Due at 11:59pm on 07/07/2016.

## Starter Files

Download lab05.zip. Inside the archive, you will find starter files for the questions in this lab, along with a copy of the OK autograder.

## Submission

By the end of this lab, you should have submitted the lab with python3 ok --submit. You may submit more than once before the deadline; only the final submission will be graded.

• Questions 1, 2, and 3 must be completed in order to receive credit for this lab. Starter code for questions 2 and 3 is in lab05.py.
• Questions 4 through 11 (Coding) are optional. It is recommended that you complete these problems on your own time. Starter code for these questions is in lab05_extra.py.

# 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.

## List Comprehension

List comprehensions are a compact and powerful way of creating new lists out of sequences. Let's work with them directly:

>>> [i**2 for i in [1, 2, 3, 4] if i%2 == 0]
[4, 16]

is equivalent to

>>> lst = []
>>> for i in [1, 2, 3, 4]:
...     if i % 2 == 0:
...         lst += [i**2]
>>> lst
[4, 16]

The general syntax for a list comprehension is

[<expression> for <element> in <sequence> if <conditional>]

The syntax is designed to read like English: "Compute the expression for each element in the sequence if the conditional is true."

Note: The if clause in a list comprehension is optional.

Python has many built-in types of sequences: lists, ranges, and strings, to name a few. In this lab, we instead construct our own type of sequence called a linked list. A linked list is a simple type of sequence that is comprised of multiple links that are connected.

Each link is a pair where the first element is an item in the linked list, and the second element is another link.

• Constructors:

• empty: The empty linked list.
• Selectors

• first(s): Returns the first element in the given linked list s.
• rest(s): Returns the rest of the linked list s.
• Other

We can construct the Linked list shown above by using the constructors. The first element of this Linked list is 12 while the rest is another Linked list that contains 99 and 37:

>>> first(x)
12
>>> first(rest(x))
99
>>> first(rest(rest(x)))
37

Note: Notice that we can just use link(37) instead link(37, empty). This is because the second argument of the link constructor has a default argument of empty.

# Required Questions

## What Would Python Display?

### Question 1: WWPD: Lists?

What would Python display? Try to figure it out before you type it into the interpreter!

Use OK to test your knowledge with the following "What Would Python Display?" questions:

python3 ok -q lists -u
>>> [x*x for x in range(5)]
______
[0, 1, 4, 9, 16]
>>> [n for n in range(10) if n % 2 == 0]
______
[0, 2, 4, 6, 8]
>>> ones = [1 for i in ["hi", "bye", "you"]] >>> ones + [str(i) for i in [6, 3, 8, 4]]
______
[1, 1, 1, '6', '3', '8', '4']
>>> [i+5 for i in [n for n in range(1,4)]]
______
[6, 7, 8]
>>> [i**2 for i in range(10) if i < 3]
______
[0, 1, 4]
>>> lst = ['hi' for i in [1, 2, 3]] >>> print(lst)
______
['hi', 'hi', 'hi']
>>> lst + [i for i in ['1', '2', '3']]
______
['hi', 'hi', 'hi', '1', '2', '3']

## Coding Practice

### Question 2: Coordinates

Implement a function coords that takes a function fn, a sequence seq, and a lower and upper bound on the output of the function. coords then returns a list of coordinate pairs (lists) such that:

• Each (x, y) pair is represented as [x, fn(x)]
• The x-coordinates are elements in the sequence
• The result contains only pairs whose y-coordinate is within the upper and lower bounds (inclusive)

See the doctest for examples.

Note: your answer can only be one line long. You should make use of list comprehensions!

def coords(fn, seq, lower, upper):
"""
>>> seq = [-4, -2, 0, 1, 3]
>>> fn = lambda x: x**2
>>> coords(fn, seq, 1, 9)
[[-2, 4], [1, 1], [3, 9]]
"""
"*** YOUR CODE HERE ***" return ______
return [[x, fn(x)] for x in seq if lower <= fn(x) <= upper]

Use OK to test your code:

python3 ok -q coords

### Question 3: Is Sorted?

Implement the is_sorted(lst) function, which returns True if the linked list lst is sorted in increasing from left to right. If two adjacent elements are equal, the linked list can still be considered sorted.

def is_sorted(lst):
"""Returns True if the linked list is sorted.

>>> is_sorted(lst1)
True
>>> is_sorted(lst2)
False
>>> is_sorted(lst3)
True
"""
if lst == empty or rest(lst) == empty:: return True elif first(lst) > first(rest(lst)): return False return is_sorted(rest(lst))

Use OK to test your code:

python3 ok -q is_sorted

# Optional Questions

## Coding Practice

Note: The following questions are in lab05_extra.py.

### Question 4: Sum

Write a function that takes in a linked list lst and a function fn which is applied to each number in lst and returns the sum. If the linked list is empty, the sum is 0.

