# Discussion 6: Orders of Growth, Midterm Review disc06.pdf

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# Efficiency

When we talk about the efficiency of a function, we are often interested in the following: as the size of the input grows, how does the runtime of the function change? And what do we mean by runtime?

Example 1: `square(1)` requires one primitive operation: multiplication. `square(100)` also requires one. No matter what input `n` we pass into `square`, it always takes a constant number of operations (1). In other words, this function has a runtime complexity of Θ(1).

As an illustration, check out the table below:

input function call return value operations
1 `square(1)` 1*1 1
2 `square(2)` 2*2 1
... ... ... ...
100 `square(100)` 100*100 1
... ... ... ...
n `square(n)` n*n 1

Example 2: `factorial(1)` requires one multiplication, but `factorial(100)` requires 100 multiplications. As we increase the input size of n, the runtime (number of operations) increases linearly proportional to the input. In other words, this function has a runtime complexity of Θ(`n`).

As an illustration, check out the table below:

input function call return value operations
1 `factorial(1)` 1*1 1
2 `factorial(2)` 2*1*1 2
... ... ... ...
100 `factorial(100)` 100*99*...*1*1 100
... ... ... ...
n `factorial(n)` n*(n-1)*...*1*1 n

Here are some general guidelines for finding the order of growth for the runtime of a function:

• If the function is recursive or iterative, you can subdivide the problem as seen above:

• Count the number of recursive calls/iterations that will be made in terms of input size `n`.
• Find how much work is done per recursive call or iteration in terms of input size `n`.
• The answer is usually the product of the above two, but be sure to pay attention to control flow!
• If the function calls helper functions that are not constant-time, you need to take the runtime of the helper functions into consideration.
• We can ignore constant factors. For example `1000000n` and `n` steps are both linear.
• We can also ignore smaller factors. For example if `h` calls `f` and `g`, and `f` is Quadratic while `g` is linear, then `h` is Quadratic.
• For the purposes of this class, we take a fairly coarse view of efficiency. All the problems we cover in this course can be grouped as one of the following:

• Constant: the amount of time does not change based on the input size. Rule: `n --> 2n` means `t --> t`.
• Logarithmic: the amount of time changes based on the logarithm of the input size. Rule: `n --> 2n` means `t --> t + k`.
• Linear: the amount of time changes with direct proportion to the size of the input. Rule: `n --> 2n` means `t --> 2t`.
• Quadratic: the amount of time changes based on the square of the input size. Rule: `n --> 2n` means `t --> 4t`.
• Exponential: the amount of time changes with a power of the input size. Rule: `n --> n + 1` means `t --> 2t`.

## Questions

### Q1: The First Order...of Growth

What is the efficiency of `rey`?

``````def rey(finn):
poe = 0
while finn >= 2:
poe += finn
finn = finn / 2
return``````
Logarithmic, because our while loop iterates at most log(`finn`) times, due to `finn` being halved in every iteration. This is commonly known as Θ(log(`finn`)) runtime. Another way of looking at this if you duplicate the input, we only add a single iteration to the time, which also indicates logarithmic.

What is the efficiency of `mod_7`?

``````def mod_7(n):
if n % 7 == 0:
return 0
else:
return 1 + mod_7(n - 1)``````
Constant, since at worst it will require 6 recursive calls to reach the base case. This is commonly known as a Θ(1) runtime.

# Midterm Review

This section is far longer than a typical discussion, and it is recommended that you also use it as a problem bank for your midterm studies! Best of luck, you got this!!

## Reverse Environment Diagrams

### Q2: Who - What - When

Fill in the lines below so that the execution of the program would lead to the environment diagram below. You may not use any numbers in any blanks.

Click here for the diagram that goes along with this problem, go to page 2 of the pdf

Run in 61A Code
Solution
``````def what(f):    def when(x):        return f(x)    return whendef who(n):
def where(k):        return 2 * k + n
return wherey = 3
what(who(y))(4)``````

### Q3: Spring 2021: YipYip Book

The following environment diagram was generated by a program:

In this series of questions, you'll fill in the blanks of the program that follows so that its execution matches the environment diagram. You may want to fill in the blanks in a different order; feel free to answer the questions in whatever order works for you.

