Discussion 3: Recursion, Tree Recursion

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The vitamin for Discussion 3 is worth 1 point and located here.


A recursive function is a function that is defined in terms of itself. Consider this recursive factorial function:

def factorial(n):
    if n == 0 or n == 1:
        return 1
        return n * factorial(n - 1)

Although we haven’t finished defining factorial, we are still able to call it since the function body is not evaluated until the function is called. When n is 0 or 1, we just return 1. This is known as the base case, and it prevents the function from infinitely recursing. Now we can compute factorial(2) in terms of factorial(1), and factorial(3) in terms of factorial(2), and factorial(4) – well, you get the idea.

There are three common steps in a recursive definition:

  1. Figure out your base case: The base case is usually the simplest input possible to the function. For example, factorial(0) is 1 by definition. You can also think of a base case as a stopping condition for the recursion. If you can’t figure this out right away, move on to the recursive case and try to figure out the point at which we can’t reduce the problem any further.
  2. Make a recursive call with a simpler argument: Simplify your problem, and assume that a recursive call for this new problem will simply work. This is called the “leap of faith”. For factorial, we reduce the problem by calling factorial(n - 1).
  3. Use your recursive call to solve the full problem: Remember that we are assuming the recursive call works. With the result of the recursive call, how can you solve the original problem you were asked? For factorial, we just multiply (n − 1)! by n.

Another way to understand recursion is by separating out two things: "internal correctness" and not running forever (known as "halting").

A recursive function is internally correct if it is always does the right thing assuming that every recursive call does the right thing.

Consider this alternative recursive factorial:

def factorial(n): # WRONG!
    if n == 2:
        return n
    return n * factorial(n-1)

It is internally correct, since 2! = 2 and n! = n ∗ (n − 1)! are both true statements.

However, that factorial does not halt on all inputs, since factorial(1) results in a call to factorial(0), and then to factorial(-1) and so on.

A recursive function is correct if and only if it is both internally correct and halts for valid inputs; but you can check each property separately. The "recursive leap of faith" is temporarily placing yourself in a mindset where you only check internal correctness.

Q1: Warm Up: Recursive Multiplication

These exercises are meant to help refresh your memory of topics covered in lecture and/or lab this week before tackling more challenging problems.

Write a function that takes two numbers m and n and returns their product. Assume m and n are positive integers. Use recursion, not mul or *!

Hint: 5 * 3 = 5 + (5 * 2) = 5 + 5 + (5 * 1).

For the base case, what is the simplest possible input for multiply?

For the recursive case, what does calling multiply(m - 1, n) do? What does calling multiply(m, n - 1) do? Do we prefer one over the other?

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Q2: Recursion Environment Diagram

Draw an environment diagram for the following code:

def rec(x, y):
    if y > 0:
        return x * rec(x, y - 1)
    return 1

rec(3, 2)
Global frame
f1: [parent=]
Return value
f2: [parent=]
Return value
f3: [parent=]
Return value

Imagine you were writing the documentation for this function. Come up with a line that describes what the function does:

Note: This problem is meant to help you understand what really goes on when we make the "recursive leap of faith". However, when approaching or debugging recursive functions, you should avoid visualizing them in this way for large or complicated inputs, since the large number of frames can bes quite unwieldy and confusing. Instead, think in terms of the three step process - base case, recursive call, solving the full problem.

Q3: Merge Numbers

Write a procedure merge(n1, n2) which takes numbers with digits in decreasing order and returns a single number with all of the digits of the two, in decreasing order. Any number merged with 0 will be that number (treat 0 as having no digits). Use recursion.

Hint: If you can figure out which number has the smallest digit out of both, then we know that the resulting number will have that smallest digit, followed by the merge of the two numbers with the smallest digit removed.

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Q4: Recursive Hailstone

Recall the hailstone function from Homework 1. First, 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.

Hint: When taking the recursive leap of faith, consider both the return value and side effect of this function.

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Q5: 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.

