Homework 5 Solutions

Solution Files

You can find the solutions in hw05.py.

Required Questions

Q1: Max Product

Implement max_product, which takes a list of numbers and returns the maximum product that can be formed by multiplying together non-consecutive elements of the list. Assume that all numbers in the input list are greater than or equal to 1.

First try multiplying the first element by the max_product of everything after the first two elements (skipping the second element because it is consecutive with the first), then try skipping the first element and finding the max_product of the rest. To find which of these options is better, use max. 
def max_product(s):
    """Return the maximum product of non-consecutive elements of s.

    >>> max_product([10, 3, 1, 9, 2])   # 10 * 9
    90
    >>> max_product([5, 10, 5, 10, 5])  # 5 * 5 * 5
    125
    >>> max_product([])                 # The product of no numbers is 1
    1
    """

    if s == []:
        return 1
    if len(s) == 1:
        return s[0]
    else:
        return max(s[0] * max_product(s[2:]), max_product(s[1:]))
        # OR
        return max(s[0] * max_product(s[2:]), s[1] * max_product(s[3:]))

Use Ok to test your code:

python3 ok -q max_product

This solution begins with the idea that we either include s[0] in the product or not:
  • If we include s[0], we cannot include s[1].
  • If we don't include s[0], we can include s[1].

The recursive case is that we choose the larger of:

  • multiplying s[0] by the max_product of s[2:] (skipping s[1]) OR
  • just the max_product of s[1:] (skipping s[0])

Here are some key ideas in translating this into code:

  • The built-in max function can find the larger of two numbers, which in this case come from two recursive calls.
  • In every case, max_product is called on a list of numbers and its return value is treated as a number.

An expression for this recursive case is:

max(s[0] * max_product(s[2:]), max_product(s[1:]))

Since this expression never refers to s[1], and s[2:] evaluates to the empty list even for a one-element list s, the second base case (len(s) == 1) can be omitted if this recursive case is used.

The recursive solution above explores some options that we know in advance will not be the maximum, such as skipping both s[0] and s[1]. Alternatively, the recursive case could be that we choose the larger of:

  • multiplying s[0] by the max_product of s[2:] (skipping s[1]) OR
  • multiplying s[1] by the max_product of s[3:] (skipping s[0] and s[2])

An expression for this recursive case is:

max(s[0] * max_product(s[2:]), s[1] * max_product(s[3:]))

Q2: Super Mario

Mario needs to jump over a sequence of Piranha plants called level. level is represented as a string of empty spaces (' '), indicating no plant, and P's ('P'), indicating the presence of a plant. Mario only moves forward and can either step (move forward one spot) or jump (move forward two spots) from each position. How many different ways can Mario traverse a level without stepping or jumping into a Piranha plant ('P')? Assume that every level begins with an empty space (' '), where Mario starts, and ends with an empty space (' '), where Mario must end up.

Hint: You can access the ith character in a string s by using s[i]. For example:

>>> s = 'abcdefg'
>>> s[0]
'a'
>>> s[2]
'c'

Hint: You can find the total number of characters in a string using the built-in len function:

>>> s = 'abcdefg'
>>> len(s)
7
>>> len('')
0
def mario_number(level):
    """Return the number of ways that Mario can perform a sequence of steps
    or jumps to reach the end of the level without ever landing in a Piranha
    plant. Assume that every level begins and ends with a space.

    >>> mario_number(' P P ')   # jump, jump
    1
    >>> mario_number(' P P  ')   # jump, jump, step
    1
    >>> mario_number('  P P ')  # step, jump, jump
    1
    >>> mario_number('   P P ') # step, step, jump, jump or jump, jump, jump
    2
    >>> mario_number(' P PP ')  # Mario cannot jump two plants
    0
    >>> mario_number('    ')    # step, jump ; jump, step ; step, step, step
    3
    >>> mario_number('    P    ')
    9
    >>> mario_number('   P    P P   P  P P    P     P ')
    180
    """

    def ways(n):
        if n == len(level)-1:
            return 1
        if n >= len(level) or level[n] == 'P':
            return 0
        return ways(n+1) + ways(n+2)
    return ways(0)

Use Ok to test your code:

python3 ok -q mario_number

Check Your Score Locally

You can locally check your score on each question of this assignment by running

python3 ok --score

This does NOT submit the assignment! When you are satisfied with your score, submit the assignment to Gradescope to receive credit for it.

Submit Assignment

Submit this assignment by uploading any files you've edited to the appropriate Gradescope assignment. Lab 00 has detailed instructions.