) days in \( g \) months, for an average month length of \( p = f/g \) days. The relationship between the running day number \( s \) and the running month number \( m \) is a step function: after a (varying) number of days, the month number increases by one. Such a step of 1 that occurs just before day \( a \) and repeats itself every \( f \) days can be obtained through the formula

\begin{equation} \dfloorratio{s − a}{f} + 1 = \dfloorratio{s + f − a}{f} \end{equation}

The \( + 1 \) arranges that the result is 0 for \( s \lt a \) and 1 for \( s ≥ a \).

When there are \( g \) months, each with its own \( a_i \) (for \( i \) from 0 through \( g − 1 \)), then their combined effect is

\begin{equation} \sum_{i=0}^{g−1} \dparen{\dfloorratio{s − a_i}{f} + 1} = g + \sum_{i=0}^{g−1} \dfloorratio{s − a_i}{f} = \sum_{i=0}^{g−1} \dfloorratio{s + f − a_i}{f} \end{equation}

As usual we want that month \( m = 0 \) begins when \( s = 0 \), so then we must have \( a_0 = 0 \).

If the successive month lengths are \( L_i \), then the first step is at \( a_0 = 0 \), the second one is at \( a_1 = a_0 + L_0 = L_0 \), the third one is at \( a_2 = a_1 + L_1 = L_0 + L_1 \), and in general

\begin{equation} a_i = \sum_{j=0}^{i-1} L_j \end{equation}

Then

\begin{equation} m = \sum_{i=0}^{g−1} \dfloorratio{s + f − \sum_{j=0}^{i−1} L_j}{f} \label{eq:willekeurigynaarx} \end{equation}

If we go in the other direction then we also have a staircase, but now days and months exchange roles as length and height of the steps. With that, the formula for the running day number \( σ \) of the first day of month \( m \) is:

\begin{equation} σ = \sum_{i=0}^{g−1} L_i \dfloorratio{m + g − 1 − i}{g} \end{equation}

The formula to go from running month number \( m \) and day number \( d \) in the current month to running day number \( s \) (all beginning at 0) is then (with \( i \) running from 0 through \( g − 1 \))

\begin{equation} s = σ + d = \sum_{i=0}^{g−1} m_i \dfloorratio{m + g − 1 − i}{g} + d \label{willekeurigxnaary} \end{equation}

These formulas are, if you write out the summation, a lot longer than the formulas that we found earlier for calendars with internal patterns, so it is convenient if you recognize such patterns for a calendar, but not every calendar has such patterns.

The simple calendar from section 12.3 has \( s = \dfloor{pm + v} + d \). To have that calendar begin month \( m \) at running day \( s \) we need \( s = \dfloor{pm + v} \), from which follows \( s ≤ pm + b \lt s + 1 \), hence (for \( m \gt 0 \)) \( (s − b)/m ≤ p \lt (s + 1 − b)/m \). We know that \( 0 ≤ b \lt 1 \), so

\[ \frac{s − 1}{m} \lt \frac{s − b}{m} \le p \lt \frac{s + 1 − b}{p} \le \frac{s + 1}{p} \]

so

\begin{equation} \frac{s − 1}{m} \lt p \lt \frac{s + 1}{m} \label{eq:beperkp} \end{equation}

Not every \( p \) that meets these restrictions yields a simple calendar, but if a \( p \) does not meet these restrictions for at least one of its months, then there is certainly no simple calendar for these month lengths.

12.9. Combinations of straight lines

It is rare for a calendar to be fully defined by just one period \( p \), so usually you have to combine several periods. Let's assume that we have formulas for two periods:

\begin{align} y_1 \| = D_1(\{ x_1, z_1 \}) \\ y_2 \| = D_2(\{ x_2, z_2 \}) \end{align}

We don't use \( m \), \( d \), and \( s \) here because we may want to use different units of time than months and days. \( x \) counts the larger unit, and \( y \) and \( z \) count the smaller unit. \( y \) is the running number of smaller units, and \( z \) is the number of smaller units since the beginning of the most recent larger unit. Mapping \( D_1 \) says that you can calculate \( y_1 \) from \( x_1 \) and \( z_1 \), and likewise for mapping \( D_2 \).

In the other direction we have

\begin{align} \{ x_1, z_1 \} \| = U_1(y_1) \\ \{ x_2, z_2 \} \| = U_2(y_2) \end{align}

where \( U \) and \( D \) are each other's inverse:

\begin{align} y \| = D(U(y)) \\ \{ x, z \} \| = U(D(\{ x, z \})) \end{align}

To go from calendar date to running day number we first apply \( D_1 \) and then \( D_2 \). There are two ways to combine these: We can equate \( y_1 \) to \( z_2 \) or to \( x_2 \). In the first case, the two periods have the same leap unit, for example days. In the second case, the two periods have different leap units (for example, sometimes an extra day for the first period, and sometimes an extra month for the second period). If the larger period needs no leap rules at all, then you can choose which combination method to use.

Let's call the first case the "flat" combination, because the leap units remain the same. Let's call the second case the "stepped" combination, because the leap unit goes a step higher.

The combination of months and years in most (perhaps all) solar calendars (such as the Gregorian calendar, the Julian calendar, and the Egyptian calendar) is flat, i.e. of the first kind. A month can be a day shorter or longer than another month (and exactly one month can be a lot shorter), and a year can be a day shorter or longer than another year. The number of days in a year does not depend on the months, because the rules to calculate the length of the year depend on the year number but not on the month number.

