fubar@squid.UUCP (01/28/88)
From squid!fubar Thu Jan 28 04:22 CDT 1988 remote from occrsh Subject: day of week *** *** PLEASE send replies to ihnp4!occrsh!squid!david *** > Here's a formula given in the book An Introduction to the Theory of Numbers, >by Niven and Zuckerman. > > Let M be the number of the month, defined so that Jan = 11, Feb = 12, >Mar = 1, Apr = 2, ..., Dec = 10. Let N be the number of the day within the >month. Let C be the hundreds in the year, and let Y be the rest of the year. >Let L be 1 for a leap year and 0 for a non-leap year. > >The formula > >(N + [2.6M - 0.2] + Y + [Y / 4] + [C / 4] - 2C - (1 + L)[M / 11]) mod 7 >yields 0 for Sunday, 1 for Monday, ..., 6 for Saturday. > >(Square brackets denote the greatest integer function.) So: Thursday, Jan 28, 1988, should yield 4: N = 28 (number of day within month) M = 11 (Jan = 11) C = 19 (Hundreds w/in year) Y = 88 (rest of year??) L = 1 (1988 is leap year) = (N + [2.6M - 0.2] + Y + [Y/4] + [C/4] - 2C - (1+L)[M/11]) mod 7 = (28 + [(2.6*11)-0.2] + 88 + [88/4] + [19/4] - (2*19) - (1+1)*[11/11]) mod 7 = (28 + [28.4] + 88 + [22] + [4.75] - 38 - 2) mod 7 = (28 + 29 + 88 + 22 + 5 - 38 - 2) mod 7 = 132 mod 7 = 6 ?? What is the correct formula for this? David Drexler uucp: {ihnp4,cbosgd,moss,uokmax}!occrsh!squid!david fidonet: 147/1 -or- 19/1 sneakernet: POB 1214, Bethany OK 73008 MaBellNet: [405] 848-8868 (voice) [405] 728-2463 (data - bbs)
wnp@killer.UUCP (Wolf Paul) (02/01/88)
In article <141900013@occrsh.ATT.COM> fubar@squid.UUCP writes:
]
]From squid!fubar Thu Jan 28 04:22 CDT 1988 remote from occrsh
]Subject: day of week
]
]***
]*** PLEASE send replies to ihnp4!occrsh!squid!david
]***
]
]>(Square brackets denote the greatest integer function.)
]So:
]Thursday, Jan 28, 1988, should yield 4:
] N = 28 (number of day within month)
] M = 11 (Jan = 11)
] C = 19 (Hundreds w/in year)
] Y = 88 (rest of year??)
] L = 1 (1988 is leap year)
] = (N + [2.6M - 0.2] + Y + [Y/4] + [C/4] - 2C - (1+L)[M/11]) mod 7
] = (28 + [(2.6*11)-0.2] + 88 + [88/4] + [19/4] - (2*19) - (1+1)*[11/11]) mod 7
] = (28 + [28.4] + 88 + [22] + [4.75] - 38 - 2) mod 7
] = (28 + 29 + 88 + 22 + 5 - 38 - 2) mod 7
] = 132 mod 7
] = 6 ??
]
]What is the correct formula for this?
Seems to me the formula is ok, but your calculations are not. The greatest
integer in 28.4 is 28, not 29; in 4.75 it is 4, not 5.
Thus, the last few lines of your calculation above should read:
= (28 + 28 + 88 + 22 + 4 - 38 - 2) mod 7
= 130 mod 7
= 4
"greatest integer" is not the same as rounding up to the next highest integer
:-) ...
Wolf Paul
ihnp4!killer!wnpjulian@uhccux.UUCP (Julian Cowley) (02/02/88)
Here is a formula that I picked out of an old HP-2000 calendar program.
Basically it figures out how many days there are since January 1, 1900.
As that date was a Monday, the result modulo 7 starts on that day of
the week. This should work for all of the years in this century, with
the exception of 1900 because that year was not a leap year.
Sample C program:
char *daystr[7] = {
"Monday",
"Tuesday",
"Wednesday",
"Thursday",
"Friday",
"Saturday",
"Sunday" };
main()
{
int day, year, month, date;
puts("Enter the date as mm/dd/yy");
scanf("%d/%d/%d", &month, &date, &year);
day = ((year-1)*365 + (year-1)/4 + (int) ((month-1)*30.57+0.5) -
(1+(year%4 != 0))*(month > 2) + date) % 7;
printf("That day was a %s\n", daystr[day]);
}
Cheers.
////// ///// Julian Cowley, University of Hawaii | "It's not the size of
// // // ...!ihnp4!sdcsvax!nosc!uhccux!julian | the disk that matters,
\\ // // julian@uhccux.uhcc.hawaii.edu | it's the way it's
\X/ ///// julian@uhccux.bitnet | formatted..."davidsen@steinmetz.steinmetz.UUCP (William E. Davidsen Jr) (02/03/88)
There has been so much commentary on the topic of "what day of the
week," that I am posting a short program to provide the information. It
returns a pseudo-julian date which may be used to uniquely identify any
date from Jan 1 1980 to Feb 29, 2100. One procedure takes the month,
day, year, and returns a unique number, the other does the inverse. The
number is such that num%7 = day_of_week, and 0 means Sunday.
