The monthshift function (starting on march 1st):     Torsten Sillke, 1999-03-15

 The step function: s(month)

 8 +    x    x    x    x    x    x    x    x    x    x    x    x
   |
   |
 7 +    x    x    x    x    x    x    x    x    x    x    x    o_A
   |
   |
 6 +    x    x    x    x    x    x    x    x    x    x_D  o    o
   |
   |
 5 +    x    x    x    x    x    x    x    x    o    o    o    o
   |
   |
 4 +    x    x    x    x    x    x    o    o    o    o    o    o
   |
   |
 3 +    x    x    x    x    x_B  o    o    o    o    o    o    o
   |
   |
 2 +    x    x    x    o    o    o    o    o    o    o    o    o
   |
   |
 1 +    x    o_C  o    o    o    o    o    o    o    o    o    o
   |
   |
 0 +----o----+----+----+----+----+----+----+----+----+----+----+-> month
   0    1    2    3    4    5    6    7    8    9   10   11   12
       mar  apr  may  jun  jul  aug  sep  oct  nov  dec  jan  feb
        0    1    1    2    2    3    4    4    5    5    6    7   s(month)
        0    31   61   92  122  153  184  214  245  275  306  337  monthshift

   s(month) = monthshift - 30*(month-1)

 We are looking for lines that separate the 'o' from the 'x'.
 So, which lines are above A and C but below B and D?

 Let  s = floor( a * month + b )

 Constraints:
 (A)  12 a + b >= 7
 (B)   5 a + b <  3
 (C)   2 a + b >= 1
 (D)  10 a + b <  6

 The admissable parameter (a,b) are inside the quadrangle

 ( 4/7,  1/7)   line AB
 ( 3/5,   0 )   line BD
 ( 5/8, -1/4)   line CD
 ( 3/5, -2/5)   line AC

 The two edges on the right are excluded the left ones are included.

 For b = 0 we get the section 7/12 <= a < 3/5.

 Now we can select a nice fraction.
 Zeller's choise:  a = 3/5, b = -2/5.
 I prefere: a = 19/32, b = 0 as the division is faster on a binary computer.


Reference:

 - Chr. Zeller, Kalender-Formeln, Acta Mathematica, 9 (1887) 131-136

 - A. D. Bradley, The Day of the Week for Gregorian Calendars  ?? 82-87

 - Keith & Craver; The ultimate perpetual calendar?, JoRM 22:4 (1990) 280-282
   day of the week as a 44 character expression in C. (illegal use of --)
   The following 45 character C expression by Keith is correct.
   dow(y,m,d) { return (d+=m<3?y--:y-2,23*m/9+d+4+y/4-y/100+y/400)%7; }

 - A. W. Butkewitsch, M. S. Selikson - Ewige Kalender, Teubner (Leipzig) 1974
   Kleine Naturwissenschaftliche Bibliothek, Bd. 23, p98-113
   ISBN 3-322-00393-0

 - Heinz Bachmann, Kalenderarithmetik, 2nd Ed.,
   Juris Verlag, Z"urich, 1984, ISBN 3-260-05035-0, p26-42