Due by 11:59pm on Friday, 2/26


Download hw03.zip. Inside the archive, you will find a file called hw03.py, along with a copy of the OK autograder.

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. See Lab 0 for 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:

Required questions


The following three problems were optional in lab. If you haven't already done so, complete them (otherwise, copy your result from lab).

Question 1: Flatten

Write a function flatten that takes a (possibly deep) list and "flattens" it. For example:

>>> lst = [1, [[2], 3], 4, [5, 6]]
>>> flatten(lst)
[1, 2, 3, 4, 5, 6]

Hint: you can check if something is a list by using the built-in type function. For example,

>>> type(3) == list
>>> type([1, 2, 3]) == list
def flatten(lst):
    """Returns a flattened version of lst.

    >>> flatten([1, 2, 3])     # normal list
    [1, 2, 3]
    >>> x = [1, [2, 3], 4]      # deep list
    >>> flatten(x)
    [1, 2, 3, 4]
    >>> x = [[1, [1, 1]], 1, [1, 1]] # deep list
    >>> flatten(x)
    [1, 1, 1, 1, 1, 1]
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q flatten

Question 2: Interleave

Write interleave(s0, s1), which takes two linked lists and produces a new linked list with elements of s0 and s1 interleaved. In other words, the resulting list should have the first element of the s0, the first element of s1, the second element of s0, the second element of s1, and so on.

If the two lists are not the same length, then the leftover elements of the longer list should still appear at the end.

def interleave(s0, s1):
    """Interleave linked lists s0 and s1 to produce a new linked

    >>> evens = link(2, link(4, link(6, link(8, empty))))
    >>> odds = link(1, link(3, empty))
    >>> print_link(interleave(odds, evens))
    1 2 3 4 6 8
    >>> print_link(interleave(evens, odds))
    2 1 4 3 6 8
    >>> print_link(interleave(odds, odds))
    1 1 3 3
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q interleave

Question 3: Merge

Write a function merge that takes 2 sorted lists lst1 and lst2, and returns a new list that contains all the elements in the two lists in sorted order.

def merge(lst1, lst2):
    """Merges two sorted lists.

    >>> merge([1, 3, 5], [2, 4, 6])
    [1, 2, 3, 4, 5, 6]
    >>> merge([], [2, 4, 6])
    [2, 4, 6]
    >>> merge([1, 2, 3], [])
    [1, 2, 3]
    >>> merge([5, 7], [2, 4, 6])
    [2, 4, 5, 6, 7]
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q merge


Question 4: Mergesort

Mergesort is a type of sorting algorithm. It follows a naturally recursive procedure:

  • Break the input list into equally-sized halves
  • Recursively sort both halves
  • Merge the sorted halves.

Using your merge function from the previous question, implement mergesort.

Challenge: Implement mergesort itself iteratively, without using recursion.

def mergesort(seq):
    """Mergesort algorithm.

    >>> mergesort([4, 2, 5, 2, 1])
    [1, 2, 2, 4, 5]
    >>> mergesort([])     # sorting an empty list
    >>> mergesort([1])   # sorting a one-element list
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q mergesort


The following problems use the same tree data abstraction as lecture, but for brevity, we've renamed make_tree as tree. The code you write should not apply the [] operation to trees directly; that's an abstraction barrier violation. The print_tree function is provided for convenience:

# An alternative implementation of the tree data abstraction #

def tree(label, children=[]):
    for branch in children:
        assert is_tree(branch), 'children must be trees'
    return (label, children)

def label(tree):
    return tree[0]

def children(tree):
    return tree[1]

def is_tree(tree):
    if type(tree) is not tuple or len(tree) != 2 \
           or (type(tree[1]) is not list and type(tree[1]) is not tuple):
        return False
    for branch in children(tree):
        if not is_tree(branch):
            return False
    return True

def is_leaf(tree):
    return not children(tree)

def print_tree(t, indent=0):
    """Print a representation of this tree in which each node is
    indented by two spaces times its depth from the label.

