Lab 2 Solutions

Solution Files

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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.

Lambda Expressions

Lambda expressions are expressions that evaluate to functions by specifying two things: the parameters and a return expression.

lambda <parameters>: <return expression>

While both lambda expressions and def statements create function objects, there are some notable differences. lambda expressions work like other expressions; much like a mathematical expression just evaluates to a number and does not alter the current environment, a lambda expression evaluates to a function without changing the current environment. Let's take a closer look.

lambda def
Type Expression that evaluates to a value Statement that alters the environment
Result of execution Creates an anonymous lambda function with no intrinsic name. Creates a function with an intrinsic name and binds it to that name in the current environment.
Effect on the environment Evaluating a lambda expression does not create or modify any variables. Executing a def statement both creates a new function object and binds it to a name in the current environment.
Usage A lambda expression can be used anywhere that expects an expression, such as in an assignment statement or as the operator or operand to a call expression. After executing a def statement, the created function is bound to a name. You should use this name to refer to the function anywhere that expects an expression.
Example
# A lambda expression by itself does not alter
# the environment
lambda x: x * x

# We can assign lambda functions to a name
# with an assignment statement
square = lambda x: x * x
square(3)

# Lambda expressions can be used as an operator
# or operand
negate = lambda f, x: -f(x)
negate(lambda x: x * x, 3)
def square(x):
    return x * x

# A function created by a def statement
# can be referred to by its intrinsic name
square(3)

Environment Diagrams

Environment diagrams are one of the best learning tools for understanding lambda expressions and higher order functions because you're able to keep track of all the different names, function objects, and arguments to functions. We highly recommend drawing environment diagrams or using Python tutor if you get stuck doing the WWPD problems below. For examples of what environment diagrams should look like, try running some code in Python tutor. Here are the rules:

Assignment Statements

  1. Evaluate the expression on the right hand side of the = sign.
  2. If the name found on the left hand side of the = doesn't already exist in the current frame, write it in. If it does, erase the current binding. Bind the value obtained in step 1 to this name.

If there is more than one name/expression in the statement, evaluate all the expressions first from left to right before making any bindings.

def Statements

  1. Draw the function object with its intrinsic name, formal parameters, and parent frame. A function's parent frame is the frame in which the function was defined.
  2. If the intrinsic name of the function doesn't already exist in the current frame, write it in. If it does, erase the current binding. Bind the newly created function object to this name.

Call expressions

Note: you do not have to go through this process for a built-in Python function like max or print.

  1. Evaluate the operator, whose value should be a function.
  2. Evaluate the operands left to right.
  3. Open a new frame. Label it with the sequential frame number, the intrinsic name of the function, and its parent.
  4. Bind the formal parameters of the function to the arguments whose values you found in step 2.
  5. Execute the body of the function in the new environment.

Lambdas

Note: As we saw in the lambda expression section above, lambda functions have no intrinsic name. When drawing lambda functions in environment diagrams, they are labeled with the name lambda or with the lowercase Greek letter λ. This can get confusing when there are multiple lambda functions in an environment diagram, so you can distinguish them by numbering them or by writing the line number on which they were defined.

  1. Draw the lambda function object and label it with λ, its formal parameters, and its parent frame. A function's parent frame is the frame in which the function was defined.

This is the only step. We are including this section to emphasize the fact that the difference between lambda expressions and def statements is that lambda expressions do not create any new bindings in the environment.

Required Questions

What Would Python Display?

Q1: WWPD: Lambda the Free

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

python3 ok -q lambda -u

For all WWPD questions, type Function if you believe the answer is <function...>, Error if it errors, and Nothing if nothing is displayed. As a reminder, the following two lines of code will not display anything in the Python interpreter when executed:

