61A Homework 5

Due by 5pm on Friday, 10/5

Submission. See the online submission instructions. We have provided a starter file for the questions below.

Readings. Section 2.4 of the online lecture notes.

Q1. When testing software, it can be useful to count the number of times that a function is called. Define a higher-order function count_calls that returns two functions:

  1. A counted version of the original function that counts the number of times it has been called, but otherwise behaves identically to the input, and
  2. A function of zero arguments that returns the number of times that the counted function has been called.

Your implementation should not include any lists or dictionaries:

def count_calls(f):
    """A function that returns a version of f that counts calls to f and can
    report that count to how_many_calls.

    >>> from operator import add
    >>> counted_add, add_count = count_calls(add)
    >>> add_count()
    >>> counted_add(1, 2)
    >>> add_count()
    >>> add(3, 4)  # Doesn't count
    >>> add_count()
    >>> counted_add(5, 6)
    >>> add_count()
    "*** YOUR CODE HERE ***"

Q2. Write a version of make_withdraw that returns password-protected account 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):
    """Return a password-protected withdraw function.

    >>> w = make_withdraw(100, 'hax0r')
    >>> w(25, 'hax0r')
    >>> w(90, 'hax0r')
    'Insufficient funds'
    >>> w(25, 'hwat')
    'Incorrect password'
    >>> w(25, 'hax0r')
    >>> w(75, 'a')
    'Incorrect password'
    >>> w(10, 'hax0r')
    >>> w(20, 'n00b')
    'Incorrect password'
    >>> w(10, 'hax0r')
    "Your account is locked. Attempts: ['hwat', 'a', 'n00b']"
    >>> w(10, 'l33t')
    "Your account is locked. Attempts: ['hwat', 'a', 'n00b']"
    "*** YOUR CODE HERE ***"

Q3. 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):
    """Return a password-protected withdraw function that has joint access to
    the balance of withdraw.

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

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

    >>> j2(25, 'password')
    'Incorrect password'
    >>> j2(5, 'secret')
    "Your account is locked. Attempts: ['my', 'secret', 'password']"
    >>> j(5, 'secret')
    "Your account is locked. Attempts: ['my', 'secret', 'password']"
    >>> w(5, 'hax0r')
    "Your account is locked. Attempts: ['my', 'secret', 'password']"
    >>> make_joint(w, 'hax0r', 'hello')
    "Your account is locked. Attempts: ['my', 'secret', 'password']"
    "*** YOUR CODE HERE ***"


Section 2.4.8 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.

Q4. (Extra for experts) Implement the function triangle_area that 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
    "*** YOUR CODE HERE ***"

Q5. (Extra for experts) The multiplier constraint from the lecture notes 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
    "*** YOUR CODE HERE ***"

Q6. (Extra for experts) 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
    "*** YOUR CODE HERE ***"

The equation solver implementation from the lecture notes:

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)
    Contradiction detected: 77.0 vs 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)
            if val != value:
                print('Contradiction detected:', val, 'vs', 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:

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):
    constraint = {'new_val': new_value, 'forget': forget_value}
    for connector in (a, b, c):
    return constraint

from operator import add, sub, mul, truediv

def adder(a, b, c):
    """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)
    adder(v, y, f)
    constant(w, 9)
    constant(x, 5)
    constant(y, 32)