Due by 11:59pm on Saturday, 7/1

Instructions

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:

Homework questions

Several doctests refer to these one-argument functions:

``````from operator import add, mul

def square(x):
return x * x

def triple(x):
return 3 * x

def identity(x):
return x

def increment(x):
return x + 1``````

Question 1: Summation

Write a `summation(term, n)` function that returns `term(1) + ... + term(n)`

``````def summation(n, term):
"""Return the summation of the first n terms in a sequence.

n    -- a positive integer
term -- a function that takes one argument

>>> summation(3, identity) # 1 + 2 + 3
6
>>> summation(5, identity) # 1 + 2 + 3 + 4 + 5
15
>>> summation(3, square)   # 1^2 + 2^2 + 3^2
14
>>> summation(5, square)   # 1^2 + 2^2 + 3^2 + 4^2 + 5^2
55
"""
"*** YOUR CODE HERE ***"
``````

Use OK to test your code:

``python3 ok -q summation``

Question 2: Product

The `summation(term, n)` function above adds up `term(1) + ... + term(n)`. Write a similar `product(n, term)` function that returns ```term(1) * ... * term(n)```. Show how to define the factorial function in terms of `product`. Hint: try using the `identity` function for `factorial`.

``````def product(n, term):
"""Return the product of the first n terms in a sequence.

n    -- a positive integer
term -- a function that takes one argument

>>> product(3, identity) # 1 * 2 * 3
6
>>> product(5, identity) # 1 * 2 * 3 * 4 * 5
120
>>> product(3, square)   # 1^2 * 2^2 * 3^2
36
>>> product(5, square)   # 1^2 * 2^2 * 3^2 * 4^2 * 5^2
14400
"""
"*** YOUR CODE HERE ***"

# The identity function, defined using a lambda expression!
identity = lambda k: k

def factorial(n):
"""Return n factorial for n >= 0 by calling product.

>>> factorial(4)
24
>>> factorial(6)
720
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'factorial', ['Recursion', 'For', 'While'])
True
"""
"*** YOUR CODE HERE ***"
return _______
``````

Use OK to test your code:

``````python3 ok -q product
python3 ok -q factorial``````

Question 3: Make Adder with a Lambda

Implement the `make_adder` function below using a single `return` statement that returns the value of a `lambda` expression.

``````def make_adder(n):
"""Return a function that takes an argument K and returns N + K.

9
3
"""
"*** YOUR CODE HERE ***"
# return lambda ________________
``````

Use OK to test your code:

``python3 ok -q make_adder``

Question 4: Accumulate

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

``````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
15
>>> accumulate(add, 11, 5, identity) # 11 + 1 + 2 + 3 + 4 + 5
26
>>> accumulate(add, 11, 0, identity) # 11
11
>>> accumulate(add, 11, 3, square)   # 11 + 1^2 + 2^2 + 3^2
25
>>> accumulate(mul, 2, 3, square)   # 2 * 1^2 * 2^2 * 3^2
72
"""
"*** 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)
55
>>> summation_using_accumulate(5, triple)
45
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'summation_using_accumulate',
...       ['Recursion', 'For', 'While'])
True
"""
"*** YOUR CODE HERE ***"
return _______

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

>>> product_using_accumulate(4, square)
576
>>> product_using_accumulate(6, triple)
524880
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'product_using_accumulate',
...       ['Recursion', 'For', 'While'])
True
"""
"*** 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 5: 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
BASE COMBINER v1 COMBINER v2 ... COMBINER vk
(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
15
>>> filtered_accumulate(add, 11, lambda x: False, 5, identity) # 11
11
>>> filtered_accumulate(add, 0, odd, 5, identity)   # 0 + 1 + 3 + 5
9
>>> filtered_accumulate(mul, 1, greater_than_5, 5, square)  # 1 * 9 * 16 * 25
3600
>>> # Do not use while/for loops or recursion
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'filtered_accumulate',
...       ['While', 'For', 'Recursion'])
True
"""
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 6: Repeated

Implement a function `repeated` so that `repeated(f, n)(x)` returns `f(f(...f(x)...))`, where `f` is applied `n` times. That is, `repeated(f, n)` returns another function that can then be applied to another argument. 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)
8
>>> repeated(triple, 5)(1) # 3 * 3 * 3 * 3 * 3 * 1
243
>>> repeated(square, 2)(5) # square(square(5))
625
>>> repeated(square, 4)(5) # square(square(square(square(5))))
152587890625
>>> repeated(square, 0)(5)
5
"""
"*** YOUR CODE HERE ***"
``````

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

``````def compose1(f, g):
"""Return a function h, such that h(x) = f(g(x))."""
def h(x):
return f(g(x))
return h``````

Use OK to test your code:

``python3 ok -q repeated``