Due by 11:59pm on Wednesday, 7/5

## Instructions

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:

## Recursion and Tree Recursion

### 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
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'g', ['While', 'For'])
True
"""

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
>>> check(HW_SOURCE_FILE, 'g_iter', ['Recursion'])
True
"""
``````

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
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'pingpong', ['Assign', 'AugAssign'])
True
"""
``````

You may use the function `has_seven`, which returns True if a number `k` contains the digit 7 at least once.

Use OK to test your code:

``python3 ok -q pingpong``

### 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
"""
``````

Hint: you may find it helpful to refer to the implementation of `count_partitions`.

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"
``````

Use OK to test your code:

``python3 ok -q move_stack``

### Question 5: Flatten

Write a function `flatten` that takes a (possibly deep) list and "flattens" it. For example:

``````>>> lst = [1, [[2], 3], 4, [5, 6]]
>>> flatten(lst)
[1, 2, 3, 4, 5, 6]``````

Hint: you can check if something is a list by using the built-in `type` function. For example,

``````>>> type(3) == list
False
>>> type([1, 2, 3]) == list
True``````
``````def flatten(lst):
"""Returns a flattened version of lst.

>>> flatten([1, 2, 3])     # normal list
[1, 2, 3]
>>> x = [1, [2, 3], 4]      # deep list
>>> flatten(x)
[1, 2, 3, 4]
>>> x = [[1, [1, 1]], 1, [1, 1]] # deep list
>>> flatten(x)
[1, 1, 1, 1, 1, 1]
"""
``````

Use OK to test your code:

``python3 ok -q flatten``

### Question 6: Merge

Write a function `merge` that takes 2 sorted lists `lst1` and `lst2`, and returns a new list that contains all the elements in the two lists in sorted order.

``````def merge(lst1, lst2):
"""Merges two sorted lists.

>>> merge([1, 3, 5], [2, 4, 6])
[1, 2, 3, 4, 5, 6]
>>> merge([], [2, 4, 6])
[2, 4, 6]
>>> merge([1, 2, 3], [])
[1, 2, 3]
>>> merge([5, 7], [2, 4, 6])
[2, 4, 5, 6, 7]
"""
``````

Use OK to test your code:

``python3 ok -q merge``

### Question 7: Mergesort

Mergesort is a type of sorting algorithm. It follows a naturally recursive procedure:

• Break the input list into equally-sized halves
• Recursively sort both halves
• Merge the sorted halves.

Using your `merge` function from the previous question, implement `mergesort`.

Challenge: Implement mergesort itself iteratively, without using recursion.

``````def mergesort(seq):
"""Mergesort algorithm.

>>> mergesort([4, 2, 5, 2, 1])
[1, 2, 2, 4, 5]
>>> mergesort([])     # sorting an empty list
[]
>>> mergesort([1])   # sorting a one-element list
[1]
"""
``python3 ok -q mergesort``