>>> x = None
>>> x
>>> lambda x: x  # A lambda expression with one parameter x
______
<function <lambda> at ...>
>>> a = lambda x: x # Assigning the lambda function to the name a >>> a(5)
______
5
>>> (lambda: 3)() # Using a lambda expression as an operator in a call exp.
______
3
>>> b = lambda x: lambda: x # Lambdas can return other lambdas! >>> c = b(88) >>> c
______
<function <lambda> at ...
>>> c()
______
88
>>> d = lambda f: f(4) # They can have functions as arguments as well. >>> def square(x): ... return x * x >>> d(square)
______
16
>>> x = None # remember to review the rules of WWPD given above!
>>> x
>>> lambda x: x
______
Function
>>> z = 3
>>> e = lambda x: lambda y: lambda: x + y + z
>>> e(0)(1)()
______
4
>>> f = lambda z: x + z >>> f(3)
______
NameError: name 'x' is not defined
>>> higher_order_lambda = lambda f: lambda x: f(x)
>>> g = lambda x: x * x
>>> higher_order_lambda(2)(g)  # Which argument belongs to which function call?
______
Error
>>> higher_order_lambda(g)(2)
______
4
>>> call_thrice = lambda f: lambda x: f(f(f(x))) >>> call_thrice(lambda y: y + 1)(0)
______
3
>>> print_lambda = lambda z: print(z) # When is the return expression of a lambda expression executed? >>> print_lambda
______
Function
>>> one_thousand = print_lambda(1000)
______
1000
>>> one_thousand
______
# print_lambda returned None, so nothing gets displayed

Q2: WWPD: Higher Order Functions

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

python3 ok -q hof-wwpd -u

For all WWPD questions, type Function if you believe the answer is <function...>, Error if it errors, and Nothing if nothing is displayed.

>>> def even(f):
...     def odd(x):
...         if x < 0:
...             return f(-x)
...         return f(x)
...     return odd
>>> steven = lambda x: x
>>> stewart = even(steven)
>>> stewart
______
<function ...>
>>> stewart(61)
______
61
>>> stewart(-4)
______
4
>>> def cake():
...    print('beets')
...    def pie():
...        print('sweets')
...        return 'cake'
...    return pie
>>> chocolate = cake()
______
beets
>>> chocolate
______
Function
>>> chocolate()
______
sweets 'cake'
>>> more_chocolate, more_cake = chocolate(), cake
______
sweets
>>> more_chocolate
______
'cake'
>>> def snake(x, y): ... if cake == more_cake: ... return chocolate ... else: ... return x + y >>> snake(10, 20)
______
Function
>>> snake(10, 20)()
______
30
>>> cake = 'cake' >>> snake(10, 20)
______
30

Coding Practice

Q3: Lambdas and Currying

We can transform multiple-argument functions into a chain of single-argument, higher order functions by taking advantage of lambda expressions. For example, we can write a function f(x, y) as a different function g(x)(y). This is known as currying. It's useful when dealing with functions that take only single-argument functions. We will see some examples of these later on.

Write a function lambda_curry2 that will curry any two argument function using lambdas. Refer to the textbook for more details about currying.

Your solution to this problem should fit entirely on the return line. You can try writing it first without this restriction, but rewrite it after in one line.

def lambda_curry2(func):
    """
    Returns a Curried version of a two-argument function FUNC.
    >>> from operator import add, mul, mod
    >>> curried_add = lambda_curry2(add)
    >>> add_three = curried_add(3)
    >>> add_three(5)
    8
    >>> curried_mul = lambda_curry2(mul)
    >>> mul_5 = curried_mul(5)
    >>> mul_5(42)
    210
    >>> lambda_curry2(mod)(123)(10)
    3
    """
return lambda arg1: lambda arg2: func(arg1, arg2)

Use Ok to test your code:

python3 ok -q lambda_curry2
Walkthrough:

Q4: Count van Count

Consider the following implementations of count_factors and count_primes:

def count_factors(n):
    """Return the number of positive factors that n has.
    >>> count_factors(6)
    4   # 1, 2, 3, 6
    >>> count_factors(4)
    3   # 1, 2, 4
    """
    i, count = 1, 0
    while i <= n:
        if n % i == 0:
            count += 1
        i += 1
    return count

def count_primes(n):
    """Return the number of prime numbers up to and including n.
    >>> count_primes(6)
    3   # 2, 3, 5
    >>> count_primes(13)
    6   # 2, 3, 5, 7, 11, 13
    """
    i, count = 1, 0
    while i <= n:
        if is_prime(i):
            count += 1
        i += 1
    return count

def is_prime(n):
    return count_factors(n) == 2 # only factors are 1 and n

The implementations look quite similar! Generalize this logic by writing a function count_cond, which takes in a two-argument predicate function condition(n, i). count_cond returns a one-argument function that takes in n, which counts all the numbers from 1 to n that satisfy condition when called.

