1. CPSC 311 Section 502 Analysis of Algorithm
Fall 2002
Department of Computer Science
Texas A&M University

2. Announcements There will be a quiz on next Tuesday (Sep 30)
CPSC 311 O. Burchan Bayazit – Fall 02
Announcements
There will be a quiz on next Tuesday (Sep 30)
A new homework will be put to the web site today ( due date next Thursday (October 3))
Midterm Tuesday October 8th.

3. Randomize-Select Randomized-Select(A, start, end, i)
CPSC 311 O. Burchan Bayazit – Fall 02
Randomize-Select
Randomized-Select(A, start, end, i)
1. if start = end then return A[p]
2. pivot = Randomized-Partition(A, start, end)
3. if pivot = i then return A[i]
4. else if i is in [start, pivot-1] then
return Randomized-Select(A, start, pivot-1, i)
6. else return Randomized-Select(A, pivot+1, end,
i - (pivot - start + 1))

4. Running Time of Randomized-Select
CPSC 311 O. Burchan Bayazit – Fall 02
Running Time of Randomized-Select
Worst-case : unlucky with bad 0 : n - 1 partitions.
T(n) = T(n - 1) + (n) = (n2)
(same as for worst-case of QuickSort)
Best-case : really lucky T(n) = (n)
Average-case : like Quick-Sort, will be asymptotically close to best-case.

5. Selection in Linear Worst-Case Time
CPSC 311 O. Burchan Bayazit – Fall 02
Selection in Linear Worst-Case Time
Select(A, start, end, i) /* i is the ith order statistic. */
divide input array A into n/5 groups of size 5
(and one leftover group if n % 5 != 0)
find the median of each group of size 5 by insertion sorting
the groups of 5 and then picking the middle element.
call Select recursively to find x=A[k], the median of the n/5
medians.
partition array around x, splitting it into two arrays A[start, pivot-1]
and A[pivot+1, end]
if (i = k) return x
else if (i < k) then /* like Randomized-Select */
call Select(A[start, pivot-1], i)
else call Select(A[pivot+1, end], i - (pivot - start + 1))

6. Selection in Linear Worst-Case Time
CPSC 311 O. Burchan Bayazit – Fall 02
Selection in Linear Worst-Case Time
Main idea: this guarantees that x causes a "good" split (at least a constant fraction of the n elements <= x and a constant fraction > x).
3(1/2 n/5 - 2)  (3n/10) - 6
elements are > x (or < x)
"At least 1/2 of the medians found in step 2 are greater than the median of medians x. So at least half of the n/5 groups contribute 3 elements that are > x, except for the one group with < 5 elements and the group with x itself." (also holds that (3n/10) - 6 are < x)
So worst-case split has (7n/10) + 6 elements in "big" section of the problem.

7. Selection in Linear Worst-Case Time
CPSC 311 O. Burchan Bayazit – Fall 02
Selection in Linear Worst-Case Time
Running Time (each step):
1. O(n) (break into groups of 5)
2. O(n) (sorting 5 numbers and finding median is O(1) time)
3. T(n/5) (recursive call to find median of medians)
4. O(n) (partition is linear time)
5. T(7n/10 + 6) (maximum size of subproblem)
T(n) = T(n/5) + T(7n/10 + 6) + O(n)
Solve this recurrence using the substitution method. Guess T(n)  cn
 cn/5 + c(7n/10 + 6) + O(n)
 c((n/5) + 1) + 7cn/10 + 6c + O(n)
= cn - (cn/10 - 7c) + O(n)
 cn This step holds since n >= 80 implies (cn/10 -7c) is positive
Choosing big enough c makes O(n) + (cn/10 -7c)
positive, so last line holds. (Try c = 200)

8. Example Partition this array using Select
CPSC 311 O. Burchan Bayazit – Fall 02
Example
Partition this array using Select
A={12,34,0,3,22,4,17,32,3,28,43,82,25,27,34,2,19,12,5,18,20,33,16,33,21,30,3,47}

9. Example First make groups of 5 12 34 3 22 4 17 32 3 28 43 82 25 27 34
CPSC 311 O. Burchan Bayazit – Fall 02
Example
First make groups of 5 12 34 3 22 4 17 32 3 28 43 82 25 27 34 2 19 12 5 18 20 33 16 21 30 3 47

10. Example Then find medians in each group 3 12 34 22 4 3 17 32 28 25 27
CPSC 311 O. Burchan Bayazit – Fall 02
Example
Then find medians in each group 3 12 34 22 4 3 17 32 28 25 27 34 43 82 2 5 12 19 18 20 16 21 33 3 30 47

