# Lab 3: Recursion and Midterm 1 Review

*Due at 11:59pm on 09/17/2015.*

## Starter Files

Download lab03.zip. Inside the archive, you will find starter files for the questions in this lab, along with a copy of the OK autograder.

## Submission

By the end of this lab, you should have submitted the lab with
`python3 ok --submit`

. You may submit more than once before the
deadline; only the final submission will be graded.

- To receive credit for this lab, you must complete Questions 1, 6, 7, 8 in lab03.py and submit through OK.
- Questions 2, 3, 4, and 5 are to show you what some common recursion misconceptions and mistakes are. You can run Python interactively to test them out!
- Questions 8, 9, and 10 go over topics from old labs for the sake of midterm review.
- Questions 11 is
*extra practice*. It can be found in the lab03_extra.py file. It is recommended that you complete this problem on your own time.

## Recursion

A recursive function is a function that calls itself in its body, either directly or indirectly. Recursive functions have three important components:

- Base case(s), the simplest possible form of the problem you're trying to solve.
- Consider a recursive call on a smaller argument.
- Recursive case(s), where the function calls itself with a
*simpler argument*as part of the computation.

Let's look at the canonical example, `factorial`

:

```
def factorial(n):
if n == 0:
return 1
return n * factorial(n - 1)
```

We know by its definition that `0!`

is `1`

. So we choose `n = 0`

as our
base case. The recursive step also follows from the definition of
factorial, i.e. `n! = n * (n-1)!`

.

The next few questions in lab will have you writing recursive functions. Here are some general tips:

- Consider how you can solve the current problem using the solution to
a simpler version of the problem. Remember to
*trust the recursion*: assume that your solution to the simpler problem works correctly without worrying about how. - Think about what the answer would be in the simplest possible case(s). These will be your base cases - the stopping points for your recursive calls. Make sure to consider the possibility that you're missing base cases (this is a common way recursive solutions fail).
- It may help to write the iterative version first.

### Question 1: Skip Add

Write a function `skip_add`

that takes a single argument `n`

and computes the sum of every other integers between 0 and `n`

starting from n. Assume `n`

is non-negative.

```
def skip_add(n):
""" Takes a number x and returns x + x-2 + x-4 + x-6 + ... + 0.
>>> skip_add(5) # 5 + 3 + 1 + 0
9
>>> skip_add(10) # 10 + 8 + 6 + 4 + 2 + 0
30
"""
"*** YOUR CODE HERE ***"
if n <= 0:
return 0
return n + skip_add(n - 2)
```

Use OK to test your code:

`python3 ok -q skip_add`

### Question 2: Common Misconception

Fix the error with the following recursive function.

```
def count_up(n):
"""Print out all numbers up to and including n in ascending order.
>>> count_up(5)
1
2
3
4
5
"""
i = 1
if i == n:
return
print(i)
i += 1
count_up(n-1)
```

The variable `i`

resets back to 1 for each function call
and will print 1 each time. Try it out!

### Question 3: Common Misconception

Fix the error with this recursive function.

```
def skip_mul(n):
"""Return the product of n * (n - 2) * (n - 4) * ...
>>> skip_mul(5) # 5 * 3 * 1
15
>>> skip_mul(8) # 8 * 6 * 4 * 2 * 0
0
"""
if n == 0:
return 0
else:
return n * skip_mul(n - 2)
```

Consider what happens when we choose an odd number for `n`

. `skip_mul(3)`

will
return `3 * skip_mul(1)`

. `skip_mul(1)`

will return `1 * skip_mul(-1)`

. You
may see the problem now. Since we are decreasing `n`

by two at a time, we've
completed missed our base case of `n == 0`

, and we will end up recursing
indefinitely. We need to add another base case to make sure this doesn't
happen.

```
def skip_mul(n):
if n == 0:
return 0
elif n == 1:
return 1
else:
return n * skip_mul(n - 2)
```

