# Homework 2: Higher Order Functions

*Due by 11:59pm on Tuesday, 7/9*

## 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: 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
```

### Q2: 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
"""
"*** 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
```

### Q3: 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`

### Q4: Num Sevens

Write a recursive function `num_sevens`

that takes a positive integer `n`

and
returns the number of times the digit 7 appears in `n`

.

*Use recursion - the tests will fail if you use any assignment statements.*

```
def num_sevens(n):
"""Returns the number of times 7 appears as a digit of n.
>>> num_sevens(3)
0
>>> num_sevens(7)
1
>>> num_sevens(7777777)
7
>>> num_sevens(2637)
1
>>> num_sevens(76370)
2
>>> num_sevens(12345)
0
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'num_sevens',
... ['Assign', 'AugAssign'])
True
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q num_sevens`

### Q5: Ping-pong

The ping-pong sequence counts up starting from 1 and is always either counting
up or counting down. At element `k`

, the direction switches if `k`

is a
multiple of 7 or contains the digit 7. The first 30 elements of the ping-pong
sequence are listed below, with direction swaps marked using brackets at the
7th, 14th, 17th, 21st, 27th, and 28th elements:

`1 2 3 4 5 6 [7] 6 5 4 3 2 1 [0] 1 2 [3] 2 1 0 [-1] 0 1 2 3 4 [5] [4] 5 6`

Implement a function `pingpong`

that returns the nth element of the ping-pong
sequence *without using any assignment statements*.

You may use the function `num_sevens`

, which you defined in the previous question.

Hint: If you're stuck, first try implementing`pingpong`

using assignment statements and a`while`

statement. Then, to convert this into a recursive solution, write a helper function that has a parameter for each variable that changes values in the body of the while loop.

```
def pingpong(n):
"""Return the nth element of the ping-pong sequence.
>>> pingpong(7)
7
>>> pingpong(8)
6
>>> pingpong(15)
1
>>> pingpong(21)
-1
>>> pingpong(22)
0
>>> pingpong(30)
6
>>> pingpong(68)
2
>>> pingpong(69)
1
>>> pingpong(70)
0
>>> pingpong(71)
1
>>> pingpong(72)
0
>>> pingpong(100)
2
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'pingpong', ['Assign', 'AugAssign'])
True
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q pingpong`

### Q6: Count change

Once the machines take over, the denomination of every coin will be a power of two: 1-cent, 2-cent, 4-cent, 8-cent, 16-cent, etc. There will be no limit to how much a coin can be worth.

Given a positive integer `amount`

, a set of coins makes change for `amount`

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

. For example, the following
sets make change for `7`

:

- 7 1-cent coins
- 5 1-cent, 1 2-cent coins
- 3 1-cent, 2 2-cent coins
- 3 1-cent, 1 4-cent coins
- 1 1-cent, 3 2-cent coins
- 1 1-cent, 1 2-cent, 1 4-cent coins

Thus, there are 6 ways to make change for `7`

. Write a recursive function
`count_change`

that takes a positive integer `amount`

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

using these coins of the future.

Hint:Refer the implementation of`count_partitions`

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

```
def count_change(amount):
"""Return the number of ways to make change for amount.
>>> count_change(7)
6
>>> count_change(10)
14
>>> count_change(20)
60
>>> count_change(100)
9828
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'count_change', ['While', 'For'])
True
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q count_change`

## 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, but you may attempt them if you would like extra practice.

### Q7: Towers of Hanoi

A classic puzzle called the Towers of Hanoi is a game that consists of three
rods, and a number of disks of different sizes which can slide onto any rod.
The puzzle starts with `n`

disks in a neat stack in ascending order of size on
a `start`

rod, the smallest at the top, forming a conical shape.

The objective of the puzzle is to move the entire stack to an `end`

rod,
obeying the following rules:

- Only one disk may be moved at a time.
- Each move consists of taking the top (smallest) disk from one of the rods and sliding it onto another rod, on top of the other disks that may already be present on that rod.
- No disk may be placed on top of a smaller disk.

Complete the definition of `move_stack`

, which prints out the steps required to
move `n`

disks from the `start`

rod to the `end`

rod without violating the
rules. The provided `print_move`

function will print out the step to move a
single disk from the given `origin`

to the given `destination`

.

Hint:Draw out a few games with various`n`

on a piece of paper and try to find a pattern of disk movements that applies to any`n`

. In your solution, take the recursive leap of faith whenever you need to move any amount of disks less than`n`

from one rod to another. If you need more help, see the following hints.

`print`

effectively "collects" all the results in the terminal as long as you make sure that the moves are printed in order.
```
def print_move(origin, destination):
"""Print instructions to move a disk."""
print("Move the top disk from rod", origin, "to rod", destination)
def move_stack(n, start, end):
"""Print the moves required to move n disks on the start pole to the end
pole without violating the rules of Towers of Hanoi.
n -- number of disks
start -- a pole position, either 1, 2, or 3
end -- a pole position, either 1, 2, or 3
There are exactly three poles, and start and end must be different. Assume
that the start pole has at least n disks of increasing size, and the end
pole is either empty or has a top disk larger than the top n start disks.
>>> move_stack(1, 1, 3)
Move the top disk from rod 1 to rod 3
>>> move_stack(2, 1, 3)
Move the top disk from rod 1 to rod 2
Move the top disk from rod 1 to rod 3
Move the top disk from rod 2 to rod 3
>>> move_stack(3, 1, 3)
Move the top disk from rod 1 to rod 3
Move the top disk from rod 1 to rod 2
Move the top disk from rod 3 to rod 2
Move the top disk from rod 1 to rod 3
Move the top disk from rod 2 to rod 1
Move the top disk from rod 2 to rod 3
Move the top disk from rod 1 to rod 3
"""
assert 1 <= start <= 3 and 1 <= end <= 3 and start != end, "Bad start/end"
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q move_stack`

### Q8: Anonymous factorial

The recursive factorial function can be written as a single expression by using a conditional expression.

```
>>> fact = lambda n: 1 if n == 1 else mul(n, fact(sub(n, 1)))
>>> fact(5)
120
```

However, this implementation relies on the fact (no pun intended) that
`fact`

has a name, to which we refer in the body of `fact`

. To write a
recursive function, we have always given it a name using a `def`

or
assignment statement so that we can refer to the function within its
own body. In this question, your job is to define fact recursively
without giving it a name!

Write an expression that computes `n`

factorial using only call
expressions, conditional expressions, and lambda expressions (no
assignment or def statements). *Note in particular that you are not
allowed to use make_anonymous_factorial in your return expression.*
The

`sub`

and `mul`

functions from the `operator`

module are the only
built-in functions required to solve this problem:```
from operator import sub, mul
def make_anonymous_factorial():
"""Return the value of an expression that computes factorial.
>>> make_anonymous_factorial()(5)
120
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'make_anonymous_factorial', ['Assign', 'AugAssign', 'FunctionDef', 'Recursion'])
True
"""
return 'YOUR_EXPRESSION_HERE'
```

Use Ok to test your code:

`python3 ok -q make_anonymous_factorial`