``````def yiip(signal, bookbook):
if signal < 0:
return ______
(a)
elif signal == 0:
return float("inf")
return signal * -98

def yip(mup, pet):
if _______:
(b)
mup += 1
if _______:
(c)
return lambda al: ______
(d)
return lambda al: lambda fal: al - fal

yuiop = yiip(_____, _____)(3)(5)
(e)    (f)``````

Which of these could fill in blank (a)?

i. `lambda al: signal - al * 3` ii. `lambda al: bookbook(signal, 3)(al)` iii. `bookbook` iv. `yiip(bookbook)` v. `yip` vi. `bookbook(-3, signal)` vii. `bookbook(signal + - 3)`

Solution vi. `bookbook(-3, signal)`

Which of these could fill in blank (b)? Select all that apply.

i. `mup < 0` ii. `pet <= 0` iii. `mup == pet` iv. `mup <= 0` v. `pet == 0` vi. `pet < 0`

Solution i. `mup < 0`, ii. `pet <= 0`, iv. `mup <= 0`, vi. `pet < 0`

Which of these could fill in blank (c)? Select all that apply.

i. `mup > 0` ii. `mup == -pet` iii. `pet == -mup` iv. `mup == pet` v. `pet > 0` vi. `mup <= 0 and pet <= 0`

Solution iv. `mup == pet`, vi. `mup <= 0` and `pet <= 0`

Which of these could fill in blank (d)?

i. `al - fal ** fal` ii. `lambda fal: mup ** al + pet ** fal` iii. `mup * al + pet * fal` iv. `lambda fal: mup + pet * fal` v. `mup * pet` vi. `lambda fal: mup * al * pet * fal` vii. `mup ** al + pet ** al` viii. `lambda fal: al * mup + fal * pet`

Solution ii. `lambda fal: mup ** al + pet ** fal`

Which of these could fill in blank (e)?

i. `yiip(2 * -1)` ii. `yiip` iii. `bookbook(yip)` iv. `-2` v. `yip` vi. `signal - 2`

Solution iv. `-2`

Which of these could fill in blank (f)?

i. `yip` ii. `yip()` iii. `lambda y: y` iv. `yiip` v. `yiip()` vi. `lambda y: yiip(y)` vii. `-2`

Solution i. `yip`

## Lists and Mutability

### Q4: List Comprehension: f

Fill in the definition of `f` below such that the interpreter prints as expected. Your solution must be on one line.

``````>>> f = _________________________________________________
>>> f = f(10)
1
2
3
4
5
6
7
8
9
10``````

Solution f = lambda x: [print(i) for i in range(1, x + 1)] Alternate Solution f = lambda x: x and (f(x - 1) or print(x))

The list comprehension is the intended solution, but the recursive approach can be a fun thing to talk about with students who really want to go above and beyond.

Then, given your definition of `f`, what will be printed below? (Assuming the above lines have also been executed in the interpreter.)

``>>> f``

Solution For the list comprehension approach: [None, None, None, None, None, None, None, None, None, None] Alternate Solution For the recursive approach: None

### Q5: Deep map

Write the function `deep_map_mut` that takes a Python list and mutates all of the elements (including elements of sublists) to be the result of calling the function given, `fn`, on each element. Note that the function does not return the mutated list!

Hint: `type(a) == list` will return `True` if `a` is a list.

Run in 61A Code
Solution
``````def deep_map_mut(fn, lst):
"""Deeply maps a function over a Python list, replacing each item
in the original list object.

Does NOT create new lists by either using literal notation
([1, 2, 3]), +, or slicing.

Does NOT return the mutated list object.

>>> l = [1, 2, [3, [4], 5], 6]
>>> deep_map_mut(lambda x: x * x, l)
>>> l
[1, 4, [9, [16], 25], 36]
"""
for i in range(len(lst)):
if type(lst[i]) == list:
deep_map_mut(fn, lst[i])
else:
lst[i] = fn(lst[i])``````

## HOF and Self Reference

### Q6: Foldl

Write a function that takes in a list `s`, a function `f`, and an initial value `start`. This function will fold `s` starting at the beginning. If `s` is ```[1, 2, 3, 4, 5]``` then the function `f` is applied as follows:

``f(f(f(f(f(start, 1), 2), 3), 4), 5)``

You may assume that the function `f` takes in two parameters.

