Lab 8 Solutions
Solution Files
Topics
Consult this section if you need a refresher on the material for this lab. It's okay to skip directly to the questions and refer back here should you get stuck.
Iterators
An iterable is any object that can be iterated through, or gone through one element at a time. One construct that we've used to iterate through an iterable is a for loop:
for elem in iterable:
# do something
for
loops work on any object that is iterable. We previously described it
as working with any sequence -- all sequences are iterable, but there are other
objects that are also iterable! We define an iterable as an object on which
calling the built-in iter
function returns an iterator. An
iterator is another type of object that allows us to iterate through an
iterable by keeping track of which element is next in the sequence.
To illustrate this, consider the following block of code, which does the exact
same thing as the for
statement above:
iterator = iter(iterable)
try:
while True:
elem = next(iterator)
# do something
except StopIteration:
pass
Here's a breakdown of what's happening:
- First, the built-in
iter
function is called on the iterable to create a corresponding iterator. - To get the next element in the sequence, the built-in
next
function is called on this iterator. - When
next
is called but there are no elements left in the iterator, aStopIteration
error is raised. In the for loop construct, this exception is caught and execution can continue.
Calling iter
on an iterable multiple times returns a new iterator each time
with distinct states (otherwise, you'd never be able to iterate through a
iterable more than once). You can also call iter
on the iterator itself, which
will just return the same iterator without changing its state. However, note
that you cannot call next
directly on an iterable.
Let's see the iter
and next
functions in action with an iterable we're
already familiar with -- a list.
>>> lst = [1, 2, 3, 4]
>>> next(lst) # Calling next on an iterable
TypeError: 'list' object is not an iterator
>>> list_iter = iter(lst) # Creates an iterator for the list
>>> list_iter
<list_iterator object ...>
>>> next(list_iter) # Calling next on an iterator
1
>>> next(list_iter) # Calling next on the same iterator
2
>>> next(iter(list_iter)) # Calling iter on an iterator returns itself
3
>>> list_iter2 = iter(lst)
>>> next(list_iter2) # Second iterator has new state
1
>>> next(list_iter) # First iterator is unaffected by second iterator
4
>>> next(list_iter) # No elements left!
StopIteration
>>> lst # Original iterable is unaffected
[1, 2, 3, 4]
Since you can call iter
on iterators, this tells us that that they are also
iterables! Note that while all iterators are iterables, the converse is not
true - that is, not all iterables are iterators. You can use iterators wherever
you can use iterables, but note that since iterators keep their state, they're
only good to iterate through an iterable once:
>>> list_iter = iter([4, 3, 2, 1])
>>> for e in list_iter:
... print(e)
4
3
2
1
>>> for e in list_iter:
... print(e)
Analogy: An iterable is like a book (one can flip through the pages) and an iterator for a book would be a bookmark (saves the position and can locate the next page). Calling
iter
on a book gives you a new bookmark independent of other bookmarks, but callingiter
on a bookmark gives you the bookmark itself, without changing its position at all. Callingnext
on the bookmark moves it to the next page, but does not change the pages in the book. Callingnext
on the book wouldn't make sense semantically. We can also have multiple bookmarks, all independent of each other.
Iterable Uses
We know that lists are one type of built-in iterable objects. You may have also
encountered the range(start, end)
function, which creates an iterable of
ascending integers from start (inclusive) to end (exclusive).
>>> for x in range(2, 6):
... print(x)
...
2
3
4
5
Ranges are useful for many things, including performing some operations for a particular number of iterations or iterating through the indices of a list.
There are also some built-in functions that take in iterables and return useful results:
map(f, iterable)
- Creates an iterator overf(x)
forx
initerable
. In some cases, computing a list of the values in this iterable will give us the same result as [func(x)
forx
initerable
]. However, it's important to keep in mind that iterators can potentially have infinite values because they are evaluated lazily, while lists cannot have infinite elements.filter(f, iterable)
- Creates an iterator overx
for eachx
initerable
iff(x)
zip(iterables*)
- Creates an iterator over co-indexed tuples with elements from each of theiterables
reversed(iterable)
- Creates an iterator over all the elements in the input iterable in reverse orderlist(iterable)
- Creates a list containing all the elements in the inputiterable
tuple(iterable)
- Creates a tuple containing all the elements in the inputiterable
sorted(iterable)
- Creates a sorted list containing all the elements in the inputiterable
reduce(f, iterable)
- Must be imported withfunctools
. Apply function of two argumentsf
cumulatively to the items ofiterable
, from left to right, so as to reduce the sequence to a single value.
