Homework 1
Due by 11:59pm on Thursday, 1/25
Instructions
Download hw01.zip.
Submission: When you are done, submit with
python3 ok --submit
.
You may submit more than once before the deadline; only the final submission
will be scored. Check that you have successfully submitted your code on
okpy.org.
See Lab 0
for more instructions on submitting assignments.
Using Ok: If you have any questions about using Ok, please refer to this guide.
Readings: You might find the following references useful:
Homework Questions
Q0: Survey
Before you get started writing code, please fill out the introductory survey.
Q1: A Plus Abs B
Fill in the blanks in the following function definition for adding a
to the
absolute value of b
, without calling abs
.
from operator import add, sub
def a_plus_abs_b(a, b):
"""Return a+abs(b), but without calling abs.
>>> a_plus_abs_b(2, 3)
5
>>> a_plus_abs_b(2, -3)
5
"""
if b < 0:
f = _____
else:
f = _____
return f(a, b)
Use Ok to test your code:
python3 ok -q a_plus_abs_b
Q2: Two of Three
Write a function that takes three positive numbers and returns the sum of the squares of the two largest numbers. Use only a single line for the body of the function.
def two_of_three(a, b, c):
"""Return x*x + y*y, where x and y are the two largest members of the
positive numbers a, b, and c.
>>> two_of_three(1, 2, 3)
13
>>> two_of_three(5, 3, 1)
34
>>> two_of_three(10, 2, 8)
164
>>> two_of_three(5, 5, 5)
50
"""
return _____
Use Ok to test your code:
python3 ok -q two_of_three
Q3: Largest Factor
Write a function that takes an integer n
that is greater than 1 and
returns the largest integer that is smaller than n
and evenly divides n
.
def largest_factor(n):
"""Return the largest factor of n that is smaller than n.
>>> largest_factor(15) # factors are 1, 3, 5
5
>>> largest_factor(80) # factors are 1, 2, 4, 5, 8, 10, 16, 20, 40
40
>>> largest_factor(13) # factor is 1 since 13 is prime
1
"""
"*** YOUR CODE HERE ***"
Hint: To check if
b
evenly dividesa
, you can use the expressiona % b == 0
, which can be read as, "the remainder of dividinga
byb
is 0."
Use Ok to test your code:
python3 ok -q largest_factor
Q4: If Function vs Statement
Let's write a function that does the same thing as an if
statement.
def if_function(condition, true_result, false_result):
"""Return true_result if condition is a true value, and
false_result otherwise.
>>> if_function(True, 2, 3)
2
>>> if_function(False, 2, 3)
3
>>> if_function(3==2, 3+2, 3-2)
1
>>> if_function(3>2, 3+2, 3-2)
5
"""
if condition:
return true_result
else:
return false_result
Despite the doctests above, this function actually does not do the
same thing as an if
statement in all cases. To prove this fact,
write functions c
, t
, and f
such that with_if_statement
returns the number 1
, but with_if_function
does not (it can do
anything else):
def with_if_statement():
"""
>>> with_if_statement()
1
"""
if c():
return t()
else:
return f()
def with_if_function():
return if_function(c(), t(), f())
def c():
"*** YOUR CODE HERE ***"
def t():
"*** YOUR CODE HERE ***"
def f():
"*** YOUR CODE HERE ***"
To test your solution, open an interactive interpreter
python3 -i hw01.py
and try calling with_if_function
and with_if_statement
to check that one
returns 1 and the other doesn't.
Hint: If you are having a hard time identifying how the if statement and if function differ, first try to get them to
Q5: Hailstone
Douglas Hofstadter's Pulitzer-prize-winning book, Gödel, Escher, Bach, poses the following mathematical puzzle.
- Pick a positive integer
n
as the start. - If
n
is even, divide it by 2. - If
n
is odd, multiply it by 3 and add 1. - Continue this process until
n
is 1.
The number n
will travel up and down but eventually end at 1 (at
least for all numbers that have ever been tried -- nobody has ever
proved that the sequence will terminate). Analogously, a hailstone
travels up and down in the atmosphere before eventually landing on
earth.
This sequence of values of n
is often called a Hailstone sequence,
Write a function that takes a single argument with formal parameter
name n
, prints out the hailstone sequence starting at n
, and
returns the number of steps in the sequence:
def hailstone(n):
"""Print the hailstone sequence starting at n and return its
length.
>>> a = hailstone(10)
10
5
16
8
4
2
1
>>> a
7
"""
"*** YOUR CODE HERE ***"
Hailstone sequences can get quite long! Try 27. What's the longest you can find?
Use Ok to test your code:
python3 ok -q hailstone