The combination of months and years in a lunisolar calendar (such as the Hebrew and Babylonian calendars) can be flat with very unequal year lengths (\( r \gt 1 \)) or stepped (then usually \( r = 1 \)). A month can be a day longer or shorter than another month, and a year can be a month shorter or longer than another month. (For the flat combination, one month of the year can be much shorter.) In this way you can follow two separate (astronomical) cycles: the motion of the Sun (with the year) and the motion of the Moon (with the month). If the combination is stepped, then the length of the year depends on the length of the months, because the year is then defined in terms of a fixed number of months, not a fixed number of days.

Lunar calendars that are not lunisolar (such as the administrative Islamic calendar) usually do not have any leap rules, so then both methods can be used.

If a calendar has more than two important large periods with leap rules (for example, not just for the month and the year, but also for the century), then it is possible that some combinations are flat and others are stepped.

If we have to deal with more than one period, then it is important for translating a running day number into a calendar date to have the running numbers begin at 0 for each of those periods.

12.9.1. Flat combination

In this case \( z_2 = y_1 \), so

\begin{align} y_1 \| = D_1\dparen{\{x_1, z_1\}} \\ y_2 \| = D_2\dparen{\{x_2, z_2\}} = D_2\dparen{\{x_2, D_1\dparen{\{x_1, z_1\}}\}} \label{eq:vlak} \end{align}

and in the other direction

\begin{align} \{ x_2, z_2 \} \| = U_2(y_2) \label{eq:vlakr} \\ \{ x_1, z_1 \} \| = U_1(z_2) \end{align}

or, with a diagram

  x₂ ──────────┐
  x₁ ┐         │
  z₁ ┴ y₁ = z₂ ┴ y₂

The lowest row has the smallest calendar unit, usually days. Higher rows have greater calendar units, for example months and years. \( x_2 \) could be the year number, \( x_1 \) the month number within the year, \( z_1 \) the day number within the month, \( y_1 \) the day number within the year, and \( y_2 \) the running day number.

To calculate the calendar date from the running day number, we must first calculate the larger period (\( x_2 \), \( z_2 \)), and then the smaller period (\( x_1 \), \( z_1 \)).

12.9.2. Stepped combination

In this case \( x_2 = y_1 \), so

\begin{align} y_1 \| = D_1\dparen{\{x_1, z_1\}} \\ y_2 \| = D_2\dparen{\{x_2, z_2\}} = D_2\dparen{D_1\dparen{\{x_1, z_1\}, z_2}} \end{align}

and in the other direction

\begin{align} \{x_2, z_2\} \| = U_2(y_2) \\ \{x_1, z_1\} \| = U_1(x_2) \end{align}

or, with a diagram

  x₁ ┐
  z₁ ┴ y₁ = x₂ ┐
  z₂ ──────────┴ y₂

Here \( x_1 \) could be the year number, \( z_1 \) the month number within the year, \( z_2 \) the day number within the month, \( y_1 \) the running month number, and \( y_2 \) the running day number.

12.10. Simultaneous Cycles

Most calendars have higher and lower periods, and the number of a higher period changes much less often than the number of a lower period. After the 7th day of the 3rd month follows the 8th day of the 3rd month − the month number changes less often than the day number.

Some calendars have different periods that change simultaneously. The most common calendars often have such a case, too, in the form of weekdays and days of the month. When the next day arrives, then the day number of the month changes, but the weekday changes, too. After Monday the 7th we get Tuesday the 8th − the weekday and the day number change equally fast.

We don't need the weekday to be able to point to a unique day (30 August 2011 indicates exactly one day; for that we do not need to know that that day was a Tuesday), so the weekday does not usually play a role in calendar calculations. However, there are calendars (for example the calendars of Central America) which do use simultaneous cycles to point at particular days. We look at this type of calendar below.

Suppose that a calendar uses simultaneous periods \( p_i \) (all of them whole numbers greater than 1) for \( i \) from 1 through \( n \). The day number in each period begins at 0. We write a date in that calendar as \( \{x\} = \{x_1,x_2,…,x_n\} \), where \( x_i \) is the day number from period \( i \). If \( x_i \) is equal to \( p_i − 1 \), then the next day has \( x_i = 0 \) again.

12.10.1. From Running Day Number to Date

To translate running day number \( s \) into \( \{x\} \) we can use the following formula:

\begin{equation} x_i = ⌊s + a_i⌉_{p_i} = (s + a_i) \bmod p_i \label{eq:cycli} \end{equation}

Here \( a_i \) is the value of \( x_i \) when \( s = 0 \). That value depends on the calendar.

12.10.2. From Date to Running Day Number

It is a lot more difficult to go in the other direction. Then we have to find a \( s \) for which equation \eqref{eq:cycli} is satisfied for all \( i \), i.e.,

\begin{equation} s ≡ x_i − a_i \pmod{p_i} \end{equation}

We define

\begin{equation} c_i = x_i − a_i \end{equation}

We first look at the case \( n = 2 \). Then we need to solve \( s \) from

\begin{align} s \| ≡ c_1 \pmod{p_1} \label{eq:c1} \\ s \| ≡ c_2 \pmod{p_2} \label{eq:c2} \end{align}

There are only solutions if

\begin{equation} c_1 ≡ c_2 \pmod{g} \end{equation}

where

\begin{equation} g = \gcd(p_1, p_2) \end{equation}

is the greatest common divisor of \( p_1 \) and \( p_2 \), so we assume that that condition is met. Then there are numbers \( n_1, n_2 \) for which

\begin{equation} g = n_1 p_1 + n_2 p_2 \end{equation}

and then the solutions are

\begin{equation} s ≡ \dfrac{c_1 n_2 p_2 + c_2 n_1 p_1}{g} \pmod{\dfrac{p_1 p_2}{g}} \label{eq:cyclitoj} \end{equation}

The numerators of these ratios have a divisor equal to \( g \) so these divisions always yield whole numbers.

