# Lab 13
# YOUR NAMES & EMAIL HERE


# Discuss this with your lab partner
def upAndDown(n):
    print("Down:", n)
    if (n > 0):
        upAndDown(n-1)
    print("Up:", n)
    return 0


# find the max in the list
# non-recursive
def maxR(lst):
    return maxRhelper(lst,0)

# find the max in the list starting from index p
# recursive
# no for loops or Python library functions
def maxRhelper(lst, p):
    return 0


# find the minimum item in the list
def minR(lst):
    return minRhelper(lst,0)

# find the minimum item in the list starting from index p
# recursive
# no for loops or Python library functions
def minRhelper(lst, p):
    return 0

    
# find the number of times item v occurs in the list
def countItems(lst, v):
    return countItemsRhelper(lst,v,0)

# find the number of times item v occurs in the list
# starting from index p
# no for loops or Python library functions
def countItemsRhelper(lst, v, p):
    return 0

# Output the number of digits in a given number
# Hint: use //
# no for loops or Python library functions
def digits(n):
    return 0

# Convert the integer value to a String representing its binary value
# Example: 12022008 -> 101101110111000011111000
# Example: 10 -> 1010
#   Note: 10 is even, so rightmost bit is 0
#         divide 10 by 2, resulting in 5
#         5 is odd, so next bit is 1
#         divide 5 by 2, resulting in 2
#         2 is even, so next bit is 0
#         divide 2 by 2, resulting in 1
#         1 is odd, so next bit is 1
# Example: 1 -> 1
# Example: 0 -> 0
# Hint: Use // and % operators
# no for loops or Python library functions
def binary(n):
    return "0"

# Multiply the input values using only addition.
# no for loops or Python library functions
# Hint: 5*7 = 7 + 4*7
def multiply(m, n):
    return 0


# Ackermann's Function
# A(m,n) = n+1 if m=0
# A(m,n) = A(m-1,1) if n=0
# A(m,n) = A(m-1,A(m,n-1)) otherwise
# Note: This is a really C-R-A-Z-Y function!
# See http://mathworld.wolfram.com/AckermannFunction.html
def Ackermann(m, n):
    return 0

# Determine all the permutations of a given input string
# Example:
#   input:  "abc"
#   output: ["abc", "acb", "bac", "bca", "cab", "cba"]
# Hint: use the : array slicing operator
# recall s[i] is the ith character in the string
# s[0:i] is the substring from positions 0 through i-1 inclusive
# s[i:0] is the substring from positions i through the end of the string
# Generally recursive, but using for loops is ok
def permutations(s):
    return list()


def main():
    # Discuss this with your lab partner
    print("upAndDown - explain this output")
    upAndDown(5)
    
    print("max")
    # 9
    values = [3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8]
    print(maxR(values))
    
    print("min")
    # 1
    values = [3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8]
    print(minR(values))
    
    print("countItems")
    # Opening lines from the song "Across the Universe" by the Beatles
    text = "\
Words are flowing out like endless rain into a paper cup\n\
They slither wildly as they slip away across the universe"
    print(text)
    print(countItems(text, " ")) # 19
    print(countItems(text, "a")) # 8
    print(countItems(text, "x")) # 0
    
    print("digits")
    print(digits(0)) # 1
    print(digits(9)) # 1
    print(digits(54321)) # 5
    
    print("binary")
    print(binary(0)) # 0
    print(binary(1)) # 1
    print(binary(10)) # 1010
    print(binary(12022008)) # 101101110111000011111000
    
    print("multiply")
    print(multiply(5,7)) # 35
    print(multiply(0,0)) # 0
    print(multiply(1,1)) # 1
    
    print("Ackermann")
    print(Ackermann(3,0)) # 5
    print(Ackermann(0,2)) # 3
    print(Ackermann(3,3)) # 61
    print(Ackermann(5,5)) # discuss why this fails, then comment out this line

    print("permutations")
    # ['abcd', 'abdc', 'acbd', 'acdb', ..., 'dcba']
    perms = permutations("abcd")
    print(perms)
    
main()