Generators
We can create our own custom iterators by writing a generator function, which
returns a special type of iterator called a generator. Generator functions
have yield
statements within the body of the function instead of return
statements. Calling a generator function will return a generator object and
will not execute the body of the function.
For example, let's consider the following generator function:
def countdown(n):
print("Beginning countdown!")
while n >= 0:
yield n
n -= 1
print("Blastoff!")
Calling countdown(k)
will return a generator object that counts down from k
to 0. Since generators are iterators, we can call iter
on the resulting
object, which will simply return the same object. Note that the body is not
executed at this point; nothing is printed and no numbers are outputted.
>>> c = countdown(5)
>>> c
<generator object countdown ...>
>>> c is iter(c)
True
So how is the counting done? Again, since generators are iterators, we call
next
on them to get the next element! The first time next
is called,
execution begins at the first line of the function body and continues until the
yield
statement is reached. The result of evaluating the expression in the
yield
statement is returned. The following interactive session continues
from the one above.
>>> next(c)
Beginning countdown!
5
Unlike functions we've seen before in this course, generator functions can
remember their state. On any consecutive calls to next
, execution picks up
from the line after the yield
statement that was previously executed. Like
the first call to next
, execution will continue until the next yield
statement is reached. Note that because of this, Beginning countdown!
doesn't
get printed again.
>>> next(c)
4
>>> next(c)
3
The next 3 calls to next
will continue to yield consecutive descending
integers until 0. On the following call, a StopIteration
error will be
raised because there are no more values to yield (i.e. the end of the function
body was reached before hitting a yield
statement).
>>> next(c)
2
>>> next(c)
1
>>> next(c)
0
>>> next(c)
Blastoff!
StopIteration
Separate calls to countdown
will create distinct generator objects with their
own state. Usually, generators shouldn't restart. If you'd like to reset the
sequence, create another generator object by calling the generator function
again.
>>> c1, c2 = countdown(5), countdown(5)
>>> c1 is c2
False
>>> next(c1)
5
>>> next(c2)
5
Here is a summary of the above:
- A generator function has a
yield
statement and returns a generator object. - Calling the
iter
function on a generator object returns the same object without modifying its current state. - The body of a generator function is not evaluated until
next
is called on a resulting generator object. Calling thenext
function on a generator object computes and returns the next object in its sequence. If the sequence is exhausted,StopIteration
is raised. A generator "remembers" its state for the next
next
call. Therefore,the first
next
call works like this:- Enter the function and run until the line with
yield
. - Return the value in the
yield
statement, but remember the state of the function for futurenext
calls.
- Enter the function and run until the line with
And subsequent
next
calls work like this:- Re-enter the function, start at the line after the
yield
statement that was previously executed, and run until the nextyield
statement. - Return the value in the
yield
statement, but remember the state of the function for futurenext
calls.
- Re-enter the function, start at the line after the
- Calling a generator function returns a brand new generator object (like
calling
iter
on an iterable object). - A generator should not restart unless it's defined that way. To start over from the first element in a generator, just call the generator function again to create a new generator.
Another useful tool for generators is the yield from
statement. yield from
will yield all values from an iterator or iterable.
>>> def gen_list(lst):
... yield from lst
...
>>> g = gen_list([1, 2, 3, 4])
>>> next(g)
1
>>> next(g)
2
>>> next(g)
3
>>> next(g)
4
>>> next(g)
StopIteration
Required Questions
Iterators & Generators
Q1: Repeated
Implement repeated
,
which takes in an iterator t
and returns the first value in t
that appears k
times in a row.
Note: You can assume that the iterator
t
will have a value that appears at leastk
times in a row. If you are receiving aStopIteration
, yourrepeated
function is likely not identifying the correct value.
Your implementation should iterate through the items in a way such that
if the same iterator is passed into repeated
twice,
it should continue in the second call at the point it left off in the first.
An example of this behavior is in the doctests.
def repeated(t, k):
"""Return the first value in iterator T that appears K times in a row.
Iterate through the items such that if the same iterator is passed into
the function twice, it continues in the second call at the point it left
off in the first.
