is leap year redux | Python Fiddle

# A python function which takes in a year (2000, 1963, 2001, etc.) and returns True if the year is a leap year, 
# or False if the year is a common year
#
# http://stackoverflow.com/questions/3220163/how-to-find-leap-year-programatically-in-c has an interesting discussion
# about leap year algorithms 
#
# http://pythonfiddle.com/is-leap-year-redux by Peter Shank

def is_leap_year(year):
    
    # years not divisible by 4 (1601, 1999, etc) are common years
    if year%4 != 0:
        return False
    
    # years that ARE divisibible by 4 but are NOT divisible by 100 (1604, 2012, etc) are leap years
    elif year%100 != 0:
        return True
    
    # years divisible by both 4 and 100, but NOT divisible by 400 (1700, 1900, etc) are common years 
    elif year%400 != 0:
        return False
    
    # years failing the three tests are divisible by 400 (1600, 2000, etc.) and they are leap years
    else:
        return True
    
def test():
    test_cases = [(1600, True), 
                  (1601, False),
                  (1602, False),
                  (1603, False),
                  (1604, True),
                  (1700, False),
                  (2000, True),
                  (2012, True)]
    passes = 0
    for (args, answer) in test_cases:
        result = is_leap_year(args)
        if result != answer:
            print "Failed when argument was: ", args
        else:
            passes = passes + 1
    print str(passes) + " out of " + str(len(test_cases)) + " cases passed."

test()

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