Warning! Formula \eqref{eq:cyclitoj} only gives useful results for \( \{x\} \) that really occur in the calendar. If \( g \) is not equal to 1, then not all possible \( \{x\} \) can occur. If you enter an \( \{x\} \) that does not occur in the calendar, then you get a reasonable-looking result from the formula, but that result then still won't be valid.

12.10.3. More than Two Periods

Using equation \eqref{eq:cyclitoj}, \( s ≡ c_1 \pmod{p_1} \) and \( s ≡ c_2 \pmod{p_2} \) lead to another equation \( s ≡ C_2 \pmod{P_2} \) with

\begin{align} C_2 \| ≡ \dfrac{c_1n_2p_2 + c_2n_1p_1}{\gcd(p_1, p_2)} \\ P_2 \| ≡ \dfrac{p_1p_2}{\gcd(p_1, p_2)} \end{align}

That formula has the same form as the two formulas that we started with, so we can combine that formula in the same manner with \( s ≡ c_3 \pmod{p_3} \), and so on until we've handled all \( p_i \). We end up with a formula that again has the same form \( s ≡ C_n \pmod{P_n} \), and that is the solution of the \( n \) equations.

The procedure is then as follows. Given \( x_i \), \( p_i \), \( a_i \) for \( i \) from 1 through \( n \),

1. Assign

   \begin{align*} C_1 \| = x_1 − a_1 \\ P_1 \| = p_1 \end{align*}

2. For \( i \) from 2 through \( n \):

   1. Determine \( g_i = \gcd(P_{i−1}, p_i) \), the greatest common divisor of \( P_{i−1} \) and \( p_i \).

   2. Find \( n_i, m_i \) such that

      \[ g_i = n_i P_i + m_i p_i \]

   3. Assign

      \begin{align*} C_i \| = \dfrac{C_{i−1} m_i p_i + c_i n_i P_{i−1}}{g_i} \\ P_i \| = \dfrac{P_{i−1}p_i}{g_i} \end{align*}

3. The solution is

   \begin{equation} s ≡ C_n \pmod{P_n} \label{eq:cycle} \end{equation}

12.10.4. To One Solution

Because the date \( \{x\} \) is given as a collection of positions in repeating periods, there is an infinite number of days that fit that date. For the running day number \( s \) that corresponds to date \( \{x\} \) we find formula \eqref{eq:cycle}, which gives a solution \( \bmod P_n \). If you have found a running day number \( s \) that fits date \( \{x\} \), then you can add or subtract arbitrary multiples of \( P_n \) and then you find another \( s \) that belongs to date \( \{x\} \). How do we pick the right one?

To pick the right solution out of an infinite collection of solutions, we need to use additional information. One appropriate way is to pick the solution that is closest to a date \( J_0 \) chosen by us. One can interpret "closest to" in at least four different ways:

• the last one at or before \( s_0 \)
• the first one at or after \( s_0 \)
• the last one before \( s_0 \)
• the first one after \( s_0 \)

We investigate these cases one by one, for \( s ≡ C_n \pmod{P_n} \).

12.10.4.1. The last one at or before

We seek the last \( s \) at or before \( s_0 \) that satisfies \( s ≡ C_n \pmod{P_n} \). Then, with \( k \) a whole number,

\begin{align} s_0 − P_n \| \lt s ≤ s_0 \\ s_0 − P_n \| \lt C_n + kP_n ≤ s_0 \\ \frac{s_0 − C_n}{P_n} − 1 \| \lt k ≤ \frac{s_0 − C_n}{P_n} \end{align}

There is only one whole number \( k \) that satisfies this inequality, and that is

\begin{equation} k = \dfloorratio{s_0 − C_n}{P_n} \end{equation}

so

\begin{align} s \| = C_n + P_n \dfloorratio{s_0 − C_n}{P_n} \notag \\ \| = C_n + P_n \dparen{\frac{s_0 − C_n}{P_n} − \mod1ratio{s_0 − C_n}{P_n}} \notag \\ \| = C_n + (s_0 − C_n) − P_n \mod1ratio{s_0 − C_n}{P_n} \notag \\ \| = s_0 − \dmod{s_0 − C_n}{P_n} \notag \\ \| = s_0 − ((s_0 − C_n) \bmod P_n) \label{eq:nearestle} \end{align}

12.10.4.2. The first at or after

We seek the first \( s \) at or after \( s_0 \) that satisfies \( s ≡ C_n \pmod{P_n} \). Then, with \( k \) a whole number,

\begin{align} s_0 \| ≤ s \lt s_0 + P_n \\ s_0 \| ≤ C_n + kP_n \lt s_0 + P_n \\ \dfrac{s_0 − C_n}{P_n} \| ≤ k \lt \dfrac{s_0 − C_n}{P_n} + 1 \end{align}

There is only one whole number \( k \) that satisifies this inequality, and that number is

\begin{equation} k = \dceilratio{s_0 − C_n}{P_n} \end{equation}

so

\begin{align} s \| = C_n + P_n \dceilratio{s_0 − C_n}{P_n} \notag \\ \| = C_n + P_n \dparen{\dfrac{s_0 − C_n}{P_n} + \dom1ratio{s_0 − C_n}{P_n}} \notag \\ \| = C_n + (s_0 − C_n) + P_n \dom1ratio{s_0 − C_n}{P_n} \\ \| = s_0 + \ddom{s_0 − C_n}{P_n} \\ \| = s_0 + \dmod{C_n − s_0}{P_n} \\ \| = s_0 + ((C_n − s_0) \bmod P_n) \label{eq:nearestge} \end{align}

