1. Analysis of Bubble Sort and Loop Invariant

2. N-1 Iteration 77 12 35 42 5
101
N - 1 5 42 12 35 77
101 42 5 35 12 77
101 42 35 5 12 77
101 42 35 12 5 77
101

3. Outer loop Inner loop procedure Bubblesort(A isoftype in/out Arr_Type)
to_do, index isoftype Num
to_do <- N – 1
loop
exitif(to_do = 0)
index <- 1
exitif(index > to_do)
if(A[index] > A[index + 1]) then
Swap(A[index], A[index + 1])
endif
index <- index + 1
endloop
to_do <- to_do - 1
endprocedure // Bubblesort
To do N-1 iterations
To bubble a value
Inner loop
Outer loop

4. Bubble Sort Time Complexity
Best-Case Time Complexity
The scenario under which the algorithm will do the least amount of work (finish the fastest)
Worst-Case Time Complexity
The scenario under which the algorithm will do the largest amount of work (finish the slowest)

5. Bubble Sort Time Complexity
Called Linear Time
O(N)
Order-of-N
Best-Case Time Complexity
Array is already sorted
Need 1 iteration with (N-1) comparisons
Worst-Case Time Complexity
Need N-1 iterations
(N-1) + (N-2) + (N-3) + …. + (1) = (N-1)* N / 2
Called Quadratic Time
O(N2)
Order-of-N-square

6. Loop Invariant
It is a condition or property that is guaranteed to be correct with each iteration in the loop
Usually used to prove the correctness of the algorithm

7. Loop Invariant for Bubble Sort
By the end of iteration i  the right-most i items (largest) are sorted and in place

8. N-1 Iterations 77 12 35 42 5
101
N - 1
1st 5 42 12 35 77
101
2nd 42 5 35 12 77
101
3rd 42 35 5 12 77
101
4th 42 35 12 5 77
101
5th

9. Correctness of Bubble Sort (using Loop Invariant)
Bubble sort has N-1 Iterations
Invariant: By the end of iteration i  the right-most i items (largest) are sorted and in place
Then: After the N-1 iterations  The right-most N-1 items are sorted
This implies that all the N items are sorted

10. Correctness of Bubble Sort (using Loop Invariant)
What if we stop early (after iteration K)
Remember the Boolean “did_swap” flag
From the invariant  the right-most K items are sorted
A[N-(K-1)] < A[N-2] < A[N-1] < A[N]
Since we stopped early, then no swaps are done, Then
A[1] < A[2] < ….. < A[N-K]
By merging these two, then the entire array is sorted