C: Finding the day for a given date

Question

This is based on Sakamoto's algorithm for find-out the day of the given date if we follow the Georgian calender.

Explanation

Source

year -= month < 3 is a nice trick. It creates a "virtual year" that starts on March 1 and ends on February 28 (or 29), putting the extra day (if any) at the end of the year; or rather, at the end of the previous year. So for example, virtual year 2011 began on Mar 1 and will end on February 29, while virtual year 2012 will begin on March 1 and end on the following February 28.

By putting the added day for leap years at the end of the virtual year, the rest of the expression is massively simplified.

Let's look at the sum:

(year + year/4 - year/100 + year/400 + t[month-1] + date) % 7

There are 365 days in a normal year. That is 52 weeks plus 1 day. So the day of the week shifts by one day per year, in general. That is what the y term is contributing; it adds one to the day for each year.

But every four years is a leap year. Those contribute an extra day every four years. Thanks to the use of virtual years, we can just add y/4 to the sum to count how many leap days happen in y years. (Note that this formula assumes integer division rounds down.)

But that is not quite right, because every 100 years is not a leap year. So we have to subtract off y/100.

Except that every 400 years is a leap year again. So we have to add y/400.

Finally we just add the day of the month d and an offset from a table that depends on the month (because the month boundaries within the year are fairly arbitrary). Then take the whole thing mod 7 since that is how long a week is.

Sample Code

#include<stdio.h>

int day_of_week(int date, int month, int year)
{
  static int t[] = { 0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4 };
  int dow;

  year -= month < 3;
  dow = ( year + year/4 - year/100 + year/400 + t[month-1] + date) % 7;

  printf("\nDate: %d/%d/%d\n\n",date,month,year);

  switch (dow)
  {
    case 0:
        printf("Day = Sunday");
        break;
    case 1:
        printf("Day = Monday");
        break;
    case 2:
        printf("Day = Tuesday");
        break;
    case 3:
        printf("Day = Wednesday");
        break;
    case 4:
        printf("Day = Thursday");
        break;
    case 5:
        printf("Day = Friday");
        break;
    case 6:
        printf("Day = Saturday");
        break;
    default:
        printf("Wrong data");
  }

  return 0;
}

void main()
{
  int d,m,y;

  printf("Enter the year ");
  scanf("%d",&y);

  printf("Enter the month ");
  scanf("%d",&m);

  printf("Enter the date ");
  scanf("%d",&d);

  day_of_week(d,m,y);
}

Output

Enter the year 2014
Enter the month 5
Enter the date 26

Date: 26/5/2014

Day = Monday