Exercise 6
A palindrome is a word that is spelled the same backward and forward, like “noon” and “redivider”. Recursively, a word is a palindrome if the first and last letters are the same and the middle is a palindrome.
You can use the first, last, and middle helper functions defined in Think Python, or do the string slices inside your function directly. Be sure to think about your base cases.
Write a function called is_palindrome that takes a string argument and returns True if it is a palindrome and False otherwise. Remember that you can use the built-in function len to check the length of a string.
Write some unit tests for your function (optionally using doctest) to show that it works as intended.
# Helper functions
def first(word):
return word[0]
def last(word):
return word[-1]
def middle(word):
return word[1:-1]
# Palindrome Solution 1
def is_palindrome(word):
if first(word)==last(word):
if len(word)<=2:
return True
else:
return is_palindrome (middle(word))
else:
return False
# Palindrome Solution 2
def is_palindrome(x):
if len(x) >= 3:
if x[0] == x[-1]:
return is_palindrome(x[1:-1])
else:
return False
elif x[0] == x[-1]:
return True
else:
return False
# Palindrome Solution 3
def is_palindrome(x):
"""
>>> is_palindrome("cart")
False
>>> is_palindrome("bobob")
True
>>> is_palindrome("kayak")
True
"""
for i in range(len(x)):
if x == x[::-1]:
return True
return False
# Palindrome Solution 4
def is_pallindrome(string):
string = string.lower()
if len(string)>1:
if string[0] == string[len(string)-1]:
string = string[1:len(string)-1]
return True
if string[0] != string[len(string)-1]:
return False
return is_pallindrome(string)
# Palindrome Solution 5
def is_palindrome(s):
"""
>>> is_palindrome("Go hang a salami Im a lasagna hog") # even palindrome
True
>>> is_palindrome("Ah Satan sees Natasha")
True
>>> is_palindrome("No melons no lemon") # odd palindrome
True
>>> is_palindrome("Now I see bees I lost") # not a palindrome
False
"""
s = s.strip().lower()
if len(s) <= 1:
return True
if s[0] == s[len(s)-1]:
return is_palindrome(s[1::len(s)-1])
else:
return False
# Palindrome Solution 6
def is_palindrome(word):
flag = True
if len(word) == 1:
return True
if len(word) < 3: #when len == 2
return first(word) == last(word)
return (first(word) == last(word) and is_palindrome(middle(word)))
# Palindrome Solution 7
def is_palindrome(WoRd):
""" Returns True/False for if input is a palindrome.
>>> is_palindrome('word')
False
>>> is_palindrome('racecar')
True
>>> is_palindrome('Racecar')
True
>>> is_palindrome('redivider')
True
>>> is_palindrome('l')
True
"""
word = WoRd.lower()
if len(word) < 3:
if first(word) == last(word):
return True
else:
return False
if len(word) >= 3:
if first(word) == last(word):
return is_palindrome(middle(word))
else:
return False
# Palindrome Solution 8
def palindrome(word):
"""
Recursively looks through a string and determines if it is a palindrome.
>>> palindrome('racecar')
True
>>> palindrome('abba')
True
>>> palindrome('palindrome')
False
"""
if len(word) < 2:
return True
elif word[0] == word[-1]:
return palindrome(word[1:-1])
return False
# Palindrome Solution 9
def is_palindrome(word):
'''
>>> is_palindrome("redivider")
True
>>> is_palindrome("bob")
True
>>> is_palindrome("ttta")
False
>>> is_palindrome("aaaaaaa")
True
'''
mid = middle(word)
if first(word) == last(word):
if mid[:] == mid[::-1]:
return True
return False
# Palindrome Solution 10
def is_palindrome(word):
word_list = list(word)
length = len(word_list)
if length <= 2:
return False
elif length > 2:
for letters in word_list:
if word_list[0] == word_list[-1]:
del(word_list[0])
del(word_list[-1])
if len(word_list)== 0:
return True
else:
return False
# Palindrome Solution 11
def is_palindrome(word):
forward = word
backward = word[::-1]
if forward == backward:
return True
return False
# Palindrome Solution 12
def palindrome(a):
"""
Returns Boolean assignment True or False to determine if word is palindrome.