12.10.4.3. The last one before

The last \( s \) before \( s_0 \) is one \( P_n \) before the first \( s \) at or after \( s_0 \), so

\begin{equation} s = s_0 − P_n − \dmod{s_0 − C_n}{P_n} \end{equation}

12.10.4.4. The first one after

The first \( s \) after \( s_0 \) is one \( P_n \) after the last \( s \) at or before \( s_0 \), so

\begin{align} s \| = s_0 + P_n + ⌈s_0 − C_n⌋_{P_n} \\ \| = s_0 + P_n + ⌊C_n − s_0⌉_{P_n} \end{align}

12.11. Summary

This summary is in terms of days and months, but holds also for other units of time. We distinguish four different degrees of difficulty: from 1 to 4 the formulas get more complicated but also more capable.

1. The simplest calendar has two different month lengths, with the greatest being one greater than the smallest, and allows no shifting of the pattern of month lengths. Month 0 is always a short month, and month \( g − 1 \) (the last month before everything repeats itself) is always a long month.

2. Calendar type 2 is equal to type 1, except that shifting of the pattern of month lengths is allowed.

3. Calendar type 3 is equal to type 2, but now the difference between the month lengths can be more than one.

4. Calendar type 4 is equal to type 3, but allows more than two month lengths.

The following table summarizes the calendar formulas, and uses the following definitions:

• \( q \) is the number of days in the shortest month. \( q \) is a whole number greater than 0.

• \( p \) is the average length of the infinite number of months that this calendar provides (even if in practice the calendar is used in a flat combination with another calendar so that only a limited number of months actually get used). \( p \) is greater than zero, but is usually not a whole number.

• \( f \), \( g \) are the numerator and denominator from \( p = f/g \). They are whole numbers greater than zero.

• \( h/g \), \( h_i/g_i \) is the fraction of the infinite number of months that has the non-basic month length (unequal to \( q \)). \( h, h_i, g_i \) are whole numbers greater than zero.

• \( r \), \( r_i \) is the difference between a month length and the basic month length \( q \). They are whole numbers greater than zero.

• \( t/g \), \( t_i/g_i \) indicate by how many months the pattern of month lengths should be shifted compared to the simplest case. They are a whole numbers. Only the values from 0 through \( g − 1 \) are important: value \( t + g \) has the same effect as value \( t \).

• \( γ_i = g/g_i \) is a factor that is needed when not all \( g_i \) are equal.

• \( m \) is the running month number. It is a whole number, and can be negative, zero, or positive.

• \( σ \) is the running day number of the beginning of month \( m \). It is a whole number, and can be negative, zero, or positive.

• \( ξ \) is a correction to be added to month number \( m \) to get a better estimate for the month number. This is only relevant for calendar types 3 and 4.

• \( d \) is the number of days since the beginning of the current month, \( d = s − σ \).

 #     

Calendar type 1 is like calendar type 2 but with \( t = 0 \).

For calendar types 3 and 4, the formula for \( m \) yields an upper boundary for the month number. The general procedure for finding the right month number is then:

Try the \( m \) from the formula in the table. Calculate the corresponding \( σ \) and then \( d = s − σ \) and \( ξ \). If \( ξ \) is equal to 0, then you're done. Otherwise, subtract 1 from \( m \) and try again.

If \( ρ ≤ 0 \) then you need to calculate \( ξ \) at most once. Otherwise you may need to do it more often.

If there are more that two relevant units of time, then different calendar levels must be combined.

13. Derivation for Specific Calendars

Almost all calendars distinguish between three basic periods: days, months, and years. One calendar level (straight line) links two periods, so at least two calendar levels are needed to link three periods.

In the next sections we give a diagram for each calendar calculation. We explain those diagrams based on the following example:

  j ───────────┐ ┌─ a₁ ──────────┐
               (1)               │
  m ─(−1)─ m₀ ─┘ └─ m₁ ─┐        │
  d ─(−1)─ d₀ ─────────(2)─ d₁ ─(3)─ s ─(+J₀)─ J

The calendar date with year, month, day \( j, m, d \) is on the left, and the CJDN \( J \) is on the right. The symbols between parentheses, like (1) and (+J₀), indicate a transformation. The other symbols, like a₁ and d₁, indicate intermediate results. The lines show which intermediate results from one transformation are used in the next transformation. The intermediate results with names with an a in them (like a₁) are measured in years, those with m are measured in months, and those with d, s, or J are measured in days. Names with a c in them (like c₁, not in the above diagram) are measured in centuries of 100 years each. The calculations with larger units are higher up in the diagram than those with smaller units.

The transformations for which there is a plus or minus sign between the parentheses indicate the corresponding increment or decrement when going from left (calendar date) to right (CJDN). So (+J₀) means that you add \( J_0 \). When going from right (CJDN) to left (calendar date) then you do the opposite, so then (+J₀) means that you should subtract \( J_0 \).

For a transformation without a plus or minus sign between the parentheses the number identifies the transformation. More details are given later. Most such transformations (such as transformations 2 and 3 here) transform one number into two, or two numbers into one, and correspond to one of the calendar levels described in the previous chapter. And sometimes there is a transformation (such as transforomation 1 here) that adjusts two numbers, produces two numbers from two numbers. If the number is shown between parentheses () then it represents a calendar level of types 1 or 2. If it is instead shown between square brackets [] then it represents a calendar level of types 3 or 4.

