# Lab 9 Solutions

## Solution Files

## Submission

In order to facilitate midterm studying, solutions to this lab were released with the lab. We encourage you to try out the problems and struggle for a while before looking at the solutions!
**Note: You do not need to run** `python ok --submit`

**to receive credit for this assignment.**

# Required Questions

### Q1: All Questions Are Optional

**Updated**: The following questions in this assignment are not graded, but they are highly recommended to help you prepare for the upcoming midterm. You will receive credit for this lab even if you not complete these questions.

This question has no Ok tests.

# Suggested Questions

## Recursion and Tree Recursion

### Q2: Subsequences

A subsequence of a sequence `S`

is a sequence of elements from `S`

, in the same
order they appear in `S`

, but possibly with elements missing. Thus, the lists
`[]`

, `[1, 3]`

, `[2]`

, and `[1, 2, 3]`

are some (but not all) of the
subsequences of `[1, 2, 3]`

. Write a function that takes a list and returns a
list of lists, for which each individual list is a subsequence of the original
input.

In order to accomplish this, you might first want to write a function `insert_into_all`

that takes an item and a list of lists, adds the item to the beginning of nested list,
and returns the resulting list.

```
def insert_into_all(item, nested_list):
"""Assuming that nested_list is a list of lists, return a new list
consisting of all the lists in nested_list, but with item added to
the front of each.
>>> nl = [[], [1, 2], [3]]
>>> insert_into_all(0, nl)
[[0], [0, 1, 2], [0, 3]]
"""
return [[item] + lst for lst in nested_list]
def subseqs(s):
"""Assuming that S is a list, return a nested list of all subsequences
of S (a list of lists). The subsequences can appear in any order.
>>> seqs = subseqs([1, 2, 3])
>>> sorted(seqs)
[[], [1], [1, 2], [1, 2, 3], [1, 3], [2], [2, 3], [3]]
>>> subseqs([])
[[]]
"""
if not s: return [[]] else:
subset = subseqs(s[1:]) return insert_into_all(s[0], subset) + subset
```

Use Ok to test your code:

`python3 ok -q subseqs`

### Q3: Increasing Subsequences

Just like the last question, we want to write a function that takes a list and returns a list of lists, where each individual list is a subsequence of the original input.

This time we have another condition: we only want the subsequences for which
consecutive elements are *nondecreasing*. For example, `[1, 3, 2]`

is a
subsequence of `[1, 3, 2, 4]`

, but since 2 < 3, this subsequence would *not*
be included in our result.

**Fill in the blanks** to complete the implementation of the `inc_subseqs`

function. You may assume that the input list contains no negative elements.

You may use the provided helper function `insert_into_all`

, which takes in an
`item`

and a list of lists and inserts the `item`

to the front of each list.

```
def inc_subseqs(s):
"""Assuming that S is a list, return a nested list of all subsequences
of S (a list of lists) for which the elements of the subsequence
are strictly nondecreasing. The subsequences can appear in any order.
>>> seqs = inc_subseqs([1, 3, 2])
>>> sorted(seqs)
[[], [1], [1, 2], [1, 3], [2], [3]]
>>> inc_subseqs([])
[[]]
>>> seqs2 = inc_subseqs([1, 1, 2])
>>> sorted(seqs2)
[[], [1], [1], [1, 1], [1, 1, 2], [1, 2], [1, 2], [2]]
"""
def subseq_helper(s, prev):
if not s:
return [[]] elif s[0] < prev:
return subseq_helper(s[1:], prev) else:
a = subseq_helper(s[1:], s[0]) b = subseq_helper(s[1:], prev) return insert_into_all(s[0], a) + b return subseq_helper(s, 0)
```

Use Ok to test your code:

`python3 ok -q inc_subseqs`

### Q4: Number of Trees

A **full binary tree** is a tree where each node has either 2 branches or 0 branches, but never 1 branch.

How many possible full binary tree structures exist that have exactly n leaves?

