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Program to calculate Dooms Day for a year

  • Difficulty Level : Expert
  • Last Updated : 06 Oct, 2021

Doomsday may refer to a hypothetical event according to which the end of human life is at the highest possibility. There are many algorithms written to calculate which day of a week in a year has the highest possibility of doomsday falling on that day.
All the statements are with respect to Gregorian Calendar. As the Gregorian calendar repeats itself every 400 years so a set of rules is decided for 1st 400 years only. Algorithms are derived from the calculations of John Conway, Lewis Carroll and many other mathematicians in history who worked on the calculation of Dooms Day.
To calculate Doom’s Day of a particular year following algorithm is used:-

  • Extract the last two digits of the year.(Let this be y)
  • Divide by 12 take the floor of the value.
  • Then add remainder of dividing y by 12.
  • Calculate the result when remainder of y divided by 12 is divided by 4.
  • Take the floor of the above value and then add.
  • Take remainder after dividing with 7 (mod 7).
  • Add the value of anchor day of the week starting from Sunday (Considering Sunday as 0)

The formula becomes –

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([y/12]+ y mod 12 + [y mod 12/4]) mod 7 + anchor
 



Here [ ] is Greatest Integer Function.
Anchor day changes after 100 years and repeats after every 400 years in the following way –

 

0-99 yrs --> Tuesday
100-199 yrs --> Sunday
200-299 yrs --> Friday
300-399 yr --> Wednesday

 

After this the above anchor days repeat as mentioned in the beginning of the article.
Examples:

 

Input :  2005
Output : Doomsday in the year 2005 = Monday

Input : 1800
Output : Doomsday in the year 1800 = Friday

 

Below is the implementation.

 

C++14




#include<bits/stdc++.h>
using namespace std;
 
string dooms_day(int year)
{
     
    // map to store days value of
    // anchor day can be known
    map<int, string> dict_day;
    dict_day[0] = "Sunday";
    dict_day[1] = "Monday";
    dict_day[2] = "Tuesday";
    dict_day[3] = "Wednesday";
    dict_day[4] = "Thursday";
    dict_day[5] = "Friday";
    dict_day[6] = "Saturday";
     
    // Gregorian calendar repeats
    // every 400 years
    int k = year % 400;
     
    int anchor;
     
    // Decide the anchor day
    if(k >= 0 && k < 100)
        anchor = 2;
         
    else if(k >= 100 && k < 200)
        anchor = 0;
         
    else if(k >= 200 && k < 300)
        anchor = 5;
         
    else
        anchor = 3;
         
    int y = year % 100;
     
    // Dooms day formula by Conway
    int doomsday = ((y / 12 + y % 12 +
                    (y % 12) / 4) % 7 + anchor) % 7;
     
    return dict_day[doomsday];
}
 
// Driver code
int main()
{
    int year = 1966;
     
    cout << "Doomsday in the year "
         << year << " = " << dooms_day(year);
     
    return 0;
}
 
// This code is contributed by yatinagg

Java




import java.util.*;
 
class GFG{
     
public static String dooms_day(int year)
{
     
    // map to store days value of 
    // anchor day can be known 
    HashMap<Integer, String> dict_day = new HashMap<>();
    dict_day.put(0, "Sunday");
    dict_day.put(1, "Monday");
    dict_day.put(2, "Tuesday");
    dict_day.put(3, "Wednesday");
    dict_day.put(4, "Thursday");
    dict_day.put(5, "Friday");
    dict_day.put(6, "Saturday");
       
    // Gregorian calendar repeats 
    // every 400 years 
    int k = year % 400;
       
    int anchor;
       
    // Decide the anchor day 
    if (k >= 0 && k < 100)
        anchor = 2;
           
    else if (k >= 100 && k < 200)
        anchor = 0;
           
    else if (k >= 200 && k < 300)
        anchor = 5;
           
    else
        anchor = 3;
           
    int y = year % 100;
       
    // Dooms day formula by Conway 
    int doomsday = ((y / 12 + y % 12
                   (y % 12) / 4) % 7 +
                    anchor) % 7;
       
    return dict_day.get(doomsday);
}
 
// Driver code
public static void main(String[] args)
{
    int year = 1966;
     
