This is a study guide with links to past lectures, assignments, and handouts, as well as additional practice problems to assist you in learning the concepts.
Draw environment diagrams automatically with Python Tutor.
Important: For solutions to these assignments once they have been released, see the main website
A lot of students were confused about the
return statement (more specifically
return statements allow the programmer to return a value from a function. You
can take the returned value and save it to a variable or do whatever you want
Think of it as a store. When you return something at the store, that item that you return will probably be purchased by someone else and they will use it however they want. (I know the analogy is not so great).
Two different functions will be used to illustrate the difference.
The first function example will return a string:
def some_function(): return 'I love John DeNero'
some_function returns the string,
'I love John DeNero', so I can also
bind that return value to a name.
my_love = some_function()
Now, the variable
my_love is set to the return value of
this case it is
'I love John DeNero'. So
my_love = 'I love John DeNero'.
The second function example will print a string:
def some_function(): print('I love John DeNero')
some_function() now will print
'I love John DeNero' on the
terminal and then return
None. (Recall that, if a function has no
statement, it will return
None.) This time, binding the return value to a
name has a different result.
my_love = some_function()
It will print 'I love John DeNero' on the terminal and bind
my_love = None.
It is important to note that return statements will terminate the function.
def some_function(): return 5 return 3
It will return 5 and end the function without ever returning 3.
Boolean contexts are "places in Python code where you place an expression but all that matters about that expression is whether it's true or false."
Here's an example from lecture of a function containing an
if statement with
two boolean contexts.
def absolute_value(x): """Return the absolute value of x.""" if x < 0: return -x elif x == 0: return 0 else: return x
if clause, Python asks, "Is
x < 0 a true value or a false
value?" This depends on the value of
x. Suppose we have
x = 3.
>>> x = 3 >>> x < 0 False
Python evaluates the call expression
x < 0 to
False. Now, we need to ask
the question, is
False a false value? This might seem like an unusual
question to ask, but it's actually a very subtle step.
In Python, false values include the following values (more to come):
False, 0, '', None
Everything else is a true value.
Because we know that
False is a false value, the entire expression
x < 0
is a false value in boolean context.
Let's look at another example to see why boolean context and true value/false value distinction matters. Suppose we want to evaluate the following boolean expression.
>>> print('hi') or 25
or operator contains two boolean contexts, one on each side of the
Let's evaluate this expression.
print('hi') prints the string
'hi' to the terminal and returns
None is a false value in Python, the first boolean context is false.
25 evaluates to the number 25. Because 25 is a non-zero number, it is a true
value in Python and so the second boolean context is true.
The final value that is returned from the entire
or is 25 because Python uses
the boolean context only to determine how to interpret the value in the context
or expression. 25 is not the same as
True; it just evaluates to a
true value in boolean context.
For each of the expressions below, write the output displayed by the interactive Python interpreter when the expression is evaluated. The output may have multiple lines. If an error occurs, write "Error", but include all output displayed before the error. If a function value is displayed, write "Function".
Recall: The interactive interpreter displays the value of a successfully
evaluated expression, unless it is
Assume that you have started
python3 and executed the following statements:
from math import sqrt x = 15 def square(x): return print(x * x) def multiply(x): x_new = x * 2 return x * 3
>>> print(print(3, 5))______3 5 None>>> square(x)______225>>> True and 17______17
>>> print(multiply(x))______45>>> multiply(multiply(3))______27>>> print(square(multiply(2)), 8) + 3______36 None 8 Error
x = 2 def take_quiz(x): print(x * 10) if x * 10 >= 100: return 'Good job!' return 'Go to office hours!' def office_hour(x): while x < 10: x = x + 3 return print(take_quiz(x))
>>> pow(print(0, 0), 0)______0 0 Error>>> office_hour(4)______100 Good job!
