# Homework 3: hw03.zip

Due by 11:59pm on Thursday 2/21

## 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 effort, not correctness. However, there is no partial credit; you must show substantial effort on every problem to receive any points.

## Required questions

### Q1: Has Seven

Write a recursive function `has_seven` that takes a positive integer `n` and returns whether `n` contains the digit `7`. Use recursion - the tests will fail if you use any assignment statements.

``````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(374)
True
>>> has_seven(140)
False
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'has_seven',
...       ['Assign', 'AugAssign'])
True
"""
``````

Use Ok to test your code:

``python3 ok -q has_seven``

### Q2: 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 `has_seven` from the previous problem.

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
"""
``````

Use Ok to test your code:

``python3 ok -q pingpong``
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``````

### Q3: Taxicab Distance

An intersection in midtown Manhattan can be identified by an avenue and a street, which are both indexed by positive integers. The Manhattan distance or taxicab distance between two intersections is the number of blocks that must be traversed to reach one from the other, ignoring one-way street restrictions and construction. For example, Times Square is on 46th Street and 7th Avenue. Ess-a-Bagel is on 51st Street and 3rd Avenue. The taxicab distance between them is 9 blocks (5 blocks from 46th to 51st street and 4 blocks from 7th avenue to 3rd avenue). Taxicabs cannot cut diagonally through buildings to reach their destination!

Implement `taxicab`, which computes the taxicab distance between two intersections using the following data abstraction. Hint: You don't need to know what a Cantor pairing function is; just use the abstraction.

``````def intersection(st, ave):
"""Represent an intersection using the Cantor pairing function."""
return (st+ave)*(st+ave+1)//2 + ave

def street(inter):
return w(inter) - avenue(inter)

def avenue(inter):
return inter - (w(inter) ** 2 + w(inter)) // 2

w = lambda z: int(((8*z+1)**0.5-1)/2)

def taxicab(a, b):
"""Return the taxicab distance between two intersections.

>>> times_square = intersection(46, 7)
>>> ess_a_bagel = intersection(51, 3)
>>> taxicab(times_square, ess_a_bagel)
9
>>> taxicab(ess_a_bagel, times_square)
9
"""
``````

Use Ok to test your code:

``python3 ok -q taxicab``

### Q4: Squares only (optional)

NOTE! This question is now optional, as it refers to material that will be covered in Wednesdayâ€™s lecture.

Implement the function `squares`, which takes in a list of positive integers. It returns a list that contains the square roots of the elements of the original list that are perfect squares. Try using a list comprehension.

You may find the `round` function useful.

``````>>> round(10.5)
10
>>> round(10.51)
11``````
``````def squares(s):
"""Returns a new list containing square roots of the elements of the
original list that are perfect squares.

>>> seq = [8, 49, 8, 9, 2, 1, 100, 102]
>>> squares(seq)
[7, 3, 1, 10]
>>> seq = [500, 30]
>>> squares(seq)
[]
"""
``````

Use Ok to test your code:

``python3 ok -q squares``

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

Use Ok to test your code:

``python3 ok -q count_change``

### Q6: 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.

See the following animation of the Towers of Hanoi, found on Wikimedia by user Trixx

The strategy used in Towers of Hanoi is to move all but the bottom disc to the second peg, then moving the bottom disc to the third peg, then moving all but the second disc from the second to the third peg.

One thing you don't need to worry about is collecting all the steps. `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"
``````

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!

### Q7: 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``