>>> s = iter([10, 9, 10, 9, 9, 10, 8, 8, 8, 7])
>>> repeated(s, 2)
9
>>> s2 = iter([10, 9, 10, 9, 9, 10, 8, 8, 8, 7])
>>> repeated(s2, 3)
8
>>> s = iter([3, 2, 2, 2, 1, 2, 1, 4, 4, 5, 5, 5])
>>> repeated(s, 3)
2
>>> repeated(s, 3)
5
>>> s2 = iter([4, 1, 6, 6, 7, 7, 8, 8, 2, 2, 2, 5])
>>> repeated(s2, 3)
2
"""
assert k > 1
count = 1
last_item = None
while True:
item = next(t)
if item == last_item:
count += 1
else:
last_item = item
count = 1
if count == k:
return item
Use Ok to test your code:
python3 ok -q repeated
Q2: Merge
Implement merge(incr_a, incr_b)
, which takes two iterables incr_a
and incr_b
whose
elements are ordered. merge
yields elements from incr_a
and incr_b
in sorted
order, eliminating repetition. You may assume incr_a
and incr_b
themselves do not
contain repeats, and that none of the elements of either are None
.
You may not assume that the iterables are finite; either may produce an infinite
stream of results.
You will probably find it helpful to use the two-argument version of the built-in
next
function: next(incr, v)
is the same as next(incr)
, except that instead of
raising StopIteration
when incr
runs out of elements, it returns v
.
See the doctest for examples of behavior.
def merge(incr_a, incr_b):
"""Yield the elements of strictly increasing iterables incr_a and incr_b, removing
repeats. Assume that incr_a and incr_b have no repeats. incr_a or incr_b may or may not
be infinite sequences.
>>> m = merge([0, 2, 4, 6, 8, 10, 12, 14], [0, 3, 6, 9, 12, 15])
>>> type(m)
<class 'generator'>
>>> list(m)
[0, 2, 3, 4, 6, 8, 9, 10, 12, 14, 15]
>>> def big(n):
... k = 0
... while True: yield k; k += n
>>> m = merge(big(2), big(3))
>>> [next(m) for _ in range(11)]
[0, 2, 3, 4, 6, 8, 9, 10, 12, 14, 15]
"""
iter_a, iter_b = iter(incr_a), iter(incr_b)
next_a, next_b = next(iter_a, None), next(iter_b, None)
while next_a is not None or next_b is not None:
if next_a is None or next_b is not None and next_b < next_a:
yield next_b
next_b = next(iter_b, None)
elif next_b is None or next_a is not None and next_a < next_b:
yield next_a
next_a = next(iter_a, None)
else:
yield next_a
next_a, next_b = next(iter_a, None), next(iter_b, None)
Use Ok to test your code:
python3 ok -q merge
Linked Lists & Trees
Q3: 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
Q4: Add Leaves
Implement add_d_leaves
, a function that takes in a Tree
instance t
and a number v
.
We define the depth of a node in t
to be the number of edges from the root to that node. The depth of root is therefore 0.
For each node in the tree, you should add d
leaves to it, where d
is the depth of the node. Every added leaf should have a label of v
. If the node at this depth has existing branches, you should add these leaves to the end of that list of branches.
For example, you should be adding 1 leaf with label v
to each node at depth 1, 2 leaves to each node at depth 2, and so on.
Here is an example of a tree t
(shown on the left) and the result after add_d_leaves
is applied with v
as 5.
Try drawing out the second doctest to visualize how the function is mutating
t3
.
Hint: Use a helper function to keep track of the depth!
def add_d_leaves(t, v):
"""Add d leaves containing v to each node at every depth d.
>>> t_one_to_four = Tree(1, [Tree(2), Tree(3, [Tree(4)])])
>>> print(t_one_to_four)
1
2
3
4
>>> add_d_leaves(t_one_to_four, 5)
>>> print(t_one_to_four)
1
2
5
3
4
5
5
5
>>> t1 = Tree(1, [Tree(3)])
>>> add_d_leaves(t1, 4)
>>> t1
Tree(1, [Tree(3, [Tree(4)])])
>>> t2 = Tree(2, [Tree(5), Tree(6)])
>>> t3 = Tree(3, [t1, Tree(0), t2])
>>> print(t3)
3
1
3
4
0
2
5
6
>>> add_d_leaves(t3, 10)
>>> print(t3)
3
1
3
4
10
10
10
10
10
10
0
10
2
5
10
10
6
10
10
10
"""
def add_leaves(t, d):
for b in t.branches:
add_leaves(b, d + 1)
t.branches.extend([Tree(v) for _ in range(d)])
add_leaves(t, 0)
Use Ok to test your code:
python3 ok -q add_d_leaves
add_d_leaves
on each branch to add the
leaves to the branches. However, notice that each recursive call would need to
know what the current depth is in order to add that many leaves. We could try
initializing a variable within the body of the function, but by now we know
that in order to keep track of changing values across recursive calls we should
use a helper function!
