Calculating the day of the week and much more

An Extended Approach to the Julian and the Gregorian calendar

2015-08-22T13:08:00.001+05:30

An Extended Approach to the Julian and

the Gregorian calendar

Welcome!

Calculating the day of the week is now simple.

I have written an article titled "An Extended Approach to The Julian and The Gregorian Calendar" which was published in The Mathematical Gazette, London in November 2004. Here, I have tried to explain the concepts given in the above article in detailed but lucid manner.

Many students, teachers and experts have visited and benefitted from this article. People who have used the below mentioned weekday formula and/or the algorithm in different programming languages have called it (for identification) 'Babwani's Method' or 'Babwani's Formula' or 'Babwani's Congruence' or 'Babwani's Algorithm'. You may call it by any name. But to maintain uniformity and identification, the same formulae are collectively referred as 'Babwani's Congruence' and steps to find out the years and the algorithms in general as 'Babwani's Algorithm'.

You need to refer the explanatory document wherein the concepts given in the Babwani's Congruence have been explained methodically.

Secondly, a program Calember (Calendar + Number), that calculates the weekday, dates, months and years based on the explanations given in the document, has been prepared and made available.

Remember, this is not the first time such a formula is made. Many formulae existed even before I invented this method. But this is the first time, you will discover a different approach to find weekdays, dates, months and even years when other data is provided. What's more special is the flexibility of methods that are at your disposal as it also explains alternative easy ways of doing the same calculations. It also presents algorithms that can be used to prepare your own program.

Thus you can find out the weekdays, dates, months and years by either by:

1) Pen and a paper and one of the formulae. (Also see the FAQ page)

2) Orally. (Depends on your mental ability and practice)

3) Using a computer program.

4) Calendar Slider. (See Calendar Slider page for more details)

View Babwani's Congruence

Download Babwani's Congruence

Download Calember v 2.0 Flash format

Download Calember v 2.0 Windows application format
Download Calendar Slider Flash Format
Download Calendar Slider Windows application format

Download Gregorian Calendar MS Excel format

Download Calember v 1.0 DOS format

C program to find the day of the week

2010-03-04T18:08:00.003+05:30

C program to find the day of the week

Here I have provided a copy-paste code to find the day of the week in C language. Remember this program works best using Turbo C 3.0. You may make some changes to make it work with the complier you use. I have also added some comments which is for you to understand. One can cast off all the comments without affecting the working of the program. Also remember that this is not the only way to write the C program to calculate the day of the week. You may try writing simpler programs.
The program given below is with respect to the international switchover from 4^th October 1582 to 15^th October 1582 which is also used for astronomical calculations. Make the required changes for other switchovers. Hint make changes in line number 3 and 5 in the jg function.

