# Problem 2.2

## Correctness of bubblesort

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

BUBBLESORT(A)
for i to A.length - 1
for j = A.length downto i + 1
if A[j] < A[j - 1]
exchange A[j] with A[j - 1]

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

$$A'[1] \leq A'[2] \leq \cdots \leq A'[n] \tag{2.3}$$

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

1. 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.

2. Using the termination condition of the loop invariant proved in part (2), 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.

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

### 1. What else?

We need to prove that $A'$ consists of the original elements in $A$, although in (possibly) different order.

### 2. Inner loop invariant

At the start of each iteration of the for loop of lines 2-4, (1) the subarray $A[j...]$ consists of elements that were in $A[j...]$ before entering the inner loop (possibly) in different order and (2) $A[j]$ is the smallest of those elements.

Initialization: It holds trivially, because $A[j...]$ consists of only one element and it is the last element of $A$ when execution starts the inner loop.

Maintenance: On each step, we replace $A[j]$ with $A[j - 1]$ if it is smaller. Because we're only adding the previous element and possibly swapping two values (1) holds. Since $A[j-1]$ becomes the smallest of $A[j]$ and $A[j-1]$ and the loop invariant states that $A[j]$ is the smallest one in $A[j...]$, we know that (2) holds.

Termination: After the loop terminates, we will get $j = i$. This implies that $A[i]$ is the smallest element of the subarray $A[i...]$ and array contains the original elements in some order.

### 3. Outer loop invariant

At the start of each iteration, $A[1..i-1]$ consists of sorted elements, all of which are smaller or equal to the ones in $A[i...]$.

Initialization: Initially we have the empty array, which holds trivially.

Maintenance: The invariant of the inner loop tells us that on each iteration, $A[i]$ becomes the smallest element of $A[i...]$ while the rest get shuffled. This impliest that at the end of the loop:

$$A[i] < A[k] \text{, for } i < k$$

Termination: The loop terminates with $i = n$, where $n$ is the length of the array. Substituting the $n$ for $i$ in the invariant, we have that the subarray $A[1..n]$ consists of the original elements, but in sorted order. This is the entire array, so the entire array is sorted.

### 4. Worst-case running time?

The number of comparisons is

$$n - 1, n - 2, \cdots , 1 = \frac{n(n - 1)}{2}$$

Which is a quadratic function. The swaps are at most the same ammount, which means that the worst-case complexity is $\Theta(n^2)$.

Insertion sort has the same worst-case complexity. In general, the best-case complexity of both algorithms should be $\Theta(n)$, but this implementation of bubble-sort has $\Theta(n^2)$ best-case complexity. That can be fixed by returning if no swaps happened in an iteration of the outer loop.

Furthermore, bubble-sort should be even slower than insertion sort, because the swaps imply a lot more assignments than what insertion sort does.