For those interested in combinatorics, this problem does have a closed form solution):

```
def num_trees(n):
"""How many full binary trees have exactly n leaves? E.g.,
1 2 3 3 ...
* * * *
/ \ / \ / \
* * * * * *
/ \ / \
* * * *
>>> num_trees(1)
1
>>> num_trees(2)
1
>>> num_trees(3)
2
>>> num_trees(8)
429
"""
if n == 1: return 1 return sum(num_trees(k) * num_trees(n-k) for k in range(1, n))
```

Use Ok to test your code:

`python3 ok -q num_trees`

## Generators

### Q5: Generators generator

Write the generator function `make_generators_generator`

, which takes a
zero-argument generator function `g`

and returns a generator that yields
generators. For each element `e`

yielded by the generator object returned by
calling `g`

, a new generator object is yielded that will generate entries 1
through `e`

yielded by the generator returned by `g`

.

```
def make_generators_generator(g):
"""Generates all the "sub"-generators of the generator returned by
the generator function g.
>>> def every_m_ints_to(n, m):
... i = 0
... while (i <= n):
... yield i
... i += m
...
>>> def every_3_ints_to_10():
... for item in every_m_ints_to(10, 3):
... yield item
...
>>> for gen in make_generators_generator(every_3_ints_to_10):
... print("Next Generator:")
... for item in gen:
... print(item)
...
Next Generator:
0
Next Generator:
0
3
Next Generator:
0
3
6
Next Generator:
0
3
6
9
"""
def gen(i):
for entry in g(): if i <= 0: return yield entry i -= 1 i = 1 for entry in g(): yield gen(i) i += 1
```

Use Ok to test your code:

`python3 ok -q make_generators_generator`

## Objects

### Q6: Keyboard

We'd like to create a `Keyboard`

class that takes in an arbitrary
number of `Button`

s and stores these `Button`

s in a dictionary. The
keys in the dictionary will be ints that represent the postition on the
`Keyboard`

, and the values will be the respective `Button`

. Fill out
the methods in the `Keyboard`

class according to each description,
using the doctests as a reference for the behavior of a `Keyboard`

.

```
class Button:
"""
Represents a single button
"""
def __init__(self, pos, key):
"""
Creates a button
"""
self.pos = pos
self.key = key
self.times_pressed = 0
class Keyboard:
"""A Keyboard takes in an arbitrary amount of buttons, and has a
dictionary of positions as keys, and values as Buttons.
>>> b1 = Button(0, "H")
>>> b2 = Button(1, "I")
>>> k = Keyboard(b1, b2)
>>> k.buttons[0].key
'H'
>>> k.press(1)
'I'
>>> k.press(2) #No button at this position
''
>>> k.typing([0, 1])
'HI'
>>> k.typing([1, 0])
'IH'
>>> b1.times_pressed
2
>>> b2.times_pressed
3
"""
def __init__(self, *args):
self.buttons = {} for button in args: self.buttons[button.pos] = button
def press(self, info):
"""Takes in a position of the button pressed, and
returns that button's output"""
if info in self.buttons.keys(): b = self.buttons[info] b.times_pressed += 1 return b.key return ''
def typing(self, typing_input):
"""Takes in a list of positions of buttons pressed, and
returns the total output"""
accumulate = '' for pos in typing_input: accumulate+=self.press(pos) return accumulate
```

Use Ok to test your code:

`python3 ok -q Keyboard`

## Nonlocal

### Q7: Advanced Counter

Complete the definition of `make_advanced_counter_maker`

,
which creates a function that creates counters. These counters can not
only update their personal count, but also a shared count for all
counters. They can also reset either count.

```
def make_advanced_counter_maker():
"""Makes a function that makes counters that understands the
messages "count", "global-count", "reset", and "global-reset".
See the examples below:
>>> make_counter = make_advanced_counter_maker()
>>> tom_counter = make_counter()
>>> tom_counter('count')
1
>>> tom_counter('count')
2
>>> tom_counter('global-count')
1
>>> jon_counter = make_counter()
>>> jon_counter('global-count')
2
>>> jon_counter('count')
1
>>> jon_counter('reset')
>>> jon_counter('count')
1
>>> tom_counter('count')
3
>>> jon_counter('global-count')
3
>>> jon_counter('global-reset')
>>> tom_counter('global-count')
1
"""
global_count = 0 def make_counter(): count = 0 def counter(msg): nonlocal global_count, count if msg == 'count':
count += 1
return count
elif msg == 'reset':
count = 0
elif msg == 'global-count':
global_count += 1
return global_count
elif msg == 'global-reset':
global_count = 0 return counter return make_counter
```

Use Ok to test your code:

`python3 ok -q make_advanced_counter_maker`

## Mutable Lists

### Q8: Trade

In the integer market, each participant has a list of positive integers to trade. When two participants meet, they trade the smallest non-empty prefix of their list of integers. A prefix is a slice that starts at index 0.

Write a function `trade`

that exchanges the first `m`

elements of list `first`

with the first `n`

elements of list `second`

, such that the sums of those
elements are equal, and the sum is as small as possible. If no such prefix
exists, return the string `'No deal!'`

and do not change either list. Otherwise
change both lists and return `'Deal!'`

. A partial implementation is provided.

