Analysis of Bubble Sort and
Loop Invariant
N-1
N-1 Iteration
1
42
2
35
3
12
4
77
5
5
6
101
1
35
2
12
3
42
4
5
5
77
6
101
1
12
2
35
3
5
4
42
5
77
6
101
1
12
2
5
3
35
4
42
5
77
6
101
1
5
2
12
3
35
4
42
5
77
6
101
procedure Bubblesort(A isoftype in/out Arr_Type)
to_do, index isoftype Num
to_do <- N – 1
Outer loop
loop
exitif(to_do = 0)
index <- 1
To bubble a value
loop
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
endloop
endprocedure // Bubblesort
Inner loop
To do N-1 iterations
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)
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
Called Quadratic Time
O(N2)
Order-of-N-square
• Worst-Case Time Complexity
– Need N-1 iterations
– (N-1) + (N-2) + (N-3) + …. + (1) = (N-1)* N / 2
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
Loop Invariant for Bubble Sort
• By the end of iteration i  the right-most i items
(largest) are sorted and in place
N-1
N-1 Iterations
1
42
2
35
3
12
4
77
5
5
6
101
1st
1
35
2
12
3
42
4
5
5
77
6
101
2nd
1
12
2
35
3
5
4
42
5
77
6
101
3rd
1
12
2
5
3
35
4
42
5
77
6
101
4th
1
5
2
12
3
35
4
42
5
77
6
101
5th
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
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