# 61A Homework 5

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

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

Q1. Define a function reverse_list that takes a list as an argument and returns None, but reverses the elements in the list as a side effect. Do not use any built-in list methods (especially not reverse) or slicing, but element assignment statements are fine:

```def reverse_list(s):
"""Reverse the contents of list s and return None.

>>> digits = [6, 2, 9, 5, 1, 4, 1, 3]
>>> reverse_list(digits)
>>> digits
[3, 1, 4, 1, 5, 9, 2, 6]
>>> d = digits
>>> reverse_list(d)
>>> digits
[6, 2, 9, 5, 1, 4, 1, 3]
"""
```

Hint: This question was also asked in discussion 5, so you get a freebie. You should try to implement it on your own before looking up the answer, though!

Q2. Define a function shuffle that takes a list 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 type 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.

>>> 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'
```

Q3. Write a version of the mutable make_withdraw function from lecture 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']"
"""
```

Q4. Suppose that our banking system requires the ability to make joint accounts. Define a function make_joint that takes three arguments.

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 QUESTIONS ARE EXTRA FOR EXPERTS (OPTIONAL)

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.

Q5. (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
"""
```

Q6. (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
"""
```

Q7. (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
"""
```

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)

>>> 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)