This is a study guide about orders of growth with links to past lectures, assignments, and handouts, as well as additional practice problems to assist you in learning the concepts.
Important: For solutions to these assignments once they have been released, see the main website.
Orders of growth
The order of growth of an algorithm is an approximation of the time required to run a computer program as the input size increases. The order of growth ignores the constant factor needed for fixed operations and focuses instead on the operations that increase proportional to input size. For example, a program with a linear order of growth generally requires double the time if the input doubles.
This table summarizes the most common orders of growth:
|Constant||Θ(1)||Indexing an item in a list|
|Logarithmic||Θ(lg N)||Repeatedly halving a number|
|Linear||Θ(n)||Summing a list|
|Quadratic||Θ(n^2)||Summing each pair of numbers in a list|
|Exponential||Θ(2^n)||Visiting each node in a binary tree|
When an algorithm has a constant order of growth, it means that it always takes a fixed number of steps, no matter how large the input size increases.
As an example, consider accessing the first element of a list:
first_post = posts
Even if the list grows to be a million items long, that operation will always require a single step.
We can visualize that relationship as a table:
We can also visualize it as a graph:
A constant run time is ideal, but is typically not possible for algorithms that process multiple pieces of data.
When an algorithm has a logarithmic order of growth, it increases proportionally to the logarithm of the input size.
The binary search algorithm is an example of an algorithm that runs in logarithmic time.
Here's the pseudocode:
def search_list(nums, target_num): """ Returns the index of TARGET_NUM in sorted list NUMS or -1 if not found. >>> search_list([1, 2, 3, 4], 3) 2 >>> search_list([14, 23, 37, 48, 59], 23) 1 >>> search_list([14, 23, 37, 48, 59], 47) -1 """ min_index = 1 max_index = len(nums) while min_index <= max_index: middle_index = (min_index + max_index) // 2 if target_num == nums[middle_index]: return middle_index elif target_num > nums[middle_index]: min_index = middle_index + 1 else: max_index = middle_index - 1 return -1
The algorithm uses a loop to look at multiple items in the list, but crucially, it does not look at every item in the list. Specifically, it looks at lg2(n) items, where n is the number of items in the list.
We can visualize that relationship in a table:
We can also see that as a graph:
The number of steps is definitely increasing as input size increases, but at a very slow rate.
When an algorithm has a linear order of growth, its number of steps increases in direct proportion to the input size.
The aptly-named linear search algorithm runs in linear time.
The code shows its simplicity compared to binary search:
def search_list(nums, target_num): """ Returns the index of TARGET_NUM in an unsorted list NUMS or -1 if not found. >>> search_list([3, 2, 1, 4], 3) 2 >>> search_list([14, 59, 99, 23, 37, 22], 23) 3 >>> search_list([14, 59, 99, 23, 37, 22], 47) -1 """ index = 1 while index < len(nums): if nums[index] == target_num: return index index += 1 return -1
This time, the loop looks at every item in the list. This exhaustive search is necessary to search for items in an unsorted list, since there's no way to narrow down where a particular item might be. This algorithm will always require at least as many steps as items in the list.
We can see that in table form:
Or as a graph:
When an algorithm has a quadratic order of growth, its steps increase in proportion to the input size squared.
Several list sorting algorithms run in quadratic time, like selection sort. That algorithm starts from the front of the list, then keeps finding the next smallest value in the list and swapping it with the current value.
Here's pseudocode for selection sort:
def linear_sort(nums): """Performs an in-place sorting of NUMS. >>> l = [2, 3, 1, 4] >>> linear_sort(l) >>> l [1, 2, 3, 4] """ i = 0 while i < len(nums): min_index = i j = i + 1 # Find next smallest value while j < len(nums): if nums[j] < nums[min_index]: min_index = j j += 1 # Swap if new minimum found if min_index != i: nums[i], nums[min_index] = nums[min_index], nums[i] i += 1
The important thing to notice about the algorithm is the nested loop: it loops through each items in the list, and loops again through the remaining items for each of those items. In this case, the algorithm ends up looking at
1/2 * (n^2 - n) values, where
n is the number of items in the list.