Run in 61A Code
Solution
``````from operator import add, sub, mul

def foldl(s, f, start):
"""Return the result of applying the function F to the initial value START
and the first element in S, and repeatedly applying F to this result and
the next element in S until we reach the end of the list.

>>> s = [3, 2, 1]
>>> foldl(s, sub, 0)      # sub(sub(sub(0, 3), 2), 1)
-6
6
>>> foldl(s, mul, 1)      # mul(mul(mul(1, 3), 2), 1)
6

>>> foldl([], sub, 100)   # return start if s is empty
100
"""
total = start
for number in s:
total = f(total, number)

### Q7: Announce Losses

It's Hog again! Write a commentary function `announce_losses` that takes in a player `who` and returns a commentary function that announces whenever that player loses points.

Run in 61A Code
Solution
``````def announce_losses(who, last_score=0):
"""
>>> f = announce_losses(0)
>>> f1 = f(10, 0)
>>> f2 = f1(1, 10) # Player 0 loses points due to swine swap
Oh no! Player 0 just lost 9 point(s).
>>> f3 = f2(7, 10)
>>> f4 = f3(7, 11) # Should not announce when player 0's score does not change
>>> f5 = f4(11, 12)
"""
assert who == 0 or who == 1, 'The who argument should indicate a player.'
def say(score0, score1):
if who == 0:
score = score0        elif who == 1:
score = score1        if score < last_score:            print("Oh no! Player", who, "just lost", last_score - score, "point(s).")        return announce_losses(who, score)    return say``````

## Recursion

### Q8: Pig Latin

Consider the below function `pig_latin`, which computes the pig latin equivalent of an English word

``````def pig_latin_original(w):
"""Return the Pig Latin equivalent of a lowercase English word w."""
if starts_with_a_vowel(w):
return w + 'ay'
return pig_latin_original(rest(w) + first(w))

def first(s):
"""Returns the first character of a string."""
return s[0]

def rest(s):
"""Returns all but the first character of a string."""
return s[1:]

def starts_with_a_vowel(w):
"""Return whether w begins with a vowel."""
c = first(w)
return c == 'a' or c == 'e' or c == 'i' or c == 'o' or c == 'u'``````

This code repeatedly moves a letter from the beginning of a word to the end, until the first letter is a vowel, at which point it adds on `'ay'` to the end. However, this code fails when the original word has no vowels in the set {a, e, i, o, u}, such as the word "sphynx." Write a new version of the `pig_latin` function that just adds `'ay'` to the original word if it does not contain a vowel in this set. Use only the `first`, `rest`, and `starts_with_a_vowel` functions to access the contents of a word, and use the built-in `len` function to determine its length. Do not use any loops.

Run in 61A Code
Solution
``````def pig_latin(w):
"""Return the Pig Latin equivalent of a lowercase English word w.

>>> pig_latin('pun')
'unpay'
>>> pig_latin('sphynx')
'sphynxay'
"""
return pig_latin_helper(w, 0)

def pig_latin_helper(w, rotations):
if starts_with_a_vowel(w) or rotations == len(w):
return w + 'ay'
return pig_latin_helper(rest(w) + first(w), rotations + 1)

# Alternate solution
def pig_latin2(w):
"""Return the Pig Latin equivalent of a lowercase English word w.

>>> pig_latin2('pun')
'unpay'
>>> pig_latin2('sphynx')
'sphynxay'
"""
# BEGIN SOLUTION
if starts_with_a_vowel(w) or has_no_vowels(w):
return w + 'ay'
return pig_latin(rest(w) + first(w))

def has_no_vowels(w):
if w == '':
return True
elif starts_with_a_vowel(w):
return False
return has_no_vowels(rest(w))``````

### Q9: Ten-pairs

Write a function that takes a positive integer `n` and returns the number of ten-pairs it contains. A ten-pair is a pair of digits within `n` that sums to 10. Do not use any assignment statements.

The number 7,823,952 has 3 ten-pairs. The first and fourth digits sum to 7+3=10, the second and third digits sum to 8+2=10, and the second and last digit sum to 8+2=10. Note that a digit can be part of more than one ten-pair.

Hint: Use a helper function to calculate how many times a digit appears in n.

