Lab 3: Recursion and Midterm Review
Due at 11:59pm on Friday, 02/09/2018.
Starter Files
Download lab03.zip. Inside the archive, you will find starter files for the questions in this lab, along with a copy of the Ok autograder.
Submission
By the end of this lab, you should have submitted the lab with
python3 ok --submit
. You may submit more than once before the
deadline; only the final submission will be graded.
Check that you have successfully submitted your code on
okpy.org.
- Questions 1 and 2 must be completed in order to receive credit for this lab. Starter code for these questions is in lab03.py.
- Questions 3 through 10 are for midterm review and are optional. It is recommended that you complete these problems on your own time. Starter code for these questions is in the lab03_extra.py file.
Note: In order to help you prepare for the midterm, we've provided the solutions to the problems in this lab assignment. You may consult them if you get stuck or to check your answers, but please try to solve them on your own first. You must still submit the required questions by Friday night if you want to receive credit for this lab assignment.
Topics
Consult this section if you need a refresher on the material for this lab. It's okay to skip directly to the questions and refer back here should you get stuck.
Recursion
A recursive function is a function that calls itself in its body, either directly or indirectly. Recursive functions have three important components:
- Base case(s), the simplest possible form of the problem you're trying to solve.
- Recursive case(s), where the function calls itself with a simpler argument as part of the computation.
- Using the recursive calls to solve the full problem.
Let's look at the canonical example, factorial
:
def factorial(n):
if n == 0:
return 1
return n * factorial(n - 1)
We know by its definition that 0! is 1. So we choose n == 0
as our base case.
The recursive step also follows from the definition of factorial, i.e., n
! =
n
* (n
-1)!.
The next few questions in lab will have you writing recursive functions. Here are some general tips:
- Consider how you can solve the current problem using the solution to a simpler version of the problem. Remember to trust the recursion: assume that your solution to the simpler problem works correctly without worrying about how.
- Think about what the answer would be in the simplest possible case(s). These will be your base cases - the stopping points for your recursive calls. Make sure to consider the possibility that you're missing base cases (this is a common way recursive solutions fail).
- It may help to write the iterative version first.
Required Questions
Q1: GCD
The greatest common divisor of two positive integers a
and b
is the
largest integer which evenly divides both numbers (with no remainder).
Euclid, a Greek mathematician in 300 B.C., realized that the greatest
common divisor of a
and b
is one of the following:
- the smaller value if it evenly divides the larger value, or
- the greatest common divisor of the smaller value and the remainder of the larger value divided by the smaller value
In other words, if a
is greater than b
and a
is not divisible by
b
, then
gcd(a, b) = gcd(b, a % b)
Write the gcd
function recursively using Euclid's algorithm.
def gcd(a, b):
"""Returns the greatest common divisor of a and b.
Should be implemented using recursion.
>>> gcd(34, 19)
1
>>> gcd(39, 91)
13
>>> gcd(20, 30)
10
>>> gcd(40, 40)
40
"""
"*** YOUR CODE HERE ***"
a, b = max(a, b), min(a, b)
if a % b == 0:
return b
else:
return gcd(b, a % b)
# Iterative solution, if you're curious
def gcd_iter(a, b):
"""Returns the greatest common divisor of a and b, using iteration.
>>> gcd_iter(34, 19)
1
>>> gcd_iter(39, 91)
13
>>> gcd_iter(20, 30)
10
>>> gcd_iter(40, 40)
40
"""
if a < b:
return gcd_iter(b, a)
while a > b and not a % b == 0:
a, b = b, a % b
return b
Use Ok to test your code:
python3 ok -q gcd
Q2: Hailstone
For the hailstone
function from homework 1, you 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. Repeat this process until n
is 1. Write a
recursive version of hailstone that prints out the values of the sequence and
returns the number of steps.
def hailstone(n):
"""Print out the hailstone sequence starting at n, and return the
number of elements in the sequence.
>>> a = hailstone(10)
10
5
16
8
4
2
1
>>> a
7
"""
"*** YOUR CODE HERE ***"
print(n)
if n == 1:
return 1
elif n % 2 == 0:
return 1 + hailstone(n // 2)
else:
return 1 + hailstone(3 * n + 1)
# Alternate solution with helper function
def hailstone2(n):
def hail_helper(n, count):
print(n)
if n == 1:
return count
else:
if n % 2 == 0:
return hail_helper(n // 2, count + 1)
else:
return hail_helper(3 * n + 1, count + 1)
return hail_helper(n, 1)
Use Ok to test your code:
python3 ok -q hailstone
Midterm Review
Note: The following questions are in lab03_extra.py.
