# CS 61A: Homework 5

Due by 11:59pm on Wednesday, 10/15

Submission: See Lab 1 for submission instructions. We have provided a hw5.py starter file for the questions below.

Readings: You might find the following references useful:

### Question 1

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'

### Question 2

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 integers that are equal in sum. 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.

``````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]
'Deal!'
>>> a
[4, 3, 1, 1, 4]
>>> b
[1, 1, 3, 2, 2, 7]
>>> c = [3, 3, 2, 4, 1]
'No deal!'
>>> b
[1, 1, 3, 2, 2, 7]
>>> c
[3, 3, 2, 4, 1]
'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])
if equal_prefix():
first[:m], second[:n] = second[:n], first[:m]
return 'Deal!'
else:
return 'No deal!'``````

The linked list data abstraction, used below, contains the following functions.

``````################################
# Linked list data abstraction #
################################

empty = 'empty'

"""s is a linked list if it is empty or a (first, rest) pair."""
return s == empty or (len(s) == 2 and is_link(s))

"""Construct a linked list from its first element and the rest."""
return [first, rest]

def first(s):
"""Return the first element of a linked list s."""
assert s != empty, "empty linked list has no first element."
return s

def rest(s):
"""Return the rest of the elements of a linked list s."""
assert s != empty, "empty linked list has no rest."
return s

"""Print elements of a linked list s.

1 2 3
"""
line = ''
while s != empty:
if line:
line += ' '
line += str(first(s))
s = rest(s)
print(line)``````

### Question 3

The mad scientist John Harvey Hilfinger has discovered a gene that compels people to enroll in CS 61A. You may be afflicted!

A DNA sequence is represented as a linked list of elements `A`, `G`, `C` or `T`. This discovered gene has sequence `C A T C A T`. Write a function `has_61A_gene` that takes a DNA sequence and returns whether it contains the 61A gene as a sub-sequence.

First, write a function `has_prefix` that takes two linked lists, `s` and `prefix`, and returns whether `s` starts with the elements of `prefix`. Note that `prefix` may be larger than `s`, in which case the function should return `False`.

``````def has_prefix(s, prefix):
"""Returns whether prefix appears at the beginning of linked list s.

3 4 6 6
>>> has_prefix(x, empty)
True
True
False
True
False
>>> has_prefix(x, x)
True
False
"""

Next, write a function `has_sublist` that takes two linked lists, `s` and `sublist`, and returns whether the elements of `sublist` appear in order anywhere within `s`.

``````def has_sublist(s, sublist):
"""Returns whether sublist appears somewhere within linked list s.

>>> has_sublist(empty, empty)
True
G A T T A C A
>>> has_sublist(x, empty)
True
False
False
True
True
False
"""

Finally, write `has_61A_gene` to detect `C A T C A T` within a linked list `dna` sequence.

``````def has_61A_gene(dna):
"""Returns whether linked list dna contains the CATCAT gene.

C A T G C A T
>>> has_61A_gene(dna)
False
G T A C A T C A T G
>>> has_61A_gene(dna)
True
>>> has_61A_gene(end)
False
"""

Note: Subsequence matching is a problem of importance in computational biology. CS 176 goes into more detail on this topic, including methods that handle errors in the DNA (because DNA sequencing is not 100% correct).

### Question 4

In lecture, we saw how to use functions to create mutable objects. Here, for example, is the function, `make_withdraw,` which produces a function that can withdraw money from an account:

``````def make_withdraw(balance):
"""Return a withdraw function with BALANCE as its starting balance.
>>> withdraw = make_withdraw(1000)
>>> withdraw(100)
900
>>> withdraw(100)
800
>>> withdraw(900)
'Insufficient funds'
"""
def withdraw(amount):
nonlocal balance
if amount > balance:
return 'Insufficient funds'
balance = balance - amount
return balance
return withdraw``````

Write a version of the `make_withdraw` function that returns password-protected withdraw functions. That is, `make_withdraw` should take a password argument (a string) in addition to an initial balance. The returned function should take two arguments: an amount to withdraw and a password.

A password-protected `withdraw` function should only process withdrawals that include a password that matches the original. Upon receiving an incorrect password, the function should:

1. Store that incorrect password in a list, and
2. Return the string 'Incorrect password'.

If a withdraw function has been called three times with incorrect passwords `p1`, `p2`, and `p3`, then it is locked. All subsequent calls to the function should return:

``"Your account is locked. Attempts: [<p1>, <p2>, <p3>]"``

The incorrect passwords may be the same or different:

``````def make_withdraw(balance, password):

>>> w = make_withdraw(100, 'hax0r')
>>> w(25, 'hax0r')
75
>>> w(90, 'hax0r')
'Insufficient funds'
>>> w(25, 'hwat')
>>> w(25, 'hax0r')
50
>>> w(75, 'a')
>>> w(10, 'hax0r')
40
>>> w(20, 'n00b')
>>> w(10, 'hax0r')
"Your account is locked. Attempts: ['hwat', 'a', 'n00b']"
>>> w(10, 'l33t')
"Your account is locked. Attempts: ['hwat', 'a', 'n00b']"
"""

### Question 5

Suppose that our banking system requires the ability to make joint accounts. Define a function `make_joint` that takes three arguments.

1. A password-protected `withdraw` function,
2. The password with which that `withdraw` function was defined, and
3. A new password that can also access the original account.

The `make_joint` function returns a `withdraw` function that provides additional access to the original account using either the new or old password. Both functions draw down the same balance. Incorrect passwords provided to either function will be stored and cause the functions to be locked after three wrong attempts.

Hint: The solution is short (less than 10 lines) and contains no string literals! The key is to call `withdraw` with the right password and interpret the result. You may assume that all failed attempts to withdraw will return some string (for incorrect passwords, locked accounts, or insufficient funds), while successful withdrawals will return a number.

Use `type(value) == str` to test if some `value` is a string:

``````def make_joint(withdraw, old_password, new_password):
the balance of withdraw.