11. Example Then find median of medians 3 12 34 22 4 3 17 32 28 25 27 34
CPSC 311 O. Burchan Bayazit – Fall 02
Example
Then find median of medians 3 12 34 22 4 3 17 32 28 25 27 34 43 82 2 5 12 19 18 20 16 21 33 3 30 47
12,12,17,21,34,30

12. Example Use 17 as the pivot value and partition original array 3 12 34
CPSC 311 O. Burchan Bayazit – Fall 02
Example
Use 17 as the pivot value and partition original array 3 12 34 22 4 3 17 32 28 25 27 34 43 82 2 5 12 19 18 20 16 21 33 3 30 47
12,12,17,21,34,30

13. Example After partitioning {12,0,3,4,3,2,12,5,16,3} 17
CPSC 311 O. Burchan Bayazit – Fall 02
Example
After partitioning
{12,0,3,4,3,2,12,5,16,3} 17
{34,22,32,28,43,82,25,27,34,19,18,20,33,33,21,30,47}

14. Stacks and Queries Stack operations: Push: Add an element to the stack
CPSC 311 O. Burchan Bayazit – Fall 02
Stacks and Queries
Both are dynamic sets
Stacks: Last-In-First-Out
Queries: First-In-First-Out
Stack operations:
Push: Add an element to the stack
Pop: Remove the last element from the stack

15. Example Stack Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Stack Operations
Initially Empty Stack

16. Example Stack Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Stack Operations
Push (10)
StackTop 10

17. Example Stack Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Stack Operations
Push (2)
StackTop 2 10

18. Example Stack Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Stack Operations
Push (5)
StackTop 5 2 10

19. Example Stack Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Stack Operations
Pop ()5 5
StackTop 2 10

20. Example Stack Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Stack Operations
Push (7)
StackTop 7 2 10

21. Example Stack Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Stack Operations
Push (12)
StackTop 12 7 2 10

22. Example Stack Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Stack Operations
Push (3)
StackTop 3 12 7 2 10

23. Example Stack Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Stack Operations
Pop ()3 3
StackTop 12 7 2 10

24. Example Stack Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Stack Operations
Pop ()12 12
StackTop 7 2 10

25. Implementation Push(S,x) { top[S]=top[S]+1 S[top[S]]=x }
CPSC 311 O. Burchan Bayazit – Fall 02
Implementation
Push(S,x) {
top[S]=top[S]+1
S[top[S]]=x }
Stack-Empty(S) {
if (top[S]=0) {
return TRUE
}else {
return FALSE }
Pop(S,x) {
if(Stack-Empty(S)) {
return error “underflow”
}else{
top[S]=top[S]-1
return S[top[S]+1] }

26. Implementation Initial Stack S= Top[S]=3 10 2 7 1 2 3 4 5 6 7
CPSC 311 O. Burchan Bayazit – Fall 02
Implementation
Initial Stack 1 2 3 4 5 6 7 S= 10 2 7
Top[S]=3

27. Implementation Push(12) S= Top[S]=4 10 2 7 12 1 2 3 4 5 6 7
CPSC 311 O. Burchan Bayazit – Fall 02
Implementation
Push(12) 1 2 3 4 5 6 7 S= 10 2 7 12
Top[S]=4

28. Implementation Push(3) S= Top[S]=5 10 2 7 12 3 1 2 3 4 5 6 7
CPSC 311 O. Burchan Bayazit – Fall 02
Implementation
Push(3) 1 2 3 4 5 6 7 S= 10 2 7 12 3
Top[S]=5

29. Implementation Pop ()3 S= Top[S]=4 10 2 7 12 1 2 3 4 5 6 7
CPSC 311 O. Burchan Bayazit – Fall 02
Implementation
Pop ()3 1 2 3 4 5 6 7 S= 10 2 7 12
Top[S]=4

30. Implementation Pop ()12 S= Top[S]=3 10 2 7 1 2 3 4 5 6 7
CPSC 311 O. Burchan Bayazit – Fall 02
Implementation
Pop ()12 1 2 3 4 5 6 7 S= 10 2 7
Top[S]=3

31. Example Queue Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Queue Operations
Initially Empty Queue
Queue operations:
Enqueue: Add an element to the queue
Dequeue: Remove the last element from the queue

32. Example Queue Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Queue Operations
Enqueue(10) 10
Queue Head and Tail

