Discussion 4: Recursion

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Getting Started

Say your name and the year of your life you think you enjoyed most (Kindergarten, senior year of high school, etc.).

Recursion

Many students find this discussion challenging. Everything gets easier with practice. Please help each other learn.

VERY IMPORTANT: In this discussion, don't check your answers until your whole group is sure that the answer is right. Figure things out and check your work by thinking about what your code will do. Your goal should be to have all checks pass the first time you run them! If you need help, ask.

Q1: Swipe

Implement swipe, which prints the digits of argument n, one per line, first backward then forward. The left-most digit is printed only once. Do not use while or for or str. (Use recursion, of course!)

Your Answer
Run in 61A Code
Solution
def swipe(n):
    """Print the digits of n, one per line, first backward then forward.

    >>> swipe(2837)
    7
    3
    8
    2
    8
    3
    7
    """
    if n < 10:
        print(n)
    else:
        print(n % 10)
        swipe(n // 10)
        print(n % 10)
First print the first line of the output, then make a recursive call, then print the last line of the output.

Q2: Skip Factorial

Define the base case for the skip_factorial function, which returns the product of every other positive integer, starting with n.

Your Answer
Run in 61A Code
Solution
def skip_factorial(n):
    """Return the product of positive integers n * (n - 2) * (n - 4) * ...

    >>> skip_factorial(5) # 5 * 3 * 1
    15
    >>> skip_factorial(8) # 8 * 6 * 4 * 2
    384
    """
    if n <= 2:
        return n
    else:
        return n * skip_factorial(n - 2)
If n is even, then the base case will be 2. If n is odd, then the base case will be 1. Try to write a condition that handles both possibilities.

Q3: 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. Complete this recursive version of hailstone that prints out the values of the sequence and returns the number of steps.

Your Answer
Run in 61A Code
Solution
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
    >>> b = hailstone(1)
    1
    >>> b
    1
    """
    print(n)
    if n % 2 == 0:
        return even(n)
    else:
        return odd(n)

def even(n):
    return 1 + hailstone(n // 2)

def odd(n):
    if n == 1:
        return 1
    else:
        return 1 + hailstone(3 * n + 1)
An even number is never a base case, so even always makes a recursive call to hailstone and returns one more than the length of the rest of the hailstone sequence.

An odd number might be 1 (the base case) or greater than one (the recursive case). Only the recursive case should call hailstone.

Document the Occasion

Please all fill out the attendance form (one submission per person per week).

Extra Questions

The questions below are recommended but optional. If you don't get to them, it's a great idea to schedule some time outside of lab to work on them together.

You'll need your whole discussion group for this question. At least try it out. You might have fun.

Q4: Sevens

The Game of Sevens: Players in a circle count up from 1 in the clockwise direction. (The starting player says 1, the player to their left says 2, etc.) If a number is divisible by 7 or contains a 7 (or both), switch directions. Numbers must be said on the beat at 60 beats per minute. If someone says a number when it's not their turn or someone misses the beat on their turn, the game ends.

For example, 5 people would count to 20 like this:

Player 1 says 1
Player 2 says 2
Player 3 says 3
Player 4 says 4
Player 5 says 5
Player 1 says 6  # All the way around the circle
Player 2 says 7  # Switch to counterclockwise
Player 1 says 8
Player 5 says 9  # Back around the circle counterclockwise
Player 4 says 10
Player 3 says 11
Player 2 says 12
Player 1 says 13
Player 5 says 14 # Switch back to clockwise
Player 1 says 15
Player 2 says 16
Player 3 says 17 # Switch back to counterclockwise
Player 2 says 18
Player 1 says 19
Player 5 says 20

Play a few games. Post the highest score your group reached on Discord.

Then, implement sevens which takes a positive integer n and a number of players k. It returns which of the k players says n. You may call has_seven.

An effective approach to this problem is to simulate the game, stopping on turn n. The implementation must keep track of the final number n, the current number i, the player who will say i, and the current direction that determines the next player (either increasing or decreasing). It works well to use integers to represent all of these, with direction switching between 1 (increase) and -1 (decreasing).

Your Answer
Run in 61A Code
Solution
def sevens(n, k):
    """Return the (clockwise) position of who says n among k players.

    >>> sevens(2, 5)
    2
    >>> sevens(6, 5)
    1
    >>> sevens(7, 5)
    2
    >>> sevens(8, 5)
    1
    >>> sevens(9, 5)
    5
    >>> sevens(18, 5)
    2
    """
    def f(i, who, direction):
        if i == n:
            return who
        if i % 7 == 0 or has_seven(i):
            direction = -direction
        who = who + direction
        if who > k:
            who = 1
        if who < 1:
            who = k
        return f(i + 1, who, direction)
    return f(1, 1, 1)

def has_seven(n):
    if n == 0:
        return False
    elif n % 10 == 7:
        return True
    else:
        return has_seven(n // 10)

First check if i is a multiple of 7 or contains a 7, and if so, switch directions. Then, add the direction to who and ensure that who has not become smaller than 1 or greater than k.

Q5: Is Prime

Implement is_prime that takes an integer n greater than 1. It returns True if n is a prime number and False otherwise. Try following the approach below, but implement it recursively without using a while (or for) statement.

def is_prime(n):
    assert n > 1
    i = 2
    while i < n:
        if n % i == 0:
            return False
        i = i + 1
    return True

You will need to define another "helper" function (a function that exists just to help implement this one). Does it matter whether you define it within is_prime or as a separate function in the global frame? Try to define it to take as few arguments as possible.

Your Answer
Run in 61A Code
Solution
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
    """
    def check_all(i):
        "Check whether no number from i up to n evenly divides n."
        if i == n:      # could be replaced with i > (n ** 0.5)
            return True
        elif n % i == 0:
            return False
        return check_all(i + 1)
    return check_all(2)
Define an inner function that checks whether some integer between i and n evenly divides n. Then you can call it starting with i=2:
def is_prime(n):
    def f(i):
        if n % i == 0:
            return ____
        elif ____:
            return ____
        else:
            return f(____)
    return f(2)