Lab 14 Solutions

lab14.zip

Solution Files

This lab has many files. Remember to write in lab14.scm for the Scheme questions, lab14.lark for the BNF question, and lab14.py for all other questions.

Required Questions

Scheme

Q1: Split

Implement split-at, which takes a list lst and a non-negative number n as input and returns a pair new such that (car new) is the first n elements of lst and (cdr new) is the remaining elements of lst. If n is greater than the length of lst, (car new) should be lst and (cdr new) should be nil.

scm> (car (split-at '(2 4 6 8 10) 3))
(2 4 6)
scm> (cdr (split-at '(2 4 6 8 10) 3))
(8 10)

(define (split-at lst n)
  (cond ((= n 0) (cons nil lst))
        ((null? lst) (cons lst nil))
        (else (let ((rec (split-at (cdr lst) (- n 1))))
                (cons (cons (car lst) (car rec)) (cdr rec)))))
)

Use Ok to test your code:

python3 ok -q split-at

Scheme Data Abstraction

Q2: Filter Odd Tree

Write a function filter-odd which takes a tree data abstraction and returns a new tree with all even labels replaced with nil.

Consider using the map procedure to apply a one-argument function to a list.

Below is a Scheme-ified data abstraction of the Tree class we've been working with this semester.

; Constructs tree given label and list of branches
(tree label branches)

; Returns the label of the tree
(label t)

; Returns the list of branches of the given tree
(branches t)

(define (filter-odd t)
	(cond
		((null? t) nil)
		((odd? (label t)) (tree (label t) (map filter-odd (branches t))))
		(else (tree nil (map filter-odd (branches t))))
	)
)

Use Ok to test your code:

python3 ok -q filter_odd

Programs as Data

Q3: Swap

Implement swap which takes an expression expr representing a call to some procedure and returns the same expression with its first two operands swapped if the value of the second operand is greater than the value of the first. Otherwise, it should just return the original expression. For example, (swap '(- 1 (+ 3 5) 7)) should return the expression (- (+ 3 5) 1 7) since 1 evaluates to 1, (+ 3 5) evaluates to 8, and 8 > 1. Any operands after the first two should not be evaluated during the execution of the procedure, and they should be left unchanged in the final expression. You may assume that every operand evaluates to a number and that there are always at least two operands in expr. You may want to consider using a let expression in addition to the provided procedures to help simplify your code.

(define (cddr s)
  (cdr (cdr s))
)

(define (cadr s)
  (car (cdr s))
)

(define (caddr s)
  (car (cddr s))
)

(define (swap expr)
    (let ((op (car expr))
        (first (car (cdr expr)))
        (second (caddr expr))
        (rest (cdr (cddr expr))))
        (if (> (eval second) (eval first))
            (cons op (cons second (cons first rest)))
            expr)
    )
)

Use Ok to test your code:

python3 ok -q swap

Regex

Q4: Address First Line

Write a regular expression that parses strings and returns whether it contains the first line of a US mailing address.

US mailing addresses typically contain a block number, which is a sequence of 3-5 digits, following by a street name. The street name can consist of multiple words but will always end with a street type abbreviation, which itself is a sequence of 2-5 English letters. The street name can also optionally start with a cardinal direction ("N", "E", "W", "S"). Everything should be properly capitalized.

Proper capitalization means that the first letter of each name is capitalized. It is fine to have things like "WeirdCApitalization" match.

See the doctests for some examples.

def address_oneline(text):
    """
    Finds and returns if there are expressions in text that represent the first line
    of a US mailing address.

    >>> address_oneline("110 Sproul Hall, Berkeley, CA 94720")
    True
    >>> address_oneline("What's at 39177 Farwell Dr? Is there a 39177 Nearwell Dr?")
    True
    >>> address_oneline("I just landed at 780 N McDonnell Rd, and I need to get to 1880-ish University Avenue. Help!")
    True
    >>> address_oneline("123 Le Roy Ave")
    True
    >>> address_oneline("110 Unabbreviated Boulevard")
    False
    >>> address_oneline("790 lowercase St")
    False
    """
    block_number = r"\d{3,5}"
    cardinal_dir = r"(?:[NEWS] )?" # whitespace is important!
    street = r"(?:[A-Z][A-Za-z]+ )+"
    type_abbr = r"[A-Z][a-z]{1,4}\b"    street_name = f"{cardinal_dir}{street}{type_abbr}"
    return bool(re.search(f"{block_number} {street_name}", text))

Use Ok to test your code:

python3 ok -q address_oneline

BNF

Q5: WWPD: PyCombinator

Consider this attempt at a BNF grammar for Pycombinator, a grammar which supports a subset of Python's functionality. Specifically, it is able to parse any expression with a Python arithmetic operator. The grammar is specified below:

?start: pycomb_expression

pycomb_expression: func "(" arg ("," arg)* ")"

arg: pycomb_expression | NUMBER

func: FUNCNAME

FUNCNAME: "add" | "mul" | "sub"

%ignore " "
%import common.NUMBER

Let's understand and modify the functionality of this BNF with a few questions.

