Solution Files

You can find the solutions in hw01.py.

Required Questions

Q1: Syllabus Quiz

Q2: A Plus Abs B

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.

from operator import add, sub

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 check that you didn't change the return statement!
    >>> import inspect, re
    >>> re.findall(r'^\s*(return .*)', inspect.getsource(a_plus_abs_b), re.M)
    ['return f(a, b)']
    """
    if b < 0:
        f = sub
    else:
        f = add
    return f(a, b)

Use Ok to test your code:

python3 ok -q a_plus_abs_b

Q3: 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(x, y, z):
    """Return a*a + b*b, where a and b are the two smallest members of the
    positive numbers x, y, and z.

    >>> 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
    >>> # check that your code consists of nothing but an expression (this docstring)
    >>> # a return statement
    >>> import inspect, ast
    >>> [type(x).__name__ for x in ast.parse(inspect.getsource(two_of_three)).body[0].body]
    ['Expr', 'Return']
    """
    return min(x*x+y*y, x*x+z*z, y*y+z*z)
    # Alternate solution
def two_of_three_alternate(x, y, z):
    return x**2 + y**2 + z**2 - max(x, y, z)**2

Hint: Consider using the max or min function:

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

Use Ok to test your code:

python3 ok -q two_of_three

Q4: 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
    """
    factor = n - 1
    while factor > 0:
        if n % factor == 0:
            return factor
        factor -= 1

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

Use Ok to test your code:

python3 ok -q largest_factor

Q5: If Function vs Statement

Let's try to write a function that does the same thing as an if statement.

def if_function(condition, true_result, false_result):
    """Return true_result if condition is a true value, and
    false_result otherwise.

    >>> if_function(True, 2, 3)
    2
    >>> if_function(False, 2, 3)
    3
    >>> if_function(3==2, 3+2, 3-2)
    1
    >>> if_function(3>2, 3+2, 3-2)
    5
    """
    if condition:
        return true_result
    else:
        return false_result

Despite the doctests above, this function actually does not do the same thing as an if statement in all cases. To prove this fact, write functions cond, true_func, and false_func such that with_if_statement prints the number 47, but with_if_function prints both 42 and 47.

def with_if_statement():
    """
    >>> result = with_if_statement()
    47
    >>> print(result)
    None
    """
    if cond():
        return true_func()
    else:
        return false_func()

def with_if_function():
    """
    >>> result = with_if_function()
    42
    47
    >>> print(result)
    None
    """
    return if_function(cond(), true_func(), false_func())

def cond():
    return False
def true_func():
    print(42)
def false_func():
    print(47)

Hint: If you are having a hard time identifying how an if statement and if_function differ, consider the rules of evaluation for if statements and call expressions.

Use Ok to test your code:

python3 ok -q with_if_statement
python3 ok -q with_if_function

Q6: 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.

Breaking News (or at least the closest thing to that in math). There was a recent development in the hailstone conjecture last year that shows that almost all numbers will eventually get to 1 if you repeat this process. This isn't a complete proof but a major breakthrough.

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
    """
    length = 1
    while n != 1:
        print(n)
        if n % 2 == 0:
            n = n // 2      # Integer division prevents "1.0" output
        else:
            n = 3 * n + 1
        length = length + 1
    print(n)                # n is now 1
    return length

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

Use Ok to test your code:

python3 ok -q hailstone

Submit

Make sure to submit this assignment by running:

python3 ok --submit