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Tree Recursion

Consider a function that requires more than one recursive call. A simple example is the recursive fibonacci function:

def fib(n):
    if n == 0:
        return 0
    elif n == 1:
        return 1
        return fib(n - 1) + fib(n - 2)

This type of recursion is called tree recursion, because it makes more than one recursive call in its recursive case. If we draw out the recursive calls, we see the recursive calls in the shape of an upside-down tree:

We could, in theory, use loops to write the same procedure. However, problems that are naturally solved using tree recursive procedures are generally difficult to write iteratively. It is sometimes the case that a tree recursive problem also involves iteration: for example, you might use a while loop to add together multiple recursive calls.

As a general rule of thumb, whenever you need to try multiple possibilities at the same time, you should consider using tree recursion.

Q6: Count Stair Ways

Imagine that you want to go up a flight of stairs that has n steps, where n is a positive integer. You can either take 1 or 2 steps each time. In this question, you'll write a function count_stair_ways that solves this problem. Before you code your approach, consider these questions.

How many different ways can you go up this flight of stairs?

What’s the base case for this question? What is the simplest input?

What do count_stair_ways(n - 1) and count_stair_ways(n - 2) represent?

Fill in the code for count_stair_ways:

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Q7: Count K

Consider a special version of the count_stair_ways problem, where instead of taking 1 or 2 steps, we are able to take up to and including k steps at a time. Write a function count_k that figures out the number of paths for this scenario. Assume n and k are positive.

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Exam prep

We recommending reading sections 1.7 from the textbook for these problems. We also recommend watching the June 29 and June 30 lectures on recursion and tree recursion, respectively.

Test your work! For example, for is_palindrome, you can type test(is_palindrome) in the python interpreter you get once you click Run in 61A Code to verify if you pass the doctests or not.

IMPORTANT: you may not use any string operations other than indexing, len and slicing. Specifically, you may not call reversed or index with a negative step size.

Here is a brief survey of string operations useful to this worksheet. Given a string s = 'abcdefg',

  • len(s) evaluates to the length of s, in this case 6
  • s[0] evaluates to the leftmost character of s, in this case 'a'
  • s[-1], or alternatively s[len(s) - 1], evaluates to the rightmost character of s, in this case g
  • s[a:b] evaluates to a slice of s from position a to just before position b, zero-indexed for both. So s[2:5] would give cde
  • s[1:] evaluates to a slice of s from position 1 until the end, in this case bcdefg
  • s[:-1], or alternatively s[:len(s)-1], evaluates to a slice of s from the beginning until one position before the end, in this case abcdef
  • Equality can be used as normal. For example, s == 'abcdefg' evaluates to True, and s == 'egg' evaluates to False.

Q8: 'Tis it?

Difficulty: ⭐

A palindrome is a string that remains identical when reversed. Given a string s, is_palindrome should return whether or not s is a palindrome.

IMPORTANT: Please use the template for this problem; if you have spare time, try to solve the problem using iteration without the template.

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Q9: Greatest Pals

Difficulty: ⭐⭐

A substring of s is a sequence of consecutive letters within s. Given a string s, greatest_pal should return the longest palindromic substring of s. If there are multiple palindromic substrings of greatest length, then return the leftmost one. You may use is_palindrome.

IMPORTANT: For this problem, each starter code template is just a suggestion. We recommend that you use the first, but feel free to modify it, try one of the other two or write your own if you'd like to. Comment out the other versions of the function to run doctests.

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Q10: Wait, It's All Palindromes?

Difficulty: ⭐⭐⭐

Given a string s, return the longest palindromic substring of s. If there are multiple palindromes of greatest length, then return the leftmost one. You may not use is_palindrome.

Hint: Given equivalent values a and b, max(a, b) will evaluate to a. You may also find the key argument to max helpful.

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Just for Fun

This is a challenge problem and not reflective of exam difficulty. We will not be going over this problem in examprep section, but we will be releasing solutions.

Q11: All-Ys Has Been

Difficulty: 😨

Given mystery function Y, complete fib and is_pal so that the given doctests work correctly. When Y is called on fib, it should return a function which takes a positive integer n and returns the nth Fibonacci number.

Similarly, when Y is called on is_pal_maker it should return a function is_pal that takes a string s and returns whether s is a palindrome.

Hint: You may use the ternary operator <a> if <bool-exp> else <b>, which evaluates to <a> if <bool-exp> is truthy and evaluates to <b> if <bool-exp> is false-y.

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