Q1: A Plus Abs B

Python's operator module contains two-argument functions such as add and sub for Python's built-in arithmetic operators. For example, add(2, 3) evalutes to 5, just like the expression 2 + 3.

Fill in the blanks in the following function for adding a to the absolute value of b, without calling abs. You may not modify any of the provided code other than the two blanks.

def a_plus_abs_b(a, b):
    """Return a+abs(b), but without calling abs.

    >>> a_plus_abs_b(2, 3)
    5
    >>> a_plus_abs_b(2, -3)
    5
    >>> a_plus_abs_b(-1, 4)
    3
    >>> a_plus_abs_b(-1, -4)
    3
    """
    if b < 0:
        f = _____
    else:
        f = _____
    return f(a, b)

My Solution

from operator import add, sub

def a_plus_abs_b(a, b):
    if b < 0:
        f = sub
    else:
        f = add
    return f(a, b)

Q2: Two of Three

Write a function that takes three positive numbers as arguments and returns the sum of the squares of the two smallest numbers. Use only a single line for the body of the function.

def two_of_three(i, j, k):
    """Return m*m + n*n, where m and n are the two smallest members of the
    positive numbers i, j, and k.

    >>> two_of_three(1, 2, 3)
    5
    >>> two_of_three(5, 3, 1)
    10
    >>> two_of_three(10, 2, 8)
    68
    >>> two_of_three(5, 5, 5)
    50
    """
    return _____

Hint: Consider using the max or min function:

>>> max(1, 2, 3)
3
>>> min(-1, -2, -3)
-3

My Solution

def two_of_three(i, j, k):
    return i*i + j*j + k*k - max(i,j,k)**2

Q3: Largest Factor

Write a function that takes an integer n that is greater than 1 and returns the largest integer that is smaller than n and evenly divides n.

def largest_factor(n):
    """Return the largest factor of n that is smaller than n.

    >>> largest_factor(15) # factors are 1, 3, 5
    5
    >>> largest_factor(80) # factors are 1, 2, 4, 5, 8, 10, 16, 20, 40
    40
    >>> largest_factor(13) # factor is 1 since 13 is prime
    1
    """
    "*** YOUR CODE HERE ***"

Hint: To check if b evenly divides a, use the expression a % b == 0, which can be read as, "the remainder when dividing a by b is 0."

My Solution 1

Search for factors starting from n // 2 down to 1.

def largest_factor(n):
    factor = n // 2
    while n % factor != 0:
        factor -= 1
    return factor

My Solution 2

Search only up to $\sqrt{n}$. Whenever i divides n, its paired factor n // i is the largest proper factor found.

def largest_factor(n):
    factor = 1
    i = 2
    while i * i <= n:
        if n % i == 0:
            factor = n // i
            break
        i += 1
    return factor

A slightly more efficient version avoids checking even numbers other than 2:

def largest_factor(n):
    if n % 2 == 0:
        return n // 2

    factor = 3
    while factor * factor <= n:
        if n % factor == 0:
            return n // factor
        factor += 2
    return 1

Q4: Hailstone

Douglas Hofstadter's Pulitzer-prize-winning book, Gödel, Escher, Bach, poses the following mathematical puzzle.

1. Pick a positive integer n as the start.
2. If n is even, divide it by 2.
3. If n is odd, multiply it by 3 and add 1.
4. Continue this process until n is 1. The number n will travel up and down but eventually end at 1 (at least for all numbers that have ever been tried -- nobody has ever proved that the sequence will terminate). Analogously, a hailstone travels up and down in the atmosphere before eventually landing on earth.

This sequence of values of n is often called a Hailstone sequence. Write a function that takes a single argument with formal parameter name n, prints out the hailstone sequence starting at n, and returns the number of steps in the sequence:

def hailstone(n):
    """Print the hailstone sequence starting at n and return its
    length.

    >>> a = hailstone(10)
    10
    5
    16
    8
    4
    2
    1
    >>> a
    7
    >>> b = hailstone(1)
    1
    >>> b
    1
    """
    "*** YOUR CODE HERE ***"

Hailstone sequences can get quite long! Try 27. What's the longest you can find?

Note that if n == 1 initially, then the sequence is one step long. Hint: If you see 4.0 but want just 4, try using floor division // instead of regular division /.

My Solution

def hailstone(n):
    length = 1
    print(n)
    while n != 1:
        if n % 2 == 0:
            n = n // 2
        else:
            n = 3 * n + 1
        print(n)
        length += 1
    return length

Contributors: Junyuan He

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