Due by 11:59pm on Tuesday, 9/13


Download hw03.zip. The vitamin problems can be found in the vitamin directory and the homework problems can be found in the problems directory. You must run python3 ok --submit twice: once inside the vitamin directory, and once inside the problems directory.

Submission: When you are done, submit with python3 ok --submit. You may submit more than once before the deadline; only the final submission will be scored. Check that you have successfully submitted your code on okpy.org. See Lab 0 for more instructions on submitting assignments.

Using OK: If you have any questions about using OK, please refer to this guide.

Readings: You might find the following references useful:

Several doctests use the construct_check module, which defines a function check. For example, a call such as

check("foo.py", "func1", ["While", "For", "Recursion"])

checks that the function func1 in file foo.py does not contain any while or for constructs, and is not an overtly recursive function (i.e., one in which a function contains a call to itself by name.)


For this set of problems, you must run ok from within the vitamin directory. While homework questions may be completed with a partner, please remember that vitamin questions must be completed alone.

Question 1: Has Seven

Write a function has_seven that takes a positive integer n and returns whether n contains the digit 7. Do not use any assignment statements - use recursion instead:

def has_seven(k):
    """Returns True if at least one of the digits of k is a 7, False otherwise.

    >>> has_seven(3)
    >>> has_seven(7)
    >>> has_seven(2734)
    >>> has_seven(2634)
    >>> has_seven(734)
    >>> has_seven(7777)
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q has_seven

Question 2: Summation

Write a recursive implementation of summation, which takes a positive integer n and a function term. It applies term to every number from 1 to n including n and returns the sum of the results.

def summation(n, term):

    """Return the sum of the first n terms in the sequence defined by term.
    Implement using recursion!

    >>> summation(5, lambda x: x * x * x) # 1^3 + 2^3 + 3^3 + 4^3 + 5^3
    >>> summation(9, lambda x: x + 1) # 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
    >>> summation(5, lambda x: 2**x) # 2^1 + 2^2 + 2^3 + 2^4 + 2^5
    >>> # Do not use while/for loops!
    >>> from construct_check import check
    >>> check(HW_SOURCE_FILE, 'summation',
    ...       ['While', 'For'])
    assert n >= 1
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q summation

Required questions

For this set of problems, you must run ok from within the problems directory. Remember that you may choose to work with a partner on homework questions.

Several doctests refer to these one-argument functions:

def square(x):
    return x * x

def triple(x):
    return 3 * x

def identity(x):
    return x

def increment(x):
    return x + 1

Question 3: Accumulate

Show that both summation and product are instances of a more general function, called accumulate:

from operator import add, mul

def accumulate(combiner, base, n, term):
    """Return the result of combining the first n terms in a sequence and base.
    The terms to be combined are term(1), term(2), ..., term(n).  combiner is a
    two-argument commutative function.

    >>> accumulate(add, 0, 5, identity)  # 0 + 1 + 2 + 3 + 4 + 5
    >>> accumulate(add, 11, 5, identity) # 11 + 1 + 2 + 3 + 4 + 5
    >>> accumulate(add, 11, 0, identity) # 11
    >>> accumulate(add, 11, 3, square)   # 11 + 1^2 + 2^2 + 3^2
    >>> accumulate(mul, 2, 3, square)   # 2 * 1^2 * 2^2 * 3^2
    "*** YOUR CODE HERE ***"

accumulate(combiner, base, n, term) takes the following arguments:

  • term and n: the same arguments as in summation and product
  • combiner: a two-argument function that specifies how the current term combined with the previously accumulated terms. You may assume that combiner is commutative, i.e., combiner(a, b) = combiner(b, a).
  • base: value that specifies what value to use to start the accumulation.

For example, accumulate(add, 11, 3, square) is

11 + square(1) + square(2) + square(3)

Implement accumulate and show how summation and product can both be defined as simple calls to accumulate:

def summation_using_accumulate(n, term):
    """Returns the sum of term(1) + ... + term(n). The implementation
    uses accumulate.

    >>> summation_using_accumulate(5, square)
    >>> summation_using_accumulate(5, triple)
    >>> from construct_check import check
    >>> check(HW_SOURCE_FILE, 'summation_using_accumulate',
    ...       ['Recursion', 'For', 'While'])
    "*** YOUR CODE HERE ***"
    return _______

def product_using_accumulate(n, term):
    """An implementation of product using accumulate.

    >>> product_using_accumulate(4, square)
    >>> product_using_accumulate(6, triple)
    >>> from construct_check import check
    >>> check(HW_SOURCE_FILE, 'product_using_accumulate',
    ...       ['Recursion', 'For', 'While'])
    "*** YOUR CODE HERE ***"
    return _______

Use OK to test your code:

python3 ok -q accumulate
python3 ok -q summation_using_accumulate
python3 ok -q product_using_accumulate

Question 4: Filtered Accumulate

Show how to extend the accumulate function to allow for filtering the results produced by its term argument, by implementing the filtered_accumulate function in terms of accumulate:

def filtered_accumulate(combiner, base, pred, n, term):
    """Return the result of combining the terms in a sequence of N terms
    that satisfy the predicate PRED.  COMBINER is a two-argument function.
    If v1, v2, ..., vk are the values in TERM(1), TERM(2), ..., TERM(N)
    that satisfy PRED, then the result is
    (treating COMBINER as if it were a binary operator, like +). The
    implementation uses accumulate.