Functions
Q3: WWPD: Call Expressions
Use Ok to test your knowledge with the following "What Would Python Display?" questions:
python3 ok -q call_expressions -u
For all WWPD questions, type
Function
if you believe the answer is<function...>
, >Error
if it errors, andNothing
if nothing is displayed.
>>> from operator import add
>>> def double(x):
... return x + x
...
>>> def square(y):
... return y * y
...
>>> def f(z):
... add(square(double(z)), 1)
...
>>> f(4)
______# f(4) returns None, which is a special value that the interpreter hides unless explicitly printed
>>> def foo(x, y):
... print("x or y")
... return x or y
...
>>> a = foo
______# We aren't calling foo here; we are just binding the variable a in the global frame to the function object foo
>>> b = foo()
______TypeError: foo() missing 2 required positional arguments: 'x' and 'y'
>>> c = a(print("x"), print("y"))
______x
y
x or y
>>> print(c)
______None
>>> def welcome():
... print('welcome to')
... return 'hello'
...
>>> def cs61a():
... print('cs61a')
... return 'world'
...
>>> print(welcome(), cs61a())
______welcome to
cs61a
hello world
Higher Order Functions
Q4: I Heard You Liked Functions...
Define a function cycle
that takes in three functions f1
, f2
,
f3
, as arguments. cycle
will return another function that should
take in an integer argument n
and return another function. That
final function should take in an argument x
and cycle through
applying f1
, f2
, and f3
to x
, depending on what n
was. Here's what the final function should do to x
for a few
values of n
:
n = 0
, returnx
n = 1
, applyf1
tox
, or returnf1(x)
n = 2
, applyf1
tox
and thenf2
to the result of that, or returnf2(f1(x))
n = 3
, applyf1
tox
,f2
to the result of applyingf1
, and thenf3
to the result of applyingf2
, orf3(f2(f1(x)))
n = 4
, start the cycle again applyingf1
, thenf2
, thenf3
, thenf1
again, orf1(f3(f2(f1(x))))
- And so forth.
Hint: most of the work goes inside the most nested function.
def cycle(f1, f2, f3):
"""Returns a function that is itself a higher-order function.
>>> def add1(x):
... return x + 1
>>> def times2(x):
... return x * 2
>>> def add3(x):
... return x + 3
>>> my_cycle = cycle(add1, times2, add3)
>>> identity = my_cycle(0)
>>> identity(5)
5
>>> add_one_then_double = my_cycle(2)
>>> add_one_then_double(1)
4
>>> do_all_functions = my_cycle(3)
>>> do_all_functions(2)
9
>>> do_more_than_a_cycle = my_cycle(4)
>>> do_more_than_a_cycle(2)
10
>>> do_two_cycles = my_cycle(6)
>>> do_two_cycles(1)
19
"""
"*** YOUR CODE HERE ***"
def ret_fn(n):
def ret(x):
i = 0
while i < n:
if i % 3 == 0:
x = f1(x)
elif i % 3 == 1:
x = f2(x)
else:
x = f3(x)
i += 1
return x
return ret
return ret_fn
Use Ok to test your code:
python3 ok -q cycle
Lambda expressions
Q5: Palindrome
A number is considered a palindrome if it reads the same forwards and backwards. Fill in the blanks '_' to help determine if a number is a palindrome. In the spirit of exam style questions, please do not edit any parts of the function other than the blanks.
def is_palindrome(n):
"""
Fill in the blanks '_____' to check if a number
is a palindrome.
>>> is_palindrome(12321)
True
>>> is_palindrome(42)
False
>>> is_palindrome(2015)
False
>>> is_palindrome(55)
True
"""
x, y = n, 0
f = lambda: _____
f = lambda: y * 10 + x % 10 while x > 0:
x, y = _____, f()
x, y = x // 10, f() return y == n
Use Ok to test your code:
python3 ok -q is_palindrome
Environment diagrams
Q6: Doge
Draw the environment diagram for the following code.
wow = 6
def much(wow):
if much == wow:
such = lambda wow: 5
def wow():
return such
return wow
such = lambda wow: 4
return wow()
wow = much(much(much))(wow)
Verify your solution with Python Tutor.
More Recursion Practice
Q7: Find the Bug
Find the bug with this recursive function.
def skip_mul(n):
"""Return the product of n * (n - 2) * (n - 4) * ...