#include<stdio.h>
#include<conio.h>
#include<math.h>
jg(int d, int m,int y) //Julian or Gregorian
{ int type;
if(y<1582 || (y==1582 && m<10) || (y==1582 && m==10 && d<5))
type=0; // 0 stands for Julian
else if(y>1582 || (y==1582 && m>10) || (y==1582 && m==10 && d>14))
type=1; // 1 stands for Gregorian
else
type=2; // 2 stands for neither of the two
return type;
}
fm(int d, int m,int y) //function of month
{ int fmonth,leap;
if(jg(d,m,y)==0) //for Julian calendar
{
if(y%4==0) //leap function 1 for leap & 0 for non-leap
leap=1;
else
leap=0;
}
if(jg(d,m,y)==1) //for Gregorian calendar
{
if((y%100==0) && (y%400!=0)) //leap function 1 for leap & 0 for non-leap
leap=0;
else if(y%4==0)
leap=1;
else
leap=0;
}
fmonth=3+(2-leap)*((m+2)/(2*m))+(5*m+m/9)/2; //f(m) formula
fmonth%=7; //f(m) is brought in range of 0 to 6
return fmonth;
}
c(int y) //function of century
{
int century;
century=y/100; //returns the first two digits of year
return century;
}
yy(int y) //function of year
{
int lastdigits;
lastdigits=y%100; //returns the last two digits of year
return lastdigits;
}
day_of_week(int d, int m, int y)
{
int dow,z; //dow stands for day of week
printf("\nDate: %d/%d/%d\n\n",d,m,y);
if(jg(d,m,y)==0)
{
z=y; //z is a new year variable
if(y<0)
{
z=y+9800; //z is made positive if negative, 9800 is multiple of 7
}
dow=1.25*z+fm(d,m,y)+d-2; //function of weekday for Julian
dow%=7; //remainder on division by 7
}
else
{
dow=1.25*yy(y)+fm(d,m,y)+d-2*(c(y)%4); //function of weekday for Gregorian
dow%=7; //remainder on division by 7
}
switch (dow)
{
case 0:
printf("weekday = Saturday");
break;
case 1:
printf("weekday = Sunday");
break;
case 2:
printf("weekday = Monday");
break;
case 3:
printf("weekday = Tuesday");
break;
case 4:
printf("weekday = Wednesday");
break;
case 5:
printf("weekday = Thursday");
break;
case 6:
printf("weekday = Friday");
break;
default:
printf("Incorrect data");
}
return 0;
}
void main()
{
int date,month,year;
char con='y'; //con = continue
while(con=='y' || con=='Y')
{
clrscr(); //clears screen
printf("Enter the year ");
scanf("%d",&year); //year input
printf("Enter the month ");
scanf("%d",&month); //month input
printf("Enter the date ");
scanf("%d",&date); //date input
day_of_week(date,month,year);
printf("\n\nDo you wish to continue? (Y/N) ");
scanf("%s",&con);
}
}

Calendar Slider

2009-08-22T13:04:00.002+05:30

Calendar Slider

Calendar Slider is easy and fun way of learning to calculate day of the week, month, date and year when other information is given. Here you don't require your mathematical skills or ability to understand and solve the formula. In fact you don't even require a computer to solve the calendar problem. You can make the Calendar Slider even at home on a cardboard. Calendar Slider is made of two parts: 1) The Calendar board 2) The Slider that slides over the board to get the desired result.

The Calendar Slider that I have made right now calculates problem based on Gregorian calendar only as Gregorian calendar is currently used calendar. I have provided the Calendar Slider in the Flash format and the Windows application format. You can download any one of them.

But solving the calendar problem on the computer with the help of Calendar Slider kills the very purpose of making Calendar Slider. Therefore make the Calendar Slider yourself at home using a cardboard and make it exactly as it is shown. Then without the help of the computer or formulae you can solve calendar problems. I have also included a help file that will teach you to solve the calendar problems by using Calendar Slider.

Download Calendar Slider Flash format
A very easy and interesting way to solve calendar problems without the help of computer or any formulae. Make the same at home using cardboard.

Download Calendar Slider Windows application format
A very easy and interesting way to solve calendar problems without the help of computer or any formulae. Make the same at home using cardboard.

Downloads

2009-08-22T13:03:00.001+05:30

Downloads

Download Babwani's Congruence

It is an explanatory document that explains the Babwani's Congruence. It includes the derivation of Babwani's congruence, which was published in The Mathematical Gazette, London, 2004. It is a detailed document that explains you how to find the weekday, dates, month and years both Julian and Gregorian, both BC and AD with the help of examples. The document also includes algorithm that you can use to prepare computer programs.

The above document is in Adobe Reader (.pdf) format. If you do not have Adobe Reader you can download the same from here.

Download Calember v 2.0 Flash format

A computer program that calculates the weekday, dates, months and years when other data is provided.

The above program is in Macromedia Flash (.swf) format. If you do not have flash player you can download the same from here.

Download Calember v 2.0 Windows Application format

A computer program that calculates the weekday, dates, months and years when other data is provided.

Download Calendar Slider Flash format
A very easy and interesting way to solve calendar problems without the help of computer or any formulae. Make the same at home using cardboard.

Download Calendar Slider Windows application format
A very easy and interesting way to solve calendar problems without the help of computer or any formulae. Make the same at home using cardboard.