Use Ok to test your knowledge by choosing the best answer for each of the following "What Would PyCombinator Do" questions:
python3 ok -q wwpd-bnf -u

Submit

Make sure to submit this assignment by running:

python3 ok --submit

Recommended Questions

The following problems are not required for credit on this lab but may help you prepare for the final.

Trees

Q6: Prune Min

Write a function that prunes a Tree t mutatively. t and its branches always have zero or two branches. For the trees with two branches, reduce the number of branches from two to one by keeping the branch that has the smaller label value. Do nothing with trees with zero branches.

Prune the tree in a direction of your choosing (top down or bottom up). The result should be a linear tree.

def prune_min(t):
    """Prune the tree mutatively.

    >>> t1 = Tree(6)
    >>> prune_min(t1)
    >>> t1
    Tree(6)
    >>> t2 = Tree(6, [Tree(3), Tree(4)])
    >>> prune_min(t2)
    >>> t2
    Tree(6, [Tree(3)])
    >>> t3 = Tree(6, [Tree(3, [Tree(1), Tree(2)]), Tree(5, [Tree(3), Tree(4)])])
    >>> prune_min(t3)
    >>> t3
    Tree(6, [Tree(3, [Tree(1)])])
    """
    if t.branches == []:
        return
    prune_min(t.branches[0])
    prune_min(t.branches[1])
    if (t.branches[0].label > t.branches[1].label):
        t.branches.pop(0)
    else:
        t.branches.pop(1)
    return # return statement to block alternate from running

# Alternate solution
    if t.is_leaf():
        return
    remove_ind = int(t.branches[0].label < t.branches[1].label)
    t.branches.pop(remove_ind)
    prune_min(t.branches[0])

Use Ok to test your code:

python3 ok -q prune_min

Q7: Add trees

Define the function add_trees, which takes in two trees and returns a new tree where each corresponding node from the first tree is added with the node from the second tree. If a node at any particular position is present in one tree but not the other, it should be present in the new tree as well.

Hint: You may want to use the built-in zip function to iterate over multiple sequences at once.

def add_trees(t1, t2):
    """
    >>> numbers = Tree(1,
    ...                [Tree(2,
    ...                      [Tree(3),
    ...                       Tree(4)]),
    ...                 Tree(5,
    ...                      [Tree(6,
    ...                            [Tree(7)]),
    ...                       Tree(8)])])
    >>> print(add_trees(numbers, numbers))
    2
      4
        6
        8
      10
        12
          14
        16
    >>> print(add_trees(Tree(2), Tree(3, [Tree(4), Tree(5)])))
    5
      4
      5
    >>> print(add_trees(Tree(2, [Tree(3)]), Tree(2, [Tree(3), Tree(4)])))
    4
      6
      4
    >>> print(add_trees(Tree(2, [Tree(3, [Tree(4), Tree(5)])]), \
    Tree(2, [Tree(3, [Tree(4)]), Tree(5)])))
    4
      6
        8
        5
      5
    """
    if not t1:
        return t2
    if not t2:
        return t1
    new_label = t1.label + t2.label
    t1_branches, t2_branches = list(t1.branches), list(t2.branches)
    length_t1, length_t2 = len(t1_branches), len(t2_branches)
    if length_t1 < length_t2:
        t1_branches += [None for _ in range(length_t1, length_t2)]
    elif length_t1 > length_t2:
        t2_branches += [None for _ in range(length_t2, length_t1)]
    return Tree(new_label, [add_trees(branch1, branch2) for branch1, branch2 in zip(t1_branches, t2_branches)])

Use Ok to test your code:

python3 ok -q add_trees

Objects

Let's implement a game called Election. In this game, two players compete to try and earn the most votes. Both players start with 0 votes and 100 popularity.

