Lowest Common Ancestors

Two vertices (u, v)Lowest common ancestors, lca (u, v)

Example

1

2 8 8

3 4

5 6 7

lca (5, 6) = 4lca (3, 7) = 2lca (7, 8) = 1

l(v): left most appearance of vr(v): right most appearance of vIf r(u) < l(v) thenlca (u,v) is the vertex with minimum level over the interval [r(u), l(v)]How to find [r(u), l(v)] efficiently?

Range Minima ProblemLet {s1,s2,s3,s4,s5,s6,s7,s8} = {5,10,3,4,7,1,8,2}

Given i, j, how to find min {ai...aj} in constant time?

Approach:Prefix minSuffix min

p=[5,5]s=[5,10]

p=[7,1]s=[1,1]

p=[5,5,3,3]s=[3,3,3,4]

p=[7,1,1,1]s=[1,1,2,2]

p=[3,3]s=[3,4]

p=[8,2]s=[2,2]

5 10 3 4 7 1 8 2

p=[5,5,3,3,3,1,1,1]s=[1,1,1,1,1,1,2,2]

min {ak..al} 1. Find lca w of ak and al

in a complete binary tree

Let x: left child, y: right child of w2. suffix min of k in x

3. prefix min of l in y 4. take min of 2 and 3

Example {a3..a5}w = v1

suffix min of 3rd in v2 = 3

prefix min of 5th in v3 =7

min {a3..a5} = 3

v1

v2 v3

v4 v5 v6 v7

Applications: Computing the minimum of its descendants

Example12

2 3

7 9

6 4

4

51For each vi, compute

xi = minimum aj among all its descendants vj of vi

Then xi = min {ak .. aj }, where k is preorder (vi) and j = k + descendants of (vi) -1

Complexity of Range Minima

O ( log n ) timewith O ( n ) PEs

Merging P [ i.. j]P [ j+1.. k]

P’ [ i .. k] parentcopy

min (, p[j+1])Each level of three O (1)total time O (log n) timefor constructing the P, S lists

Two part: • Initial construction• Searching

Construction:

Searching:Constant time

Breadth First Traversal

BFS: 1, 2, 8, 9, 3, 4, 5, 6, 7

1

2 8 9

3 4

5 6 7

level 0

level 1

Remove the left most edges, and right most edges from ETET’ = (v1,v2), (v2,v3),(v1,v2), (v2,v3), (v3,v2), (v2,v4), ), (v4,v5),(v4,v5), (v5,v4), (v4,v6), (v6,v4), (v4,v7), (v7,v4),v7,v4), (v4,v2),(v4,v2), (v2,v1), (v1,v8), (v8,v1), (v1,v9), (v9,v1)(v9,v1)How to detect?

Prefix maxima, and suffix maxia of edge levelIf prefix max has ai - ai-1 > 0 then i-th edge is a leftmost edge

Back Edge: “( “ Advance edge: “)”find mate of each parenthesis

Mostly well definedIf (vi,vj) is the right most edge at level k,

then its mate is leftmost edge at level k+1 (if it exists)(vk,vl) is a mate of (vi,vj) then make a link vk -> vi

(vi,vj) is the first edge, vi -> vj

Eulerian Tour ET = (v1,v2), (v2,v3), (v3,v2), (v2,v4), (v4,v5), (v5,v4), (v4,v6), (v6,v4), (v4,v7), (v7,v4), (v4,v2), (v2,v1), (v1,v8), (v8,v1), (v1,v9), (v9,v1)

Generalized Prefix Computation

• Next class: read section 4.9