Python – Find if given year is Leap Year in Georgian Calendar

Python – Check if year is Leap Year in Georgian Calendar

In this tutorial, we will write a Python function to check if the given year is a leap year in Georgian Calendar.

Following are the criteria for an year to be Leap year.

  • The year can be evenly divided by 4, is a leap year, unless:
    • The year can be evenly divided by 100, it is NOT a leap year, unless:
      • The year is also evenly divisible by 400. Then it is a leap year.

You may combine all these conditions into a single boolean expression.

Consider that we have our year in variable year.

To satisfy the first condition, “The year can be evenly divided by 4, is a leap year”, following condition has to be satisfied.

year%4==0
Run this program ONLINE

To satisfy the second condition, “The year can be evenly divided by 100, it is NOT a leap year,”, following condition has to be satisfied.

year%100==0
Run this program ONLINE

To satisfy the third condition, ” The year is also evenly divisible by 400″, following condition has to be satisfied.

year%400==0
Run this program ONLINE

By combining above three conditions, we get the following boolean expression.

year%4==0 and not (year%100==0 and not year%400==0)
Run this program ONLINE

unless in the given condition translates to and not and hence the and not operators in the above condition.

Following is the example program, in which we check if a given year is a leap year or not.

Python Program

def is_leap(year):
    return year%4==0 and not (year%100==0 and not year%400==0)

year = int(input())

print(is_leap(year))

Output

2004
True

When you run the program, input() waits for your input. Enter a number and our program shall check if the given number is a leap year or not.

Summary

In this tutorial of Python Examples, we learned how to check if given year is a leap year of Georgian Calendar or not.