Lesson 3 Exercise 1 Solution

import re 

datePattern = re.compile('\d\d\d\d-(0[1-9]|1[0-2])-(0[1-9]|[12]\d|3[01])$') 

def isValidDate(s): 
    return datePattern.match(s) != None 

Explanation: Since we are using match(…) to compare the compiled pattern in variable datePattern to the string in parameter s given to our function isValidDate(…), we don’t have to worry about additional characters before the start of the date because match(…) will always try to match the pattern to the start of the string. However, we use $ as the last character in our pattern to make sure there are no additional characters following the date. That means the pattern has the form

“…-…-…$”

where the dots have to be replaced with some regular expression notation for the year, month, and day parts. The year part is easy, since we allow for any 4-digit number here. So we can use \d\d\d\d here, or alternatively \d{4,4} (remember that \d stands for the predefined class of all digits).

For the month, we need to distinguish two cases: either it is a 0 followed by one of the digits 1-9 (but not another 0) or a 1 followed by one of the digits 0-2. We therefore write this part as a case distinction (…|…) with the left part 0[1-9] representing the first option and the second part 1[0-2] representing the second option.

For the day, we need to distinguish three cases: (1) a 0 followed by one of the digits 1-9, (2) a 1 or 2 followed by any digit, or (3) a 3 followed by a 0 or a 1. Therefore we use a case-distinction with three options (…|…|…) for this part. The first part 0[1-9] is for option (1), the second part [12]\d for option (2), and the third part 3[01] for the third option.