>>> print(take_quiz(x))______20 Go to office hours!>>> print(office_hour(x))______110 Good job! None
>>> office_hour(office_hour(x))______110 Good job! Error
>>> take_quiz(-3) and print(office_hour(x))______-30 110 Good Job! None
Q3: There can only be Wan
Hint: Draw environment diagrams to track progress! Don't just quietly recite the names to yourself, as names that sound similar can easily be mixed up!
def wan1(wan1, wan11): return wan1 == wan11 def wan11(wan11, wan1): return wan1 / wan11 one, wan, won = 1, 3, 17 def derek(wan, won, one): print(wan1(wan, wan11(won, 1))) return wan11(one, wan1)
>>> print(one or wan11(0, one))______1>>> print(wan1(one, wan11(one, one)))______True
>>> print(print(wan - won), wan1)______-14 None Function
>>> derek(1, 1, 2)______True Error
>>> derek(wan, 2, one)______False Error
Q4: Whoos hat?
s = 2 def hat(s): while s > 0: whoos(s) s = s - 1 return whoos(s) def someones(their, hat): his, her = their, hat hat = print(his or her) and her or his if hat: return 'hat' def whoos(hat): return print(hat)
>>> print(s) or 1 / 0______2 Error>>> 0 or 2 == True and print(5)______False
>>> her, cat = 'her', print('theirs') or 'his'______theirs>>> someones(her, cat)______her 'hat'
>>> hat(s)______2 1 0
>>> print(whoos(hat), someones(cat, hat))______Function his None hat
Q5: Out at the Ballgame
def announce(score, inning): hits = score * 3 % inning - 1 print('The Giants have ' + str(score) + ' runs!') if hits < 2: jumbotron_text = print('That\'s the end of the inning!') inning = inning + 1 return jumbotron_text else: score = score + hits % 3 crowd_noise = print('The Giants now have: ' + str(score)) print(str(crowd_noise)) return score def player(hits, inning): if hits < 2: print('I need to practice in the batting cage!') score = hits * 2 + 1 saying = announce(score, inning) return saying * 2
>>> print(3, print(5 % 3))______2 3 None
>>> player(0, 0)______I need to practice in the batting cage! Error
>>> player(3, 2)______The Giants have 7 runs! That's the end of the inning! Error
>>> player(5, 9) - 5______The Giants have 11 runs! The Giants now have 13 None 21
>>> player(announce(15, 8), 4)______The Giants have 15 runs! The Giants have 16 None The Giants have 33 runs! The Giants now have 35 None 70
Implement a function called
distance(x1, y1, x2, y2):
y1form an x-y coordinate pair
y2form an x-y coordinate pair
distance returns the Euclidean distance between the two points. Use the
from math import sqrt def distance(x1, y1, x2, y2): """Calculates the Euclidian distance between two points (x1, y1) and (x2, y2) >>> distance(1, 1, 1, 2) 1.0 >>> distance(1, 3, 1, 1) 2.0 >>> distance(1, 2, 3, 4) 2.8284271247461903 """"*** YOUR CODE HERE ***"return sqrt(square(x1-x2) + square(y1-y2))
Q7: Distance (3D)
Now, let us edit this program to get the distance between two
3-dimensional coordinates. Your
distance3d function should take six
arguments and compute the following:
def distance3d(x1, y1, z1, x2, y2, z2): """Calculates the 3D Euclidian distance between two points (x1, y1, z1) and (x2, y2, z2). >>> distance3d(1, 1, 1, 1, 2, 1) 1.0 >>> distance3d(2, 3, 5, 5, 8, 3) 6.164414002968976 """"*** YOUR CODE HERE ***"return sqrt(square(x1-x2) + square(y1-y2) + square(z1-z2))
harmonic, which returns the harmonic mean of two positive numbers
y. The harmonic mean of 2 numbers is 2 divided by the sum of the
reciprocals of the numbers. (The reciprocal of
def harmonic(x, y): """Return the harmonic mean of x and y. >>> harmonic(2, 6) 3.0 >>> harmonic(1, 1) 1.0 >>> harmonic(2.5, 7.5) 3.75>>> harmonic(4, 12) 6.0""""*** YOUR CODE HERE ***"return 2/(1/x + 1/y)
Python Tutor is a great visualization tool for environment diagrams. Paste in your Python code and it will generate an environment diagram you can walk through step-by-step! Use it to help you check your answers!
