Lab 7: Midterm Review

Due at 11:59pm on Friday, 07/19/2019.

Starter Files

Download Inside the archive, you will find starter files for the questions in this lab, along with a copy of the Ok autograder.


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

  • In order to facilitate midterm studying, solutions to this lab were released with the lab. We encourage you to try out the problems and struggle for a while before looking at the solutions!
  • Note: submitting this lab is entirely optional. It is not worth any credit, but will be helpful for your studying.

Required Questions


Q1: 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, and Nothing 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)
>>> def welcome():
...     print('welcome to')
...     return 'hello'
>>> def cs61a():
...     print('cs61a')
...     return 'world'
>>> print(welcome(), cs61a())
welcome to cs61a hello world

Higher Order Functions


Draw the environment diagram for the following code:

x = 5
def compose1(f, g):
    def h(x):
        return f(g(x))
    return h
d = lambda y: y * x
x = 4
result = compose1(lambda z: z - 1, d)(3)

There are 5 frames total (including the Global frame). In addition, consider the following questions:

  1. In frame f1 (the frame for compose1), the parameter f points to a function object. What is the intrinsic name of that function object, and what frame is its parent?
  2. In frame f2, what name is the frame labeled with (h or λ)? Which frame is the parent of f2?
  3. In frame f3, what name is the frame labeled with (f, g, d, or λ)? Which frame is the parent of f3? In order to compute the return value y * x, in which frame does Python find x? What is that value of x?
  4. In frame f4, what name is the frame labeled with (f, g, d, or λ)? Which frame is the parent of f3?
  5. What value is the variable result bound to in the Global frame?

You can try out the environment diagram at

  1. The intrinsic name of the function object that f points to is λ (specifically, the lambda whose parameter is z). The parent frame of this lambda is Global.
  2. f2 is labeled with the name h; the parent frame of f2 is f1, since that is where h is defined.
  3. f3 is labeled with the name λ (specifically, it is the λ that takes in a parameter y). The parent frame of f3 is Global, since that is where this lambda was defined (the line d = lambda y: y * x).
  4. f4 is labeled with the name λ (specifically, it is the λ that takes a parameter z). The parent frame of f4 is Global, since that is where the lambda is defined.

    You might think that the parent of f4 is f1, since lambda z: z - 1 is being passed into compose1. Remember, however, that operands are evaluated before the operator is applied (that is, before we create the frame f1). Since we are evaluating lambda z: z - 1 in the Global frame, its parent is Global.

  5. The variable result is bound to 11.

Recursion & Tree Recursion

Q3: Insect Combinatorics

Consider an insect in an M by N grid. The insect starts at the bottom left corner, (0, 0), and wants to end up at the top right corner, (M-1, N-1). The insect is only capable of moving right or up. Write a function paths that takes a grid length and width and returns the number of different paths the insect can take from the start to the goal. (There is a closed-form solution to this problem, but try to answer it procedurally using recursion.)


For example, the 2 by 2 grid has a total of two ways for the insect to move from the start to the goal. For the 3 by 3 grid, the insect has 6 diferent paths (only 3 are shown above).

def paths(m, n):
    """Return the number of paths from one corner of an
    M by N grid to the opposite corner.

    >>> paths(2, 2)
    >>> paths(5, 7)
    >>> paths(117, 1)
    >>> paths(1, 157)
"*** YOUR CODE HERE ***"
if m == 1 or n == 1: return 1 return paths(m - 1, n) + paths(m, n - 1)

Use Ok to test your code:

python3 ok -q paths

Q4: Number of Trees

How many different possible full binary tree (each node has two branches or 0, but never 1) structures exist that have exactly n leaves?

