# Homework 2: Higher Order Functions

*Due by 11:59pm on Friday, 2/8*

## Instructions

Download hw02.zip. Inside the archive, you will find a file called
hw02.py, along with a copy of the `ok`

autograder.

**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:

**Grading:** Homework is graded based on effort, not
correctness. However, there is no partial credit; you must show substantial
effort on every problem to receive any points.

The `construct_check`

module is used in this assignment, 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.)

## Required questions

Several doctests refer to these functions:

```
from operator import add, mul, sub
square = lambda x: x * x
identity = lambda x: x
triple = lambda x: 3 * x
increment = lambda x: x + 1
```

### Q1: Make Adder with a Lambda

Implement the `make_adder`

function, which takes in a number `n`

and returns a
function that takes in an another number `k`

and returns `n + k`

. Your
solution must consist of a single `return`

statement.

```
def make_adder(n):
"""Return a function that takes an argument K and returns N + K.
>>> add_three = make_adder(3)
>>> add_three(1) + add_three(2)
9
>>> make_adder(1)(2)
3
"""
return 'YOUR EXPRESSION HERE'
```

Use Ok to test your code:

`python3 ok -q make_adder`

### Q2: Product

The `summation(n, term)`

function from the higher-order functions lecture adds
up `term(1) + ... + term(n)`

. Write a similar function called `product`

that
returns `term(1) * ... * term(n)`

. **Do not use recursion.**

```
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
>>> product(3, increment) # (1+1) * (2+1) * (3+1)
24
>>> product(3, triple) # 1*3 * 2*3 * 3*3
162
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'product', ['Recursion'])
True
"""
"*** YOUR CODE HERE ***"
```

Now, define the factorial function in
terms of `product`

in one line.

```
def factorial(n):
"""Return n factorial for n >= 0 by calling product.
>>> factorial(4) # 4 * 3 * 2 * 1
24
>>> factorial(6) # 6 * 5 * 4 * 3 * 2 * 1
720
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'factorial', ['Recursion', 'For', 'While'])
True
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

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

### Q3: Accumulate

Let's take a look at how `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, associative 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
>>> accumulate(lambda x, y: x + y + 1, 2, 3, square)
19 #(((2 + 1^2 + 1) + 2^2 + 1) + 3^2 + 1)
"""
"*** YOUR CODE HERE ***"
```

`accumulate`

has the following parameters:

`term`

and`n`

: the same parameters as in`summation`

and`product`

`combiner`

: a two-argument function that specifies how the current term is combined with the previously accumulated terms.`base`

: value at which to start the accumulation.

For example, the result of `accumulate(add, 11, 3, square)`

is

`11 + square(1) + square(2) + square(3) = 25`

Note: You may assume that

`combiner`

is associative and commutative. That is,`combiner(a, combiner(b, c)) == combiner(combiner(a, b), c)`

and`combiner(a, b) == combiner(b, a)`

for all`a`

,`b`

, and`c`

. However, you may not assume`combiner`

is chosen from a fixed function set and hard-code the solution.

After implementing `accumulate`

, 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 ***"
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 ***"
```

Use Ok to test your code:

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

## Extra questions

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

### Q4: Make Repeater

Implement a function `make_repeater`

so that `make_repeater(f, n)(x)`

returns
`f(f(...f(x)...))`

, where `f`

is applied `n`

times. That is,
`make_repeater(f, n)`

returns another function that can then be applied to
another argument. For example, `make_repeater(square, 3)(42)`

evaluates to
`square(square(square(42)))`

. See if you can figure out a reasonable function
to return for that case. You may use either loops or recursion in your
implementation.

```
def make_repeater(f, n):
"""Return the function that computes the nth application of f.
>>> add_three = make_repeater(increment, 3)
>>> add_three(5)
8
>>> make_repeater(triple, 5)(1) # 3 * 3 * 3 * 3 * 3 * 1
243
>>> make_repeater(square, 2)(5) # square(square(5))
625
>>> make_repeater(square, 4)(5) # square(square(square(square(5))))
152587890625
>>> make_repeater(square, 0)(5) # Yes, it makes sense to apply the function zero times!
5
"""
"*** YOUR CODE HERE ***"
```

For an extra challenge, try defining

`make_repeater`

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 make_repeater`