# Homework 3

*Due by 11:59pm on Wednesday, 9/23*

## 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. See Lab 1 for 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:

## Required questions

### Question 1: 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)`

:

```
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
"""
"*** 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
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

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

### Question 2: 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. *Do not use any assignment statements; however, you
may use def statements*.

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

first using assignment and a`while`

statement. Any name that changes value will become an argument to a function in the recursive definition.

```
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
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q pingpong`

You may use the function `has_seven`

, which returns True if a number `k`

contains the digit 7 at least once.

```
def has_seven(k):
"""Returns True if at least one of the digits of k is a 7, False otherwise.
>>> has_seven(3)
False
>>> has_seven(7)
True
>>> has_seven(2734)
True
>>> has_seven(2634)
False
>>> has_seven(734)
True
>>> has_seven(7777)
True
"""
if k % 10 == 7:
return True
elif k < 10:
return False
else:
return has_seven(k // 10)
```

### Question 3: 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.

A set of coins makes change for `n`

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

. 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 function
`count_change`

that takes a positive integer `n`

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

using these coins of the future:

```
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
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q count_change`

### Question 4: 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.

```
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`

## Extra questions

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

### Question 5: 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
"""
return 'YOUR_EXPRESSION_HERE'
```

Use OK to test your code:

`python3 ok -q make_anonymous_factorial`