Try drawing environment diagrams for the following examples and predicting what Python will output:
>>> def square(x): ... return x * x >>> def double(x): ... return x + x >>> a = square(double(4)) >>> a______64
>>> x, y = 4, 3 >>> def reassign(arg1, arg2): ... x = arg1 ... y = arg2 >>> reassign(5, 6) >>> x______4>>> y______3
>>> def f(x): ... f(x) >>> print, f = f, print >>> a = f(4)______4>>> a______# Nothing shows up, because a = None>>> b = print(4)______4>>> b______# Nothing shows up, because b = None
Q10: Fix the Bug
The following snippet of code doesn't work! Figure out what is wrong and fix the bugs.
def compare(a, b): """ Compares if a and b are equal. >>> compare(4, 2) 'not equal' >>> compare(4, 4) 'equal' """ if a = b: return 'equal' return 'not equal'
a = b will cause a
SyntaxError. Instead, it should be
if a == b:
Q11: Last square
Implement the function
last_square, which takes as input a positive
integer and returns the largest perfect square less than its argument.
A perfect square is any integer multiplied by itself:
Hint: If you're stuck, try writing a function that prints out the
first 5 perfect squares using a
while statement: 1, 4, 9, 16, 25.
Then, adapt that
while statement to this question by changing the
def last_square(x): """Return the largest perfect square less than X, where X>0. >>> last_square(10) 9 >>> last_square(39) 36 >>> last_square(100) 81 >>> result = last_square(2) # Return, don't print >>> result 1>>> cases = [(1, 0), (2, 1), (3, 1), (4, 1), (5, 4), (6, 4), ... (10, 9), (17, 16), (26, 25), (36, 25), (46, 36)] >>> [last_square(s) == t for s, t in cases].count(False) 0""""*** YOUR CODE HERE ***"k = 0 while k * k < x: k = k + 1 return (k-1) * (k-1)
We iterate over perfect squares until we find the first one larger or equal to the input. The answer is then the square before that one. This solution is inefficient, but an efficient solution requires taking a square root.
An open interval is a range of numbers that does not include its end points.
For example, (10, 15) stands for all numbers that are strictly greater than 10
and strictly less than 15. Two intervals overlap if they contain any points
in common. For example (10, 15) overlaps (14, 16), but not (1, 5) or (15, 16).
The intervals (10, 10) or (10, 9) contain no numbers, since nothing is both
greater than and less than 10, or greater than 10 and less than 9. Implement
overlaps to take four numbers as arguments, representing the
bounds of two intervals, and return
True if the intervals overlap and
def overlaps(low0, high0, low1, high1): """Return whether the open intervals (LOW0, HIGH0) and (LOW1, HIGH1) overlap. >>> overlaps(10, 15, 14, 16) True >>> overlaps(10, 15, 1, 5) False >>> overlaps(10, 10, 9, 11) False >>> result = overlaps(1, 5, 0, 3) # Return, don't print >>> result True>>> [overlaps(a0, b0, a1, b1) for a0, b0, a1, b1 in ... ( (1, 4, 2, 3), (1, 4, 0, 2), (1, 4, 3, 5), (0.1, 0.4, 0.2, 0.3), ... (2, 3, 1, 4), (0, 2, 1, 4), (3, 5, 1, 4) )].count(False) 0 >>> [overlaps(a0, b0, a1, b1) for a0, b0, a1, b1 in ... ( (1, 4, -1, 0), (1, 4, 5, 6), (1, 4, 4, 5), (1, 4, 0, 1), ... (-1, 0, 1, 4), (5, 6, 1, 4), (4, 5, 1, 4), (0, 1, 1, 4), ... (5, 5, 3, 6), (5, 3, 4, 6), (5, 5, 5, 5), ... (3, 6, 5, 5), (4, 6, 5, 3), (0.3, 0.6, 0.5, 0.5) )].count(True) 0""""*** YOUR CODE HERE ***"return low1 < min(high0, high1) > low0
There are many solutions to this problem. One way to look at it is to consider
conditions under which the intervals don't overlap. Clearly for two non-empty
not to overlap, one has to come entirely before the other. This becomes
<= low0 or high0 <= low1, which when negated is
high1 > low0 and high1 >
low1. In addition, both lower bounds must be less than their respective upper
bounds (or the intervals are empty). The solution given combines these
Q13: Triangular numbers
The nth triangular number is defined as the sum of all integers from 1 to n, i.e.
