%NAME: ukdate.sty
% --- ------------------------------------------------------------------
% ---       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