The helper function should take in a tree and a depth value, and we will define
it as a function that adds d
leaves to the branches of the root node, d + 1
leaves to the branches of each of the root node's branches, and so on:
def add_leaves(t, d):
"""Adds a number of leaves to each node in t equivalent to the depth of
the node, assuming that the root node is at depth d, the children of
the root node are at depth d + 1, and so on."""
...
We don't need a parameter for v
since that value won't change and we can
access it from the parent frame. With this function defined as such, we can
call add_leaves
with arguments t
and 0 to add leaves starting at depth 0.
def add_d_leaves(t, v):
def add_leaves(t, d):
"""Adds a number of leaves to each node in t equivalent to the depth of
the node, assuming that the root node is at depth d, the children of
the root node are at depth d + 1, and so on."""
...
add_leaves(t, 0)
Inside the helper function, we can now call it recursively on each branch.
Each node's branch is one depth level greater than the node itself, so we
should update d
to d + 1
:
def add_leaves(t, d):
for b in t.branches:
add_leaves(b, d + 1)
...
Now that we've made these recursive calls, let's take a step back and look at our progress. Taking the leap of faith, we know that each recursive call should've successfully added the correct number of leaves at each node in each branch. That means that the only step left is to add the correct number of leaves to the current node!
The parameter d
tells us how many leaves to add at this node. Since we are
mutating t
to add these leaves, we need to mutate the list of t
's branches.
We know a few different ways to mutatively add elements to a list:insert
,
append
, and extend
. Which one makes most sense to use here? Well, we know that
we have to add d
elements to the end of t.branches
. Index doesn't matter
so we can rule out insert
. append
is good for adding a single element, while
extend
is useful for adding multiple elements contained in a list, so let's
use extend
!
The input to extend
should be a list, so how do we create a list with the
leaves that we need? The most concise way is with a list comprehension. To
create each leaf, we call Tree(v)
:
[Tree(v) for _ in range(d)]
Now, we just have to extend t.branches
by this list:
def add_leaves(t, d):
for b in t.branches:
add_leaves(b, d + 1)
t.branches.extend([Tree(v) for _ in range(d)])
Do we need an explicitly base case? Let's take a look at what happens when t
is a leaf. In that case, t.branches
would be an empty list, so we would not
enter the for
loop. Then, the function will extend t.branches
, which is an
empty list, by a list containing the new leaves. This is exactly the desired
result, so no base case is needed!
Efficiency
Q5: Efficiency Practice
Choose the term that fills in the blank for the functions defined below:
<function>
runs in ____
time in the length of its input.
- Constant
- Logarithmic
- Linear
- Quadratic
- Exponential
- None of these
Assume that len
runs in constant time
and all
runs in linear time in the length of its input.
Selecting an element of a list by its index requires constant time.
Constructing a range requires constant time.
def count_partitions(n, m):
"""Counts the number of partitions of a positive integer n,
using parts up to size m."""
if n == 0:
return 1
elif n < 0:
return 0
elif m == 0:
return 0
else:
with_m = count_partitions(n-m, m)
without_m = count_partitions(n, m-1)
return with_m + without_m
def is_palindrome(s):
"""Return whether a list of numbers s is a palindrome."""
return all([s[i] == s[len(s) - i - 1] for i in range(len(s))])
def binary_search(lst, n):
"""Takes in a sorted list lst and returns the index where integer n
is contained in lst. Returns -1 if n does not exist in lst."""
low = 0
high = len(lst)
while low <= high:
middle = (low + high) // 2
if lst[middle] == n:
return middle
elif n < lst[middle]:
high = middle - 1
else:
low = middle + 1
return -1
The
is_palindrome
question was reformatted from question 6(d) on fall 2019's final.