### Question 4: Common Misconception

Fix the error with the following recursive function.

```
def factorial(n):
"""Return n * (n - 1) * (n - 2) * ... * 1.
>>> factorial(5)
120
"""
if n == 0:
return 1
else:
n * factorial(n-1)
```

The result of the recursive calls is not returned.

```
def factorial(n):
if n == 0:
return 1
else:
return n * factorial(n-1)
```

### Question 5: Common Misconception

Fix the error with the following recursive function:

```
def print_up_to(n):
"""Print every natural number up to n, inclusive.
>>> print_up_to(5)
1
2
3
4
5
"""
i = 1
if i > n:
return
else:
print(i)
i += 1
print_up_to(n)
```

The function never reduces toward the base case of `i == n`

.

```
def print_up_to(n):
def helper(i):
print(i)
if i < n:
helper(i + 1)
helper(1)
```

Note:

- To keep track of info between recursive calls, we passsed the information as arguments to the function. This also allows us to get by without explicitly assigning a counter.
- The helper function doesn't have a base case written.
The base case is implicit because we have the condition
`i < n`

We will stop recursing once`i == n`

and we violate this condition.

### Question 6: GCD

The greatest common divisor of two positive integers `a`

and `b`

is the
largest integer which evenly divides both numbers (with no remainder).
Euclid, a Greek mathematician in 300 B.C., realized that the greatest
common divisor of `a`

and `b`

is one of the following:

- the smaller value if it evenly divides the larger value, OR
- the greatest common divisor of the smaller value and the remainder of the larger value divided by the smaller value

In other words, if `a`

is greater than `b`

and `a`

is not divisible by
`b`

, then

`gcd(a, b) == gcd(b, a % b)`

Write the `gcd`

function recursively using Euclid's algorithm.

```
def gcd(a, b):
"""Returns the greatest common divisor of a and b.
Should be implemented using recursion.
>>> gcd(34, 19)
1
>>> gcd(39, 91)
13
>>> gcd(20, 30)
10
>>> gcd(40, 40)
40
"""
"*** YOUR CODE HERE ***"
a, b = max(a, b), min(a, b)
if a % b == 0:
return b
else:
return gcd(b, a % b)
# Iterative solution, if you're curious
def gcd_iter(a, b):
"""Returns the greatest common divisor of a and b, using iteration.
>>> gcd_iter(34, 19)
1
>>> gcd_iter(39, 91)
13
>>> gcd_iter(20, 30)
10
>>> gcd_iter(40, 40)
40
"""
if a < b:
return gcd_iter(b, a)
while a > b and not a % b == 0:
a, b = b, a % b
return b
```

Use OK to test your code:

`python3 ok -q gcd`

### Question 7: Hailstone

For the `hailstone`

function from homework 1, you 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. Repeat this process until `n`

is 1.
Write a recursive version of hailstone that prints out the values of
the sequence and returns the number of steps.

```
def hailstone(n):
"""Print out the hailstone sequence starting at n, and return the
number of elements in the sequence.
>>> a = hailstone(10)
10
5
16
8
4
2
1
>>> a
7
"""
"*** YOUR CODE HERE ***"
print(n)
if n == 1:
return 1
elif n % 2 == 0:
return 1 + hailstone(n // 2)
else:
return 1 + hailstone(3 * n + 1)
```

Use OK to test your code:

`python3 ok -q hailstone`

## Midterm 1 Review

### Question 8: Palindrome

A number is considered a palindrome if it reads the same forwards and backwards. Fill in the blanks '_' to help determine if a number is a palindrome. In the spirit of exam style questions, please do not edit any parts of the function other than the blanks.

```
def is_palindrome(n):
"""
Fill in the blanks '_____' to check if a number
is a palindrome.
>>> is_palindrome(12321)
True
>>> is_palindrome(42)
False
>>> is_palindrome(2015)
False
>>> is_palindrome(55)
True
"""
x, y = n, 0
f = lambda: _____
f = lambda: y * 10 + x % 10 while x > 0:
x, y = _____, f()
x, y = x // 10, f() return y == n
```

Use OK to test your code:

`python3 ok -q is_palindrome`

### Question 9: WWPP: To Boolean or not to Boolean

The following WWPP is meant to review your understanding of boolean operators.