    >>> print_tree(tree(1))
    >>> print_tree(tree(1, [tree(2)]))
    >>> numbers = tree(1, [tree(2), tree(3, [tree(4), tree(5)]), tree(6, [tree(7)])])
    >>> print_tree(numbers)
    print('  ' * indent + str(label(t)))
    for child in children(t):
        print_tree(child, indent + 1)


Acknowledgements. This mobile example is based on a classic problem from Structure and Interpretation of Computer Programs, Section 2.2.2.

A mobile is a type of hanging sculpture. A binary mobile consists of two sides. Each side is a rod of a certain length, from which hangs either a weight or another mobile.

We will represent a binary mobile using the data abstractions below, which use the tree data abstraction for their representation.

  • A mobile has a left side (index 0) and a right side (index 1).
  • A side has a length and a structure, which is either a mobile or weight.
  • A weight has a size, which is a positive number.

Question 5

Implement the weight data abstraction by completing the weight constructor, the size selector, and the is_weight predicate so that a weight is represented using a tree. The total_weight example is provided to demonstrate use of the mobile, side, and weight abstractions.

def mobile(left, right):
    """Construct a mobile from a left side and a right side."""
    return tree(None, [left, right])

def sides(m):
    """Select the sides of a mobile."""
    return children(m)
def side(length, mobile_or_weight):
    """Construct a side: a length of rod with a mobile or weight at the end."""
    return tree(length, [mobile_or_weight])

def length(s):
    """Select the length of a side."""
    return label(s)

def end(s):
    """Select the mobile or weight hanging at the end of a side."""
    return children(s)[0]
def weight(size):
    """Construct a weight of some size."""
    assert size > 0
    "*** YOUR CODE HERE ***"

def size(w):
    """Select the size of a weight."""
    "*** YOUR CODE HERE ***"

def is_weight(w):
    """Whether w is a weight, not a mobile."""
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q total_weight

Question 6

Implement the balanced function, which returns whether m is a balanced mobile. A mobile is said to be balanced if the torque applied by its left side is equal to that applied by its right side (that is, if the length of the left rod multiplied by the total weight hanging from that rod is equal to the corresponding product for the right side) and if each of the submobiles hanging off its sides is balanced.

def balanced(m):
    """Return whether m is balanced.

    >>> t, u, v = examples()
    >>> balanced(t)
    >>> balanced(v)
    >>> w = mobile(side(3, t), side(2, u))
    >>> balanced(w)
    >>> balanced(mobile(side(1, v), side(1, w)))
    >>> balanced(mobile(side(1, w), side(1, v)))
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q balanced

Simplifying Expressions

Question 7

In lecture, you saw that one use of trees is in representing expressions (such as arithmetic expressions). So, for example, the expression 2 * (3 + x) can be represented as the tree

Expression example

That is, each operand is a child of the operator that applies to it. In lecture, we looked at evaluating an expression that contains only numbers and operators. For this problem, we'll work at simplifying an expression that may contain variables without necessarily evaluating it. For example, 2 * (x + 0) + y * 0 could be simplified to 2 * x.

For this problem, the only operators are *, +, and - (as strings), and the labels of leaves will either be numbers or strings containing variable names. Thus, our first example would be represented with

  tree('*', [tree(2), tree('+', [tree(3), tree('x')])])

To help you, we've defined a few useful things that may come in handy:

# Alternative names of parts of an expression tree.

def left_opnd(expr):
    return children(expr)[0]

def right_opnd(expr):
    return children(expr)[1]

def oper(expr):
    return label(expr)

# Useful constants:

ZERO = tree('0')
ONE = tree('1')

def same_expr(expr0, expr1):
    """Return true iff expression trees EXPR0 and EXPR1 are identical."""
    if oper(expr0) != oper(expr1):
        return False
    elif is_leaf(expr0):
        return True
        return same_expr(left_opnd(expr0), left_opnd(expr1)) and \
               same_expr(right_opnd(expr0), right_opnd(expr1))

def postfix_to_expr(postfix_expr):
    """Return an expression tree equivalent to POSTFIX_EXPR, a string
    in postfix ("reverse Polish") notation.  In postfix, one writes
    E1 OP E2 (where E1 and E2 are expressions and OP is an operator) as
    E1' E2' OP, where E1' and E2' are the postfix versions of E1 and E2. For
    example, '2*(3+x)' is written '2 3 x + *' and '2*3+x' is `2 3 * x +'.
    >>> print_tree(postfix_to_expr("2 3 x + *"))

def expr_to_infix(expr):
    """A string containing a standard infix denotation of the expression
    tree EXPR"""

