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

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.

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 5. return Randomized-Select(A, start, pivot-1, i) 6. else return Randomized-Select(A, pivot+1, end, i - (pivot - start + 1))

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.

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))

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.

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)

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}

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

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

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

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

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}

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

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

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

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

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

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

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

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

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

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

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

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] }

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 }

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

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

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

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

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

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

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

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

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

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

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

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

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

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