    System.out.println("Doomsday in the year " +
                        year + " = " +
                        dooms_day(year));
}
}
 
// This code is contributed divyeshrabadiya07

Python3




def dooms_day(year):
     
    # dictionary to store days
    # value of anchor day can be known
    dict_day ={ 0 : "Sunday",
               1 : "Monday",
               2 : "Tuesday",
               3 : "Wednesday",
               4 : "Thursday",
               5 : "Friday",
               6 : "Saturday" }
     
    # gregorian calendar repeats
    # every 400 years
    k = year % 400
     
    # decide the anchor day
    if(k >= 0 and k < 100):
        anchor = 2
         
    elif(k >= 100 and k < 200):
        anchor = 0
         
    elif(k >= 200 and k < 300):
        anchor = 5
         
    else:
        anchor = 3
         
    y = year % 100
     
    # dooms day formula by Conway
    doomsday = ((y//12 + y % 12 + (y % 12)//4)% 7 + anchor) % 7
     
    return dict_day[doomsday]
 
# Driver code
year = 1966
print("Doomsday in the year % s = % s"%(year,
                                        dooms_day(year)))

C#




using System;
using System.Collections.Generic;  
class GFG
{  
    static String dooms_day(int year)
    {
          
        // map to store days value of 
        // anchor day can be known 
        Dictionary<int, string> dict_day =  
                       new Dictionary<int, string>(); 
        dict_day.Add(0, "Sunday");
        dict_day.Add(1, "Monday");
        dict_day.Add(2, "Tuesday");
        dict_day.Add(3, "Wednesday");
        dict_day.Add(4, "Thursday");
        dict_day.Add(5, "Friday");
        dict_day.Add(6, "Saturday");
            
        // Gregorian calendar repeats 
        // every 400 years 
        int k = year % 400;
            
        int anchor;
            
        // Decide the anchor day 
        if (k >= 0 && k < 100)
            anchor = 2;
                
        else if (k >= 100 && k < 200)
            anchor = 0;
                
        else if (k >= 200 && k < 300)
            anchor = 5;
                
        else
            anchor = 3;
                
        int y = year % 100;
            
        // Dooms day formula by Conway 
        int doomsday = ((y / 12 + y % 12 + 
                       (y % 12) / 4) % 7 +
                        anchor) % 7;
            
        return dict_day[doomsday];
    }
   
  // Driver code
  static void Main()
  {
        int year = 1966;
      
        Console.WriteLine("Doomsday in the year " +
                            year + " = " +
                            dooms_day(year));
  }
}
 
// This code is contributed by divyesh072019

Javascript




<script>
      function dooms_day(year) {
 
        // map to store days value of
        // anchor day can be known
        var dict_day = new Map();
        dict_day.set(0, "Sunday");
        dict_day.set(1, "Monday");
        dict_day.set(2, "Tuesday");
        dict_day.set(3, "Wednesday");
        dict_day.set(4, "Thursday");
        dict_day.set(5, "Friday");
        dict_day.set(6, "Saturday");
 
        // Gregorian calendar repeats
        // every 400 years
        var k = year % 400;
 
        var anchor;
 
        // Decide the anchor day
        if (k >= 0 && k < 100)
            anchor = 2;
 
        else if (k >= 100 && k < 200)
            anchor = 0;
 
        else if (k >= 200 && k < 300)
            anchor = 5;
 
        else
            anchor = 3;
 
        var y = parseInt(year % 100);
 
        // Dooms day formula by Conway
        var doomsday = parseInt(parseInt(parseInt(y / 12) + y % 12 + parseInt((y % 12) / 4)) % 7 + anchor) % 7;
 
        return dict_day.get(doomsday);
    }
 
    // Driver code
        var year = 1966;
        document.write("Doomsday in the year " + year + " = " + dooms_day(year));
 
// This code is contributed by gauravrajput1
</script>
Output: 
Doomsday in the year 1966 = Monday

 




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