# Homework 4 Solutions

## Solution Files

You can find solutions for all questions in hw04.py.

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

### Q1: 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
"""
return abs(street(a)-street(b)) + abs(avenue(a)-avenue(b))
```

Use Ok to test your code:

`python3 ok -q taxicab`

The main focus of this problem is to get familiar with using data abstraction. With some previous problems involving abstract data types, it might have been possible to break the abstraction barrier and still solve the problem. This time around, the abstraction uses the Cantor pairing function to obfuscate the original data!

Through the power of abstraction however, you don't need to understand how the Cantor pairing function works. In truth, we could have also not told you anything about how the abstract data type was implemented. As long as you use the provided selectors, you should be able to solve the problem.

Speaking of which, the selectors give the `street`

and `avenue`

of an
intersection. If we have the street and the avenue for each intersection,
the taxicab distance is just the sum of the absolute difference of the
two.

For more information, Wikipedia has a useful visualization.

Video walkthrough: https://youtu.be/QueVasKQQBI

### Q2: Squares only

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)
[]
"""
return [round(n ** 0.5) for n in s if n == round(n ** 0.5) ** 2]
```

Use Ok to test your code:

`python3 ok -q squares`

It might be helpful to construct a skeleton list comprehension to begin with:

`[sqrt(x) for x in s if is_perfect_square(x)]`

This is great, but it requires that we have an `is_perfect_square`

function. How might we check if something is a perfect square?

- If the square root of a number is a whole number, then it is a perfect
square. For example,
`sqrt(61) = 7.81024...`

(not a perfect square) and`sqrt(49) = 7`

(perfect square). Once we obtain the square root of the number, we just need to check if something is a whole number. The

`is_perfect_square`

function might look like:`def is_perfect_square(x): return is_whole(sqrt(x))`

- One last piece of the puzzle: to check if a number is whole, we just
need to see if it has a decimal or not. The way we've chosen to do it in
the solution is to compare the original number to the round version
(thus removing all decimals), but a technique employing floor division
(
`//`

) or something else entirely could work too.

We've written all these helper functions to solve this problem, but they are actually all very short. Therefore, we can just copy the body of each into the original list comprehension, arriving at the solution we finally present.

Video walkthrough: https://youtu.be/YwLFB9paET0

### Q3: 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
>>> check(HW_SOURCE_FILE, 'g', ['While', 'For'])
True
"""
if n in (1, 2, 3):
return n
return g(n-1) + 2*g(n-2) + 3*g(n-3)
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
"""
if n == 1 or n == 2 or n == 3:
return n
a, b, c = 1, 2, 3
while n > 3:
a, b, c = b, c, c + 2*b + 3*a
n = n - 1
return c
```

Use Ok to test your code:

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

This is an example of how a function might be easier to write recursively
versus iteratively. Since we have defined the `g`

function in terms of
older versions of itself, the solution tends very naturally towards
recursion.

The iterative solution is trickier, since we can only track a finite amount of state at a given time. We need to pick our variables carefully so that we have just the information we need to calculate the next step. In a sense, this problem is very similar to the Fibonacci sequence (assuming we start at the 0th Fibonacci number):

```
f(n) = n, if n <= 2
f(n) = f(n-1) + f(n-2), if n > 2
```

As you may recall, the solution for Fibonacci carried two variables around for the two previous values.

```
def fib_iter(n):
prev, curr = 0, 1
while n > 0:
prev, curr = curr, prev + curr
return prev
```

Since the `g`

function depends on the three previous values, it might make
sense that we might have to track three values instead!

Consider the three previous values `old`

, `older`

, and `oldest`

. To do an
update, the `older`

value ages and becomes `oldest`

, the `old`

value ages
and becomes `older`

. Finally, `old`

gets the new value which is derived
from the three previous values: `old + 2 * older + 3 * oldest`

.

Video walkthrough: https://youtu.be/pltx7u2kGGE

### 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 `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
"""
def constrained_count(amount, smallest_coin):
if amount == 0:
return 1
if smallest_coin > amount:
return 0
without_coin = constrained_count(amount, smallest_coin * 2)
with_coin = constrained_count(amount - smallest_coin, smallest_coin)
return without_coin + with_coin
return constrained_count(amount, 1)
```

Use Ok to test your code:

`python3 ok -q count_change`

This is remarkably similar to the `count_partitions`

problem, with a
few minor differences:

- A maximum partition size
`m`

is not given, so we need to create a helper function that takes in two arguments and also create another helper function to find the max coin. - Partition size is not linear, but rather multiples of two. To get the next partition you need to divide by two instead of subtracting one.

One other implementation detail here is that we enforce a *maximum* partition
size, rather than a *minimum* coin. Many students attempted to
start at 1 and work there way up. That will also work, but is less similar
to `count_partitions`

. As long as there is some ordering on the coins being enforced, we ensure we cover all the combinations of coins without any duplicates.

See the walkthrough for a more thorough explanation and a visual of the recursive calls. Video walkthrough: https://youtu.be/EgZJPNFnoxM.

### 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.

```
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"
if n == 1:
print_move(start, end)
else:
other = 6 - start - end
move_stack(n-1, start, other)
print_move(start, end)
move_stack(n-1, other, end)
```

Use Ok to test your code:

`python3 ok -q move_stack`

To solve the Towers of Hanoi problem for `n`

disks, we need to do three
steps:

- Move everything but the last disk (
`n-1`

disks) to someplace in the middle (not the start nor the end rod). - Move the last disk (a single disk) to the end rod. This must occur after step 1 (we have to move everything above it away first)!
- Move everything but the last disk (the disks from step 1) from the middle on top of the end rod.

We take advantage of the fact that the recursive function `move_stack`

is
guaranteed to move `n`

disks from `start`

to `end`

while obeying the rules
of Towers of Hanoi. The only thing that remains is to make sure that we
have set up the playing board to make that possible.

Since we move a disk to end rod, we run the risk of `move_stack`

doing an
improper move (big disk on top of small disk). But since we're moving the
biggest disk possible, nothing in the `n-1`

disks above that is bigger.
Therefore, even though we do not explicitly state the Towers of Hanoi
constraints, we can still carry out the correct steps.

Video walkthrough: https://youtu.be/VwynGQiCTFM

## Extra questions

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

### 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
>>> check(HW_SOURCE_FILE, 'make_anonymous_factorial', ['Assign', 'AugAssign', 'FunctionDef', 'Recursion'])
True
"""
return (lambda f: lambda k: f(f, k))(lambda f, k: k if k == 1 else mul(k, f(f, sub(k, 1))))
# Alternate solution:
# return (lambda f: f(f))(lambda f: lambda x: 1 if x == 0 else x * f(f)(x - 1))
```

Use Ok to test your code:

`python3 ok -q make_anonymous_factorial`