Before I get a million postings, yes I have a real julian date routine.
This one works and returns values which will fit in a short as long as I
live.
/*
* JULI - pseudo julian date package
*
* Convert any post-1980 date to an integer day number and back to
* day, month, year. This will work up to feb 29, 2100.
*/
static months[12][2] = {
{0, 0}, {31, 31}, {59, 60}, {90, 91},
{120, 121}, {151, 152}, {181, 182}, {212, 213},
{243, 244}, {273, 274}, {304, 305}, {334, 335}};
int jdate (month, day, year)
int month, day, year;
{
register int iy, ly, lyf;
iy = year - 1980;
ly = (iy+3) / 4; /* leap year count */
lyf = (iy%4 == 0);
return (365*iy + ly + months[month-1][lyf] + day);
}
/*
* Convert day number to date
*/
unjuli (jd, month, day, year)
int jd; /* julian day */
int *month, *day, *year;
{
register int
lyf, /* leap year (this year) flag */
dd, mm, /* calculated day and month */
yy, ly; /* incremental years, leapyears */
/* first guess as to year */
yy = jd / 365;
ly = (yy+3) / 4;
while ((dd = jd - 365*yy - ly) <= 0) {
/* correct for dates near the end of a year */
yy--;
ly = (yy+3) / 4;
}
/* find the month */
lyf = (yy%4 == 0);
mm = 0;
while (months[mm][lyf] < dd && ++mm < 12);
/* return values */
*year = yy + 1980;
*month = mm;
*day = dd - months[mm-1][lyf];
}
/*
* Test routine for julian package
*/
#ifdef TEST
#include <stdio.h>
main () {
int m, d, y, jd;
for ( ; ; ) {
printf ("Enter month, day, year: ");
scanf ("%d%d%d", &m, &d, &y);
if (y < 1900) y += 1900;
jd = jdate (m, d, y);
printf ("Juliam for %d/%d/%d = %d\n", m, d, y, jd);
unjuli (jd, &m, &d, &y);
printf ("Converted back = %d/%d/%d\n", m, d, y);
}
}
#endif
--
bill davidsen (wedu@ge-crd.arpa)
{uunet | philabs | seismo}!steinmetz!crdos1!davidsen
"Stupidity, like virtue, is its own reward" -mehirayama@suvax1.UUCP (Pat Hirayama) (02/04/88)
In article <141900013@occrsh.ATT.COM>, fubar@squid.UUCP writes: > *** > *** PLEASE send replies to ihnp4!occrsh!squid!david > > > Here's a formula given in the book An Introduction to the Theory of Numbers, > >by Niven and Zuckerman. > > Let M be the number of the month, defined so that Jan = 11, Feb = 12, > >Mar = 1, Apr = 2, ..., Dec = 10. Let N be the number of the day within the > >month. Let C be the hundreds in the year, and let Y be the rest of the year. > >Let L be 1 for a leap year and 0 for a non-leap year. > > > > > >(N + [2.6M - 0.2] + Y + [Y / 4] + [C / 4] - 2C - (1 + L)[M / 11]) mod 7 > >yields 0 for Sunday, 1 for Monday, ..., 6 for Saturday. > > > >(Square brackets denote the greatest integer function.) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ > > Thursday, Jan 28, 1988, should yield 4: > > N = 28 (number of day within month) > M = 11 (Jan = 11) > C = 19 (Hundreds w/in year) > Y = 88 (rest of year??) > L = 1 (1988 is leap year) > > = (N + [2.6M - 0.2] + Y + [Y/4] + [C/4] - 2C - (1+L)[M/11]) mod 7 > = (28 + (2.6*11)-0.2] + 88 + [88/4] + [19/4] - (2*19) - (1+1)*[11/11]) mod 7 > = (28 + [28.4] + 88 + [22] + [4.75] - 38 - 2) mod 7 > = (28 + 29 + 88 + 22 + 5 - 38 - 2) mod 7 > = 6 ?? > > What is the correct formula for this? David, You are ignoring the underlined statement indicated above. If you consider this, you will realize the following: Given, your initial data values given above. Let X=Number you are seeking... X = (N + [2.6M - 0.2] + Y + [Y/4] + [C/4] - 2C - (1+L)[M/11]) mod 7 = (28 + (2.6*11)-0.2] + 88 + [88/4] + [19/4] - (2*19) - (1+1)*[11/11]) mod 7 = (28 + [28.4] + 88 + [22] + [4.75] - 38 - 2) mod 7 = (28 + 28 + 88 + 22 + 4 - 38 - 2) mod 7 = 4 Which is Thursday and the answer that you expect to get. ------------------------------------------------------------------------------ Pat Hirayama UUCP: ...!uw-entropy!dataio!suvax1!hirayama Seattle University USNail: 28625-47th Pl S, Auburn, WA 98001-1140 Class of 1988 Phone: (206) 946-0833 ". . . the starships of the Federation are the physical, tangible manifestations of humanity's stubborn insistence that life does indeed mean something." Spock Final Frontier ------------------------------------------------------------------------------