Path: utzoo!utgpu!jarvis.csri.toronto.edu!mailrus!cs.utexas.edu!uunet!mcsun!cernvax!cui!ugun2b!ugsc2a!fisher
From: fisher@sc2a.unige.ch (Markus Fischer)
Newsgroups: comp.binaries.ibm.pc.d
Subject: Re: Looking for D.O.W. algorithm
Message-ID: <125@sc2a.unige.ch>
Date: 18 Sep 89 17:37:38 GMT
References: <6470@hubcap.clemson.edu>
Organization: University of Geneva, Switzerland
Lines: 74

In article <6470@hubcap.clemson.edu>, chrisb@hubcap.clemson.edu (Chris) writes:
>   I am looking for an algorithm which gives the day of the week (M,T,W,Th,...)
> for a given date... (9,15,1989) Sept 15, 1989 --- would be Friday. 

Here you go:

The method is simple: for jan. and feb., add 12 to `month' and substract 1 to
`year' (in fact, the year was always meant to start with march ! ever noticed
that sept. is 7, oct. 8, nov. 9, and dec. is 10 [as in decimal] ? :-).
Then take year*365.25 + (month+1)*30.6 + day, this gives you an arbitrary
`number of days'.
As this works only when even years come every four years, the range is
1.3.1900 - 28.2.2100.  So the best is to substract the value you'd get for the
fist sunday after the (non-existing) 29.2.1900, namely the 4.3.1900.

Easy!

Ah yes, the number modulus 7 gives you the day of the week (0 for sundays).

Finally, here's some C-code as an example:

--- file day.c ---
#include <stdio.h>
#include <stdlib.h>

long
days (int d, int m, int y)
/*
 * This function returns the number of days since the March 4th, 1900.
 *
 * Parameters: 3 ints representing the day, month and year.
 *             no checking is done: any int is accepted !
 * Range:      4.03.1989 - 27.2.2100 (there will be no 28.2.2100)
 *             For a longer range add the usual tests for missing even years.
 */
{
	return (
		+ (long)((float)(y-(m<3)) * 365.25)  /* y->d (-1 for jan. and feb.) */
		+ (long)((float)((m+9)%12+4) * 30.6) /* m->d (+12 "   "    "   "  ) */
		+ (long)d                            /* add d :-) */
		- (long)694101                       /* val of 4.03.1989 (a sunday) */
	);
}

main (int argc, char** argv)
{
	int day, month, year;
	long n;
	char *dow[7] = {"sunday","monday","tuesday","wednesday",
	                "thursday","friday","saturday"};

	if (argc<4)
		printf("Enter 3 ints as parameter \n");
	else {
		day = atoi(argv[1]);
		month = atoi(argv[2]);
		year = atoi(argv[3]);
		n = days (day, month, year);
		printf (" On %s %d.%02d.%04d, %ld days have elapsed since"
		        " sunday 4.03.1989.\n",
		        dow[(unsigned)n%7], day,month,year, n);
	}
}
--- EOF ---

Hope this helps !

Markus Fischer                -|--|--|--|--|--|--I   Department of Anthropology
                        -|--|--|--|--|--|--|-(#)-I   University of Geneva
      -|--|--|--|--|--|--|--|--|-(#)-|-(#)(#)(_)-I   CH-1227  Carouge (GE)
   -&-(_)-|--|--|-(#)-&--|-(#)(#)(_)(#)-&-(_)(#)-I   Switzerland
            -|--|--|--|--|-(#)(_)-|-(_)(_)(_)(#)-I
black (#) to kill ! --|--|-(#)(_)(_)(_)(#)(#)(_)(_)  fisher@sc2a.unige.ch
=+==+==+==+==+==+==+==+==+==+==+==+==+==+==+=(#)=+   fisher@cgeuge52.bitnet