**Solutions:** You can find the file with solutions for all
questions here.

## Lists Review

### Question 1: Adding matrices

Write a function that adds two matrices together using list comprehensions. The function should take in two 2D lists of the same dimensions. Try to implement this in one line!

```
def add_matrices(x, y):
"""
>>> matrix1 = [[1, 3],
... [2, 0]]
>>> matrix2 = [[-3, 0],
... [1, 2]]
>>> add_matrices(matrix1, matrix2)
[[-2, 3], [3, 2]]
>>> matrix4 = [[ 1, -2, 3],
... [-4, 5, -6]]
>>> matrix5 = [[-1, 2, -3],
... [ 4, -5, 6]]
>>> add_matrices(matrix4, matrix5)
[[0, 0, 0], [0, 0, 0]]
"""
return [[x[i][j] + y[i][j] for j in range(len(x[0]))]
for i in range(len(x))]
```

Use OK to test your code:

`python3 ok -q add_matrices`

## Higher Order Functions

### Question 2: Mul_by_num

Using higher order functions, complete the `mul_by_num`

function. This
function should take an argument and return a one argument function
that multiplies any value passed to it by the original number.

```
def mul_by_num(factor):
"""
Returns a function that takes one argument and
returns the product of factor and that argument.
>>> x = mul_by_num(5)
>>> y = mul_by_num(2)
>>> x(3)
15
>>> y(-4)
-8
"""
def f(x):
return factor*x
return f
```

Use OK to test your code:

`python3 ok -q mul_by_num`

### Question 3: This Question is so Derivative

Define a function `make_derivative`

that returns a function: the derivative of a
function `f`

. Assuming that `f`

is a single-variable mathematical function, its
derivative will also be a single-variable function. When called with a number
`a`

, the derivative will estimate the slope of `f`

at point `(a, f(a))`

.

Recall that the formula for finding the derivative of `f`

at point `a`

is:

where `h`

approaches 0. We will approximate the derivative by choosing a very
small value for `h`

. The closer `h`

is to 0, the better the estimate of the
derivative will be.

```
def make_derivative(f):
"""Returns a function that approximates the derivative of f.
Recall that f'(a) = (f(a + h) - f(a)) / h as h approaches 0. We will
approximate the derivative by choosing a very small value for h.
>>> def square(x):
... # equivalent to: square = lambda x: x*x
... return x*x
>>> derivative = make_derivative(square)
>>> result = derivative(3)
>>> round(result, 3) # approximately 2*3
6.0
"""
h=0.00001
def derivative(x):
return (f(x + h) - f(x)) / h
return derivative
```

Use OK to test your code:

`python3 ok -q make_derivative`

### Question 4: 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."""
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."""
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 ```
mystery_function(n,
i)
```

. `count_cond`

returns a count of all the numbers from 1 to `n`

that satisfy
`mystery_function`

.

Note: A predicate function is a function that returns a boolean (`True`

or `False`

).

```
def count_cond(mystery_function, n):
"""
>>> def divisible(n, i):
... return n % i == 0
>>> count_cond(divisible, 2) # 1, 2
2
>>> count_cond(divisible, 4) # 1, 2, 4
3
>>> count_cond(divisible, 12) # 1, 2, 3, 4, 6, 12
6
>>> def is_prime(n, i):
... return count_cond(divisible, i) == 2
>>> count_cond(is_prime, 2) # 2
1
>>> count_cond(is_prime, 3) # 2, 3
2
>>> count_cond(is_prime, 4) # 2, 3
2
>>> count_cond(is_prime, 5) # 2, 3, 5
3
>>> count_cond(is_prime, 20) # 2, 3, 5, 7, 11, 13, 17, 19
8
"""
i, count = 1, 0
while i <= n:
if mystery_function(n, i):
count += 1
i += 1
return count
```

Use OK to test your code:

`python3 ok -q count_cond`

### Question 5: 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 the 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.

*Hint 2*: given `n`

, how many function calls are made on `x`

?

*Hint 3*: for help with how to cycle through the
functions (i.e., how to go back to applying `f1`

as your outermost
function call when `n = 4`

), consider looking at
this python tutor demo
which has similar cycling behaviour.

```
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
```

Use OK to test your code:

`python3 ok -q cycle`

### Question 6: The Word Guessing Game

Write a higher order function called `store_word`

that takes in a secret word. It will return the length of the secret word and another function, `guess_word`

, that the user can use to try to guess the secret word.

Assume that when the user tries to guess the secret word, they will only guess words that are equal in length to the secret word. The user can pass their guess into the `guess_word`

function, and it will return a list where every element in the list is a boolean, True or False, indicating whether the letter at that index matches the letter in the secret word!

```
def store_word(secret):
"""
>>> word_len, guess_word = store_word("cake")
>>> word_len
4
>>> guess_word("corn")
[True, False, False, False]
>>> guess_word("come")
[True, False, False, True]
>>> guess_word("cake")
[True, True, True, True]
>>> word_len, guess_word = store_word("pop")
>>> word_len
3
>>> guess_word("ate")
[False, False, False]
>>> guess_word("top")
[False, True, True]
>>> guess_word("pop")
[True, True, True]
"""
def guess_word(guess):
return [secret[i] == guess[i] for i in range(len(guess))]
return len(secret), guess_word
```

Use OK to test your code:

`python3 ok -q store_word`

## Submit

Make sure to submit this assignment by running:

`python3 ok --submit`