Let's look at the month \( m \). Going from left to right, we first encounter (−1) so 1 is subtracted and yields \( m_0 \). That goes into transformation 1 together with \( j \) which yield \( a_1 \) (measured in years) and \( m_1 \) (measured in months). \( m_1 \) and \( d_0 \) go into transformation 2 and yield \( d_1 \). That one goes into transformation 3 together with \( a_1 \) and yields \( s \). Then \( J_0 \) is added, and we find the end result \( J \).

You can also read the diagram from right to left. We begin with CJDN \( J \) on the right-hand side. Going toward the left we first encounter (+J₀) so we must subtract \( J_0 \) (because we're going from right to left) and get \( s \). That goes into transformation 3 which yields \( a_1 \) and \( d_1 \). That \( d_1 \) goes into transformation 2, yielding \( d_0 \) and \( m_1 \). \( m_1 \) and \( a_1 \) go into transformation 1 which yield \( m_0 \) and \( j \). Then we reach (−1) so we must add 1 (because we're going from right to left) and then we find \( m \).

In the calculations that follow we sometimes use intermediate results that aren't shown in the diagrams. We indicate them using \( α_i \) if they are measured in years, \( μ_i \) if they are measured in months, \( δ_i \) if they are measured in days, and \( ω_i \) if they are measured in a different unit. Those names can be used for different things in the opposite calendar directions, so \( α_1 \) in the description of how to calculate the running day number from the calendar date can refer to something different than \( α_1 \) in the description of how to calculate the calendar date from the running day number.

13.1. The Julian Calendar

13.1.1. From Julian Date to CJDN (1)

In the Julian calendar, every three regular years of 365 days are followed by a leap year of 366 days. This period of four years contains 3×365 + 366 = 1461 days. Each period of four years has the same sequence of months with their month lengths.

All months have 30 or 31 days, except for February which has 28 or 29 days. That extra-short month February is a bit of a problem; because of it, we have to deal not with two but with four different month lengths. We work around the problem by regarding February as the last calculation month of the calculation year. Then March is calculation month 0 of the calculation year, and the following February is calculation month 11 of the same calculation year. We combine a calendar level between months and days with a calendar level between years and days and then we use that second calendar level to set the length of the year such that the last calculation month (February) can be shorter than 30 days.

The calculations are schematically as shown in the following diagram.

  j ───────────┐ ┌─ a₁ ──────────┐
               (1)               │
  m ─(−1)─ m₀ ─┘ └─ m₁ ─┐        │
  d ─(−1)─ d₀ ─────────(2)─ d₁ ─(3)─ s ─(+J₀)─ J

The summary of the calculation levels (as in the previous chapter) is

1. First we shift months and days such that the first ones have number 0. January 1 then corresponds to \( m_0 = 0, d_0 = 0 \).

   \begin{align} m_0 \| = m − 1 \\ d_0 \| = d − 1 \end{align}

2. Transformation 1 shifts the month number and year number such that March instead of January is the first calculation month (month \( m_1 = 0 \)) of the calculation year \( a_1 \).

   \begin{align} \{ α_1, m_1 \} \| = \Div(m_0 − 2, 12) \\ a_1 \| = j + α_1 \end{align}

   \({j}\)	\({m}\)	\({a_0}\)	\({a_1}\)	\({m_1}\)
   0	1	−1	−1	10
   0	2	−1	−1	11
   0	3	0	0	0
   0	10	0	0	7
   0	11	0	0	8
   0	12	0	0	9
   1	1	−1	0	10
   1	2	−1	0	11
   1	3	0	1	0

3. Transformation 2 calculates the day number \( d_1 \) since the beginning of the calculation year from the calculation month \( m_1 \) and the day number \( d_0 \) since the beginning of the calculation month:

   \begin{equation} d_1 = \dfloorratio{153 m_1 + 2}{5} + d_0 = 30 m_1 + \dfloorratio{3 m_1 + 2}{5} + d_0 \end{equation}

   This is a calendar level of type 2, with \( f = 153, g = 5, q = 30, h = 3, t = 2 \).

   \({m}\)	\({d}\)	\({m_1}\)	\({d_0}\)	\({d_1}\)	\({L}\)
   1	1	10	0	306	31
   1	2	10	1	307
   1	31	10	30	336
   2	1	11	0	337	?
   2	2	11	1	338
   2	28	11	27	364
   2	29	11	28	365
   3	1	0	0	0	31
   4	1	1	0	31	30
   5	1	2	0	61	31
   6	1	3	0	92	30
   7	1	4	0	122	31
   8	1	5	0	153	31
   9	1	6	0	184	30
   10	1	7	0	214	31
   11	1	8	0	245	30
   12	1	9	0	275	31
   12	31	9	30	305	30

   In the preceding table, \( L \) indicates the length of the month. The length of the last calculation month (February) is shown as "?" because it is determined by the length of the year, which isn't always the same.

4. Transformation 3 calculates the running day number \( s \) since the beginning of calculation year 0 from the calculation year \( a_1 \) and the day number \( d_1 \) since the beginning of the calculation year:

   \begin{equation} s = \dfloorratio{1461 a_1}{4} + d_1 \end{equation}

   This is a calendar level of type 2, with \( f = 1461, g = 4, q = 365, h = 1, t = 0 \).

   \({a_1}\)	\({s}\)	\({L}\)
   0	0	365
   1	365	365
   2	730	365
   3	1095	366
   4	1461

   In the preceding table, \( L \) is the length of the year: 3 years of 365 days each, followed by a year of 366 days ― the leap year. After those 4 years the pattern repeats itself.

