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ajiten

03-29-2024 02:44 AM

Please tell the error in code below, as seems as perfectly fine as it's alternative.

Please tell why the below code is considered incorrect?

Code:

def is_leap(year):

    leap = False

    

    # Write your logic here

    if year%400==0 :

        leap = True

    elif year%4 == 0 and year%100 != 0:

        leap = True

    return leap



year = int(input())

print(is_leap(year))

If asked for, can provide link of where it has been marked as incorrect.

The alternative provided is :

Code:

def is_leap(year):

  leap = False



  if (year % 4 == 0) and (year % 100 != 0): 

      # Note that in your code the condition will be true if it is not..

      leap = True

  elif (year % 100 == 0) and (year % 400 != 0):

      leap = False

  elif (year % 400 == 0):

      # For some reason here you had False twice

      leap = True

  else:

      leap = False



  return leap

But, I see nothing gained, except redundant code.

Also, ran the stated (earlier) code, and gives correct results.

grail

03-29-2024 08:17 AM

Please provide why the first code is incorrect?

teckk

03-29-2024 08:35 AM

First all, this is python.

Both of those work.

Code:

python

Python 3.11.8 (main, Feb 12 2024, 14:50:05) [GCC 13.2.1 20230801] on linux

Type "help", "copyright", "credits" or "license" for more information.

>>> def is_leap(year):

...    leap = False

...    

...    # Write your logic here

...    if year%400==0 :

...        leap = True

...    elif year%4 == 0 and year%100 != 0:

...        leap = True

...    return leap

... 

>>> year = int(input())

2010

>>> print(is_leap(year))

False

Second example has mixed indents.(Which wont work in python)

Code:

>>> def is_leap(year):

...    leap = False

...    

...    if (year % 4 == 0) and (year % 100 != 0): 

...        # Note that in your code the condition will be true if it is not..

...        leap = True

...    elif (year % 100 == 0) and (year % 400 != 0):

...        leap = False

...    elif (year % 400 == 0):

...        # For some reason here you had False twice

...        leap = True

...    else:

...        leap = False

...        

...    return leap

... 

>>> year = int(input())

2010

>>> print(is_leap(year))

False

boughtonp

03-29-2024 09:48 AM

The Python code for determining a leap year is "calendar.isleap(year)"

Anything else is redundant code.

If this is homework, then a correct answer should start by pointing out the built-in function, then solving it anyway, along with with a full set of test cases showing how the function handles all edge cases correctly.

The "alternative provided" is stupidly over-verbose but also looks like it is a correction of different code - so maybe the teacher sent you an updated version of someone else's attempt. Have you tried asking the teacher?

ajiten

03-30-2024 05:34 AM

Quote:

Originally Posted by grail (Post 6492689)

Please provide why the first code is incorrect?

I don't know, the source says. If you allow, can quote the source.

ajiten

03-30-2024 05:35 AM

Quote:

Originally Posted by boughtonp (Post 6492703)

The python code is:

Code:

def isleap(year):

    """Return True for leap years, False for non-leap years."""

    return year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)

grail

03-30-2024 07:09 AM

I would suggest that if you are doing a course then it is expecting specific answers.
Whilst your version may work and still be correct, it would appear the source material required the answer provided

sundialsvcs

03-30-2024 04:59 PM

I actually prefer your version, as long as it can be proved to be correct using an appropriate set of test cases. I have doubts about the second with regard to the "% 400" case, since "% 100" is tested first, and it would also "pass" the "400 case." Very importantly, yours does not do that.

However, I would have written the final bit as: "else return false." I would not bother to have a "leap" variable – just "return true/false" – because you never subsequently test that variable.

The resulting code is clear: the most-restrictive case first, each one directly "returns" true or false, and the final "else" exists to very-clearly show the catch-all case.

But, such a function then needs to be rigorously tested against a set of appropriately-chosen dates which test both the expected results and the "edge cases."

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