Use OK to test your knowledge with the following "What Would Python Print?" questions:

`python3 ok -q boolean -u`

```
>>> ((0 or 5) and (False or '') and 1/0)
______''
>>> fi, foo = max, min
>>> def switcheroo(foo):
... foo = fi
... return 10
>>> foo(switcheroo(foo), 5)
______5
```

### Question 10: WWPP: Lambda Trivia

The following WWPP questions will review your understanding of lambdas. Python Tutor may be helpful in understanding what is happening.

Use OK to test your knowledge with the following "What Would Python Print?" questions:

`python3 ok -q lambda -u`

Hint: Remember for all WWPP questions, enter`Function`

if you believe the answer is`<function ...>`

and`Error`

if it errors.

```
>>> x = 5
>>> x = lambda x: lambda x: lambda y: 3+x
>>> x(3)(5)(7)
______8
>>> you = 'fly'
>>> yo, da = lambda do: you, lambda can: True
>>> yo = yo('jedi')
>>> da = (3 and 4 and 5 and (False or ' you shall'))
>>> yo+da
______'fly you shall'
>>> annoying = lambda yell: annoying
>>> annoying('aiya')('aaa')('aaaa')('aaaaa')('aaaaaa')
______Function
>>> more = ' productive'
>>> lip_service = lambda say: print(say)
>>> useful_person = lambda do: do + more
>>> lip_service('blah blah blah') + useful_person('be')
______blah blah blah
Error
```

## Extra Questions

Questions in this section are not required for submission. However, we encourage you to try them out on your own time for extra practice.

### Question 11: Interleaved Sum

Recall that the `summation`

function computes the sum of a sequence of
terms from 1 to `n`

:

```
def summation(n, term):
"""Return the sum of the first n terms of a sequence.
>>> summation(5, lambda x: pow(x, 3))
225
"""
total, k = 0, 1
while k <= n:
total, k = total + term(k), k + 1
return total
```

Write a function `interleaved_sum`

that similarly computes the sum of a
sequence of terms from 1 to `n`

, but uses different function to compute
the terms for odd and even numbers. Do so without using any loops or
testing in any way if a number is odd or even. (You may test if a
number is equal to 0, 1, or `n`

.)

Hint: Use recursion and a helper function!

```
def interleaved_sum(n, odd_term, even_term):
"""Compute the sum odd_term(1) + even_term(2) + odd_term(3) + ..., up
to n.
>>> # 1 + 2^2 + 3 + 4^2 + 5
... interleaved_sum(5, lambda x: x, lambda x: x*x)
29
"""
"*** YOUR CODE HERE ***"
return interleaved_helper(n, odd_term, even_term, 1)
def interleaved_helper(n, term0, term1, k):
if k == n:
return term0(k)
return term0(k) + interleaved_helper(n, term1, term0, k+1)
# Alternate solution
def interleaved_sum2(n, odd_term, even_term):
"""Compute the sum odd_term(1) + even_term(2) + odd_term(3) + ..., up
to n.
>>> # 1 + 2^2 + 3 + 4^2 + 5
... interleaved_sum2(5, lambda x: x, lambda x: x*x)
29
"""
total, term0, term1 = interleaved_helper2(n, odd_term, even_term)
return total
def interleaved_helper2(n, odd_term, even_term):
if n == 1:
return odd_term(1), even_term, odd_term
else:
total, term0, term1 = interleaved_helper2(n-1, odd_term,
even_term)
return total + term0(n), term1, term0
```

Use OK to test your code:

`python3 ok -q interleaved_sum`