The two players alternate turns, and the first player starts. Each turn, the current player chooses an action. There are two types of actions:

The player can debate, and either gain or lose 50 popularity. If the player has popularity p1 and the other player has popularity p2, then the probability that the player gains 50 popularity is max(0.1, p1 / (p1 + p2)) Note that the max causes the probability to never be lower than 0.1.
The player can give a speech. If the player has popularity p1 and the other player has popularity p2, then the player gains p1 // 10 votes and popularity and the other player loses p2 // 10 popularity.

The game ends when a player reaches 50 votes, or after a total of 10 turns have been played (each player has taken 5 turns). Whoever has more votes at the end of the game is the winner!

Q8: Player

First, let's implement the Player class. Fill in the debate and speech methods, that take in another Player other, and implement the correct behavior as detailed above. Here are two additional things to keep in mind:

In the debate method, you should call the provided random function, which returns a random float between 0 and 1. The player should gain 50 popularity if the random number is smaller than the probability described above, and lose 50 popularity otherwise.
Neither players' popularity should ever become negative. If this happens, set it equal to 0 instead.

### Phase 1: The Player Class
class Player:
    """
    >>> random = make_test_random()
    >>> p1 = Player('Hill')
    >>> p2 = Player('Don')
    >>> p1.popularity
    100
    >>> p1.debate(p2)  # random() should return 0.0
    >>> p1.popularity
    150
    >>> p2.popularity
    100
    >>> p2.votes
    0
    >>> p2.speech(p1)
    >>> p2.votes
    10
    >>> p2.popularity
    110
    >>> p1.popularity
    135


    >>> # Additional correctness tests
    >>> p1.speech(p2)
    >>> p1.votes
    13
    >>> p1.popularity
    148
    >>> p2.votes
    10
    >>> p2.popularity
    99
    >>> for _ in range(4):  # 0.1, 0.2, 0.3, 0.4
    ...     p1.debate(p2)
    >>> p2.debate(p1)
    >>> p2.popularity
    49
    >>> p2.debate(p1)
    >>> p2.popularity
    0
    """
    def __init__(self, name):
        self.name = name
        self.votes = 0
        self.popularity = 100

    def debate(self, other):
        prob = max(0.1, self.popularity / (self.popularity + other.popularity))
        if random() < prob:
            self.popularity += 50
        else:
            self.popularity = max(0, self.popularity - 50)
    def speech(self, other):
        self.votes += self.popularity // 10
        self.popularity += self.popularity // 10
        other.popularity -= other.popularity // 10
    def choose(self, other):
        return self.speech

Use Ok to test your code:

python3 ok -q Player

Q9: Game

Now, implement the Game class. Fill in the play method, which should alternate between the two players, starting with p1, and have each player take one turn at a time. The choose method in the Player class returns the method, either debate or speech, that should be called to perform the action.

In addition, fill in the winner method, which should return the player with more votes, or None if the players are tied.

### Phase 2: The Game Class
class Game:
    """
    >>> p1, p2 = Player('Hill'), Player('Don')
    >>> g = Game(p1, p2)
    >>> winner = g.play()
    >>> p1 is winner
    True


    >>> # Additional correctness tests
    >>> winner is g.winner()
    True
    >>> g.turn
    10
    >>> p1.votes = p2.votes
    >>> print(g.winner())
    None
    """
    def __init__(self, player1, player2):
        self.p1 = player1
        self.p2 = player2
        self.turn = 0

    def play(self):
        while not self.game_over():
            if self.turn % 2 == 0:
                curr, other = self.p1, self.p2
            else:
                curr, other = self.p2, self.p1
            curr.choose(other)(other)
            self.turn += 1        return self.winner()

    def game_over(self):
        return max(self.p1.votes, self.p2.votes) >= 50 or self.turn >= 10

    def winner(self):
        if self.p1.votes > self.p2.votes:
            return self.p1
        elif self.p2.votes > self.p1.votes:
            return self.p2
        else:
            return None

Use Ok to test your code:

python3 ok -q Game

Q10: New Players

The choose method in the Player class is boring, because it always returns the speech method. Let's implement two new classes that inherit from Player, but have more interesting choose methods.

Implement the choose method in the AggressivePlayer class, which returns the debate method if the player's popularity is less than or equal to other's popularity, and speech otherwise. Also implement the choose method in the CautiousPlayer class, which returns the debate method if the player's popularity is 0, and speech otherwise.