Use Ok to test your understanding:
python3 ok -q efficiency_practice -u
count_partitions: exponential.
count_partitions
takes two inputs n
and m
so the runtime depends on both inputs. However, when determining what the order of growth is, we're mostly concerned with what happens to the runtime for larger values of n
and m
.
The conditional statements such as if n == 0
are logical comparisons that take constant time. Therefore, what affects our runtime the most are the number of recursive calls we to count_partitions
. For every function call to count_partitions
we make two more recursive calls; one that uses m
as a partition, and one that does not.
Visualizing the tree of recursive calls for this function, we'll see that the number of function calls at each level doubles. Thus, the runtime for count_partitions
is exponential.
is_palindrome: linear.
We're interested in seeing how the function runs in relation to
the length of its input, which is len(s)
.
In is_palindrome
, it takes constant time to calculate len(s)
and then to construct a range from 0 to len(s)
(exclusive).
For each element in this range, we will select two elements from s
(s[i]
and s[len(s)-i-1]
) and compare them, which takes some
constant time for each element.
Once we've done this for all the elements,
we will have built up the input list to all
in linear time in relation
to the length of is_palindrome
's input.
We assume that all
runs in linear time in the length of its input,
which is the length of the list we've just built and the same as
the length of is_palindrome
's input.
Overall, is_palindrome
will therefore take linear time
in relation to the length of its input.
binary_search: logarithmic.
The binary_search
algorithm's runtime is dependent on the length of lst
because we are searching through it to find some integer n
.
Again, assuming len(lst)
and logical comparisons take constant time to calculate, we look at how many iterations of the while loop we might go through before completing the function call. First we check the element at the middle of our sorted list. If lst[middle]
is not equal to n
, the function determines whether lst[middle]
is less than or greater than n
and narrows our search down to either the lower half or upper half of lst
, respectively.
For every iteration of the while loop, binary_search
cuts down the number of elements in the input lst
we are searching through by half. Thus, the maximum number of iterations for the while loop before returning an index or -1
is log(len(lst)) in base 2. binary_search
takes logarithmic time in relation to the length of its input lst
.
Submit
Make sure to submit this assignment by running:
python3 ok --submit
Extra Practice
Recursion and Tree Recursion
Q6: Subsequences
A subsequence of a sequence S
is a subset of elements from S
, in the same
order they appear in S
. Consider the list [1, 2, 3]
. Here are a few of it's
subsequences []
, [1, 3]
, [2]
, and [1, 2, 3]
.
Write a function that takes in a list and returns all possible subsequences of that list. The subsequences should be returned as a list of lists, where each nested 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 each nested list,
and returns the resulting list.
def insert_into_all(item, nested_list):
"""Return a new list consisting of all the lists in nested_list,
but with item added to the front of each. You can assume that
nested_list is a list of lists.
>>> 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):
"""Return a nested list (a list of lists) of all subsequences of S.
The subsequences can appear in any order. You can assume S is a list.
>>> 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
Q7: Non-Decreasing 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 non_decrease_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 non_decrease_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 = non_decrease_subseqs([1, 3, 2])
>>> sorted(seqs)
[[], [1], [1, 2], [1, 3], [2], [3]]
>>> non_decrease_subseqs([])
[[]]
>>> seqs2 = non_decrease_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 non_decrease_subseqs
Mutable Lists
Q8: 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.
To interleave two sequences s0
and s1
is to create a new sequence such that the new sequence contains (in this order) the first element of s0
, the first element of s1
, the second element of s0
, the second element of s1
, and so on. If the two lists are not the same length, then the leftover elements of the longer list should still appear at the end.
Note: If you're running into an issue where the special heart / diamond / spades / clubs symbols are erroring in the doctests, feel free to copy paste the below doctests into your file as these don't use the special characters and should not give an "illegal multibyte sequence" error.
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 = ['H', 'D', 'S', 'C']
>>> cards = [card(n) + suit for n in range(1,14) for suit in suits]
>>> cards[:12]
['AH', 'AD', 'AS', 'AC', '2H', '2D', '2S', '2C', '3H', '3D', '3S', '3C']
>>> cards[26:30]
['7S', '7C', '8H', '8D']
>>> shuffle(cards)[:12]
['AH', '7S', 'AD', '7C', 'AS', '8H', 'AC', '8D', '2H', '8S', '2D', '8C']
>>> shuffle(shuffle(cards))[:12]
['AH', '4D', '7S', '10C', 'AD', '4S', '7C', 'JH', 'AS', '4C', '8H', 'JD']
>>> cards[:12] # Should not be changed
['AH', 'AD', 'AS', 'AC', '2H', '2D', '2S', '2C', '3H', '3D', '3S', '3C']
"""
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
Generators & Iterators
Q9: Pairs (generator)
Write a generator function pairs
that takes a list and yields all the
possible pairs of elements from that list.
def pairs(lst):
"""
>>> type(pairs([3, 4, 5]))
<class 'generator'>
>>> for x, y in pairs([3, 4, 5]):
... print(x, y)
...
