% --- ------------------------------------------------------------------
% --- English-style Date Macros for LaTeX, version 1.00--0.
% --- ------------------------------------------------------------------
%
% --- \today (replaces American-style macro of same name) generates
% --- today's date in the form Thursday 4th October 1066.
% --- The `th' is raised, reduced in size and underlined in the
% --- same font style (family) as the rest of the date.
%
% --- \st generates a raised, reduced, underlined `st', as in 1st.
% --- \nd generates a raised, reduced, underlined `nd', as in 2nd.
% --- \rd generates a raised, reduced, underlined `rd', as in 3rd.
% --- \th generates a raised, reduced, underlined `th', as in 4th.
%
% --- \dayofweek generates the day of the week, based on TeX's values
% --- of \day, \month and \year.
%
% --- \phaseofmoon generates the current phase of the moon, again based
% --- on TeX's values for \day, \month and \year.
%
% --- ------------------------------------------------------------------
% --- Adrian F. Clark (alien@uk.ac.essex.ese) 26-Oct-1988 10:05:50
% --- ------------------------------------------------------------------
% --- Counters. Note that we use the same registers as TeX holds other
% --- things in (e.g., \count0 holds the page number). This requires
% --- that \@savestyle, \@setstyle, \dayofweek and \phaseofmoon perform
% --- all their register manipulations within a group. This may seem
% --- a bit messy, but it saves having eight registers permanently set
% --- aside just for date calculation.
\def\@cent{\count0 } % century number (1979 == 20)
\def\@diy{\count1 } % day in the year
\def\@dow{\count2 } % gets day of the week
\def\@epact{\count3 } % age of the moon on Jan. 1
\def\@golden{\count4 } % Moon's golden number
\def\@leap{\count5 } % leap year fingaler
\def\@x{\count6 } % temp register
\def\@y{\count7 } % another temp register
% --- A replacement for the ``plain'' TeX and LaTeX \today macro, to
% --- to output the date in English-style.
% --- They ensure the smaller text comes out in the right font by saving
% --- the font family before reducing the size, then restoring it. (This
% --- was suggested by Leslie Lamport.) Of course, it requires that the
% --- font in use when today is invoked has a sensible family.
\def\@up#1{{\@savestyle\thinspace$^{\underline{\hbox{%
\scriptsize\@setstyle#1\fam=-1 }}}$}}
\def\st{\@up{st}}
\def\nd{\@up{nd}}
\def\rd{\@up{rd}}
\def\th{\@up{th}}
% --- Macros to save and restore the font family.
\def\@savestyle{\count0=\the\fam}
\def\@setstyle{\ifcase\count0\rm\or\mit\or\cal\or\rm% what's family 3?
\or\it\or\sl\or\bf\or\tt\fi}
% --- The date, English style (e.g. Thursday 4th October 1066).
\def\today{\dayofweek\ \number\day\ifcase\day
\or\st\or\nd\or\rd\or\th\or\th\or\th\or\th\or\th\or\th\or\th
\or\th\or\th\or\th\or\th\or\th\or\th\or\th\or\th\or\th\or\th
\or\st\or\nd\or\rd\or\th\or\th\or\th\or\th\or\th\or\th\or\th\or\st\fi
\space\ifcase\month\or January\or February\or March\or April\or May\or
June\or July\or August\or September\or October\or November\or December\fi
\space\number\year}
% --- The day of the week ("Sunday", etc.) is inserted into the text
% --- by \dayofweek. (This uses registers \@dow, \@leap, \@x and \@y.)
% --- I acquired this from elsewhere; I don't know who wrote it.
\def\dayofweek{{%
% leap = year + (month - 14)/12;
\@leap=\month \advance\@leap by -14 \divide\@leap by 12
\advance\@leap by \year
% dow = (13 * (month + 10 - (month + 10)/13*12) - 1)/5
\@dow=\month \advance\@dow by 10
\@y=\@dow \divide\@y by 13 \multiply\@y by 12
\advance\@dow by -\@y \multiply\@dow by 13
\advance\@dow by -1 \divide\@dow by 5
% dow += day + 77 + 5 * (leap % 100)/4
\advance\@dow by \day \advance\@dow by 77
\@x=\@leap \@y=\@x \divide\@y by 100 \multiply\@y by 100 \advance\@x by -\@y
\multiply\@x by 5 \divide\@x by 4 \advance\@dow by \@x
% dow += leap / 400
\@x=\@leap \divide\@x by 400 \advance\@dow by \@x
% dow -= leap / 100 * 2;
% dow = (dow % 7)
\@x=\@leap \divide\@x by 100 \multiply\@x by 2 \advance\@dow by -\@x
\@x=\@dow \divide\@x by 7 \multiply\@x by 7 \advance\@dow by -\@x
\ifcase\@dow Sunday\or Monday\or Tuesday\or Wednesday\or
Thursday\or Friday\or Saturday\fi}}
%
% --- Likewise, \phaseofmoon inserts the phase of the moon into the
% --- text. This was written by the same person as \dayofweek.