5. And then we add to \( s \) the CJDN of the beginning of calculation year 0 (March 1 of Year 0), which is 1721118:

   \begin{equation} J = s + J_0 = s + 1721118 \end{equation}

We can combine the preceding steps into

\begin{align} \{ a_2, m_1 \} \| = \Div(m ― 3, 12) \\ J \| = \dfloorratio{1461\dparen{j + a_2}}{4} + \dfloorratio{153m_1 + 2}{5} + d + 1721117 \end{align}

13.1.2. From CJDN to Julian Date (1)

We run through the opposite procedure from the one described above, with the same calendar levels as above.

  j ───────────┐ ┌─ a₁ ──────────┐
               (1)               │
  m ─(−1)─ m₀ ─┘ └─ m₁ ─┐        │
  d ─(−1)─ d₀ ─────────(2)─ d₁ ─(3)─ s ─(+J₀)─ J

The summary of the calculation levels (as in the previous chapter) is

1. First we calculate the running day number \( s \) by subtracting from the CJDN \( J \) of the date of interest the CJDN of the start of calculation year 0.

   \begin{equation} s = J − J_0 = J − 1721118 \end{equation}

2. Transformation 3 calculates from the running day number \( s \) the calculation year \( a_1 \) and the day number \( d_1 \) since the beginning of that calculation year.

   \begin{align} \{ a_1, ω_1 \} \| = \Div(4s + 3, 1461) \\ d_1 \| = \dfloorratio{ω_1}{4} \end{align}

3. Transformation 2 calculates calculation month \( m_1 \) and day number \( d_0 \) since the beginning of the calculation month from day number \( d_1 \) since the beginning of the calculation year.

   \begin{align} \{ m_1, ω_2 \} \| = \Div(5d_1 + 2, 153) \\ d_0 \| = \dfloorratio{ω_2}{5} \end{align}

4. Transformation 1 shifts the month number and year number such that January and not March is month 0.

   \begin{align} \{ α_1, m_0 \} \| = \Div(m_1 + 2, 12) \\ j \| = a_1 + α_1 \end{align}

5. And finally we shift the month and day numbers such that they begin at 1 instead of 0.

   \begin{align} m \| = m_0 + 1 \\ d \| = d_0 + 1 \end{align}

We can condense this a bit:

\begin{align} \{ a_1, ω_1 \} \| = \Div(4J − 6884469, 1461) \\ \{ m_1, ω_2 \} \| = \Div\dparen{5\dfloorratio{ω_1}{4} + 2, 153} \\ \{ α_1, m_0 \} \| = \Div(m_1 + 2, 12) \\ j \| = a_1 + α_1 \\ m \| = m_0 + 1 \\ d \| = \dfloorratio{ω_2}{5} + 1 \end{align}

13.1.3. From Julian Date to CJDN (2)

We can also use a stepped combination of calendars; then we need a calendar level between a running month number and a running day number, taking into account leap years. In that case there is no longer a higher calendar level that will shorten a too-long last month, so the formulas must be correct also for very large running month numbers.

The calculations are schematically as follows:

  j ────────────┐
  m ─(−1)─ m₀ ─(1)─ m₁ ─┐
  d ─(−1)─ d₀ ─────────[2]─ s ─(+J₀)─ J

The summary of the calendar levels is

1. First we shift months and days such that the first ones have number 0. January 1 then corresponds to \( m_0 = 0, d_0 = 0 \).

   \begin{align} m_0 \| = m − 1 \\ d_0 \| = d − 1 \end{align}

2. Transformation 1 calculates the running month number \( m_1 \) from the year \( j \) and the month number \( m_0 \) since the beginning of the year. The formula is very simple:

   \begin{equation} m_1 = 12j + m_0 \end{equation}

   This is a calendar level of type 1, with \( f = 12, g = 0, q = 12, h = 0, t = 0 \). Because \( t = 0 \) it is also a calendar level of type 1.

3. Transformation 2 must calculate the running day number \( s \) from the running month number \( m_1 \). Of the 12 months of a year, 7 are long, with 31 days each, so we need a formula like \( \dfloor{(7(m_1 + a))/12} \) (which repeats itself every 12 months), but what should be the value of \( a \)?

   With \( a = 0 \) we get the values

   \({m_1}\)	\({\dfloorratio{7m_1}{12}}\)	∆
   0	0	0
   1	0	1
   2	1	0
   3	1	1
   4	2	0
   5	2	1
   6	3	1
   7	4	0
   8	4	1
   9	5	0
   10	5	1
   11	6	1
   12	7

   Each time that the value increases by one, the preceding month was a long one, so these values mean that the sequence of short (S) and long (L) months was: S L S L S L L S L S L L, but we're searching for L S L S L S L L S L S L, because July and August (the 7th and 8th months) are both long. We get that if we shift the pattern by −1 month (one to the left), so \( a = −1 \). Then \( t = \dmod{fa}{g} = \dmod{−7}{12} = 5 \), so we find \( \dfloor{(7m_1 + 5)/12} \).

   Then all months are good, except for February (\( m_1 = 1 \)), because that month gets 30 days but should have only 28 days in a regular year or 29 days in a leap year. We subtract 2 days from every February using the formula \( −2\dfloor{(m_1 + 10)/12} \). We add 1 day back in for the first of every four years by using \( \dfloor{(m_1 + 46)/48} \). All in all we find

   \begin{equation} s = 30m_1 + \dfloorratio{7m_1 + 5}{12} − 2\dfloorratio{m_1 + 10}{12} + \dfloorratio{m_1 + 46}{48} + d_0 \end{equation}

   This is calendar type 4 with

   \( q = 30 \), \( g = 48 \), \( \{ r_1, h_1, t_1, g_1 \} = \{ 1, 7, 5, 12 \} \), \( \{r_2, h_2, t_2, g_2 \} = \{ −2, 1, 10, 12 \} \), \( \{ r_3, h_3, t_3, g_3 \} = \{ 1, 1, 46, 48\} \).

