# Debugging example - fix this!
#
def is_reverse(word1, word2):
    if len(word1) != len(word2):
        return False
    i = 0
    j = len(word2) 

    while j > 0:
        # Use print line below to help debug ...
        #print("i = ", i, "j = ", j)
        if word1[i] != word2[j]:
            return False
        i = i + 1
        j = j - 1

    return True


# for loops and range
#
# how do we do the equivalent of loopChars (from lec8loopchars.py)
# with a for loop and range.  loopChars2 doesn't give us access to the
# index.
#
def loopCharsWithIndex(inputString):
    for currentIndex in range(len(inputString)):
        currentChar = inputString[currentIndex]
        print(currentChar, currentIndex)
    return
        
##########
#
# Questions like some that might be on quiz 1


def ex1a(n1, n2):
    return (not (n1 > 5)) or (n2 <= 10)

# Give simple expresssions for a, b, c, d, e so that ex1a and ex1b are equivalent.
# Your answers MUST NOT include any AND, OR, or NOT operations. They must
# be simple comparisions between numbers and/or variables, or return statements.



def ex1b(n1, n2):
    # if ---a---
    #    ---b---
    # if ---c---
    #    ---d---
    # ---e---
    if n2 <= 10:
        return True
    if n1 <= 5:
        return True
    return False
    

# Do f1a and f1b return the same thing for every possible input string?
# If not, provide example input for which they return diffrent results.
#
def f1a(string1):
    i = 0
    newString = ""
    while i < len(string1):
        newString = newString + string1[i] + string1[i]
        i = i + 2
    return newString

def f1b(string1):
    i = len(string1) - 1
    newString = ""
    while i >= 0:
        newString = string1[i] + string1[i] + newString
        i = i - 2 
    return newString