Implement the function simplify on these trees. Given an expression tree expr, this function returns a new expression tree, simplified from expr by applying the following rules:

  • the expressions E * 0 and 0 * E, where E here can be any expression tree, are replaced by 0.
  • the expressions E * 1 and 1 * E are replaced by E.
  • the expressions E + 0, 0 + E, and E - 0 are replaced by E.
  • the expression E - E (where the two operands are identical trees) is replaced by 0.

These simplifications may cause a cascade, as in y * (x - (0 + x)) which simplifies to y * (x - x), then to y * 0, and then to 0. In order for that to work, you must be careful to simplify operands before simplifying the whole expression.

def simplify(expr):
    """EXPR must be an expression tree involving the operators
    '+', '*', and '-' in inner nodes; numbers and strings (standing for
    variable names) in leaves.  Returns an equivalent, simplified version
    of EXPR.
    >>> def simp(postfix_expr):
    ...     return expr_to_infix(simplify(postfix_to_expr(postfix_expr)))
    >>> simp("x y + 0 *")
    >>> simp("0 x y + *")
    >>> simp("x y + 0 +")
    '(x + y)'
    >>> simp("0 x y + +")
    '(x + y)'
    >>> simp("x y + 1 *")
    '(x + y)'
    >>> simp("1 x y + *")
    '(x + y)'
    >>> simp("x y + x y + -")
    >>> simp("x y y - + x - a b * *")
    >>> simp("x y 3 * -")
    '(x - (y * 3))'
    >>> simp("x y 0 + 3 * -")
    '(x - (y * 3))'
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q simplify

Extra questions

Extra questions are not worth extra credit and are entirely optional. They are designed to challenge you to think creatively!

Question 8

The well-known Eight Queens Problem is to place eight chess queens on an 8x8 chessboard in such a way that none of them attack any of the others. Queens in chess can move (and attack) any number of squares horizontally, vertically, or diagonally. Your problem is to complete the place_queens function to find such a configuration of N queens for a board of any size NxN (if the configuration exists). This function returns a list containing, for each row, the position of the queen in that row (the conditions of the problem guarantee that there must be one queen in each row), or None if no such configuration exists.

def place_queens(size):
    """Return a list. p, of length SIZE in which p[r] is the column in
    which to place a queen in row r (0 <= r < SIZE) such that no two
    queens are attacking each other.  Return None if there is no such
    >>> place_queens(2) == None
    >>> place_queens(3) == None
    >>> check_board(4, place_queens(4))
    >>> check_board(8, place_queens(8))
    >>> check_board(14, place_queens(14))
    "*** YOUR CODE HERE ***"

def check_board(n, cols):
    """Check that COLS is a valid solution to the N-queens problem
    (N == len(COLS)).  COLS has the format returned by place_queens."""
    if cols is None:
        return False
    if n != len(cols):
        return False
    if set(cols) != set(range(n)):
        return False
    if n != len(set([ r + c for r, c in enumerate(cols) ])):
        return False
    if n != len(set([ r - c for r, c in enumerate(cols) ])):
        return False
    return True

def print_board(cols):
    """Print a board, COLS, returned by place_queens (as a list of column
    positions of queens for each row)."""
    if cols is None:
        print("No solution")
        for c in cols:
            print("- " * c + "Q " + "- " * (len(cols) - c - 1))

> print_board(place_queens(5))
Q - - - - 
- - Q - - 
- - - - Q 
- Q - - - 
- - - Q - 

Use OK to test your code:

python3 ok -q place_queens