Hint:You can mutate a slice of a list usingslice assignment. To do so, specify a slice of the list`[i:j]`

on the left-hand side of an assignment statement and another list on the right-hand side of the assignment statement. The operation will replace the entire given slice of the list from`i`

inclusive to`j`

exclusive with the elements from the given list. The slice and the given list need not be the same length.`>>> a = [1, 2, 3, 4, 5, 6] >>> b = a >>> a[2:5] = [10, 11, 12, 13] >>> a [1, 2, 10, 11, 12, 13, 6] >>> b [1, 2, 10, 11, 12, 13, 6]`

Additionally, recall that the starting and ending indices for a slice can be left out and Python will use a default value.

`lst[i:]`

is the same as`lst[i:len(lst)]`

, and`lst[:j]`

is the same as`lst[0:j]`

.

```
def trade(first, second):
"""Exchange the smallest prefixes of first and second that have equal sum.
>>> a = [1, 1, 3, 2, 1, 1, 4]
>>> b = [4, 3, 2, 7]
>>> trade(a, b) # Trades 1+1+3+2=7 for 4+3=7
'Deal!'
>>> a
[4, 3, 1, 1, 4]
>>> b
[1, 1, 3, 2, 2, 7]
>>> c = [3, 3, 2, 4, 1]
>>> trade(b, c)
'No deal!'
>>> b
[1, 1, 3, 2, 2, 7]
>>> c
[3, 3, 2, 4, 1]
>>> trade(a, c)
'Deal!'
>>> a
[3, 3, 2, 1, 4]
>>> b
[1, 1, 3, 2, 2, 7]
>>> c
[4, 3, 1, 4, 1]
"""
m, n = 1, 1
equal_prefix = lambda: sum(first[:m]) == sum(second[:n]) while m < len(first) and n < len(second) and not equal_prefix(): if sum(first[:m]) < sum(second[:n]): m += 1
else:
n += 1
if equal_prefix():
first[:m], second[:n] = second[:n], first[:m]
return 'Deal!'
else:
return 'No deal!'
```

Use Ok to test your code:

`python3 ok -q trade`

### Q9: Shuffle

Define a function `shuffle`

that takes a sequence with an even number of
elements (cards) and creates a new list that interleaves the elements
of the first half with the elements of the second half.

```
def card(n):
"""Return the playing card numeral as a string for a positive n <= 13."""
assert type(n) == int and n > 0 and n <= 13, "Bad card n"
specials = {1: 'A', 11: 'J', 12: 'Q', 13: 'K'}
return specials.get(n, str(n))
def shuffle(cards):
"""Return a shuffled list that interleaves the two halves of cards.
>>> shuffle(range(6))
[0, 3, 1, 4, 2, 5]
>>> suits = ['♡', '♢', '♤', '♧']
>>> cards = [card(n) + suit for n in range(1,14) for suit in suits]
>>> cards[:12]
['A♡', 'A♢', 'A♤', 'A♧', '2♡', '2♢', '2♤', '2♧', '3♡', '3♢', '3♤', '3♧']
>>> cards[26:30]
['7♤', '7♧', '8♡', '8♢']
>>> shuffle(cards)[:12]
['A♡', '7♤', 'A♢', '7♧', 'A♤', '8♡', 'A♧', '8♢', '2♡', '8♤', '2♢', '8♧']
>>> shuffle(shuffle(cards))[:12]
['A♡', '4♢', '7♤', '10♧', 'A♢', '4♤', '7♧', 'J♡', 'A♤', '4♧', '8♡', 'J♢']
>>> cards[:12] # Should not be changed
['A♡', 'A♢', 'A♤', 'A♧', '2♡', '2♢', '2♤', '2♧', '3♡', '3♢', '3♤', '3♧']
"""
assert len(cards) % 2 == 0, 'len(cards) must be even'
half = len(cards) // 2 shuffled = []
for i in range(half): shuffled.append(cards[i]) shuffled.append(cards[half+i]) return shuffled
```

Use Ok to test your code:

`python3 ok -q shuffle`

## Linked Lists

### Q10: Insert

Implement a function `insert`

that takes a `Link`

, a `value`

, and an
`index`

, and inserts the `value`

into the `Link`

at the given `index`

.
You can assume the linked list already has at least one element. Do not
return anything -- `insert`

should mutate the linked list.