    >>> filtered_accumulate(add, 0, lambda x: True, 5, identity)  # 0 + 1 + 2 + 3 + 4 + 5
    >>> filtered_accumulate(add, 11, lambda x: False, 5, identity) # 11
    >>> filtered_accumulate(add, 0, odd, 5, identity)   # 0 + 1 + 3 + 5
    >>> filtered_accumulate(mul, 1, greater_than_5, 5, square)  # 1 * 9 * 16 * 25
    >>> # Do not use while/for loops or recursion
    >>> from construct_check import check
    >>> check(HW_SOURCE_FILE, 'filtered_accumulate',
    ...       ['While', 'For', 'Recursion'])
    def combine_if(x, y):
        "*** YOUR CODE HERE ***"
    return accumulate(combine_if, base, n, term)

def odd(x):
    return x % 2 == 1

def greater_than_5(x):
    return x > 5

filtered_accumulate(combiner, base, pred, n, term) takes the following arguments:

  • combiner, base, term and n: the same arguments as accumulate.
  • pred: a one-argument predicate function applied to the values of term(k), k from 1 to n. Only values for which pred returns a true value are combined to form the result. If no values satisfy pred, then base is returned.

For example, filtered_accumulate(add, 0, is_prime, 11, identity) would be

0 + 2 + 3 + 5 + 7 + 11

for a suitable definition of is_prime.

Implement filtered_accumulate by defining the combine_if function. Exactly what this function does is something for you to discover. Do not write any loops or recursive calls to filtered_accumulate.

Use OK to test your code:

python3 ok -q filtered_accumulate

Question 5: Repeated

Previously, you implemented the function repeated(f, n, x), where:

  • f was a one-argument function
  • n was a non-negative integer
  • x was an argument for f

repeated(f, n, x) returned the result of composing f n times on x, i.e., f(f(...f(x)...)). Now let's write a higher-order version of this function, repeated(f, n).

The new repeated, instead of returning the result directly, returns function that, when given the argument x, will compute f(f(...f(x)...)). For example, repeated(square, 3)(42) evaluates to square(square(square(42))). Yes, it makes sense to apply the function zero times! See if you can figure out a reasonable function to return for that case.

def repeated(f, n):
    """Return the function that computes the nth application of f.

    >>> add_three = repeated(increment, 3)
    >>> add_three(5)
    >>> repeated(triple, 5)(1) # 3 * 3 * 3 * 3 * 3 * 1
    >>> repeated(square, 2)(5) # square(square(5))
    >>> repeated(square, 4)(5) # square(square(square(square(5))))
    >>> repeated(square, 0)(5)
    "*** YOUR CODE HERE ***"

For an extra challenge, try defining repeated using compose1 and accumulate in a single one-line return statement.

Use OK to test your code:

python3 ok -q repeated

Extra questions

Extra questions are not worth extra credit and are entirely optional. They are designed to challenge you to think creatively!

Question 6: Quine

Write a one-line program that prints itself, using only the following features of the Python language:

  • Number literals
  • Assignment statements
  • String literals that can be expressed using single or double quotes
  • The arithmetic operators +, -, *, and /
  • The built-in print function
  • The built-in eval function, which evaluates a string as a Python expression
  • The built-in repr function, which returns an expression that evaluates to its argument

You can concatenate two strings by adding them together with + and repeat a string by multipying it by an integer. Semicolons can be used to separate multiple statements on the same line. E.g.,

>>> c='c';print('a');print('b' + c * 2)

Hint: Explore the relationship between single quotes, double quotes, and the repr function applied to strings.

A program that prints itself is called a Quine. Place your solution in the multi-line string named quine.

Note: No tests will be run on your solution to this problem.

Question 7: Church numerals

The logician Alonzo Church invented a system of representing non-negative integers entirely using functions. The purpose was to show that functions are sufficient to describe all of number theory: if we have functions, we do not need to assume that numbers exist, but instead we can invent them.

Your goal in this problem is to rediscover this representation known as Church numerals. Here are the definitions of zero, as well as a function that returns one more than its argument:

def zero(f):
    return lambda x: x

def successor(n):
    return lambda f: lambda x: f(n(f)(x))

First, define functions one and two such that they have the same behavior as successor(zero) and successsor(successor(zero)) respectively, but do not call successor in your implementation.

Next, implement a function church_to_int that converts a church numeral argument to a regular Python integer.

Finally, implement functions add_church, mul_church, and pow_church that perform addition, multiplication, and exponentiation on church numerals.

def one(f):
    """Church numeral 1: same as successor(zero)"""
    "*** YOUR CODE HERE ***"

def two(f):
    """Church numeral 2: same as successor(successor(zero))"""
    "*** YOUR CODE HERE ***"

three = successor(two)

def church_to_int(n):
    """Convert the Church numeral n to a Python integer.

    >>> church_to_int(zero)
    >>> church_to_int(one)
    >>> church_to_int(two)
    >>> church_to_int(three)
    "*** YOUR CODE HERE ***"

def add_church(m, n):
    """Return the Church numeral for m + n, for Church numerals m and n.

    >>> church_to_int(add_church(two, three))
    "*** YOUR CODE HERE ***"

def mul_church(m, n):
    """Return the Church numeral for m * n, for Church numerals m and n.

    >>> four = successor(three)
    >>> church_to_int(mul_church(two, three))
    >>> church_to_int(mul_church(three, four))
    "*** YOUR CODE HERE ***"

def pow_church(m, n):
    """Return the Church numeral m ** n, for Church numerals m and n.

    >>> church_to_int(pow_church(two, three))
    >>> church_to_int(pow_church(three, two))
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q church_to_int
python3 ok -q add_church
python3 ok -q mul_church
python3 ok -q pow_church