How to calculate the Easter date for a given year using Gauss’ Algorithm

Given an integer Y, denoting a year, as input, the task is to find the date of Easter for that year.

Examples:

Input: Y = 2020
Output: 2020-04-12
Explanation:
In 2020, Easter was on 12 April
Hence, Easter date will be 2020-04-12

Input: Y = 1991
Output: 1991-03-30
Explanation:
In 1991, Easter was on 30 March
Hence, Easter date will be 1991-03-30

Approach: Gauss Easter Algorithm is used to easily calculate the date of Easter for a year. This algorithm was first thought of by Carl Friedrich Gauss. There is no proper explanation on how Gauss landed up with this algorithm but the implementation of this algorithm has proved to be very accurate.



A detailed explanation of the Gauss Easter Algorithm is as follows:

  1. First, calculate the location of the year Y in the Metonic cycle.

    A = Y mod 19

  2. Now, find the number of leap days according to Julian calendar.

    B = Y mod 4

  3. Then, let’s take into account that the non-leap year is one day longer than 52 weeks.

    C = Y mod 7

  4. M depends on the century of year Y. For 19th century, M = 23. For 21st century, M = 24 and so on.
    It is calculated using following relations:

    P = floor (Y / 100)
    Q = ((13 + 8 * P) / 25)
    M = (15 – Q + P – (P / 4)) mod 30

  5. The difference between the number of leap days between the Julian and the Gregorian calendar is given by:

    N = (4 + P – (P / 4)) mod 7

  6. The number of days to be added to March 21 to find the date of the Paschal Full Moon is given by:

    D = (19*A + M) mod 30

  7. And, the number of days from the Paschal full moon to the next Sunday is given by:

    E = (N + 2*B + 4*C + 6*D) mod 7

  8. Therefore, using D ans E, the date of Easter Sunday is going to be March (22 + D + E). If this number comes out to be greater than 31, then we move to April.
  9. Now the lunar month is not exactly 30 days but a little less than 30 days. So to nullify this inconsistency, following cases are followed:

    if (D == 29) and (E == 6)
    return “April 19”
    else if (D == 28) and (E == 6)
    return “April 18”

Below is the implementation of the above approach:

C++

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// C++ program for the
// above approach
  
#include <iostream>
#include <math.h>
  
using namespace std;
  
// Function calculates and prints
// easter date for given year Y
void gaussEaster(int Y)
{
    float A, B, C, P, Q,
        M, N, D, E;
  
    // All calculations done
    // on the basis of
    // Gauss Easter Algorithm
    A = Y % 19;
    B = Y % 4;
    C = Y % 7;
    P = (float)floor(Y / 100);
    Q = (float)floor((13 + 8 * P) / 25);
    M = (int)(15 - Q + P - P / 4) % 30;
    N = (int)(4 + P - P / 4) % 7;
    D = (int)(19 * A + M) % 30;
    E = (int)(2 * B + 4 * C + 6 * D + N) % 7;
    int days = (int)(22 + D + E);
  
    // A corner case,
    // when D is 29
    if ((D == 29) && (E == 6)) {
        cout << Y << "-04-19";
        return;
    }
    // Another corner case,
    // when D is 28
    else if ((D == 28) && (E == 6)) {
        cout << Y << "-04-18";
        return;
    }
    else {
        // If days > 31, move to April
        // April = 4th Month
        if (days > 31) {
            cout << Y << "-04-"
                 << (days - 31);
            return;
        }
        else {
            // Otherwise, stay on March
            // March = 3rd Month
            cout << Y << "-03-"
                 << days;
            return;
        }
    }
}
  
// Driver Code
int main()
{
    int Y = 2020;
    gaussEaster(Y);
  
    return 0;
}

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Java

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// Java program for the
// above approach
  
import java.util.Date;
import java.util.Scanner;
  
// Function calculates and prints
// easter date for given year Y
public class GaussEaster {
  
    static void gaussEaster(int Y)
    {
        float A, B, C, P, Q,
            M, N, D, E;
  
        // All calculations done
        // on the basis of
        // Gauss Easter Algorithm
        A = Y % 19;
        B = Y % 4;
        C = Y % 7;
        P = (float)Math.floor(Y / 100);
        Q = (float)Math.floor(
            (13 + 8 * P) / 25);
        M = (15 - Q + P - P / 4) % 30;
        N = (4 + P - P / 4) % 7;
        D = (19 * A + M) % 30;
        E = (2 * B + 4 * C + 6 * D + N) % 7;
        int days = (int)(22 + D + E);
  
        // A corner case,
        // when D is 29
        if ((D == 29) && (E == 6)) {
            System.out.println(Y + "-04"
                               + "-19");
            return;
        }
        // Another corner case,
        // when D is 28
        else if ((D == 28) && (E == 6)) {
            System.out.println(Y + "-04"
                               + "-18");
            return;
        }
        else {
  
            // If days > 31, move to April
            // April = 4th Month
            if (days > 31) {
                System.out.println(Y + "-04-"
                                   + (days - 31));
                return;
            }
            // Otherwise, stay on March
            // March = 3rd Month
            else {
                System.out.println(Y + "-03-"
                                   + days);
                return;
            }
        }
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int Y = 2020;
        gaussEaster(Y);
    }
}

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Output:

2020-04-12

Time complexity: O(1)
Auxiliary Space: O(1)

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