The LCA Problem Revisited Michael A.Bender & Martin Farach-Colton

Outline Definitions. Reduction from LCA to RMQ. Trivial Solutions for RMQ Faster solution for sub-problem of RMQ. Solution for general RMQ.

LCA - Lowest Common Ancestor LCAT(u,v) The LCA of nodes u and v in a tree T is the shared ancestor of u and v that is located farthest from the root. u v

RMQ - Range Minimum Query RMQA(i,j) For indices i and j between 1 and n, query RMQA(i,j) returns the index of the smallest element in the sub array A[i…j]. A[0] A[1] A[2] A[3] A[4] A[5] A[6] A[7] A[8] A[9] 1 2 34 7 19 10 12 13 16 RMQA(3,7) = 4

Complexity Notation Preprocessing Time Query Time

LCA - Trivial Algorithm For each pair of vertices, follow both paths toward the root until the first shared vertex is found. Complexity =

Reduction From LCA to RMQ If there is an -time solution for RMQ, then there is an -time solution for LCA. The reduction - We build three array from the input tree T. 1. E[1,…,2n-1] stores the nodes visited in an Euler Tour of T. 2. L[1,…,2n-1], where L[i] is the level of E[i] in T. 3. R[1,…,n], where R[i] is the index of the first occurrence of the node i in the array E.

Example 1 4 5 6 7 9 2 3 8 E: 0 1 2 1 3 1 0 4 0 5 6 5 7 8 7 5 9 5 0 L: 0 1 2 1 2 1 0 1 0 1 2 1 2 3 2 1 2 1 0 R: 0 1 2 4 7 9 10 12 13 16

Computing LCAT(u,v) The nodes in the Euler Tour between the first visits to u and to v are E[R[u],…,R[v]]. The shallowest node is at index RMQL(R[u],R[v]). The node E[RMQL(R[u],R[v])] is LCAT(u,v).

Example - LCAT(6,9) 1 4 5 6 7 9 2 3 8 E[10,…,16] E: 0 1 2 1 3 1 0 4 0 5 6 5 7 8 7 5 9 5 0 L: 0 1 2 1 2 1 0 1 0 1 2 1 2 3 2 1 2 1 0 R: 0 1 2 4 7 9 10 12 13 16 RMQL(10,16) = 11 LCAT(6,9) = E[11]=5 R[6] R[9]

Complexity Analysis Preprocessing Array Construction : O(n). Preprocessing of the array L : f(2n-1). Query Three Array references : O(1). RMQ query in L : g(2n-1). Overall :

Naive Solutions for RMQ From now on, we will focus on the RMQ problem. Computing the RMQ for every pair of indices - Trivial dynamic programming -

ST Algorithm for RMQ Preprocessing Time: O(nlogn). an a1 ... ... i i+2j-1-1 i+2j-1 an a1 ... ...

ST Algorithm for RMQ Cont. Arbitrary RMQ(i,j) Query Time: O(1). ST Algorithm Complexity: i j 2k elements an a1 ... ... 2k elements

RMQ Solution A’: block min value … ... B : block min index … ... ... A’[i] A’[2n/logn] A’: block min value … ... B[0] B[i] B[2n/logn] B : block min index … ... ... ... ... ... ... A

RMQ Solution Cont. Size(A’) = 2n / logn. ST_Preprocessing(A’) ST(A’)=

RMQ Solution Cont. Arbitrary RMQ(i,j) Query We should compute the following values: 1. The minimum from i to the end of its block. 2. The minimum of all the blocks in between i’s block and j’s block. 3. The minimum from the beginning of j’s block to j. A[i] A[j] ... 1 2 3

RMQ Solution Cont. ST Preprocessing of a block Per Block All Blocks A[i] A[j] ... 1 2 3

Observation Let two arrays X & Y such that 3 4 5 6 5 4 5 6 5 4 1 2 3 2 1 2 3 2 1 2 3 2 1 +1 +1 +1 -1 -1 +1 +1 -1 -1

 1RMQ Block size = 1 sequence block size = Number of possible sequences = Preprocessing all possible 1 blocks: Determining which table to use for each block: O(n). Overall time complexity =

General RMQ Solution Reduction from RMQ to LCA. Claim: If there is an -time solution for LCA, then there is an -time solution for RMQ.

Cartesian Tree A[0] A[1] A[2] A[3] A[4] A[5] A[6] A[7] A[8] A[9] 10 25 22 34 7 19 10 12 26 16 Given an array, we can build a Cartesian tree. The root of a Cartesian tree is the minimum element of the array. The root is labeled with the position of this minimum.

Cartesian Tree 10 25 22 34 7 19 10 12 26 16 4 A[0] A[1] A[2] A[3] A[4]

Cartesian Tree 10 25 22 34 7 19 10 12 26 16 4 6 A[0] A[1] A[2] A[3]

Cartesian Tree 10 25 22 34 7 19 10 12 26 16 4 6 7 A[0] A[1] A[2] A[3]

Cartesian Tree 10 25 22 34 7 19 10 12 26 16 4 6 7 9 A[0] A[1] A[2]

Cartesian Tree 10 25 22 34 7 19 10 12 26 16 4 6 7 9 8 A[0] A[1] A[2]

Cartesian Tree 10 25 22 34 7 19 10 12 26 16 4 6 5 7 9 8 A[0] A[1] A[2]

Cartesian Tree 10 25 22 34 7 19 10 12 26 16 4 6 5 7 9 8 A[0] A[1] A[2] 6 5 7 9 8

Cartesian Tree A[0] A[1] A[2] A[3] A[4] A[5] A[6] A[7] A[8] A[9] 10 25 22 34 7 19 10 12 26 16 4 6 2 5 7 1 3 9 8

Linear Cartesian Tree Construction 10 25 22 34 7 19 10 12 26 16 Suppose Ci is the Cartesian tree of A[0,…,i]. The i+1 node belongs to the rightmost path of Ci+1. We climb up the rightmost path of Ci until finding the position where i+1 node belongs.

Linear Cartesian Tree Construction 10 25 22 34 7 19 10 12 26 16 10

Linear Cartesian Tree Construction 10 25 22 34 7 19 10 12 26 16 10 25 1

Linear Cartesian Tree Construction 10 25 22 34 7 19 10 12 26 16 10 22 2 25 1

Linear Cartesian Tree Construction 10 25 22 34 34 34 7 7 19 10 12 26 16 10 22 2 34 25 1 3 3

Linear Cartesian Tree Construction 10 25 22 34 7 19 10 12 26 16 7 4 10 22 2 34 25 1 3

Linear Cartesian Tree Construction 10 25 22 34 7 19 10 12 26 16 4 6 2 5 7 1 3 9 8

Reduction from RMQ to LCA If there is an -time solution for LCA, then there is an -time solution for RMQ. Let A be the input array to the RMQ problem. Let C be the Cartesian tree of A. RMQA(i,j) = LCAC(i,j)

Linear Cartesian Tree Construction 10 25 22 34 7 19 10 12 26 16 4 6 2 5 7 1 3 9 8