Download Gregorian Calendar MS Excel format

A computer program that calculates the weekday, dates, months and years when other data is provided. The above program is applicable only for the Gregorian calendar.

Download Calember v 1.0 DOS format

A computer program that calculates the weekday, dates of the month, months and years when other data is provided.

Frequently Asked Questions

2009-08-11T20:49:00.012+05:30

Frequently Asked Questions

1: Could you possibly give me a formula to calculate the weekday, which I can put into my HP calculator?

In order to calculate the weekday using a standard calculator try the following method:

To calculate the weekday for 21st May 2006 (or any date on or between 1/3/1900 and 28/2/2100)

d = 21
m = 5
y = 2006

∴ f(m) = f(5) = 1 (from table number 3)

The formula is:

$\frac{1.25\times y+f\left( m \right)+d-1}{7}$

$=\frac{1.25\times 2006+1+21-1}{7}$

$=\frac{\text{2528}.\text{5}}{7}$

=361.21428571428571428571428571429

Now subtract 361 from the above number (as it is the integer part coming before the decimal point) and then multiply by 7 and round it off to two decimal places.

You will get 1.50

Ignore the decimal part and only retain the integer part to get 1 which corresponds to Sunday ( 0 = Sat, 1 = Sun, 2 = Mon, ..)

2: What are these Heap years, Seap years, etc. What is the significance of its classification?

The names of these years have been specially coined by me. Like we know that the Olympic games are held in Leap years. That is in the years divisible by 4. But what about the other years. For instance, the FIFA (soccer) World Cup are held only in Seap years. Means the years which have remainder 2 when divided by 4. Similarly, the Cricket World Cup is held in Teap years with the exception of 1992 and 1996 World Cup.

Secondly, this classification is especially useful and imperative when we wish to find the years that satisfy a particular condition.

3: Is it true that everyone celebrate their 28th birthday on the same weekday on which they were born?

About 79 % of the people celebrate their 28th birthday on the weekday they were born. Refer chapter number 7 to know the 5-6-11-6 cycle. Which means that your birthday falls on the same weekday of your birth in the cycle of 5 years, 6 years, 11 years and 6 years and again 5 years.... depending upon the type of year of your birth. Therefore, one can infer that one's birthday repeats on the same weekday of birth after 28 years. This takes place if the years follow the pattern of 1 leap year after every three non-leap years. (One in every four years)

But this cycle is obstructed by the Heap year. (Heap years are those years which is divisible by 100 but not divisible by 400 E.g. 1800, 1900, 2100, 2200, etc.) A Heap year though looks like a Leap year because of its divisibility by 4, does not have 29 days in February. Therefore if a person is born in the year 1887 will not have his/her 28th birthday on the same weekday of his/her birth. Because he/she goes across through the Heap year where one day (29th February 1900) was missing thus disturbing the entire 5-6-11-6 cycle.

Thus any one born on or in between 29 Feb 1872 to 28 Feb 1900 will not have his/her 28th birthday on the same weekday.

However, one born on or in between 1 Mar 1900 to 28 Feb 2172 will have his/her 28th birthday on the same weekday. This includes every one of us. Thus, for practical purposes (for this present time) we all (about 100 %) have our 28th birthdays on the same weekday of our birth. This is because 2000 was not a Heap year but a normal Leap year.

4: How is the Babwani's Congruence useful in our everyday life?

We use calendars in our everyday life and thus Babwani's Congruence can be used to a great extent. One can orally calculate the weekday of any given date especially the one in the present year.

For example, if we are currently in 2007, we calculate in advance what is x\[\left\lfloor 1.25\times 2007 \right\rfloor =\left\lfloor \text{2508}\text{.75} \right\rfloor \,=\left\lfloor 2.75 \right\rfloor =2\text{(mod7)}\]

Now correlate the above mentioned formula as

$day=\left\lfloor 1.25\times y \right\rfloor +f\left( m \right)+d-1$

$=2+f\left( m \right)+d-1$

$=1+f\left( m \right)+d$

Now use this formula, remembering the table number 3 showing f(m), calculate in seconds the weekday.