"""Applies a function FN to each number in LST and returns the sum
of the resulting values

>>> square = lambda x: x * x
>>> double = lambda y: 2 * y
30
44
"""
if lst == empty: return 0 return fn(first(lst)) + sum_linked_list(rest(lst), fn) # Iterative Solution def sum_linked_list(lst, fn): sum = 0 while lst != empty: sum += fn(first(lst)) lst = rest(lst) return sum

Use OK to test your code:

### Question 5: Change

Write a function that takes in a linked list, lst, and two elements, s and t. The function returns lst but with all instances of s replaced with t.

def change(lst, s, t):
"""Returns a link matching lst but with all instances of s
replaced by t. If s does not appear in lst, then return lst

>>> new = change(lst, 3, 1)
1 2 1
>>> newer = change(new, 1, 2)
2 2 2
2 2 2
"""
if lst == empty: return lst if first(lst) == s: return link(t, change(rest(lst), s, t)) return link(first(lst), change(rest(lst), s, t))

Use OK to test your code:

python3 ok -q change

### Question 6: Link to List

Write a function link_to_list that takes a linked list and converts it to a Python list.

Hint: To check if a linked list is empty, you can use lst == empty. Also, you can combine two Python lists using +.

"""Return a list that contains the values inside of linked_lst

[]
[1, 2, 3]
"""

Use OK to test your code:

### Question 7: Insert

Implement the insert function that creates a copy of the original list with an item inserted at the specific index. If the index is greater than the current length, you should insert the item at the end of the list. Review your solution for change if you are stuck.

Hint: This will be much easier to implement using recursion, rather than using iteration!

Note: Remember we are not actually inserting the item into the original linked list. Instead, we are creating a copy of the original linked list, but with the provided item added at the specified index. The original linked list stays the same.

def insert(lst, item, index):
"""Returns a link matching lst but with the given item inserted at the
specified index. If the index is greater than the current length, the item
is appended to the end of the list.

>>> new = insert(lst, 9001, 1)
1 9001 2 3
>>> newer = insert(new, 9002, 15)
1 9001 2 3 9002
"""
if lst == empty: return link(item, empty) elif index == 0: return link(item, lst) else: return link(first(lst), insert(rest(lst), item, index-1))

Use OK to test your code:

python3 ok -q insert

### Question 8: Interleave

Write interleave(s0, s1), which takes two linked lists and produces a new linked list with elements of s0 and s1 interleaved. In other words, the resulting list should have the first element of the 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.

def interleave(s0, s1):
list.

1 2 3 4 6 8
2 1 4 3 6 8
1 1 3 3
"""
# Recursive version if s0 == empty: return s1 elif s1 == empty: return s0 return link(first(s0), link(first(s1), interleave(rest(s0), rest(s1)))) # Iterative version def interleave(s0, s1): interleaved = empty while s0 != empty and s1 != empty: interleaved = link(first(s1), link(first(s0), interleaved)) s0, s1 = rest(s0), rest(s1) remaining = s1 if s0 == empty else s0 while remaining != empty: interleaved = link(first(remaining), interleaved) remaining = rest(remaining) return reverse_iterative(interleaved) def reverse_iterative(s): rev_list = empty while s != empty: rev_list = link(first(s), rev_list) s = rest(s) return rev_list

Use OK to test your code:

python3 ok -q interleave

### Question 9: Filter

Implement a filter_list function that takes a linked list lst and returns a new linked list only containing elements from lst that satisfy predicate. Remember, recursion is your friend!

def filter_list(predicate, lst):
"""Returns a link only containing elements in lst that satisfy
predicate.

>>> new = filter_list(lambda x : x % 2 == 0, lst)
50 80
"""
if lst == empty: return lst elif predicate(first(lst)): return link(first(lst), filter_list(predicate, rest(lst))) else: return filter_list(predicate, rest(lst))

Use OK to test your code:

python3 ok -q filter_list

### Question 10: Reverse

Write iterative and recursive functions that reverse a given linked list, producing a new linked list with the elements in reverse order. Use only the link constructor and first and rest selectors to manipulate linked lists. (You may write and use helper functions.)

def reverse_iterative(s):
"""Return a reversed version of a linked list s.

>>> reversed_primes = reverse_iterative(primes)
7 5 3 2
"""
rev_list = empty while s != empty: rev_list = link(first(s), rev_list) s = rest(s) return rev_list
return reverse_helper(s, empty) def reverse_helper(s, tail): if s == empty: return tail return reverse_helper(rest(s), link(first(s), tail))

Use OK to test your code:

python3 ok -q reverse_iterative
python3 ok -q reverse_recursive

### Question 11: Kth to Last

Implement the kth_last(lst, k) function, which returns the element that is k positions from the last element.

def kth_last(lst, k):
"""Return the kth to last element of `lst`.