# Lab 2: Higher-Order Functions, Lambda Expressions lab02.zip

Due by 11:59pm on Wednesday, September 8.

## Starter Files

Download lab02.zip. Inside the archive, you will find starter files for the questions in this lab, along with a copy of the Ok autograder.

# 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?

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

### Q1: WWPD: Lambda the Free

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

``python3 ok -q lambda -u``

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``

``````>>> 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 first writing a solution without the restriction, and then rewriting it into one line after.

If the syntax check isn't passing: Make sure you've removed the line containing `"***YOUR CODE HERE***"` so that it doesn't get treated as part of the function for the syntax check.

``````def lambda_curry2(func):
"""
Returns a Curried version of a two-argument function FUNC.
>>> from operator import add, mul, mod
8
>>> curried_mul = lambda_curry2(mul)
>>> mul_5 = curried_mul(5)
>>> mul_5(42)
210
>>> lambda_curry2(mod)(123)(10)
3
"""
"*** YOUR CODE HERE ***"
return ______
``````

Use Ok to test your code:

``python3 ok -q lambda_curry2``

### 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 = 1
count = 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 = 1
count = 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
"""
"*** YOUR CODE HERE ***"
``````

Use Ok to test your code:

``python3 ok -q count_cond``

# Environment Diagram Practice

There is no Ok submission for this component.

However, we still encourage you to do these problems on paper to develop familiarity with Environment Diagrams, which might appear in an alternate form on the exam.

### Q5: Make Adder

Draw the environment diagram for the following code:

``````n = 9
return lambda k: k + 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. What name is frame `f2` 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 check your work with the Online Python Tutor, but try drawing the environment diagram on your own first.

### Q6: Lambda the Environment Diagram

Draw the environment diagram for the following code and predict what Python will output.

You can check your work with the Online Python Tutor, but try drawing the environment diagram on your own first.

``````a = lambda x: x * 2 + 1
def b(b, x):
return b(x + a(x))
x = 3
x = b(a, x)``````

## Submit

Make sure to submit this assignment by running:

``python3 ok --submit``

# Optional Questions

### Q7: 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. Try to use the `composer` function defined below for more HOF practice.

``````def composer(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 = composer(square, add_one)   # (x + 1)^2
>>> a1(4)
25
>>> mul_three = lambda x: x * 3      # multiplies 3 to x
>>> a2 = composer(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
"""
"*** YOUR CODE HERE ***"
``````

Use Ok to test your code:

``python3 ok -q composite_identity``

### Q8: 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.

...     return x + 1
>>> def times2(x):
...     return x * 2
...     return x + 3
>>> identity = my_cycle(0)
>>> identity(5)
5
>>> add_one_then_double = my_cycle(2)
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
"""
"*** YOUR CODE HERE ***"
``````

Use Ok to test your code:

``python3 ok -q cycle``