# CLRS Problem 2.2

2-2 Correctness of bubblesort

# Correctness of bubblesort

Bubblesort is a popular, but inefficient, sorting algorithm. It works by repeatedly swapping adjacent elements that are out of order.

for i = 1 to A.length - 1
for j = A.length downto i + 1
if A[j] < A[j-1]
swap(A[j], A[j-1])


a.) Let $A’$ denote the output of BUBBLESORT(A). To prove that BUBBLESORT is correct, we need to prove that it terminates and that

where $n = A.length$. In order to show that BUBBLESORT actually sorts, what else do we need to prove?

b.) State precisely a loop invariant for the for loop in lines 2–4, and prove that this loop invariant holds. Your proof should use the structure of the loop invariant proof presented in this chapter.

c.) Using the termination condition of the loop invariant proved in part (b), state a loop invariant for the for loop in lines 1–4 that will allow you to prove inequality (2.3). Your proof should use the structure of the loop invariant proof presented in this chapter.

d.) What is the worst-case running time of bubblesort? How does it compare to the running time of insertion sort?

# 1.) What else do we need to prove?

We need to prove that

contains the same elements in $A$, but possibly in a different order.

# 2.) Inner loop invariant

At the beginning of every inner loop iteration, $A[j]$ will be the smallest element in the subarray $A[j, \cdots]$ and $A[j, \cdots]$ will consist of the original elements of $A[j, \cdots]$ possibly in a different order. Put differently, at the end of every inner loop iteration $A[j-1]$ will be the smallest element in the subarray $A[j-1, \cdots]$ and $A[j-1, \cdots]$ will consist of the original elements of $A[j-1, \cdots]$, possibly in a different order.

## Initialization

Initialization trivially holds, as the subarray $A[j, \cdots]$ consists only of a single element (the very last of array $A$). $A[j]$ is therefore the smallest element of this array.

## Maintenance

Every iteration we introduce $A[j-1]$ to the subarray $A[j, \cdots]$, lengthening it by $1$. We swap $A[j]$ with $A[j-1]$ if $A[j]$ is the smaller of the two. At the end of every iteration, $A[j-1]$ consists of the smallest element in the subarray $A[j-1, \cdots]$.

## Termination

The inner loop terminates when $j = i$. This suggests that after the loop terminates, $A[i]$ is the smallest element in the subarray $A[i, \cdots]$ and $A[i, \cdots]$ consists of the original elements in $A[i, \cdots]$, possibly in a different order.

# 3.) Outer loop invariant

Using the termination condition of the inner loop invariant, state the loop invariant for the outer loop.

At the beginning of each loop iteration, the subarray $A[1, \cdots, i-1]$ is in sorted order and consists only of elements smaller than those in the subarray $A[i, \cdots]$.

## Initialization

Initialization trivially holds, as the subarray $A[1, \cdots, i-1]$ is empty which by definition is in sorted order.

## Maintenance

Due to the inner loop invariant, $A[i]$ becomes the smallest element in the subarray $A[i, \cdots]$ and is less than or equal to all of the elements in the subarray $A[i+1, \cdots]$.

## Termination

The outer loop terminates when $i = A.length$. This suggests that the subarray $A[1, \cdots, i-1]$ is in sorted order, where each element is less than or equal to elements in the subarray $A[i, \cdots]$ which only consists of the final element when we substitute $A.length \text{ for } i$. Thus, the array $A[1, \cdots, A.length]$ is sorted.

# 4.) Worst case running time of bubblesort

The number of comparisons bubblesort will make is:

Bubblesort will also make at most this many swaps. This makes the worst case of bubblesort $\Theta(n^2)$, the same as insertion sort. Ideally, we could make the best case of bubblesort $\Theta(n)$ by creating some boolean flag that tells us whether we swapped any values after the first iteration of the outer loop. If no swaps occur, we can duck out early.

Written on July 2, 2016