1 + 2 + ... + n
The closed-form formula for the nth triangular number is
(n + 1) * n / 2
triangular_sum, which takes an integer
n and returns the sum of the
n triangular numbers, while printing each of the triangular numbers
between 1 and the
nth triangular number.
def triangular_sum(n): """ >>> t_sum = triangular_sum(5) 1 3 6 10 15 >>> t_sum 35 """"*** YOUR CODE HERE ***"count = 1 t_sum = 0 while count <= n: t_number = count * (count + 1) // 2 print(t_number) t_sum += t_number count += 1 return t_sum
Q14: Same hailstone
same_hailstone, which returns whether positive integer arguments
b are part of the same hailstone sequence. A hailstone sequence is
defined in Homework 1 as the following:
- Pick a positive integer
nas the start.
nis even, divide it by 2.
nis odd, multiply it by 3 and add 1.
- Continue this process until
def same_hailstone(a, b): """Return whether a and b are both members of the same hailstone sequence. >>> same_hailstone(10, 16) # 10, 5, 16, 8, 4, 2, 1 True >>> same_hailstone(16, 10) # order doesn't matter True >>> result = same_hailstone(3, 19) # return, don't print >>> result FalseExtra tests: >>> same_hailstone(19, 3) False >>> same_hailstone(4858, 61) True >>> same_hailstone(7, 6) False""""*** YOUR CODE HERE ***"return in_hailstone(a, b) or in_hailstone(b, a) def in_hailstone(a, b): """Return whether b is in hailstone sequence of a.""" while a > 1: if a == b: return True elif a % 2 == 0: a = a // 2 else: a = a * 3 + 1 return False
Q15: Pi Fraction
Complete the implementation of
pi_fraction, which takes a positive number
gap and prints the fraction that is no more than
gap away from
pi and has
the smallest possible positive integer denominator. See the doctests for the
format of the printed output.
Hint: If you want to find the nearest integer to a number, use the built-in
round function. It's possible to solve this problem without using
You may change the starter implementation if you wish.
from math import pi def pi_fraction(gap): """Print the fraction within gap of pi that has the smallest denominator. >>> pi_fraction(0.01) 22 / 7 = 3.142857142857143 >>> pi_fraction(1) 3 / 1 = 3.0 >>> pi_fraction(1/8) 13 / 4 = 3.25 >>> pi_fraction(1e-6) 355 / 113 = 3.1415929203539825>>> pi_fraction(1e-3) 201 / 64 = 3.140625 >>> pi_fraction(1/32) 19 / 6 = 3.1666666666666665""" numerator, denominator = 3, 1"*** YOUR CODE HERE ***"while abs(numerator/denominator-pi) > gap: denominator = denominator + 1 numerator = round(pi * denominator)print(numerator, '/', denominator, '=', numerator/denominator)
This implementation repeatedly increases
denominator until the nearest
pi is within