### Phase 3: New Players
class AggressivePlayer(Player):
    """
    >>> random = make_test_random()
    >>> p1, p2 = AggressivePlayer('Don'), Player('Hill')
    >>> g = Game(p1, p2)
    >>> winner = g.play()
    >>> p1 is winner
    True


    >>> # Additional correctness tests
    >>> p1.popularity = p2.popularity
    >>> p1.choose(p2) == p1.debate
    True
    >>> p1.popularity += 1
    >>> p1.choose(p2) == p1.debate
    False
    >>> p2.choose(p1) == p2.speech
    True
    """
    def choose(self, other):
        if self.popularity <= other.popularity:
            return self.debate
        else:
            return self.speech
class CautiousPlayer(Player):
    """
    >>> random = make_test_random()
    >>> p1, p2 = CautiousPlayer('Hill'), AggressivePlayer('Don')
    >>> p1.popularity = 0
    >>> p1.choose(p2) == p1.debate
    True
    >>> p1.popularity = 1
    >>> p1.choose(p2) == p1.debate
    False


    >>> # Additional correctness tests
    >>> p2.choose(p1) == p2.speech
    True
    """
    def choose(self, other):
        if self.popularity == 0:
            return self.debate
        else:
            return self.speech

Use Ok to test your code:

python3 ok -q AggressivePlayer
python3 ok -q CautiousPlayer

Lists

Q11: Intersection - Summer 2015 MT1 Q4

Implement intersection(lst_of_lsts), which takes a list of lists and returns a list of distinct elements that appear in all the lists in lst_of_lsts. If no number appears in all of the lists, return the empty list. You may assume that lst_of_lsts contains at least one list.

def intersection(lst_of_lsts):
    """Returns a list of distinct elements that appear in every list in
    lst_of_lsts.

    >>> lsts1 = [[1, 2, 3], [1, 3, 5]]
    >>> intersection(lsts1)
    [1, 3]
    >>> lsts2 = [[1, 4, 2, 6], [7, 2, 4], [4, 4]]
    >>> intersection(lsts2)
    [4]
    >>> lsts3 = [[1, 2, 3], [4, 5], [7, 8, 9, 10]]
    >>> intersection(lsts3)         # No number appears in all lists
    []
    >>> lsts4 = [[3, 3], [1, 2, 3, 3], [3, 4, 3, 5]]
    >>> intersection(lsts4)         # Return list of distinct elements
    [3]
    """
    elements = []
    for elem in lst_of_lsts[0]:
        condition = elem not in elements
        for lst in lst_of_lsts[1:]:
            if elem not in lst:
                condition = False
        if condition:
            elements = elements + [elem]
    return elements

Use Ok to test your code:

python3 ok -q intersection

Q12: Deck of cards

Write a list comprehension that will create a deck of cards, given a list of suits and a list of ranks. Each element in the list will be a card, which is represented by a 2-element list of the form [suit, rank].

def deck(suits, ranks):
    """Creates a deck of cards (a list of 2-element lists) with the given
    suits and ranks. Each element in the returned list should be of the form
    [suit, rank].

    >>> deck(['S', 'C'], [1, 2, 3])
    [['S', 1], ['S', 2], ['S', 3], ['C', 1], ['C', 2], ['C', 3]]
    >>> deck(['S', 'C'], [3, 2, 1])
    [['S', 3], ['S', 2], ['S', 1], ['C', 3], ['C', 2], ['C', 1]]
    >>> deck([], [3, 2, 1])
    []
    >>> deck(['S', 'C'], [])
    []
    """
    return [[suit, rank] for suit in suits
                           for rank in ranks]

Use Ok to test your code:

python3 ok -q deck

Linked Lists

Q13: O!-Pascal - Fall 2017 Final Q4

Pasal's Triangle is perhaps familiar to you from the diagram below, which shows the first five rows.

Pascal's Triangle

Every square is the sum of the two squares above it (as illustrated by the arrows showing here the value 4 comes from), unless it doesn't have two squares above it, in whih case its value is 1.

Given a linked list that represents a row in Pasal's triangle, return a linked list that will represent the row below it.

Diagonal

def pascal_row(s):
    """
    >>> a = Link.empty
    >>> for _ in range(5):
    ...     a = pascal_row(a)
    ...     print(a)
    <1>
    <1 1>
    <1 2 1>
    <1 3 3 1>
    <1 4 6 4 1>
    """
    if s is Link.empty:
        return Link(1)
    start = Link(1)
    last, current = start, s
    while current.rest is not Link.empty:
        last.rest = Link(current.first + current.rest.first)
        last, current = last.rest, current.rest
    last.rest = Link(1)
    return start

Use Ok to test your code:

python3 ok -q pascal_row