3 3
3 4
3 5
4 3
4 4
4 5
5 3
5 4
5 5
"""
for i in lst:
for j in lst:
yield i, j
Use Ok to test your code:
python3 ok -q pairs
Q10: Pairs (iterator)
Important note (March 14): This question is not in scope for this semester (Spring 2022), as it involves
__iter__
and__next__
.
Now write an iterator that does the same thing. You are only allowed to use a linear amount of space - so computing a list of all of the possible pairs is not a valid answer. Notice how much harder it is - this is why generators are useful.
class PairsIterator:
"""
>>> for x, y in PairsIterator([3, 4, 5]):
... print(x, y)
...
3 3
3 4
3 5
4 3
4 4
4 5
5 3
5 4
5 5
"""
def __init__(self, lst):
self.lst = lst
self.i = 0
self.j = 0
def __next__(self):
if self.i == len(self.lst):
raise StopIteration
result = (self.lst[self.i], self.lst[self.j])
if self.j == len(self.lst) - 1:
self.i += 1
self.j = 0
else:
self.j += 1
return result
def __iter__(self):
return self
Use Ok to test your code:
python3 ok -q PairsIterator
Trees
Q11: Long Paths
Implement long_paths
, which returns a list of all paths in a tree with
length at least n
. A path in a tree is a linked list of node values that
starts with the root and ends at a leaf. Each subsequent element must be from a
child of the previous value's node. The length of a path is the number of
edges in the path (i.e. one less than the number of nodes in the path).
Paths are listed in order from left to right. See the doctests for some
examples.
def long_paths(tree, n):
"""Return a list of all paths in tree with length at least n.
>>> t = Tree(3, [Tree(4), Tree(4), Tree(5)])
>>> left = Tree(1, [Tree(2), t])
>>> mid = Tree(6, [Tree(7, [Tree(8)]), Tree(9)])
>>> right = Tree(11, [Tree(12, [Tree(13, [Tree(14)])])])
>>> whole = Tree(0, [left, Tree(13), mid, right])
>>> for path in long_paths(whole, 2):
... print(path)
...
<0 1 2>
<0 1 3 4>
<0 1 3 4>
<0 1 3 5>
<0 6 7 8>
<0 6 9>
<0 11 12 13 14>
>>> for path in long_paths(whole, 3):
... print(path)
...
<0 1 3 4>
<0 1 3 4>
<0 1 3 5>
<0 6 7 8>
<0 11 12 13 14>
>>> long_paths(whole, 4)
[Link(0, Link(11, Link(12, Link(13, Link(14)))))]
"""
paths = []
if n <= 0 and tree.is_leaf():
paths.append(Link(tree.label))
for b in tree.branches:
for path in long_paths(b, n - 1):
paths.append(Link(tree.label, path))
return paths
Use Ok to test your code:
python3 ok -q long_paths
Linked Lists
Q12: Flip Two
Write a recursive function flip_two
that takes as input a
linked list s
and mutates s
so that every pair of values in the linked list
is flipped.
def flip_two(s):
"""
>>> one_lnk = Link(1)
>>> flip_two(one_lnk)
>>> one_lnk
Link(1)
>>> lnk = Link(1, Link(2, Link(3, Link(4, Link(5)))))
>>> flip_two(lnk)
>>> lnk
Link(2, Link(1, Link(4, Link(3, Link(5)))))
"""
# Recursive solution:
if s is Link.empty or s.rest is Link.empty:
return
s.first, s.rest.first = s.rest.first, s.first
flip_two(s.rest.rest)
# For an extra challenge, try writing out an iterative approach as well below!
return # separating recursive and iterative implementations
# Iterative approach
while s is not Link.empty and s.rest is not Link.empty:
s.first, s.rest.first = s.rest.first, s.first
s = s.rest.rest
Use Ok to test your code:
python3 ok -q flip_two