33. Example Queue Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Queue Operations
Enqueue(2) 10 2
Head
Tail

34. Example Queue Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Queue Operations
Enqueue(5) 10 2 5
Tail
Head

35. Example Queue Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Queue Operations
Dequeue ()10 2 5
Head
Tail

36. Example Queue Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Queue Operations
Enqueue(7) 2 5 7
Tail
Head

37. Example Queue Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Queue Operations
Enqueue(12) 2 5 7 12
Tail
Head

38. Example Queue Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Queue Operations
Dequeue ()2 5 7 12 3
Head
Tail

39. Example Queue Operations
CPSC 311 O. Burchan Bayazit – Fall 02
Example Queue Operations
Dequeue ()5 7 12 3
Head
Tail

40. Implementation Enqueue(Q,x) { Q[tail[Q]]=x if (tail[Q]=length[Q]) {
CPSC 311 O. Burchan Bayazit – Fall 02
Implementation
Enqueue(Q,x) {
Q[tail[Q]]=x
if (tail[Q]=length[Q]) {
tail[Q]=1
}else {
tail[Q]=tail[Q]+1 }
Dequeue(Q) {
x=Q[head[Q]]=x
if (tail[Q]=length[Q]) {
tail[Q]=1
}else {
tail[Q]=tail[Q]+1 }

41. Implementation Initially Queue Q= 10 2 7 1 2 3 4 5 6 7 Tail[Q]=6
CPSC 311 O. Burchan Bayazit – Fall 02
Implementation
Initially Queue 1 2 3 4 5 6 7 Q= 10 2 7
Tail[Q]=6
Head[Q]=4

42. Implementation Dequeue ()10 Q= 10 2 7 1 2 3 4 5 6 7 Head[Q]=5
CPSC 311 O. Burchan Bayazit – Fall 02
Implementation
Dequeue ()10 1 2 3 4 5 6 7 Q= 10 2 7
Head[Q]=5
Tail[Q]=6

43. Implementation Enqueue (12) Q= 10 2 7 12 1 2 3 4 5 6 7 Head[Q]=5
CPSC 311 O. Burchan Bayazit – Fall 02
Implementation
Enqueue (12) 1 2 3 4 5 6 7 Q= 10 2 7 12
Head[Q]=5
Tail[Q]=7

44. Implementation Enqueue (3) Q= 3 10 2 7 12 1 2 3 4 5 6 7 Tail[Q]=1
CPSC 311 O. Burchan Bayazit – Fall 02
Implementation
Enqueue (3) 1 2 3 4 5 6 7 Q= 3 10 2 7 12
Tail[Q]=1
Head[Q]=5

45. Implementation Dequeue ()2 Q= 3 10 2 7 12 1 2 3 4 5 6 7 Tail[Q]=1
CPSC 311 O. Burchan Bayazit – Fall 02
Implementation
Dequeue ()2 1 2 3 4 5 6 7 Q= 3 10 2 7 12
Tail[Q]=1
Head[Q]=6

46. Implementation Dequeue ()2 Q= 3 10 2 7 12 1 2 3 4 5 6 7 Tail[Q]=1
CPSC 311 O. Burchan Bayazit – Fall 02
Implementation
Dequeue ()2 1 2 3 4 5 6 7 Q= 3 10 2 7 12
Tail[Q]=1
Head[Q]=7

47. CPSC 311 O. Burchan Bayazit – Fall 02
Linked list
head 12 3 1
NULL

48. Double-Linked list head 12 3 1 prev next key
CPSC 311 O. Burchan Bayazit – Fall 02
Double-Linked list
prev
next
key
head
NULL 12 3 1
NULL

49. Double-Linked list Insert 20 head 20 12 3 1
CPSC 311 O. Burchan Bayazit – Fall 02
Double-Linked list
Insert 20
head
NULL 20 12 3 1
NULL

50. Double-Linked list Remove 3 head 20 12 1
CPSC 311 O. Burchan Bayazit – Fall 02
Double-Linked list
Remove 3
head
NULL 20 12 1
NULL

51. Implementation List-Search(L,k) { x=head[L]
CPSC 311 O. Burchan Bayazit – Fall 02
Implementation
List-Search(L,k) {
x=head[L]
while (x  NULL and key[x]  k) {
x=next[x] }
return x
List-Delete(L,x) {
if (prev[x]  NULL) {
next[prev[x]]=next[x]
else head[L]=next[x] }
if (next[x] ]  NULL) {
prev[next[x]]=prev[x]
List-Insert(L,k) {
next[x]=head[L]
If (head[L]  NULL ) {
prev[head[L]]=x }
head[L]=x
prev[x]=NULL

52. Rooted Trees Regular Tree Any number of children
CPSC 311 O. Burchan Bayazit – Fall 02
Rooted Trees
Regular Tree
Any number of children

53. Rooted Trees Binary Tree At most 2 children
CPSC 311 O. Burchan Bayazit – Fall 02
Rooted Trees
Binary Tree
At most 2 children

54. Rooted Trees Left-Child Right Sibling
CPSC 311 O. Burchan Bayazit – Fall 02
Rooted Trees
Left-Child Right Sibling