For those interested in combinatorics, this problem does have a closed form solution):

def num_trees(n):
    """How many full binary trees have exactly n leaves? E.g.,

    1   2        3       3    ...
    *   *        *       *
       / \      / \     / \
      *   *    *   *   *   *
              / \         / \
             *   *       *   *

    >>> num_trees(1)
    >>> num_trees(2)
    >>> num_trees(3)
    >>> num_trees(8)

"*** YOUR CODE HERE ***"
if n == 1: return 1 return sum(num_trees(k) * num_trees(n-k) for k in range(1, n))

Use Ok to test your code:

python3 ok -q num_trees

Tree Practice

Q5: Pruning Leaves

Define a function prune_leaves that given a tree t and a tuple of values vals, produces a version of t with all its leaves that are in vals removed. Do not attempt to try to remove non-leaf nodes and do not remove leaves that do not match any of the items in vals. Return None if pruning the tree results in there being no nodes left in the tree.

def prune_leaves(t, vals):
    """Return a modified copy of t with all leaves that have a label
    that appears in vals removed.  Return None if the entire tree is
    pruned away.

    >>> t = tree(2)
    >>> print(prune_leaves(t, (1, 2)))
    >>> numbers = tree(1, [tree(2), tree(3, [tree(4), tree(5)]), tree(6, [tree(7)])])
    >>> print_tree(numbers)
    >>> print_tree(prune_leaves(numbers, (3, 4, 6, 7)))
"*** YOUR CODE HERE ***"
if is_leaf(t) and (label(t) in vals): return None new_branches = [] for b in branches(t): new_branch = prune_leaves(b, vals) if new_branch: new_branches += [new_branch] return tree(label(t), new_branches)

Use Ok to test your code:

python3 ok -q prune_leaves

Sequences and Mutability

Q6: WWPD: Mutability?

What would Python display? Try to figure it out before you type it into the interpreter!

Use Ok to test your knowledge with the following "What Would Python Display?" questions:

python3 ok -q mutability -u
>>> lst = [5, 6, 7, 8]
>>> lst.append(6)
>>> lst
[5, 6, 7, 8, 6]
>>> lst.insert(0, 9) >>> lst
[9, 5, 6, 7, 8, 6]
>>> x = lst.pop(2) >>> lst
[9, 5, 7, 8, 6]
>>> lst.remove(x) >>> lst
[9, 5, 7, 8]
>>> a, b = lst, lst[:] >>> a is lst
>>> b == lst
>>> b is lst
>>> pokemon = {'pikachu': 25, 'dragonair': 148, 'mew': 151}
>>> pokemon['pikachu']
>>> len(pokemon)
>>> pokemon['jolteon'] = 135 >>> pokemon['mew'] = 25 >>> len(pokemon)
>>> 'mewtwo' in pokemon
>>> 'pikachu' in pokemon
>>> 25 in pokemon
>>> 151 in pokemon
>>> pokemon['ditto'] = pokemon['jolteon'] >>> pokemon['ditto']

Q7: Dict to List

Fill in the blanks in the code to complete the implementation of the dict_to_lst function, which takes in a dictionary d and returns a list. The resulting list should contain all the (key, value) pairs in the input dictionary as two-element tuples, where the pairs with smaller values come first.

Note: The .items() method returns a collection of (key, value) pairs for a dictionary:

>>> for pair in {1: 2, 3: 4, 5: 6}.items():
...     print(pair)
(1, 2)
(3, 4)
(5, 6)
def dict_to_lst(d):
    """Returns a list containing all the (key, value) pairs in d as two-element
    tuples ordered by increasing value.

    >>> nums = {1: 5, 2: 3, 3: 4}
    >>> dict_to_lst(nums)
    [(2, 3), (3, 4), (1, 5)]
    >>> words = {'first': 'yes', 'second': 'no', 'third': 'perhaps'}
    >>> dict_to_lst(words)
    [('second', 'no'), ('third', 'perhaps'), ('first', 'yes')]
    result = []
    for _ in range(len(d)):
pair = min(d.items(), key=______________________)
pair = min(d.items(), key=lambda item: item[1])
return result
# Alternate solution # This solution is the ideal way to solve this problem, but we haven't learned # how to use the sorted function so don't worry if you don't understand it. def dict_to_list2(d): return sorted(d.items(), key=lambda pair: pair[1])

Use Ok to test your code:

python3 ok -q dict_to_lst