1 / 11

Pipelined Merging of Two sorted list in a constant time (Cole’s Algorithm)

Pipelined Merging of Two sorted list in a constant time (Cole’s Algorithm). Leaves contain the value Internal nodes merge at each time by updating the values L v : the sequence of values of descendants of v Q v (j): At time j, a sorted sequence v has. An increasing subsequence of L v

jaden
Télécharger la présentation

Pipelined Merging of Two sorted list in a constant time (Cole’s Algorithm)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Pipelined Merging of Two sorted list in a constant time(Cole’s Algorithm) • Leaves contain the value • Internal nodes merge at each time by updating the values • Lv: the sequence of values of descendants of v • Qv(j): At time j, a sorted sequence v has. An increasing subsequence of Lv When Qv(j) = Lv, then node v is complete. • All leaf nodes are complete. • At step j+1, if v’s parent is not complete at j-th step, it sends Rv(j) and Qv(j) to its parent. • Qv(j) = merge Rw(j) and Rz(j), where w and z are children of v • How to compute R? If w is not complete at j-1 step, Rw(j) consists of every 4-th elements of Qw(j-1). If w is complete after j step, (i) Rw(j+1) consists of every 4-th elements of Qw(j) (ii) Rw(j+2) consists of every 2nd elements of Qw(j) (iii) Rw(j+3) = Qw(j) • If w and z becomes complete at the j-th step, then v becomes complete at j+3 step • => total complexity 3logn • How to merge Rw(j) and Rz(j) in constant time?

  2. Merging two samples in constant time • Two sequences S and T. • Predecessor of x in S: the largest element T smaller than x. • Example: S={1,3,4,9}, T={2,5,6,7} pred(3:T) = 2, pred(4:T) = 2, pred(5:S) = 4. • If each element of S and T know the position of its pred in T and S, => S and T can be merged in constant time using |S| + |T| PEs. • How to find the pred of Rw(j) and Rz(j) ? => Inductively. 1.Rw(j-1) and Rz(j-1) know their predecessors, and two sequence merged to Qv(j-1) . 2. each element in Rw(j-1) finds its pred in Qw(j-1) in constant time and its pred in Rw(j) in constant time. Note that no more than 4 elements of Rw(j-1) have the same pred in Rw(j) Each element in Rw(j) finds its pred in Rw(j-1) 3. Same for Rz. 4. With these pred knowledge, Rw(j) can determine their pred in Rz(j) in cons time.

  3. v Rw(j-1) Rz (j-1) Qw (j-1) w z Rank(x,S): number of elements in S smaller than x Rv(j) consists of every 4-th elements of Qv(j-1). update rank Rw(j), Rw (j-1) ranks are obtained Rz(j), Rz (j-1) ranks are obtained Ranks of Qw(j) : Qw(j-1)

  4. A1 A2 Ai A ……. X X1 X2 Xi B ……. B1 B2 Bi X = (x1,...,xs) rank(X:A) = (r1,...,rs) rank(X,B) = (t1,...,ts) How to compute rank(a,B)? Suppose a is in Ai Note that at most 4 elements are in Ai rank(a:B) = ti + rank(a,Bi) But |Bi| <= 4 so rank(a,Bi), and rank(a,B) in constant time

  5. Example of merging using ranks and predecessors A1 A2 A3 A4 -10 -5 -2 0 4 5 10 12 20 22 26 -4 1 2 6 17 18 19 21 24 25 B2 B3 B4 B1 C: 0 6 12 21 C is a sample Suppose that pred(a,C) and pred(b,C) are known for all a in A and b in B How to compute rank(x:AB)? AB is the merged list of A and B pred (4:C) = 0 rank(0,A) = 3 rank (0:B) = 1 rank(4,B) = rank (0:B) + rank(4,B2) = 1+rank (4,B2) = 3 rank (4:AB) = rank(4,A) + rank(4,B) = 4 + 3 = 7

  6. Example (sorting using Cole’s merging) Q(t) arrays are shown R(t) = 4th ele of Q(t-1) Qv(t) = Rw(t) + Rz(t) t=6 15 : 16 t=7 8,15 : 9,16 t=8 5,8,13,15 : 3,9,11,16 t=9 1,5,7,8,10,13,14,15 : 2,3,4,6,9,11,12,16 t=4 15 : 14 11 : 16 7,15 : 10,14 6,11 : 4,16 t=5 5,7,8,15 : 1,10,13,14 2,6,9,11 : 3,4,12,16 (1,5,7,8,10,13,14,15) (2,3,4,6,9,11,12,16) t=6 t=2 15 : 8 14 : 13 9 : 11 16 : 12 t=3 5,15 : 7,8 10,14 : 1,13 6,9 : 2,11 3,16 : 4,12 (5 7 8 15) ( 1 10 13 14) (2 6 9 11) (3 4 12 16) t=1 15:5 7,8 10,14 13,1 6,9 2,11 16,3 4,12 15 5 7 8 14 10 13 1 6 9 2 11 16 3 4 12

  7. Complexity • Number of steps: O(logn) • At each step: • For each data not in the lowest level i, there are 4 data at level i+1 • n + n/4 + n/16 + ... = O(n) => Only O(n) data are involved • O(n) processors are involved

  8. Reconstructing Tree • Given Inorder and preorder, how to reconstruct • the tree (find the parent)?

  9. 6 F 4 D H 8 B E G J 10 7 5 2 A C I K 11 9 3 1 Inorder A B C D E F G H I J K 1 2 3 4 5 6 7 8 9 10 11 Inorder Preorder sequence (IP sequence) Preorder F D B A C E H G J I K IP 6 4 2 1 3 5 8 7 10 9 11 lm: index of the largest elem (if any) in IP such that lm < m and IP(lm) < IP(m) rm: index of the smallest elem (if any) in IP such that rm < m and IP(rm) > IP(m)

  10. Let IP(x): Inorder number of node x. IP(T) is the smallest node which is larger than IP(L) and the largest one among those smaller than IP(R) T R L lm, rm • If lm > rm then IP(lm) appears after IP(rm) => IP(m) is a right child of IP(lm) else IP(m) is a left child of IP(rm)

  11. Example Inorder A B C D E F G H I J K m 1 2 3 4 5 6 7 8 9 10 11 IP(m) 6 4 2 1 3 5 8 7 10 9 11 F D B A C E H G J I K IP(lm) -- -- -- -- 2 4 6 6 8 8 1 node B D F F H H J lm 3 2 1 1 7 7 9 IP(rm) -- 6 4 2 4 6 - 8 - 10 - node (F) D B D F - H - J - rm 1 2 3 2 1 - 7 - 9 - f(m) = m, y(m) = IP(m) * defined: i*j = i provided IP(i) > IP(j) = j otherwise D(m) gives lm Operation *: max f(m) = y(m) = IP(m) D(m) gives IP(lm) Operation *: min f(m) = IP(m), y(m) = -IP(m) D(m) gives IP(rm)

More Related