def count_cond(condition):
    """Returns a function with one parameter N that counts all the numbers from
    1 to N that satisfy the two-argument predicate function Condition, where
    the first argument for Condition is N and the second argument is the
    number from 1 to N.

    >>> count_factors = count_cond(lambda n, i: n % i == 0)
    >>> count_factors(2)   # 1, 2
    2
    >>> count_factors(4)   # 1, 2, 4
    3
    >>> count_factors(12)  # 1, 2, 3, 4, 6, 12
    6

    >>> is_prime = lambda n, i: count_factors(i) == 2
    >>> count_primes = count_cond(is_prime)
    >>> count_primes(2)    # 2
    1
    >>> count_primes(3)    # 2, 3
    2
    >>> count_primes(4)    # 2, 3
    2
    >>> count_primes(5)    # 2, 3, 5
    3
    >>> count_primes(20)   # 2, 3, 5, 7, 11, 13, 17, 19
    8
    """
def counter(n): i, count = 1, 0 while i <= n: if condition(n, i): count += 1 i += 1 return count return counter

Use Ok to test your code:

python3 ok -q count_cond

Q5: Both Paths

Let a path be some sequence of directions, starting with S for start, and followed by a sequence of U and Ds representing up and down directions along the path. For example, the path SUDDDUUU represents the path up, down, down, down, up, up, up.

Your task is to implement the function both_paths, which prints out the path so far (at first just S), and then returns two functions, each of which keeps track of a branch down or up. This is probably easiest to understand with an example, which can be found in the doctest of both_paths as seen below.

Note about default arguments: Python allows certain arguments to be given default values. For example, the function

def root(x, degree=2):
    return x ** (1 / degree)

can be called either with or without the degree argument, since it has a default value of 2. For example

>>> root(64)
8
>>> root(64, 3)
4

In the given skeleton, we give the default argument sofar.

Hint: you can return multiple things from a function

>>> def func(x):
...     return x * 2, x * 4
>>> x, y = func(5)
>>> print(x, y)
10 20

 

Another hint: if you get a RecursionError, you probably accidentally called both_paths too early. How is a statement like x = func different from another statement like x = func(5)?

def both_paths(sofar="S"):
    """
    >>> up, down = both_paths()
    S
    >>> upup, updown = up()
    SU
    >>> downup, downdown = down()
    SD
    >>> _ = upup()
    SUU
    """
print(sofar) def up(): return both_paths(sofar + "U") def down(): return both_paths(sofar + "D") return up, down

Use Ok to test your code:

python3 ok -q both_paths

Environment Diagram Practice

There is no submission for this component. However, we still encourage you to do these problems on paper to develop familiarity with Environment Diagrams, which will appear on the exam.

Q6: Make Adder

Draw the environment diagram for the following code:

n = 9
def make_adder(n):
    return lambda k: k + n
add_ten = make_adder(n+1)
result = add_ten(n)

There are 3 frames total (including the Global frame). In addition, consider the following questions:

  1. In the Global frame, the name add_ten points to a function object. What is the intrinsic name of that function object, and what frame is its parent?
  2. In frame f2, what name is the frame labeled with (add_ten or λ)? Which frame is the parent of f2?
  3. What value is the variable result bound to in the Global frame?

You can try out the environment diagram at tutor.cs61a.org. To see the environment diagram for this question, click here.

  1. The intrinsic name of the function object that add_ten points to is λ (specifically, the lambda whose parameter is k). The parent frame of this lambda is f1.
  2. f2 is labeled with the name λ the parent frame of f2 is f1, since that is where λ is defined.
  3. The variable result is bound to 19.