Run in 61A Code
Solution
``````def ten_pairs(n):
"""Return the number of ten-pairs within positive integer n.

>>> ten_pairs(7823952)
3
>>> ten_pairs(55055)
6
>>> ten_pairs(9641469)
6
"""
if n < 10:
return 0
else:
return ten_pairs(n//10) + count_digit(n//10, 10 - n%10)

def count_digit(n, digit):
"""Return how many times digit appears in n.

>>> count_digit(55055, 5)
4
"""
if n == 0:
return 0
else:
if n%10 == digit:
return count_digit(n//10, digit) + 1
else:
return count_digit(n//10, digit)``````

### Q10: Num Splits

Given a list of numbers `s` and a target difference `d`, write a function `num_splits` that calculates how many different ways are there to split `s` into two subsets, such that the sum of the first is within `d` of the sum of the second. The number of elements in each subset can differ.

You may assume that the elements in `s` are distinct and that `d` is always non-negative.

Note that the order of the elements within each subset does not matter, nor does the order of the subsets themselves. For example, given the list `[1, 2, 3]`, you should not count `[1, 2], [3]` and `[3], [1, 2]` as distinct splits.

Hint: If the number you return is too large, you may be double-counting somewhere. If the result you return is off by some constant factor, it will likely be easiest to simply divide/subtract away that factor.

Run in 61A Code
Solution
``````def num_splits(s, d):
"""Return the number of ways in which s can be partitioned into two
sublists that have sums within d of each other.

>>> num_splits([1, 5, 4], 0)  # splits to [1, 4] and [5]
1
>>> num_splits([6, 1, 3], 1)  # no split possible
0
>>> num_splits([-2, 1, 3], 2) # [-2, 3], [1] and [-2, 1, 3], []
2
>>> num_splits([1, 4, 6, 8, 2, 9, 5], 3)
12
"""
def difference_so_far(s, difference):
if not s:
if abs(difference) <= d:
return 1
else:
return 0
element = s[0]
s = s[1:]
return difference_so_far(s, difference + element) + difference_so_far(s, difference - element)
return difference_so_far(s, 0)//2``````

## Trees

### Q11: Pruning Leaves

Define a function `prune_leaves` that given a tree `t` and a tuple of values `vals`, produces a version of `t` with all its leaves that are in `vals` removed. Do not attempt to try to remove non-leaf nodes and do not remove leaves that do not match any of the items in `vals`. Return `None` if pruning the tree results in there being no nodes left in the tree.

Run in 61A Code
Solution
``````def prune_leaves(t, vals):
"""Return a modified copy of t with all leaves that have a label
that appears in vals removed.  Return None if the entire tree is
pruned away.

>>> t = tree(2)
>>> print(prune_leaves(t, (1, 2)))
None
>>> numbers = tree(1, [tree(2), tree(3, [tree(4), tree(5)]), tree(6, [tree(7)])])
>>> print_tree(numbers)
1
2
3
4
5
6
7
>>> print_tree(prune_leaves(numbers, (3, 4, 6, 7)))
1
2
3
5
6
"""
if is_leaf(t) and (label(t) in vals):
return None
new_branches = []
for b in branches(t):
new_branch = prune_leaves(b, vals)
if new_branch:
new_branches += [new_branch]
return tree(label(t), new_branches)``````

### Q12: Hailstone Tree

We can represent the hailstone sequence as a tree in the figure below, showing the route different numbers take to reach 1. Remember that a hailstone sequence starts with a number `n`, continuing to `n/2` if `n` is even or `3n+1` if `n` is odd, ending with 1. Write a function `hailstone_tree(n, h)` which generates a tree of height `h`, containing hailstone numbers that will reach `n`.

Hint: A node of a hailstone tree will always have at least one, and at most two branches (which are also hailstone trees). Under what conditions do you add the second branch?

Run in 61A Code
Solution
``````def hailstone_tree(n, h):
"""Generates a tree of hailstone numbers that will reach N, with height H.
>>> print_tree(hailstone_tree(1, 0))
1
>>> print_tree(hailstone_tree(1, 4))
1
2
4
8
16
>>> print_tree(hailstone_tree(8, 3))
8
16
32
64
5
10
"""
if h == 0:        return tree(n)    branches = [hailstone_tree(n * 2, h - 1)]    if (n - 1) % 3 == 0 and ((n - 1) // 3) % 2 == 1 and (n - 1) // 3 > 1:        branches += [hailstone_tree((n - 1) // 3, h - 1)]    return tree(n, branches)

def print_tree(t):
def helper(i, t):
print("    " * i + str(label(t)))
for b in branches(t):
helper(i + 1, b)
helper(0, t)``````