Homework 1 Solutions

Solution Files

You can find the solutions in hw01.py.


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Required Questions

Q1: A Plus Abs B

Python's operator module defines binary functions for Python's intrinsic arithmetic operators. For example, calling operator.add(2,3) is equivalent to calling the expression 2 + 3; both will return 5.

Note that when the operator module is imported into the namespace, like at the top of hw01.py, you can just call add(2,3) instead of operator.add(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 = sub
else:
f = add
return f(a, b)

Use Ok to test your code:

python3 ok -q a_plus_abs_b

If b is positive, we add the numbers together. If b is negative, we subtract the numbers. Therefore, we choose the operator add or sub based on the sign of 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 min(i*i+j*j, i*i+k*k, j*j+k*k) # Alternate solution def two_of_three_alternate(i, j, k): return i**2 + j**2 + k**2 - max(i, j, k)**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

We use the fact that if x>y and y>0, then square(x)>square(y). So, we can take the min of the sum of squares of all pairs. The min function can take an arbitrary number of arguments.

Alternatively, we can do the sum of squares of all the numbers. Then we pick the largest value, and subtract the square of that.

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
    """
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

Iterating from n-1 to 1, we return the first integer that evenly divides n. This is guaranteed to be the largest factor of n.

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
    """
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?

Note that if n == 1 initially, then the sequence is one step long.
Hint: Recall the different outputs from using regular division / and floor division //

Use Ok to test your code:

python3 ok -q hailstone

Curious about hailstones or hailstone sequences? Take a look at these articles:

  • Check out this article to learn more about how hailstones work!
  • In 2019, there was a major development in understanding how the hailstone conjecture works for most numbers!

We keep track of the current length of the hailstone sequence and the current value of the hailstone sequence. From there, we loop until we hit the end of the sequence, updating the length in each step.

Note: we need to do floor division // to remove decimals.

Submit

Make sure to submit this assignment by running:

python3 ok --submit

If you completed all problems correctly, you should see that your score is 6.0 in terminal. Each homework assignment counts for 2 points, so in this case you will receive the full 2 points for homework. Remember that every incorrect question costs you 1 point, so a 5.0/6.0 on this assignment will translate to a 1.0/2.0 homework grade for this assignment.