# countingOddNumbers.py  P. Conrad for CS5nm, 11/10/08
#   Three different ways to write isEven.
#   All of these will produce the same results


# The shortest version


#  consumes: n, an integer
#  produces: True if n is even, otherwise False
def isEven(n):
    return (n%2==0)

#  consumes: n, an integer
#  produces: True if n is odd, otherwise False
def isOdd(n):
    return not isEven(n)

# consumes: myList, an list of integers
# produces: a count of how many integers in the list are odd

def countOdd(myList):
    if (len(myList)==0):
        return 0
    elif (len(myList==1) and isOdd(myList[0])):
        return 1
    elif (len(myList==1) and not isOdd(myList[0])):
        return 0
    else:  # split the list into the first and the rest, and add them together
        return(countOdd([myList[0]]) + countOdd(myList[1:])

# This next version (howManyOdd) works the same way, except
# the last line.    Keep in mind that both countOdd (above) and
# howManyOdd both take a "list of integers".   We can't pass in a single
# integer and expect the function to work.  So, we can't pass in myList[0]
# which is of type int.  We have to pass in [myList[0]], which is a list
# with a single item in it.   Another way is myList[0:1] which is a list with
# a single item in it (namely the item that is in position [0]).
               
# consumes: myList, an list of integers
# produces: a count of how many integers in the list are odd        
def howManyOdd(myList):
    if (len(myList)==0):
        return 0
    elif (len(myList==1) and isOdd(myList[0])):
        return 1
    elif (len(myList==1) and not isOdd(myList[0])):
        return 0
    else:  # split the list into first and rest, and add them together
        return(howManyOdd(myList[0:1]) + howManyOdd(myList[1:])