This table shows how many items it'd examine for lists of increasing sizes:
Here's what that looks like in graph form:
Both the table and the graph show that the number of steps for this algorithm increases at a much faster rate than the previous ones.
When an algorithm has a superpolynomial order of growth, its number of steps increases faster than a polynomial function of the input size.
An algorithm often requires superpolynomial time when it must look at every permutation of values. For example, consider an algorithm that generates all possible numerical passwords for a given password length.
For a password length of 2, it generates 100 passwords:
00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99
That algorithm requires at least 10^2 steps, since there are 10 possibilities for each digit (0-9) and it must try out every possibility for each of the 2 digits.
For any given password length of n, the algorithm requires 10^n steps. That run time increases incredibly quickly, since each additional digit requires 10 times the number of steps.
This table shows how fast that grows for just the first 5 digits:
Here's what that looks like as a graph:
All together now
Now that we've seen examples of possible run times for algorithms, let's compare them on a graph:
That graph makes it even more clear that there's a definite difference in these run times, especially as input size increases. Since modern computer programs increasingly deal with large data sets (like from millions of users or sensors), the run time efficiency matters quite a bit.
Q1: Is Prime
What is the order of growth of
is_prime in terms of
def is_prime(n): for i in range(2, n): if n % i == 0: return False return True
Choose one of:
- None of these
Explanation: In the worst case, n is prime, and we have to execute the loop n - 2 times. Each iteration takes constant time (one conditional check and one return statement). Therefore, the total time is (n - 2) x constant, or simply linear.
Q2: Growth: Bar
What is the order of growth of
bar in terms of
def bar(n): i, sum = 1, 0 while i <= n: sum += biz(n) i += 1 return sum def biz(n): i, sum = 1, 0 while i <= n: sum += i**3 i += 1 return sum
Choose one of:
- None of these
The body of the while loop in
bar is executed n times.
Each iteration, one call to
biz(n) is made. Note that n never changes,
so this call takes the same time to run each iteration.
Taking a look at
biz, we see that there is another while loop. Be careful
to note that although the term being added to
sum is cubed (
i itself is only incremented by 1 in each iteration.
This tells us that this while loop also executes n times, with each iteration
taking constant time, so the total time of
biz(n) is n x constant, or linear.
Knowing that each call to
biz(n) takes linear time,
we can conclude that each iteration of the while loop in
bar is linear.
Therefore, the total runtime of
bar(n) is quadratic.
fooin terms of
nis the length of
lst? Assume that slicing a list and calling
lenon a list can both be done in constant time.
def foo(lst, i): mid = len(lst) // 2 if mid == 0: return lst elif i > 0: return foo(lst[mid:], -1) else: return foo(lst[:mid], 1)
Explanation: A single recursive call is made in the body of
foo on half the
input list (either the first half or the second half depending on the input
i). The base case is executed when the list either is empty or has only
one element. We start with an n element list and halve the list until there
is at most 1 element, which means there will be log(n) total calls. Each
call, constant work is done if we ignore the recursive call. The total runtime
is then log(n) * θ(1).
Note: We simplified this problem by assuming that slicing a list takes constant time. In reality, this operation is a bit more nuanced and may take linear time. As an additional exercise, try determining the order of growth of this function if we assuming slicing takes linear time.
Describe the order of growth of the function below.
def bonk(n): sum = 0 while n >= 2: sum += n n = n / 2 return sum
Choose one of:
- None of these
Explanation: As we increase the value of
n, the amount of time needed to evaluate a call to
bonk scales logarithmically. Let's use the number of iterations of our while loop to illustrate an example. When
n = 1, our loop iterates 0 times. When
n = 2, our loop iterates 1 time. When
n = 4, we have 2 iterations. And when
n = 8, a call to
bonk(8) results in 3 iterations of this while loop. As the value of the input scales by a factor of 2, the number of iterations increases by 1. This indicates that this function runtime has a logarithmic order of growth.