E.g. 10th April 2007; day = 1 + 10 + 6 (mod 7) = 17 (mod 7) = 3 = Tuesday. (remember f(4) = 6)

26th September 2007; day = 1 + 26 + 5 (mod 7) = 32 (mod 7) = 4 = Wednesday. (remember f(9) = 5)

Secondly, consider a situation where the year 2007 is about to get over and you don't possess a calendar for the year 2008. So temporarily, till you get a new calendar of 2008; which calendar month of 2007 can you use instead of January 2008?

The answer is May 2007 can be used temporarily used. How? Let us see.

Let us generalize the current year (not necessarily 2007) as y and the next year as (y + 1).

The weekday of 1st January of the year (y + 1) = the weekday of 1st (unknown month) of the year y

$\left\lfloor 1.25\times \left( y+1 \right) \right\rfloor +1+f\left( 1 \right)-1=\left\lfloor 1.25\times y \right\rfloor +1+f\left( m \right)-1$

$\therefore \left\lfloor 1.25y \right\rfloor +\left\lfloor 1.25 \right\rfloor +1+0-1=\left\lfloor 1.25y \right\rfloor +1+f\left( m \right)-1$

$\therefore f\left( m \right)=\left\lfloor 1.25 \right\rfloor $

$\therefore f\left( m \right)=1$

but f(5) = 1

∴ the required month is May.

This is true irrespective of the nature (leap or non-leap) of the current or the upcoming year. Try it out.

Similarly, you can find many other personalized applications.

5: Which weekday cannot come at the end of the century?

There can be two cases to answer your questions.

CASE 1: Many (including me) consider 31st December xx00 to be the last day of the century (e.g. 31st December 2000) and the next day to be the first day of the next century.

CASE 2: While others consider 31st December xx99 to be the last day of the century (e.g. 31st December 1999) and the next day to be the first day of the next century.

You may directly refer the case you are interested in. However both are solved in the same manner, of course the answers will differ.

CASE 1:

Let us first find the weekday that can be the last day of the century (31st December xx00).

According to the congruence formula, substituting the values: d = 31, m = 12, c = c, y = 0

$w=\left\lfloor \frac{5y}{4} \right\rfloor +f\left( m \right)+d-2\times \bmod \left( c,4 \right)$

$=\left\lfloor \frac{5\times 0}{4} \right\rfloor +f\left( 12 \right)+31-2\times \bmod \left( c,4 \right)$

$=0+5+31-2\times \bmod \left( c,4 \right)$

$=36-2\times \bmod \left( c,4 \right)$

$=1-2\times \bmod \left( c,4 \right)\left( \bmod 7 \right)$

Now, be whatever the value of c, mod (c,4) can have only four values { 0, 1, 2, 3 )

w = 1 – 2 × 0 = 1 = Sunday (mod 7)

or w = 1 – 2 × 1 = – 1 = 6 = Friday (mod 7)

or w = 1 – 2 × 2 = – 3 = 4 = Wednesday (mod 7)

or w = 1 – 2 × 3 = – 5 = 2 = Monday (mod 7)

So except for the above weekdays, the other weekdays cannot be the last day of the century. So the answer is Tuesday, Thursday and Saturday cannot be the last day of any Gregorian century.

CASE 2:

Let us first find the weekday that can be the last day of the century (31st December xx99).

According to the congruence formula, substituting the values: d = 31, m = 12, c = c, y = 99

$w=\left\lfloor \frac{5y}{4} \right\rfloor +f\left( m \right)+d-2\times \bmod \left( c,4 \right)$

$=\left\lfloor \frac{5\times 99}{4} \right\rfloor +f\left( 12 \right)+31-2\times \bmod \left( c,4 \right)$

$=\left\lfloor 123.75 \right\rfloor +5+31-2\times \bmod \left( c,4 \right)$

$=123+5+31-2\times \bmod \left( c,4 \right)$

$=159-2\times \bmod \left( c,4 \right)$

$=5-2\times \bmod \left( c,4 \right)\left( \bmod 7 \right)$

Now, be whatever the value of c, mod (c,4) can have only four values { 0, 1, 2, 3 )

w = 5 – 2 × 0 = 5 = Thursday (mod 7)

or w = 5 – 2 × 1 = 3 = Tuesday (mod 7)

or w = 5 – 2 × 2 = 1 = Sunday (mod 7)

or w = 5 – 2 × 3 = – 1 = 6 = Friday (mod 7)

So except for the above weekdays, the other weekdays cannot be the last day of the century. So the answer is Monday, Wednesday, Saturday cannot be the last day of any Gregorian century.