4. And then we add the CJDN of the beginning of the year 0 (January 1st, Year 0), and that is 1721058.

   \begin{equation} J = s + J_0 = s + 1721058 \end{equation}

This condenses to

\begin{align} m_1 \| = 12j + m − 1 \\ J \| = 30m_1 + \dfloorratio{7m_1 + 5}{12} − 2\dfloorratio{m_1 + 10}{12} + \dfloorratio{m_1 + 46}{48} + d + 1721057 \end{align}

13.1.4. From Julian Day Number to Julian Date (2)

The calculations are schematically as follows:

  j ────────────┐
  m ─(−1)─ m₀ ─(1)─ m₁ ─┐
  d ─(−1)─ d₀ ─────────[2]─ s ─(+J₀)─ J

The summary of the calendar levels is

We run through the above procedure in the opposite direction. First we calculate the running day number \( s \) by subtracting the CJDN \( J_0 \) of the beginning of year 0 from the CJDN \( J \) of the sought calendar date:

\begin{equation} s = J − J_0 = J − 1721058 \end{equation}

Transformation 2 produces the running month number \( m_1 \) and the day number \( d_0 \) since the beginning of the month from the running day number \( s \).

We have

\begin{align} μ_1 \| = \dfloorratio{gs + g\dparen{\sum_{r_i \gt 0} r_i} − \dparen{\sum_i γ_i r_i t_i} − 1}{f} \notag \\ \| = \dfloorratio{48s + 48×2 − (4×1×5 + 4×−2×10 + 1×1×46) − 1}{1461} \notag \\ \| = \dfloorratio{48s + 109}{1461} \notag \\ δ_1 \| = s − 30μ_1 − \dfloorratio{7μ_1 + 5}{12} + 2\dfloorratio{μ_1 + 10}{12} − \dfloorratio{μ_1 + 46}{48} \end{align}

Because \( ρ ≤ 0 \) we can use the procedure with the fixed steps. Then

\begin{align} m_1 \| = μ_1 + \dfloorratio{δ_1}{32} \\ d_0 \| = s − 30m_1 − \dfloorratio{7m_1 + 5}{12} + 2\dfloorratio{m_1 + 10}{12} − \dfloorratio{m_1 + 46}{48} \end{align}

Transformation 1 goes from running month number \( m_1 \) to year number \( j \) and month number \( m_0 \) since the beginning of the year. This is a calendar level of type 1.

\begin{equation} \{ j, m_0 \} = \Div(m_1, 12) \end{equation}

And finally we shift the month and day numbers such that they begin at 1 instead of 0.

\begin{align} m \| = m_0 + 1 \\ d \| = d_0 + 1 \end{align}

13.2. The Gregorian Calendar

13.2.1. From Gregorian Date to CJDN (1)

The Gregorian calendar is equal to the Julian calendar, except for the rules that say when the month of February has 29 days instead of 28 days. In the Julian calendar, February has 29 days in every fourth year, when the year number is evenly divisible by 4. In the Gregorian calendar the same rule holds, except that February has 28 days when the year number is evenly divisible by 100, except if the year number is evenly divisible by 400, when February has 29 days after all.

To approximate Gregorian years with a straight line with a single pattern, the leap values have to return after every \( ⌊Q⌋ \) or \( ⌈Q⌉ \) periods (for a well-chosen \( Q \); see equation \eqref{eq:Q}). Unfortunately, the leap year rules of the Gregorian Calendar do not meet that demand, because they mean that the time between two leap years is sometimes 4 years and sometimes 8 years.

The following \( p \) might be considered:

• 146097 days in 400 years → \( p = f/g = 146097/400 \). This yields 85 sequences of a leap year every 4 years, but also 12 sequences of a leap year every 5 years. That does not fit the Gregorian Calendar, so this straight line is not useful for calculating the Gregorian Calendar.

• 146097 days in 100 periods of 4 years → \( p = f/g = 146097/100 \). This yields two sequences of 33 periods (of 4 years) with one fewer leap day at the end (1460 days instead of 1461 days), and one sequence of 34 periods. The Gregorian Calendar has two sequences of 25 periods (of 4 years) with one fewer leap day at the end, and then a sequence of 50 periods with one fewer leap day at the end. This one is not suitable, either.

• 36524 days in 100 years → \( p = f/g = 36524/100 \). This yields 20 sequences of a leap year every 4 years, but also 4 sequences of a leap year every 5 years. This one is not suitable.

• 146097 days in 4 periods of 100 years → \( p = f/g = 146097/4 \). This yields 3 sequences of 36524 days and 1 sequence of 36525 days, and that fits the Gregorian Calendar.

• 36525 days in 100 years → \( p = f/g = 36525/100 \). This yields 25 sequences with a leap year every 4 years. That fits, except that we sometimes need to stop already after 36524 days.