>>> palindrome('DogGOD')
True
>>> palindrome('RACE car')
True
>>> palindrome('allen')
False
>>> palindrome('PAul ruvOLo')
False
>>> palindrome('stanley YELnats')
True
"""
a = a.lower() #converts all characters in a to lowercase
a = a.rstrip() #strips all spaces from a
if len(a) <= 1:
return True
elif first(a) != last(a):
return False
return palindrome(middle(a))
#what do you know, it works
#fun with palindromes
palindrome('Borrow or Rob') #a moral dilemma
palindrome('Murder for a jar of red rum') #creepiest childhood game
palindrome('Yo Banana Boy') #it's Gabe
palindrome('satan oscillate my metallic sonatas') #that's unfortunate
palindrome('SolO GIGOLOS') #I am having altogether to much fun with this
palindrome('dammit Im mad')#lol not really but I bet whoever's reading this is
True
Exercise 6.4
A number $a$ is a power of $b$ if it is divisible by $b$ and $a/b$ is a power of $b$. Write a function called is_power that takes parameters a and b and returns True if a is a power of b. Note: you will have to think about the base case.
# is_power solution 1
def is_power(a, b):
if a < b:
return False
if a == b:
return True
else:
return is_power(a / b, b)
# is_power solution 2
def is_power(a,b):
if a == b:
return True
elif a > b:
if a%b == 0:
return is_power(a//b,b)
else:
return False
else:
return False
# is_power solution 3
def is_power(a,b):
if a>b:
if a%b == 0:
return True
if a%b != 0:
return False
return is_power(a/b,b)
# is_power solution 4
def is_power(a,b):
if a is b:
return True
elif a%b == 0:
return is_power(a//b,b)
else:
return False
# is_power solution 5
def is_power(a,b):
"""
>>> is_power(4,2)
True
>>> is_power(3,2)
False
"""
if a == 1:
return True
return a%b == 0 and is_power(a/b, b)
# is_power solution 6
def is_power(a,b):
"""returns whether a is a power of b.
doctests test for a is zeroeth power, a is first power, a is nth power, and a is not power.
>>> is_power(1, 70)
True
>>> is_power(5, 5)
True
>>> is_power(128, 2)
True
>>> is_power(132, 2)
False
"""
if a == 1:
return True
elif a % b == 0:
return is_power(a//b,b)
else:
return False
# is_power solution 7
def is_power(a,b):
""" Returns True/False for if a is a power of b.
>>> is_power(3,2)
True
>>> is_power(1,5)
True
>>> is_power(6,6)
True
>>> is_power(2,0)
False
>>> is_power(6,1)
False
>>> is_power(6,3)
False
"""
if a == 0 or b == 0:
return False
elif b == 1:
if a == 1:
return True
else:
return False
if a==1 or (a/b)==1:
result = True
elif a % b == 0 and a!=0:
result = is_power((a//b),b)
else:
result = False
return result
# is_power solution 8
def is_power(a, b):
"""
Returns if A is a power of b.
>>> power(10, 2)
False
>>> power(32, 2)
True
"""
if not (a % b == 0):
return False
if a//b == 1:
return True
else:
return power(a//b, b)
# is_power solution 9
def is_power(a,b):
if a == b:
return True
test1 = a%b
a = a//b
if test1 == 0 and is_power(a,b):
return True
else:
return False
# is_power solution 10
def is_power(a, b):
if a % b == 0:
if (a//b) % b == 0:
return True
return False
# is_power solution 11
def is_power(a, b):
if a%b != 0:
return False
else:
return True
c = a//b
return is_power(c, b)
# is_power solution 12
def is_power(a,b):
if a%b == 0:
if a == b:
return True
a = a//b
return is_power(a,b)
else:
return False