In case your question refers to the Julian century, then there is no weekday that doesn’t come at the end of the century.

Contact me

2009-08-11T20:38:00.004+05:30

Contact me

You can post your comments and views in the comments column at the end of the article.

If you have a question, first have a look in the FAQ page. If you have a different question, feel free to ask me at sohaelbabwani@gmail.com. I will try to clear your doubts at the earliest.

Feel free to express your opinion/constructive criticism/suggestion at sohaelbabwani@gmail.com.

No spam or junk emails please.

About me

2009-08-11T20:36:00.001+05:30

About me

Hi! I am Sohael Sadruddin Bābwāni (pronounced as baa-b-waa-nee). I live in Mumbai, India. I am a teacher as well as a student of Mathematics. I did my Bachelor's degree in Commerce in 2001. I like to do Math being my favourite subject; and also its allied subjects like Physics, Astronomy, Geography etc. where even a bit of Math is used.

I have made a formula to calculate the weekday if the date, month and year of any given date (both AD and BC) is given. The same formula was published in The Mathematical Gazette, London in November 2004 (vol. 88, no. 513 pages 569-573). A method already existed to find the same which is known by the name of Zeller's Algorithm; (made by Christian Zeller of Germany in 1886); which was not known to me till I made my formula independently. While going through the Zeller's Algorithm, I found some limitations in it; as it was slightly complicated and could not find date, month and year where other information is given. I wrote an article explaining and proving the same and it was considered to be published.

For instance, in Zeller's algorithm the months are numbered from 3 for March to 14 for February. The year is assumed to begin in March; this means, for example, that January 1995 is to be treated as month 13 of 1994. But in my method of finding the weekday, the months are numbered properly 1 for January to 12 for December.

I had made this formula independently way back in 1996. But I only got to know about the Mathematical Gazette in the year 2002. I submitted the formula in September 2002 and it was approved (of course after a few revisions) in March 2003 and published in November 2004.

This is not the first and the last formula used to find the weekday. The best example is the Zeller's Algorithm in 1886. However, I didn't know about it till I happen to search the net in 2002. Fortunately, my approach was different and I also had elaborated the ways to find the date, month and the year. Thereby, I qualified to get it published.

A word from me

Many believe that the elementary algebra and geometry learnt in the school or college is burdensome and is of no use. But, I firmly believe that though everyone may not be dealing in solving the sums directly, we use these concepts in our everyday life at most times unknowingly and indirectly. But whether you use a computer or an mp3 player or even a house or a bridge, you need math and logic. It is important therefore to teach the students the practical importance of learning math, logic, algebra and geometry.

For example, the calendar formula we are dealing in has many practical applications. One can orally calculate the weekday of any given date especially the one in the same year. Suppose the year 2007 is about to get over and you don't possess a calendar for the year 2008. So temporarily, till you get the calendar of 2008; which calendar month of 2007 can you use instead of January 2008? Using the calendar formula you can calculate that it is May of 2007 that you can use temporarily for January 2008.

Feel free to express your opinion/constructive criticism/suggestion at sohaelbabwani@gmail.com.

Calculating the day of the week and much more

An Extended Approach to the Julian and the Gregorian calendar

An Extended Approach to the Julian andthe Gregorian calendar

C program to find the day of the week

Calendar Slider

Calendar Slider

Downloads

Downloads

Frequently Asked Questions

Frequently Asked Questions

Contact me

Contact me

About me

About me

An Extended Approach to the Julian and

the Gregorian calendar