>>> skip_mul(5) # 5 * 3 * 1
15
>>> skip_mul(8) # 8 * 6 * 4 * 2
384
"""
if n == 2:
return 2
else:
return n * skip_mul(n - 2)
When you find it, run this to check your answer:
python3 ok -q skip_mul_ok -u
Then, fix the code in lab03_extra.py
and run this to check your code:
python3 ok -q skip_mul
Consider what happens when we choose an odd number for n
. skip_mul(3)
will
return 3 * skip_mul(1)
. skip_mul(1)
will return 1 * skip_mul(-1)
. You
may see the problem now. Since we are decreasing n
by two at a time, we've
completed missed our base case of n == 2
, and we will end up recursing
indefinitely. We need to add another base case to make sure this doesn't
happen.
def skip_mul(n):
if n == 1:
return 1
elif n == 2:
return 2
else:
return n * skip_mul(n - 2)
Q8: Is Prime
Write a function is_prime
that takes a single argument n
and returns True
if n
is a prime number and False
otherwise. Assume n > 1
. We implemented
this in Discussion 1 iteratively, now time to do it recursively!
Hint: You will need a helper function! Remember helper functions are useful if you need to keep track of more variables than the given parameters, or if you need to change the value of the input.
def is_prime(n):
"""Returns True if n is a prime number and False otherwise.
>>> is_prime(2)
True
>>> is_prime(16)
False
>>> is_prime(521)
True
"""
"*** YOUR CODE HERE ***"
def helper(i):
if i > (n ** 0.5): # Could replace with i == n
return True
elif n % i == 0:
return False
return helper(i + 1)
return helper(2)
Use Ok to test your code:
python3 ok -q is_prime
Q9: Interleaved Sum
Recall that the summation
function computes the sum of a sequence of
terms from 1 to n
:
def summation(n, term):
"""Return the sum of the first n terms of a sequence.
>>> summation(5, lambda x: pow(x, 3))
225
"""
total, k = 0, 1
while k <= n:
total, k = total + term(k), k + 1
return total
Write a function interleaved_sum
that similarly computes the sum of a
sequence of terms from 1 to n
, but uses different functions to compute
the terms for odd and even numbers. Do so without using any loops or
testing in any way if a number is odd or even. (You may test if a
number is equal to 0, 1, or n
.)
Hint: Use recursion and a helper function!
def interleaved_sum(n, odd_term, even_term):
"""Compute the sum odd_term(1) + even_term(2) + odd_term(3) + ..., up
to n.
>>> # 1 + 2^2 + 3 + 4^2 + 5
... interleaved_sum(5, lambda x: x, lambda x: x*x)
29
"""
"*** YOUR CODE HERE ***"
def interleaved_helper(term0, term1, k):
if k == n:
return term0(k)
return term0(k) + interleaved_helper(term1, term0, k+1)
return interleaved_helper(odd_term, even_term, 1)
# Alternate solution
def interleaved_sum2(n, odd_term, even_term):
"""Compute the sum odd_term(1) + even_term(2) + odd_term(3) + ..., up
to n.
>>> # 1 + 2^2 + 3 + 4^2 + 5
... interleaved_sum2(5, lambda x: x, lambda x: x*x)
29
"""
total, term0, term1 = interleaved_helper2(n, odd_term, even_term)
return total
def interleaved_helper2(n, odd_term, even_term):
if n == 1:
return odd_term(1), even_term, odd_term
else:
total, term0, term1 = interleaved_helper2(n-1, odd_term, even_term)
return total + term0(n), term1, term0
# Alternate solution using odd_term and even_term separately
def interleaved_sum3(n, odd_term, even_term):
def interleaved_helper3(i, fn):
if i > n:
return 0
return fn(i) + interleaved_helper3(i + 2, fn)
return interleaved_helper3(0, even_term) + interleaved_helper3(1, odd_term)
Use Ok to test your code:
python3 ok -q interleaved_sum
Q10: Ten-pairs
Write a function that takes a positive integer n
and returns the
number of ten-pairs it contains. A ten-pair is a pairs of digits
within n
that sum to 10. Do not use any assignment statements.
The number 7,823,952 has 3 ten-pairs. The first and fourth digits sum to 7+3=10, the second and third digits sum to 8+2=10, and the second and last digit sum to 8+2=10:
Hint: Use a helper function to calculate how many times a digit appears in n.
def ten_pairs(n):
"""Return the number of ten-pairs within positive integer n.
>>> ten_pairs(7823952)
3
>>> ten_pairs(55055)
6
>>> ten_pairs(9641469)
6
"""
"*** YOUR CODE HERE ***"
if n < 10:
return 0
else:
return ten_pairs(n//10) + count_digit(n//10, 10 - n%10)
def count_digit(n, digit):
"""Return how many times digit appears in n.
>>> count_digit(55055, 5)
4
"""
if n == 0:
return 0
else:
if n%10 == digit:
return count_digit(n//10, digit) + 1
else:
return count_digit(n//10, digit)
Use Ok to test your code:
python3 ok -q ten_pairs