Q7: Lambda the Environment Diagram

Try drawing an environment diagram for the following code and predict what Python will output.

You do not need to submit or unlock this question through Ok. Instead, you can check your work with the Online Python Tutor, but try drawing it yourself first!

>>> a = lambda x: x * 2 + 1
>>> def b(b, x):
...     return b(x + a(x))
>>> x = 3
>>> b(a, x)
______
21 # Interactive solution: https://goo.gl/Lu99QR

Submit

Make sure to submit this assignment by running:

python3 ok --submit

Optional Questions

Q8: Composite Identity Function

Write a function that takes in two single-argument functions, f and g, and returns another function that has a single parameter x. The returned function should return True if f(g(x)) is equal to g(f(x)). You can assume the output of g(x) is a valid input for f and vice versa. You may use the compose1 function defined below.

def compose1(f, g):
    """Return the composition function which given x, computes f(g(x)).

    >>> add_one = lambda x: x + 1        # adds one to x
    >>> square = lambda x: x**2
    >>> a1 = compose1(square, add_one)   # (x + 1)^2
    >>> a1(4)
    25
    >>> mul_three = lambda x: x * 3      # multiplies 3 to x
    >>> a2 = compose1(mul_three, a1)    # ((x + 1)^2) * 3
    >>> a2(4)
    75
    >>> a2(5)
    108
    """
    return lambda x: f(g(x))

def composite_identity(f, g):
    """
    Return a function with one parameter x that returns True if f(g(x)) is
    equal to g(f(x)). You can assume the result of g(x) is a valid input for f
    and vice versa.

    >>> add_one = lambda x: x + 1        # adds one to x
    >>> square = lambda x: x**2
    >>> b1 = composite_identity(square, add_one)
    >>> b1(0)                            # (0 + 1)^2 == 0^2 + 1
    True
    >>> b1(4)                            # (4 + 1)^2 != 4^2 + 1
    False
    """
def identity(x): return compose1(f, g)(x) == compose1(g, f)(x) return identity # Alternative solution return lambda x: f(g(x)) == g(f(x))

Use Ok to test your code:

python3 ok -q composite_identity

Q9: I Heard You Liked Functions...

Define a function cycle that takes in three functions f1, f2, f3, as arguments. cycle will return another function that should take in an integer argument n and return another function. That final function should take in an argument x and cycle through applying f1, f2, and f3 to x, depending on what n was. Here's what the final function should do to x for a few values of n:

  • n = 0, return x
  • n = 1, apply f1 to x, or return f1(x)
  • n = 2, apply f1 to x and then f2 to the result of that, or return f2(f1(x))
  • n = 3, apply f1 to x, f2 to the result of applying f1, and then f3 to the result of applying f2, or f3(f2(f1(x)))
  • n = 4, start the cycle again applying f1, then f2, then f3, then f1 again, or f1(f3(f2(f1(x))))
  • And so forth.

Hint: most of the work goes inside the most nested function.

def cycle(f1, f2, f3):
    """Returns a function that is itself a higher-order function.

    >>> def add1(x):
    ...     return x + 1
    >>> def times2(x):
    ...     return x * 2
    >>> def add3(x):
    ...     return x + 3
    >>> my_cycle = cycle(add1, times2, add3)
    >>> identity = my_cycle(0)
    >>> identity(5)
    5
    >>> add_one_then_double = my_cycle(2)
    >>> add_one_then_double(1)
    4
    >>> do_all_functions = my_cycle(3)
    >>> do_all_functions(2)
    9
    >>> do_more_than_a_cycle = my_cycle(4)
    >>> do_more_than_a_cycle(2)
    10
    >>> do_two_cycles = my_cycle(6)
    >>> do_two_cycles(1)
    19
    """
def ret_fn(n): def ret(x): i = 0 while i < n: if i % 3 == 0: x = f1(x) elif i % 3 == 1: x = f2(x) else: x = f3(x) i += 1 return x return ret return ret_fn
Video Walkthorough: https://youtu.be/BkAxfl2fy-g

Use Ok to test your code:

python3 ok -q cycle