% --- The routine calculates the year's epact (the age of the moon on Jan 1.),
% --- adds this to the number of days in the year, and calculates the phase
% --- of the moon for this date. It returns the phase as a string, e.g.,
% --- "new", "full", etc.
%
% --- In the algorithm:
% --- diy is the day of the year - 1 (i.e., Jan 1 is day 0).
% --- golden is the number of the year in the Mentonic cycle, used to
% --- determine the position of the calender moon.
% --- epact is the age of the calender moon (in days) at the beginning
% --- of the year. To calculate epact, two century-based
% --- corrections are applied:
% --- Gregorian: (3 * cent)/4 - 12
% --- is the number of years such as 1700, 1800 when
% --- leap year was not held.
% --- Clavian: (((8 * cent) + 5) / 25) - 5
% --- is a correction to the Mentonic cycle of about
% --- 8 days every 2500 years. Note that this will
% --- overflow 16 bits in the year 409600. Beware.
% --- The algorithm is accurate for the Gregorian calender only.
%
% --- The magic numbers used in the phase calculation are:
% --- 29.5 The moon's period in days.
% --- 177 29.5 scaled by 6
% --- 22 (29.5 / 8) scaled by 6 (this gets the phase)
% --- 11 ((29.5 / 8) / 2) scaled by 6
%
% --- Theoretically, this should yield a number in the range 0 .. 7. However,
% --- two days per year, things don't work out too well.
%
% --- Epact is calculated by the algorithm given in Knuth vol. 1 (Calculation
% --- of Easter). See also the article on Calenders in the Encyclopaedia
% --- Britannica and Knuth's algorithm in CACM April 1962, page 209.
%
\def\phaseofmoon{{%
\@diy=\day \advance\@diy by \ifcase\month % Jan 1 == 0
-1\or -1\or 30\or 58\or 89\or 119\or 150\or % Jan .. Jun
180\or 211\or 241\or 272\or 303\or 333\fi % Jul .. Dec
% if ((month > 2) && ((year % 4 == 0) &&
% ((year % 400 == 0) || (year % 100 != 0))))
% diy++; /* Leapyear fixup */
\ifnum \month>2
\@x=\year \@y=\@x \divide\@y by 4 \multiply\@y by 4 \advance\@x by -\@y
\ifnum \@x=0 % month > 2 and maybe leapyear
\@x=\year \@y=\@x \divide\@y by 400
\multiply\@y by 400 \advance\@x by -\@y
\ifnum \@x=0 % 2000 is a leap year
\advance\@diy by 1 % so it's one day later
\else % not 2000, check other '00's
\@x=\year \@y=\@x \divide\@y by 100
\multiply\@y by 100 \advance\@x by -\@y
\ifnum \@x>0 % not some other '00' year
\advance\@diy by 1 % it's still one day later
\fi % not odd century
\fi % not 2000-type century
\fi % not leapish year
\fi % not march or later
% cent = (year / 100) + 1; /* Century number */
% golden = (year % 19) + 1; /* Golden number */
\@cent=\year \divide\@cent by 100 \advance\@cent by 1
\@golden=\year
\@y=\year \divide\@y by 19 \multiply\@y by 19 \advance\@golden by -\@y
\advance\@golden by 1
% epact = ((11 * golden) + 20 /* Golden number */
% + (((8 * cent) + 5) / 25) - 5 /* 400 year cycle */
% - (((3 * cent) / 4) - 12)) % 30; /* Leap year correction */
\@epact=11 \multiply\@epact by \@golden
\advance\@epact by 20
\@x=8 \multiply\@x by \@cent \advance\@x by 5
\divide\@x by 25 \advance\@x by -5
\advance\@epact by \@x
\@x=3 \multiply\@x by \@cent \divide\@x by 4 \advance\@x by -12
\advance\@epact by -\@x
\@y=\@epact \divide\@y by 30 \multiply\@y by 30 \advance\@epact by -\@y
% if (epact <= 0)
% epact += 30; /* Age range is 1 .. 30 */
% if ((epact == 25 && golden > 11) || epact == 24)
% epact++;
\ifnum \@epact<0
\advance\@epact by 30
\fi
\ifnum \@epact=25
\ifnum \@golden>11
\advance \@epact by 1
\fi
\else
\ifnum \@epact=24
\advance \@epact by 1
\fi
\fi
%
% --- Calculate the phase, using the magic numbers defined above.
% --- Note that phase may be equal to 8 (== 0) on two days of the year
% --- due to the way the algorithm was implemented.
% --- phase = (((((diy + epact) * 6) + 11) % 177) / 22) & 7;
%
\@x=\@diy \advance\@x by \@epact \multiply\@x by 6 \advance\@x by 11
\@y=\@x \divide\@y by 177 \multiply\@y by 177 \advance\@x by -\@y
\divide\@x by 22
\ifcase\@x new\or waxing crescent\or in its first quarter\or
waxing gibbous\or full\or waning gibbous\or
in its last quarter\or waning crescent\or new\fi}}
% --- End of ukdate.sty