# Homework 2: Higher-Order Functions

*Due by 11:59pm on Wednesday, July 3*

## 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 the assignment by uploading all code files you've edited to Gradescope. 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 Gradescope. 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
correctness. Each incorrect problem will decrease the total score by one point.
**This homework is out of 2 points.**

# Required Questions

## Getting Started Videos

These videos may provide some helpful direction for tackling the coding problems on this assignment.

To see these videos, you should be logged into your berkeley.edu email.

Several doctests refer to these functions:

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

## Higher-Order Functions

### Q1: Product

Write a function called `product`

that returns the product of the first `n`

terms of a sequence.
Specifically, `product`

takes in an integer `n`

and `term`

, a single-argument function that determines a sequence.
(That is, `term(i)`

gives the `i`

th term of the sequence.)
`product(n, term)`

should return `term(1) * ... * term(n)`

.

```
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 to produce the term
>>> 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
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q product`

### Q2: Accumulate

Let's take a look at how `product`

is an instance of a more
general function called `accumulate`

, which we would like to implement:

```
def accumulate(fuse, start, n, term):
"""Return the result of fusing together the first n terms in a sequence
and start. The terms to be fused are term(1), term(2), ..., term(n).
The function fuse 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 (fuse is never used)
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
>>> # 2 + (1^2 + 1) + (2^2 + 1) + (3^2 + 1)
>>> accumulate(lambda x, y: x + y + 1, 2, 3, square)
19
"""
"*** YOUR CODE HERE ***"
```

`accumulate`

has the following parameters:

`fuse`

: a two-argument function that specifies how the current term is fused with the previously accumulated terms`start`

: value at which to start the accumulation`n`

: a non-negative integer indicating the number of terms to fuse`term`

: a single-argument function;`term(i)`

is the`i`

th term of the sequence

Implement `accumulate`

, which fuses the first `n`

terms of the sequence defined
by `term`

with the `start`

value using the `fuse`

function.

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

is

```
add(11, add(square(1), add(square(2), square(3)))) =
11 + square(1) + square(2) + square(3) =
11 + 1 + 4 + 9 = 25
```

Assume that

`fuse`

is commutative,`fuse(a, b) == fuse(b, a)`

, and associative,`fuse(fuse(a, b), c) == fuse(a, fuse(b, c))`

.

Then, implement `summation`

(from lecture) and `product`

as one-line calls to
`accumulate`

.

Important:Both`summation_using_accumulate`

and`product_using_accumulate`

should be implemented with a single line of code starting with`return`

.

```
def summation_using_accumulate(n, term):
"""Returns the sum: term(1) + ... + term(n), using accumulate.
>>> summation_using_accumulate(5, square) # square(0) + square(1) + ... + square(4) + square(5)
55
>>> summation_using_accumulate(5, triple) # triple(0) + triple(1) + ... + triple(4) + triple(5)
45
>>> # This test checks that the body of the function is just a return statement.
>>> import inspect, ast
>>> [type(x).__name__ for x in ast.parse(inspect.getsource(summation_using_accumulate)).body[0].body]
['Expr', 'Return']
"""
return ____
def product_using_accumulate(n, term):
"""Returns the product: term(1) * ... * term(n), using accumulate.
>>> product_using_accumulate(4, square) # square(1) * square(2) * square(3) * square()
576
>>> product_using_accumulate(6, triple) # triple(1) * triple(2) * ... * triple(5) * triple(6)
524880
>>> # This test checks that the body of the function is just a return statement.
>>> import inspect, ast
>>> [type(x).__name__ for x in ast.parse(inspect.getsource(product_using_accumulate)).body[0].body]
['Expr', 'Return']
"""
return ____
```

Use Ok to test your code:

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

### Q3: Make Repeater

Implement the function `make_repeater`

which takes a one-argument function `f`

and a positive integer `n`

. It returns a one-argument function, where
`make_repeater(f, n)(x)`

returns the value of `f(f(...f(x)...))`

in which `f`

is
applied `n`

times to `x`

. For example, `make_repeater(square, 3)(5)`

squares 5
three times and returns 390625, just like `square(square(square(5)))`

.

```
def make_repeater(f, n):
"""Returns 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, 3)(5) # square(square(square(5)))
390625
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q make_repeater`

## Recursion

### Q4: Digit Distance

For a given integer, the *digit distance* is the sum of the absolute differences between
consecutive digits. For example:

- The digit distance of
`6`

is`0`

. - The digit distance of
`61`

is`5`

, as the absolute value of`6 - 1`

is`5`

. - The digit distance of
`71253`

is`12`

(`6 + 1 + 3 + 2`

).

Write a function that determines the digit distance of a given positive integer. You must use recursion or the tests will fail.