Note: If the index is out of bounds, you can raise an`IndexError`

with:`raise IndexError`

```
def insert(link, value, index):
"""Insert a value into a Link at the given index.
>>> link = Link(1, Link(2, Link(3)))
>>> print(link)
<1 2 3>
>>> insert(link, 9001, 0)
>>> print(link)
<9001 1 2 3>
>>> insert(link, 100, 2)
>>> print(link)
<9001 1 100 2 3>
>>> insert(link, 4, 5)
IndexError
"""
if index == 0: link.rest = Link(link.first, link.rest) link.first = value # line not needed elif link.rest is Link.empty: raise IndexError else:
insert(link.rest, value, index - 1)
# iterative solution
def insert_iter(link, value, index):
while index > 0 and link.rest is not Link.empty:
link = link.rest
index -= 1
if index == 0:
link.rest = Link(link.first, link.rest)
link.first = value
else:
raise IndexError
```

Use Ok to test your code:

`python3 ok -q insert`

### Q11: Deep Linked List Length

A linked list that contains one or more linked lists as elements is called a
*deep* linked list. Write a function `deep_len`

that takes in a (possibly deep)
linked list and returns the *deep length* of that linked list. The deep length of
a linked list is the total number of non-link elements in the list, as well as the
total number of elements contained in all contained lists. See the function's doctests
for examples of the deep length of linked lists.

Hint:Use`isinstance`

to check if something is an instance of an object.

```
def deep_len(lnk):
""" Returns the deep length of a possibly deep linked list.
>>> deep_len(Link(1, Link(2, Link(3))))
3
>>> deep_len(Link(Link(1, Link(2)), Link(3, Link(4))))
4
>>> levels = Link(Link(Link(1, Link(2)), \
Link(3)), Link(Link(4), Link(5)))
>>> print(levels)
<<<1 2> 3> <4> 5>
>>> deep_len(levels)
5
"""
if lnk is Link.empty: return 0
elif not isinstance(lnk, Link): return 1
else:
return deep_len(lnk.first) + deep_len(lnk.rest)
```

Use Ok to test your code:

`python3 ok -q deep_len`

### Q12: Linked Lists as Strings

Kevin and Jerry like different ways of displaying the linked list
structure in Python. While Kevin likes box and pointer diagrams,
Jerry prefers a more futuristic way. Write a function
`make_to_string`

that returns a function that converts the
linked list to a string in their preferred style.

*Hint*: You can convert numbers to strings using the `str`

function,
and you can combine strings together using `+`

.

```
>>> str(4)
'4'
>>> 'cs ' + str(61) + 'a'
'cs 61a'
```

```
def make_to_string(front, mid, back, empty_repr):
""" Returns a function that turns linked lists to strings.
>>> kevins_to_string = make_to_string("[", "|-]-->", "", "[]")
>>> jerrys_to_string = make_to_string("(", " . ", ")", "()")
>>> lst = Link(1, Link(2, Link(3, Link(4))))
>>> kevins_to_string(lst)
'[1|-]-->[2|-]-->[3|-]-->[4|-]-->[]'
>>> kevins_to_string(Link.empty)
'[]'
>>> jerrys_to_string(lst)
'(1 . (2 . (3 . (4 . ()))))'
>>> jerrys_to_string(Link.empty)
'()'
"""
def printer(lnk):
if lnk is Link.empty: return empty_repr else:
return front + str(lnk.first) + mid + printer(lnk.rest) + back return printer
```

Use Ok to test your code:

`python3 ok -q make_to_string`

## Trees

### Q13: Prune Small

Complete the function `prune_small`

that takes in a `Tree`

`t`

and a
number `n`

and prunes `t`

mutatively. If `t`

or any of its branches
has more than `n`

branches, the `n`

branches with the smallest labels
should be kept and any other branches should be *pruned*, or removed,
from the tree.

```
def prune_small(t, n):
"""Prune the tree mutatively, keeping only the n branches
of each node with the smallest label.
>>> t1 = Tree(6)
>>> prune_small(t1, 2)
>>> t1
Tree(6)
>>> t2 = Tree(6, [Tree(3), Tree(4)])
>>> prune_small(t2, 1)
>>> t2
Tree(6, [Tree(3)])
>>> t3 = Tree(6, [Tree(1), Tree(3, [Tree(1), Tree(2), Tree(3)]), Tree(5, [Tree(3), Tree(4)])])
>>> prune_small(t3, 2)
>>> t3
Tree(6, [Tree(1), Tree(3, [Tree(1), Tree(2)])])
"""
while len(t.branches) > n: largest = max(t.branches, key=lambda x: x.label) t.branches.remove(largest) for b in t.branches: prune_small(b, n)
```

Use Ok to test your code:

`python3 ok -q prune_small`