1. COMPUTER 2430 Object Oriented Programming and Data Structures I

2. Selection Sort Bubble Sort Insertion Sort . . .
Sorting Algorithms
Selection Sort Bubble Sort Insertion Sort . . .

3. Algorithm Evaluation
How to compare the sorting algorithms? Can we design a better algorithm? Is there a better/best algorithm? When the problem size (number of elements) becomes larger and larger. Implement algorithms and compare running times. Compare the orders of the algorithms, the big Os.

4. Sorting Algorithms
How to compare the orders of the sorting algorithms? For an array of n elements (n is very large!) How many times is the comparison done as a function of n?

5. Selection Sort
n: 9 12 7 34 15 5 22 14 21 17
# of comparisons: 8 5 7 34 15 12 22 14 21 17
# of comparisons: 7 5 7 34 15 12 22 14 21 17
# of comparisons: 6 5 7 12 15 34 22 14 21 17
# of comparisons: 5 5 7 12 14 34 22 15 21 17
# of comparisons: 4 5 7 12 14 15 22 34 21 17
# of comparisons: 3 5 7 12 14 15 17 34 21 22
# of comparisons: 2 5 7 12 14 15 17 21 34 22
# of comparisons: 1 5 7 12 14 15 17 21 22 34
Don’t do it!

6. Selection Sort n: 9 Number of comparisons12 7 34 15 5 22 14 21 17
n: 9
Number of comparisons
= 36

7. Selection Sort Pass i First of j Last of j # of Comparisons 1 2 . n-4
for i = 0 to (n - 2) minIndex = i for j = i + 1 to (n – 1) if a[j] < a[minIndex] minIndex = j swap a[i], a[minIndex] How many comparisons for each i?
Pass i First of j Last of j # of Comparisons 1 2 .
n-4
n-3
n-2 1 2 3 .
n-3
n-2
n-1
n-1 .
n-1
n-2
n-3 . 3 2 1

8. Counting: What is the sum?
(n-3) + (n–2) + (n-1)
S = (n-3)+(n–2)+(n-1)
S = (n-1)+(n-2)+(n-3)
2*S = n + n + n n + n + n
= n * (n-1)
S = n * (n – 1) / 2
= (first + last) * (number of items) / 2 +

9. Selection Sort
for i = 0 to (n - 2) minIndex = i for j = i + 1 to (n – 1) if a[j] < a[minIndex] minIndex = j swap a[i], a[minIndex] Total # of comparisons: (n-1) + (n–2) + (n-3) = ((n-1) + 1) * (n-1) / 2 = n * (n-1) / 2 = (n2 – n) / 2 = O(n2) i 1 2 …
n-4
n-3
n-2
# Comparisons
n-1 3

10. Selection Sort
Total number of comparisons (n2 – n) / 2 The Big O of Selection Sort O(n2) Ignore all less significant terms Ignore the coefficient of the most significant term

11. Bubble Sort Pass i First of j Last of j # of Comparisons 1 2 . n-4 n-3
for i = 0 to (n - 2) for j = (n - 1) down to (i + 1) if a[j] < a[j - 1] then Swap a[j], a[j - 1] How many comparisons for each i?
Pass i First of j Last of j # of Comparisons 1 2 .
n-4
n-3
n-2
n-1 . 1 2 3 .
n-3
n-2
n-1
n-1
n-2
n-3 . 3 2 1

12. Bubble Sort
for i = 0 to (n - 2) for j = (n - 1) down to (i + 1) if a[j] < a[j - 1] then Swap a[j], a[j - 1] Total number of comparisons (n-1) + (n–2) + (n-3) = (n2 – n) / 2 The Big O of Bubble Sort O(n2)

13. Insertion Sort
for i = 1 to (n - 1) j = i while j > 0 and a[j] < a[j-1] Swap a[j], a[j - 1] j—- Total number of comparisons The best case: elements are sorted already n – 1 Big O: O(n)?

14. Insertion Sort
for i = 1 to (n - 1) j = i while j > 0 and a[j] < a[j-1] Swap a[j], a[j - 1] j—- Total number of comparisons The worse case: elements in reverse order (n-3) + (n-2) + (n-1) = (n2 – n) / 2 The Big O: O(n2)

15. Insertion Sort
for i = 1 to (n - 1) j = i while j > 0 and a[j] < a[j-1] Swap a[j], a[j - 1] j—- Total number of comparisons The average case: ( (n-3) + (n-2) + (n-1)) / 2 = ((n2 – n) / 2) / 2 The Big O: O(n2)

16. Sorting Algorithms
For an array of n elements How many times is the comparison done as a function of n? O(n2) for all three algorithms How many times is swap done as a function of n? Selection sort: O(n) Bubble sort: O(n2) Insertion sort: O(n2)

17. Test 2: Stack and User
public Additive sumThemAll(AdditiveStack s, int stackSize) { Additive sum = new Additive(); AdditiveStack ts = new AdditiveStack(stackSize); while (!s.isEmpty()) Additive temp = s.pop(); sum = sum.plus(temp); ts.push(temp); } while (!ts.isEmpty()) s.push( ts.pop() ); return sum;

18. Test 2: Stack and User
public Additive sumThemAll(AdditiveStack s, int stackSize) { Additive sum = new Additive(); // Make sure ts is large enough AdditiveStack ts = new AdditiveStack(stackSize); while (!s.isEmpty()) Additive temp = s.pop(); sum = sum.plus(temp); ts.push(temp); } while (!ts.isEmpty()) // We need the loop s.push( ts.pop() ); // This does not work: s = ts; return sum;

19. Test 2: Queue and Implementer
public class Queue { private Object items[]; int front, rear, count; // size is items.length, not count! public Queue (int size) items = new Object[size]; front = rear = count = 0; } public boolean find(Object target) . . .

20. Test 2: Queue and Implementer
public class Queue { private Object items[]; int front, rear, count; public boolean find(Object target) for (int i = 0; i < count; i++) int index = (front + i) % items.length; if (items[index].equals(target)) return true; } return false;

21. Test 2: Queue and User
Queue myQueue = new Queue (size); Object obj = nextTarget(); boolean found = peeker(myQueue, size, obj); if (found) { } else

22. Test 2: Queue and User
public boolean peeker(Queue q, int queueSize, object target) { Queue tq = new Queue(queueSize); boolean isThere = false; while (!q.isEmpty()) Object temp = q.remove(); if (temp.equals(target)) isThere = true; tq.add(temp); } while (!tq.isEmpty()) q.add(tq.remove); return isThere;

23. Test 2: Queue and User
public boolean peeker(Queue q, int queueSize, object target) { Queue tq = new Queue(queueSize); // Make sure tq is large enough boolean isThere = false; while (!q.isEmpty()) Object temp = q.remove(); if (temp.equals(target)) isThere = true; tq.add(temp); } while (!tq.isEmpty()) // We need the loop! q.add(tq.remove); return isThere;

24. Expected Learning Outcomes
Develop software using elementary data structures.
Design, implement, and test programs using classes, inheritance, and polymorphism.
Compare and contrast algorithm efficiency at a very basic level.
Write module tests, system tests, and develop test specifications.
Perform simple object-oriented analysis and design.
Work in a small team (two or three people) on the analysis, design, implementation, and testing of a project.

25. Prog 5 Punch in and out in SE Tool Let me know of any issues
With the tool
With your partners
Prog5 score could be different for students in the same group

26. Lab 11
Due today by 11 pm