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From: Thomas Wolf <t_wolf@angelfire.com>
Newsgroups: comp.lang.ada
Subject: Re: Zeller's Algorithm
Date: Mon, 29 Jul 2002 10:29:09 +0200
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Date: 2002-07-29T10:29:09+02:00
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paul.a.storm@lmco.com wrote:
> 
> I am still confused as to why the Zeller algorithm wouldn't work.  Since
> the algorithm has been
> around since 1875 I figure it has proven out.  Although I am still
> cogitating on whether it
> takes into account for centennial non-leap years(1800,1900, etc.) and
> quatra-centennial leap years(i.e. 2000).

Yes it does.

> It bugs me when I can't figure out the reason for some code not
> working.  Know what I mean?

Zeller's *algorithm* does work for dates in the Gregorian calendar.
It's just the poor, buggy *implementation* posted at AdaPower that
doesn't work.

> Did you try to compile the program from the link I gave and did you have
> any problems like I had?

No, I didn't try to run that program. It is flawed in two ways:

1. The "if Month < 3" stuff should be moved out of GetValidDate into
   UseZeller.
2. UseZeller shouldn't add 1 at the end.

For a working implementation of Zeller's algorithm, try this:

   type Weekday is (Mon, Tue, Wed, Thu, Fri, Sat, Sun);
   --  ISO starts the week on a monday, so let's do the same here...

   function Day_Of_Week
     (Date : in Ada.Calendar.Time)
     return Weekday
   is
     --  Returns the day of the week. Works for dates of the 
     --  Gregorian calendar.

      Day   : Ada.Calendar.Day_Number;
      Month : Ada.Calendar.Month_Number;
      Year  : Ada.Calendar.Year_Number;
      Secs  : Ada.Calendar.Day_Duration;

      M, Y, Century, Z : Integer;

   begin
      --  This is Zeller's formula.
      Ada.Calendar.Split (Date, Year, Month, Day, Secs);
      if Month < 3 then
         M := Integer (Month) + 10;
         Y := Integer (Year) - 1;
      else
         M := Integer (Month) - 2;
         Y := Integer (Year);
      end if;
      Century := Y / 100; Y := Y mod 100;
      Z :=
        (Integer (Day) + (13 * M - 1) / 5 + Y + Y / 4 +
         Century / 4 - 2 * Century) mod 7;
      --  The left operand of the "mod" may be negative, but "mod 7"
      --  does the right thing and gives us a positive remainder in
      --  the range 0 .. 6.
      --  
      --  Now 0 is Sunday, 1 is Monday, and so on. Map this such that
      --  0 is Mon, 1 is Tue, and 6 is Sun.
      return Weekday'Val ((Z + 6) mod 7);
   end Day_Of_Week;

I'll make available this function in a future release of
my Util subsystem.

-- 
-----------------------------------------------------------------
Thomas Wolf                          e-mail: t_wolf@angelfire.com