# Homework 2 Solutions

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

## Questions

### Question 1: Falling Factorial

Let's write a function `falling`

, which is a "falling" factorial
that takes two arguments, `n`

and `k`

, and returns the product of `k`

consecutive numbers, starting from `n`

and working downwards.

If `k`

is larger than n, only multiply up to n consecutive numbers!

```
def falling(n, k):
"""Compute the falling factorial of n to depth k.
>>> falling(6, 3) # 6 * 5 * 4
120
>>> falling(4, 3) # 4 * 3 * 2
24
>>> falling(4, 1) # 4
4
>>> falling(4, 0)
1
"""
total, stop = 1, n-k
while n > stop:
total, n = total*n, n-1
return total
```

Use OK to test your code:

`python3 ok -q falling`

### Question 2: Hailstone

Complete this question using iteration!

Douglas Hofstadter's Pulitzer-prize-winning book, *GĂ¶del, Escher,
Bach*, poses the following mathematical puzzle:

- Pick a positive integer
`n`

as the start. - If
`n`

is even, divide it by 2. - If
`n`

is odd, multiply it by 3 and add 1. - Continue this process until
`n`

is 1.

The sequence of values of `n`

is often called a Hailstone sequence,
because hailstones also travel up and down in the atmosphere before
falling to earth. Write a function that takes a single argument with
formal parameter name `n`

, prints out the hailstone sequence starting
at `n`

, and returns the number of steps in the sequence:

```
def hailstone(n):
"""Print the hailstone sequence starting at n and return its
length.
>>> a = hailstone(10)
10
5
16
8
4
2
1
>>> a
7
"""
s = 1
print(n)
while n != 1:
if n % 2 == 0:
n = n // 2
else:
n = n * 3 + 1
print(n)
s = s + 1
return s
```

Hailstone sequences can get quite long! Try 27. What's the longest you can find?

Use OK to test your code:

`python3 ok -q hailstone`

### Question 3: Classify the elements

Complete the function `odd_even`

that classifies an number as either `'odd'`

or `'even'`

and the function `classify`

that takes in a list and applies `odd_even`

to all elements in the list.

```
def odd_even(x):
"""Classify a number as odd or even.
>>> odd_even(4)
'even'
>>> odd_even(3)
'odd'
"""
if (x % 2) == 0:
return 'even'
else:
return 'odd'
def classify(s):
"""
Classify all the elements of a sequence as odd or even
>>> classify([0, 1, 2, 4])
['even', 'odd', 'even', 'even']
"""
return [odd_even(x) for x in s]
```

Use OK to test your code:

`python3 ok -q odd_even`

Use OK to test your code:

`python3 ok -q classify`

### Question 4: Deep List

Implement the function `deep_list`

, which takes in a list, and returns a new list which contains only elements of the original list that are also lists. Use a list comprehension.

```
def deep_list(seq):
"""Returns a new list containing elements of the original list that are lists.
>>> seq = [49, 8, 2, 1, 102]
>>> deep_list(seq)
[]
>>> seq = [[500], [30, 25, 24], 8, [0]]
>>> deep_list(seq)
[[500], [30, 25, 24], [0]]
>>> seq = ["hello", [12, [25], 24], 8, [0]]
>>> deep_list(seq)
[[12, [25], 24], [0]]
"""
return [n for n in seq if type(n) is list]
```

Use OK to test your code:

`python3 ok -q deep_list`

### Question 5: arange

Implement the function `arange`

, which behaves just like np.arange(start, end, step) from Data 8. You only need to support positive values for step.

```
def arange(start, end, step=1):
"""
arange behaves just like np.arange(start, end, step).
You only need to support positive values for step.
>>> arange(1, 3)
[1, 2]
>>> arange(0, 25, 2)
[0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24]
>>> arange(999, 1231, 34)
[999, 1033, 1067, 1101, 1135, 1169, 1203]
"""
return [n for n in range(start, end, step)]
```

Use OK to test your code:

`python3 ok -q arange`

### Question 6: 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 ```
condition(n,
i)
```

. `count_cond`

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

that satisfy
`condition`

.

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

or `False`

).

```
def count_cond(condition, 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 condition(n, i):
count += 1
i += 1
return count
```

Use OK to test your code:

`python3 ok -q count_cond`

### Question 7: Match and Apply

Sometimes when we are given a dataset, we need to alter it for specific values. For example, say we have a table with one column being people's names and the other being the price they have to pay.

We can use a list of pairs for this:

`[["Jessica", 5], ["Andrew", 9], ["Alex", 2], ["Amir", 11], ["John", 3], ["Lyric", 2]]`

The first value in each pair is the name, the second is the price.

Now, let's say we want to give a discount to specific people. We have a discount function that we want to apply to the person's price. Now, we need a function that will only apply the discount function to specific people.

Implement `match_and_apply(pairs, function)`

:

`pairs`

is a list of pairs.`function`

is some function

`match_and_apply`

returns a function such that when the function is given an input that
matches the first of a pair, returns the result of applying `function`

to the second value in the pair.

```
def match_and_apply(pairs, function):
"""
>>> pairs = [[1, 2], [3, 4], [5, 6], [7, 8], [9, 0]]
>>> def square(num):
... return num**2
>>> func = match_and_apply(pairs, square)
>>> result = func(3)
>>> result
16
>>> result = func(1)
>>> result
4
>>> result = func(7)
>>> result
64
>>> result = func(15)
>>> print(result)
None
"""
def foo(num):
for pair in pairs:
if pair[0] == num:
return function(pair[1])
return None
return foo
```

Use OK to test your code:

`python3 ok -q match_and_apply`