# Homework 3: Higher-Order Functions, Self Reference, Recursion, Tree Recursion

*Due by 11:59pm on Tuesday, July 7*

## Instructions

Download hw03.zip. Inside the archive, you will find a file called
hw03.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
correctness. Each incorrect problem will decrease the total score by one point. There is a homework recovery policy as stated in the syllabus.
**This homework is out of 3 points.**

# Required questions

### Q1: Compose

Write a higher-order function `composer`

that returns two functions, `func`

and `func_adder`

.
`func`

is a one-argument function that applies all of the functions that have been composed so far to it. The functions are applied with the most recent function being applied first (see doctests and examples).
`func_adder`

is used to add more functions to our composition, and when called on another function `g`

, `func_adder`

should return a new `func`

, and a new `func_adder`

.

If no parameters are passed into `composer`

, the `func`

returned is the identity function.

For example:

```
>>> add_one = lambda x: x + 1
>>> square = lambda x: x * x
>>> times_two = lambda x: x + x
>>> f1, func_adder = composer()
>>> f1(1)
1
>>> f2, func_adder = func_adder(add_one)
>>> f2(1)
2 # 1 + 1
>>> f3, func_adder = func_adder(square)
>>> f3(3)
10 # 1 + (3**2)
>>> f4, func_adder = func_adder(times_two)
>>> f4(3)
37 # 1 + ((2 * 3) **2)
```

Hint:Your`func_adder`

should return two arguments`func`

and`func_adder`

. What function do we know that returns`func`

and`func_adder`

?

```
def composer(func=lambda x: x):
"""
Returns two functions -
one holding the composed function so far, and another
that can create further composed problems.
>>> add_one = lambda x: x + 1
>>> mul_two = lambda x: x * 2
>>> f, func_adder = composer()
>>> f1, func_adder = func_adder(add_one)
>>> f1(3)
4
>>> f2, func_adder = func_adder(mul_two)
>>> f2(3) # should be 1 + (2*3) = 7
7
>>> f3, func_adder = func_adder(add_one)
>>> f3(3) # should be 1 + (2 * (3 + 1)) = 9
9
"""
def func_adder(g):
"*** YOUR CODE HERE ***"
return func, func_adder
```

Use Ok to test your code:

`python3 ok -q composer`

### Q2: G function

A mathematical function `G`

on positive integers is defined by two
cases:

```
G(n) = n, if n <= 3
G(n) = G(n - 1) + 2 * G(n - 2) + 3 * G(n - 3), if n > 3
```

Write a recursive function `g`

that computes `G(n)`

. Then, write an
iterative function `g_iter`

that also computes `G(n)`

:

Hint:The`fibonacci`

example in the tree recursion lecture is a good illustration of the relationship between the recursive and iterative definitions of a tree recursive problem.

```
def g(n):
"""Return the value of G(n), computed recursively.
>>> g(1)
1
>>> g(2)
2
>>> g(3)
3
>>> g(4)
10
>>> g(5)
22
>>> from construct_check import check
>>> # ban iteration
>>> check(HW_SOURCE_FILE, 'g', ['While', 'For'])
True
"""
"*** YOUR CODE HERE ***"
def g_iter(n):
"""Return the value of G(n), computed iteratively.
>>> g_iter(1)
1
>>> g_iter(2)
2
>>> g_iter(3)
3
>>> g_iter(4)
10
>>> g_iter(5)
22
>>> from construct_check import check
>>> # ban recursion
>>> check(HW_SOURCE_FILE, 'g_iter', ['Recursion'])
True
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

```
python3 ok -q g
python3 ok -q g_iter
```

### Q3: Missing Digits

Write the recursive function `missing_digits`

that takes a number `n`

that is sorted in increasing order (for example, `12289`

is valid but `15362`

and `98764`

are not). It returns the number of missing digits in `n`

. A missing digit is a number between the first and last digit of `n`

of a that is not in `n`

.
*Use recursion - the tests will fail if you use while or for loops.*

```
def missing_digits(n):
"""Given a number a that is in sorted, increasing order,
return the number of missing digits in n. A missing digit is
a number between the first and last digit of a that is not in n.
>>> missing_digits(1248) # 3, 5, 6, 7
4
>>> missing_digits(1122) # No missing numbers
0
>>> missing_digits(123456) # No missing numbers
0
>>> missing_digits(3558) # 4, 6, 7
3
>>> missing_digits(35578) # 4, 6
2
>>> missing_digits(12456) # 3
1
>>> missing_digits(16789) # 2, 3, 4, 5
4
>>> missing_digits(19) # 2, 3, 4, 5, 6, 7, 8
7
>>> missing_digits(4) # No missing numbers between 4 and 4
0
>>> from construct_check import check
>>> # ban while or for loops
>>> check(HW_SOURCE_FILE, 'missing_digits', ['While', 'For'])
True
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q missing_digits`

### Q4: 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 `total`

, a set of coins makes change for `total`

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

. 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 `total`

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

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

Watch the hints video below for somewhere to start:

Use Ok to test your code:

`python3 ok -q count_change`

## Submit

Make sure to submit this assignment by running:

`python3 ok --submit`

# Just for Fun Questions

The first question below used to be required (but caused students lots of trouble)! You're welcome to try it. The second question demonstrates that it's possible to write recursive functions without assigning them a name in the global frame.

### Q5: 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 ***"
```

Watch the hints video below for somewhere to start:

Use Ok to test your code:

`python3 ok -q move_stack`

### Q6: 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
>>> # ban any assignments or recursion
>>> 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`