Hint:There are multiple valid ways of solving this problem! If you're stuck, try writing out an iterative solution first, and then convert your iterative solution into a recursive one.

```
def digit_distance(n):
"""Determines the digit distance of n.
>>> digit_distance(3)
0
>>> digit_distance(777)
0
>>> digit_distance(314)
5
>>> digit_distance(31415926535)
32
>>> digit_distance(3464660003)
16
>>> from construct_check import check
>>> # ban all loops
>>> check(HW_SOURCE_FILE, 'digit_distance',
... ['For', 'While'])
True
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q digit_distance`

### Q5: Interleaved Sum

Write a function `interleaved_sum`

, which takes in a number `n`

and
two one-argument functions: `odd_func`

and `even_func`

. It applies `odd_func`

to every odd number and `even_func`

to every even number from 1 to `n`

*inclusive*
and returns the sum.

For example, executing `interleaved_sum(5, lambda x: x, lambda x: x * x)`

returns `1 + 2*2 + 3 + 4*4 + 5 = 29`

.

Important:Implement this function without using any loops or directly testing if a number is odd or even -- aka modulos (`%`

) are not allowed! Instead of directly checking whether a number is even or odd, start with 1, which you know is an odd number.

Hint:Introduce an inner helper function that takes an odd number`k`

and computes an interleaved sum from`k`

to`n`

(including`n`

).

```
def interleaved_sum(n, odd_func, even_func):
"""Compute the sum odd_func(1) + even_func(2) + odd_func(3) + ..., up
to n.
>>> identity = lambda x: x
>>> square = lambda x: x * x
>>> triple = lambda x: x * 3
>>> interleaved_sum(5, identity, square) # 1 + 2*2 + 3 + 4*4 + 5
29
>>> interleaved_sum(5, square, identity) # 1*1 + 2 + 3*3 + 4 + 5*5
41
>>> interleaved_sum(4, triple, square) # 1*3 + 2*2 + 3*3 + 4*4
32
>>> interleaved_sum(4, square, triple) # 1*1 + 2*3 + 3*3 + 4*3
28
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'interleaved_sum', ['While', 'For', 'Mod']) # ban loops and %
True
>>> check(HW_SOURCE_FILE, 'interleaved_sum', ['BitAnd', 'BitOr', 'BitXor']) # ban bitwise operators, don't worry about these if you don't know what they are
True
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q interleaved_sum`

### Q6: Count Coins

Given a positive integer `total`

, a set of coins makes change for `total`

if
the sum of the values of the coins is `total`

.
Here we will use standard US Coin values: 1, 5, 10, 25.
For example, the following sets make change for `15`

:

- 15 1-cent coins
- 10 1-cent, 1 5-cent coins
- 5 1-cent, 2 5-cent coins
- 5 1-cent, 1 10-cent coins
- 3 5-cent coins
- 1 5-cent, 1 10-cent coin

Thus, there are 6 ways to make change for `15`

. Write a **recursive** function
`count_coins`

that takes a positive integer `total`

and returns the number of
ways to make change for `total`

using coins.

You can use *either* of the functions given to you:

`next_larger_coin`

will return the next larger coin denomination from the input, i.e.`next_larger_coin(5)`

is`10`

.`next_smaller_coin`

will return the next smaller coin denomination from the input, i.e.`next_smaller_coin(5)`

is`1`

.- Either function will return
`None`

if the next coin value does not exist

There are two main ways in which you can approach this problem.
One way uses `next_larger_coin`

, and another uses `next_smaller_coin`

.
It is up to you which one you want to use!

Important:Use recursion; the tests will fail if you use loops.

Hint:Refer to the implementation of`count_partitions`

for an example of how to count the ways to sum up to a final value with smaller parts. If you need to keep track of more than one value across recursive calls, consider writing a helper function.

```
def next_larger_coin(coin):
"""Returns the next larger coin in order.
>>> next_larger_coin(1)
5
>>> next_larger_coin(5)
10
>>> next_larger_coin(10)
25
>>> next_larger_coin(2) # Other values return None
"""
if coin == 1:
return 5
elif coin == 5:
return 10
elif coin == 10:
return 25
def next_smaller_coin(coin):
"""Returns the next smaller coin in order.
>>> next_smaller_coin(25)
10
>>> next_smaller_coin(10)
5
>>> next_smaller_coin(5)
1
>>> next_smaller_coin(2) # Other values return None
"""
if coin == 25:
return 10
elif coin == 10:
return 5
elif coin == 5:
return 1
def count_coins(total):
"""Return the number of ways to make change using coins of value of 1, 5, 10, 25.
>>> count_coins(15)
6
>>> count_coins(10)
4
>>> count_coins(20)
9
>>> count_coins(100) # How many ways to make change for a dollar?
242
>>> count_coins(200)
1463
>>> from construct_check import check
>>> # ban iteration
>>> check(HW_SOURCE_FILE, 'count_coins', ['While', 'For'])
True
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q count_coins`

## Check Your Score Locally

You can locally check your score on each question of this assignment by running

`python3 ok --score`

**This does NOT submit the assignment!** When you are satisfied with your score, submit the assignment to Gradescope to receive credit for it.

# Submit Assignment

Submit this assignment by uploading any files you've edited **to the appropriate Gradescope assignment.** Lab 00 has detailed instructions.

## Exam Practice

Here are some related questions from past exams for you to try. These are optional. There is no way to submit them.

Note that exams from Spring 2020, Fall 2020, and Spring 2021 gave students access to an interpreter, so the question format may be different than other years. Regardless, the questions below are good problems to try

withoutaccess to an interpreter.

- Fall 2019 MT1 Q3: You Again [Higher-Order Functions]
- Spring 2021 MT1 Q4: Domain on the Range [Higher-Order Functions]
- Fall 2021 MT1 Q1b: tik [Functions and Expressions]