We can now use the 146097/4 line to find the 100-year periods, and then use the 36525/100 line to find the years within the 100-year period, and finally (just like for the Julian Calendar) use the 153/5 line to find the months within the year. 1 March of year 0 in the Gregorian Calendar corresponds to \( J_0 = 1721120 \). The calculations are schematically as follows:

                         ┌── c₁ ──────────┐
  j ───────────┐ ┌─ a₁ ─(2)─ a₂ ─┐        │
               (1)               │        │
  m ─(−1)─ m₀ ─┘ └─ m₁ ─┐        │        │
  d ─(−1)─ d₀ ─────────(3)─ d₁ ─(4)─ d₂ ─(5)─ s ─(+J₀)─ J

The summary of the calendar levels is

The steps to get from the Gregorian calendar to the CJDN are now:

1. First we shift months and days such that the first ones have number 0. January 1 then corresponds to \( m_0 = 0, d_0 = 0 \).

   \begin{align} m_0 \| = m − 1 \\ d_0 \| = d − 1 \end{align}

2. Transformation 1 shifts the month number and year number such that March instead of January is the first calculation month (month \( m_1 = 0 \)) of the calculation year \( a_1 \), just like for the Julian calendar (13.1.1).

   \begin{align} \{ α_1, m_1 \} \| = \Div(m_0 − 2, 12) \\ a_1 \| = j + α_1 \end{align}

3. Transformation 2 calculates the calculation century number \( c_1 \) and year number \( a_1 \) since the beginning of the calculation century from the calculation year number \( a_1 \).

   \begin{equation} \{ c_1, a_2 \} = \Div(a_1, 100) \end{equation}

4. Transformation 3 calculates the day number \( d_1 \) since the beginning of the calculation year from the calculation month \( m_1 \) and the day number \( d_0 \) since the beginning of the calculation month:

   \begin{equation} d_1 = \dfloorratio{153 m_1 + 2}{5} + d_0 = 30 m_1 + \dfloorratio{3 m_1 + 2}{5} + d_0 \end{equation}

5. Transformation 4 calculates the day number \( d_2 \) since the beginning of the calculation century from the year number \( a_2 \) since the beginning of the calculation century and the day number \( d_1 \) since the beginning of the calculation year.

   \begin{equation} d_2 = \dfloorratio{36525 a_2}{100} + d_1 \end{equation}

6. Transformation 5 calculates the running day number \( s \) from the calculation century number \( c_1 \) and the day number \( d_2 \) since the beginning of the calculation century.

   \begin{equation} s = \dfloorratio{146097 c_1}{4} + d_2 \end{equation}

7. And then we add to \( s \) the CJDN of the beginning of calculation year 0 (March 1 of Year 0), which is 1721120:

   \begin{equation} J = s + J_0 = s + 1721120 \end{equation}

Combined and condensed, this yields:

\begin{align} \{ α_1, m_1 \} \| = \Div(m − 3, 12) \\ \{ c_1, a_2 \} \| = \Div(j + α_1, 100) \\ J \| = \dfloorratio{146097 c_1}{4} + \dfloorratio{36525 a_2}{100} + \dfloorratio{153 m_1 + 2}{5} + d + 1721119 \end{align}

13.2.2. From CJDN to Gregorian Date (1)

Now we go in the opposite direction again, with the same calendar levels as before.

                         ┌── c₁ ──────────┐
  j ───────────┐ ┌─ a₁ ─(2)─ a₂ ─┐        │
               (1)               │        │
  m ─(−1)─ m₀ ─┘ └─ m₁ ─┐        │        │
  d ─(−1)─ d₀ ─────────(3)─ d₁ ─(4)─ d₂ ─(5)─ s ─(+J₀)─ J

The summary of the calendar levels is

1. First we subtract the CJDN that goes with running day number 0 from the CJDN to get the running day number:

   \begin{equation} s = J − J_0 = J − 1721120 \end{equation}

2. Transformation 5 produces from the running day number \( s \) the calculation century \( c_1 \) and the day number \( d_2 \) since the beginning of the calculation century:

   \begin{align} \{ c_1, ω_1 \} \| = \Div(4s + 3, 146097) \\ d_2 \| = \dfloorratio{ω_1}{4} \end{align}

3. Transformation 4 produces from the day number \( d_2 \) since the beginning of the calculation century the year number \( a_2 \) since the beginning of the calculation century and the day number \( d_1 \) since the beginning of the calculation year.

   \begin{align} \{ a_2, ω_2 \} \| = \Div(100d_2 + 99, 36525) \\ d_1 \| = \dfloorratio{ω_2}{100} \end{align}

4. Transformation 3 separates the day number \( d_1 \) since the beginning of the calculation year into the calculation month number \( m_1 \) since the beginning of the calculation year and the day number \( d_0 \) since the beginning of the calculation month.

   \begin{align} \{ m_1, ω_3 \} \| = \Div(5d_1 + 2, 153) \\ d_0 \| = \dfloorratio{ω_3}{5} \end{align}

5. Transformation 2 combines the calculation century number \( c_1 \) and the calculation year number \( a_2 \) since the beginning of the calculation century into the calculation year number \( a_1 \).

   \begin{equation} a_1 = 100c_1 + a_2 \end{equation}

6. Transformation 1 shifts the calculation month number \( m_1 \) and calculation year number \( a_1 \) such that January instead of March is the first month of the year.

   \begin{align} \{ α_1, m_0 \} \| = \Div(m_1 + 2, 12) \\ j \| = a_1 + α_1 \end{align}

7. And then we shift the month number and the day numbers such that the first ones have number 1 instead of 0.

   \begin{align} m \| = m_0 + 1 \\ d \| = d_0 + 1 \end{align}

This condenses a little bit, to:

\begin{align} \{ c_1, ω_1 \} \| = \Div(4J − 6884477, 146097) \\ \{ a_2, ω_2 \} \| = \Div\dparen{100\dfloorratio{ω_1}{4} + 99, 36525} \\ \{ m_1, ω_3 \} \| = \Div\dparen{5\dfloorratio{ω_2}{100} + 2, 153} \\ \{ α_1, m_0 \} \| = \Div(m_1 + 2, 12) \\ j \| = 100c_1 + a_2 + α_1 \\ m \| = m_0 + 1 \\ d \| = \dfloorratio{ω_3}{5} + 1 \end{align}