>>> w = make_withdraw(100, 'hax0r')
>>> w(25, 'hax0r')
75
>>> make_joint(w, 'my', 'secret')
>>> j = make_joint(w, 'hax0r', 'secret')
>>> w(25, 'secret')
>>> j(25, 'secret')
50
>>> j(25, 'hax0r')
25
>>> j(100, 'secret')
'Insufficient funds'

>>> j2 = make_joint(j, 'secret', 'code')
>>> j2(5, 'code')
20
>>> j2(5, 'secret')
15
>>> j2(5, 'hax0r')
10

>>> j2(5, 'secret')
>>> j(5, 'secret')
>>> w(5, 'hax0r')
>>> make_joint(w, 'hax0r', 'hello')
"""

ALL FOLLOWING PROBLEMS ARE CHALLENGE PROBLEMS (OPTIONAL)

Section 2.4.9 describes a system for solving equations with multiple free parameters using constraint programming, a declarative style of programming that asserts constraints and then applies a general method of constraint satisfaction. The following questions ask you to extend that system. The code for the system appears at the end of this homework.

### Question 6: Challenge Problem (optional)

Implement the function `triangle_area`, which defines a relation among three connectors, the base `b`, height `h`, and area `a` of a triangle, so that `a = 0.5 * b * h`:

``````def triangle_area(a, b, h):
"""Connect a, b, and h so that a is the area of a triangle with base b and
height h.

>>> a, b, h = [connector(n) for n in ('area', 'base', 'height')]
>>> triangle_area(a, b, h)
>>> a['set_val']('user', 75.0)
area = 75.0
>>> b['set_val']('user', 15.0)
base = 15.0
height = 10.0
"""

### Question 7: Challenge Problem (optional)

The `multiplier` constraint from the readings is insufficient to model equations that include squared quantities because constraint networks must not include loops. Implement a new constraint `squarer` that represents the squaring relation:

``````def squarer(a, b):
"""The constraint that a*a=b.

>>> x, y = connector('X'), connector('Y')
>>> s = squarer(x, y)
>>> x['set_val']('user', 10)
X = 10
Y = 100
>>> x['forget']('user')
X is forgotten
Y is forgotten
>>> y['set_val']('user', 16)
Y = 16
X = 4.0
"""

Use your `squarer` constraint to build a constraint network for the Pythagorean theorem: `a` squared plus `b` squared equals `c` squared:

``````def pythagorean(a, b, c):
"""Connect a, b, and c into a network for the Pythagorean theorem:
a*a + b*b = c*c

>>> a, b, c = [connector(name) for name in ('A', 'B', 'C')]
>>> pythagorean(a, b, c)
>>> a['set_val']('user', 5)
A = 5
>>> c['set_val']('user', 13)
C = 13
B = 12.0
"""

Here is the equation solver implementation from the readings:

``````def connector(name=None):
"""A connector between constraints.

>>> celsius = connector('Celsius')
>>> fahrenheit = connector('Fahrenheit')
>>> converter(celsius, fahrenheit)

>>> celsius['set_val']('user', 25)
Celsius = 25
Fahrenheit = 77.0

>>> fahrenheit['set_val']('user', 212)

>>> celsius['forget']('user')
Celsius is forgotten
Fahrenheit is forgotten

>>> fahrenheit['set_val']('user', 212)
Fahrenheit = 212
Celsius = 100.0
"""
informant = None  # The source of the current val
constraints = []  # A list of connected constraints

def set_value(source, value):
nonlocal informant
val = connector['val']
if val is None:
informant, connector['val'] = source, value
if name is not None:
print(name, '=', value)
inform_all_except(source, 'new_val', constraints)
else:
if val != value:

def forget_value(source):
nonlocal informant
if informant == source:
informant, connector['val'] = None, None
if name is not None:
print(name, 'is forgotten')
inform_all_except(source, 'forget', constraints)
connector = {'val': None,
'set_val': set_value,
'forget': forget_value,
'has_val': lambda: connector['val'] is not None,
'connect': lambda source: constraints.append(source)}

return connector

def inform_all_except(source, message, constraints):
"""Inform all constraints of the message, except source."""
for c in constraints:
if c != source:
c[message]()

def ternary_constraint(a, b, c, ab, ca, cb):
"""The constraint that ab(a,b)=c and ca(c,a)=b and cb(c,b)=a."""
def new_value():
av, bv, cv = [connector['has_val']() for connector in (a, b, c)]
if av and bv:
c['set_val'](constraint, ab(a['val'], b['val']))
elif av and cv:
b['set_val'](constraint, ca(c['val'], a['val']))
elif bv and cv:
a['set_val'](constraint, cb(c['val'], b['val']))
def forget_value():
for connector in (a, b, c):
connector['forget'](constraint)
constraint = {'new_val': new_value, 'forget': forget_value}
for connector in (a, b, c):
connector['connect'](constraint)
return constraint

from operator import add, sub, mul, truediv

"""The constraint that a + b = c."""
return ternary_constraint(a, b, c, add, sub, sub)

def multiplier(a, b, c):
"""The constraint that a * b = c."""
return ternary_constraint(a, b, c, mul, truediv, truediv)

def constant(connector, value):
"""The constraint that connector = value."""
constraint = {}
connector['set_val'](constraint, value)
return constraint

def converter(c, f):
"""Connect c to f to convert from Celsius to Fahrenheit."""
u, v, w, x, y = [connector() for _ in range(5)]
multiplier(c, w, u)
multiplier(v, x, u)