STRING MANIPULATION,
GUESS-and-CHECK,
APPROXIMATIONS,
BISECTION
(download slides and follow along on repl.it!)
COMP100 LECTURE 3
COMP100 LECTURE 3 1
LAST TIME
strings
branching – if/elif/else
while loops
for loops
COMP100 LECTURE 3
2
THIS WEEK
string manipulation
guess and check algorithms
approximate solutions
bisection method
COMP100 LECTURE 3
3
STRINGS
think of as a sequence of case sensitive characters
can compare strings with ==, >, < etc.
len() is a function used to retrieve the length of
the string in the parentheses
s = "abc"
len(s) evaluates to 3
COMP100 LECTURE 3
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STRINGS
square brackets used to perform indexing into a string to
get the value at a certain index/position
s = "abc"
index: 0 1 2 indexing always starts at 0
index: -3 -2 -1 last element always at index -1
s[0] evaluates to "a"
s[1] evaluates to "b"
s[2] evaluates to "c"
s[3] trying to index out of bounds, error
s[-1] evaluates to "c"
s[-2] evaluates to "b"
s[-3] evaluates to "a"
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STRINGS
can slice strings using [start:stop:step]
if give two numbers, [start:stop], step=1 by default
you can also omit numbers and leave just colons
s = "abcdefgh"
s[3:6] evaluates to "def", same as s[3:6:1]
s[3:6:2] evaluates to "df"
s[::] evaluates to "abcdefgh", same as s[0:len(s):1]
s[::-1] evaluates to "hgfedbca", same as s[-1:-(len(s)+1):-1]
s[4:1:-2]evaluates to "ec"
COMP100 LECTURE 3
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STRINGS
can slice strings using [start:stop:step]
if give two numbers, [start:stop], step=1 by default
you can also omit numbers and leave just colons
s = "abcdefgh"
s[3:6] evaluates to "def", same as s[3:6:1]
s[3:6:2] evaluates to "df"
s[::] evaluates to "abcdefgh", same as s[0:len(s):1]
s[::-1] evaluates to "hgfedbca", same as s[-1:-(len(s)+1):-1]
s[4:1:-2]evaluates to "ec"
COMP100 LECTURE 3
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STRINGS
strings are “immutable” – cannot be modified
s = "hello"
s[0] = 'y' gives an error
"hello"
COMP100 LECTURE 3
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STRINGS
strings are “immutable” – cannot be modified
s = "hello"
s[0] = 'y' gives an error
s = 'y'+s[1:len(s)] is allowed,
s bound to new object
"hello"
"yello"
COMP100 LECTURE 3
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for LOOPS RECAP
for loops have a loop variable that iterates over a set of
values
for var in range(4): var iterates over values 0,1,2,3
<expressions> expressions inside loop executed
with each value for var
for var in range(4,6): var iterates over values 4,5
<expressions>
range is a way to iterate over numbers, but a for loop variable
can iterate over any set of values, not just numbers!
COMP100 LECTURE 3
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STRINGS AND LOOPS
these two code snippets do the same thing
bottom one is more “pythonic”
s = "abcdefgh"
for index in range(len(s)):
if s[index] == 'i' or s[index] == 'u':
print("There is an i or u")
for char in s:
if char == 'i' or char == 'u':
print("There is an i or u")
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CODE EXAMPLE:
ROBOT CHEERLEADERS
an_letters = "aefhilmnorsxAEFHILMNORSX"
word = input("I will cheer for you! Enter a word: ")
times = int(input("Enthusiasm level (1-10): "))
i = 0
while i < len(word):
char = word[i]
if char in an_letters:
print("Give me an " + char + "! " + char)
else:
print("Give me a " + char + "! " + char)
i += 1
print("What does that spell?")
for i in range(times):
print(word, "!!!")
COMP100 LECTURE 3
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CODE EXAMPLE:
ROBOT CHEERLEADERS
an_letters = "aefhilmnorsxAEFHILMNORSX"
word = input("I will cheer for you! Enter a word: ")
times = int(input("Enthusiasm level (1-10): "))
i = 0
while i < len(word): for char in word:
char = word[i]
if char in an_letters:
print("Give me an " + char + "! " + char)
else:
print("Give me a " + char + "! " + char)
i += 1
print("What does that spell?")
for i in range(times):
print(word, "!!!")
COMP100 LECTURE 3
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EXERCISE
s1 = "mit u rock"
s2 = "i rule mit"
if len(s1) == len(s2):
for char1 in s1:
for char2 in s2:
if char1 == char2:
print("common letter")
break
COMP100 LECTURE 3
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GUESS-AND-CHECK
the process below also called exhaustive enumeration
given a problem…
you are able to guess a value for solution
you are able to check if the solution is correct
keep guessing until finding the solution
or guessed all possible values
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GUESS-AND-CHECK
– cube root
cube = 8
for guess in range(cube+1):
if guess**3 == cube:
print("Cube root of", cube, "is", guess)
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GUESS-AND-CHECK
– cube root
cube = 8
for guess in range(abs(cube)+1):
if guess**3 >= abs(cube):
break
if guess**3 != abs(cube):
print(cube, 'is not a perfect cube')
else:
if cube < 0:
guess = -guess
print('Cube root of '+str(cube)+' is '+str(guess))
COMP100 LECTURE 3 14
APPROXIMATE SOLUTIONS
good enough solution
start with a guess and increment by some small value
keep guessing if |guess3-cube| >= epsilon
for some small epsilon
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APPROXIMATE SOLUTIONS
good enough solution
start with a guess and increment by some small value
keep guessing if |guess3-cube| >= epsilon
for some small epsilon
decreasing increment size slower program
increasing epsilon less accurate answer
COMP100 LECTURE 3
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APPROXIMATE SOLUTION
– cube root
cube = 27
epsilon = 0.01
guess = 0.0
increment = 0.0001
num_guesses = 0
while abs(guess**3 - cube) >= epsilon: and guess <= cube :
guess += increment
num_guesses += 1
print('num_guesses =', num_guesses)
if abs(guess**3 - cube) >= epsilon:
print('Failed on cube root of', cube)
else:
print(guess, 'is close to the cube root of', cube)
COMP100 LECTURE 3
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BISECTION SEARCH
half interval each iteration
new guess is halfway in between
to illustrate, let’s play a game!
GUESS
GUESS
GUESS
6.0001 LECTURE 3 17
BISECTION SEARCH
– cube root
cube = 27
epsilon = 0.01
num_guesses = 0
low = 0
high = cube
guess = (high + low)/2.0
while abs(guess**3 - cube) >= epsilon:
if guess**3 < cube :
low = guess
else:
high = guess
guess = (high + low)/2.0
num_guesses += 1
print 'num_guesses =', num_guesses
print guess, 'is close to the cube root of', cube
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BISECTION SEARCH
CONVERGENCE
search space
◦ first guess: N/2
◦ second guess: N/4
◦ kth guess: N/2k
guess converges on the order of log2N steps
bisection search works when value of function varies
monotonically with input
code as shown only works for positive cubes > 1 – why?
challenges modify to work with negative cubes!
modify to work with x < 1!
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x<1
if x < 1, search space is 0 to x but cube root is
greater than x and less than 1
modify the code to choose the search space
depending on value of x
COMP100 LECTURE 3
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