CS 111 f21 — Introduction to Sorting

1 Introduction

What is sorting and where do we encounter it?
Having sorted data may make future operations much more efficient

2 Why Study Sorting?

Not to implement—standard libraries implement, see list.sort()
Excellent context for practicing analysis and design decisions
Very practical: you will be expected to know it

3 How would you sort a list?

3.1 Bubble Sort

go through the list swapping out of order elements
on the first pass, largest element "bubbles" up to the end
repeat this process, once you make a pass with no swaps, the list is sorted

3.2 Insertion Sort

keep a "sorted list" and an "unsorted list"
go through each element of the unsorted list and insert it into the appropriate position in the sorted list

3.3 Selection Sort

find the smallest element, swap it to the beginning
repeat with finding the second smallest, and so on
"selects" the smallest unsorted element

3.4 Bogosort

randomize the list until it ends up sorted

4 Insertion Sort

pseudocode:

for i from 1 to n - 1
    find where to insert element i into the sorted portion of the list (indexes 0 to i - 1)
    insert element i and shift other elements over

slides

4.1 Worst-Case Analysis

Just carefully count up the steps:

when \(i=1\), 1 comparison + 1 shift
when \(i=2\), 2 comparisons + 2 shifts
when \(i=3\), 3 comparisons + 3 shifts
…
when \(i=n-1\), \(n-1\) comparisons + \(n-1\) shifts

sum of \(1\) through \(n-1\) is \(\frac{n(n-1)}{2}\)

\(\frac{n(n-1)}{2}\) comparisons + \(\frac{n(n-1)}{2}\) shifts

\(\frac{(n^2 - n)}{2} + \frac{(n^2 - n)}{2}\)

\(n^2 - n\)
\(O(n^2)\)

5 Selection Sort

pseudocode:

for i from 0 to n - 2
    find index of the smallest element, j, in range of indexes i to n - 1
    swap elements at i and j

slides

5.1 Worst-Case Analysis

Just carefully count up the steps:

when \(i=0\), \(n-1\) comparison + \(1\) swap
when \(i=1\), \(n-2\) comparisons + \(1\) swap
when \(i=2\), \(n-3\) comparisons + \(1\) swap
…
when \(i=n-2\), \(1\) comparison + \(1\) swap

\(\frac{n(n-1)}{2}\) comparisons + \(n\) swaps

\(\frac{(n^2 - n)}{2} + n\)

\(\frac{n^2}{2} - \frac{n}{2} + n\)

\(\frac{n^2}{2} + \frac{n}{2}\)

\(O(n^2)\)

6 Analysis Practice

def contains(nums, x):
    for num in nums:
        if num == x:
        return True
    return False

This will be \(O(n)\) since it loops over each list element once (i.e., goes around the loop \(n\) times for a list with \(n\) elements)

def max(a, b):
    """assume a and b are numbers"""
    if a >= b:
        return a
    return b

This will be \(O(1)\) because the number of operations does not depend on the input

def count_digits(x):
    digits = 0
    while x >= 1:
        digits += 1
    x = x / 10
    return digits

This will be \(O(\log_{10}n)\) because \(\log_{10}n\) is how many times the while loop will iterate when \(x=n\)