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

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.

LCAT(u,v)

u v

RMQ - Range Minimum Query

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].

RMQA(i,j)

0 1634 13 7 19 10 12 1 2

RMQA(3,7) = 4

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

Complexity Notation

)n(g),n(f

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 = )1(O),n(O 3

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.

)(),( ngnf

)1()12(),()12( OngnOnf

8

Example

0

1

2 3 6 9

54

7

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)0

1

2 3 6 9

5

8

4

7

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

R[6] R[9]

E[10,…,16]

RMQL(10,16) = 11

LCAT(6,9) = E[11]=5

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 : )1()12(),()12( OngnOnf

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 -

)1(),( 3 OnO

)1(),( 2 OnO

ST Algorithm for RMQPreprocessing

Time: O(nlogn).

a1 ... ... an

i i+2j-1i+2j-1-1

)lognj1 n,i(1 j]M[i,matrix thecompute willWe

][min)12,(],[12

kAiiRMQjiM jiki

jA

]1j,2M[i Otherwise

1]-jM[i, ]]1j,2A[M[i1]]-jA[M[i,

]j,i[M1-j

1-j

ST Algorithm for RMQ Cont.

Arbitrary RMQ(i,j) Query

Time: O(1). ST Algorithm Complexity:

a1an... ...

i j2k elements

2k elements

1ij2rmaxk r

k,12jM Otherwise

ki,M k,12jMAk,iMA

)j,i(RMQk

k

)1(O),nlogn(O

RMQ Solution

… ...B[0] B[2n/logn]B[i]

... ... .........

A

… ...A’[0] A’[2n/logn]A’[i]

A’: block min value

B : block min index

blocksn log

2n

nlog2

1 size of blocks

n log

2n intopartition

Size(A’) = 2n / logn.

ST_Preprocessing(A’)

ST(A’)=

RMQ Solution Cont.

)n(Onlog

)nloglogn2(logn2

n log

2nlog

n log

2n

)1(O),n(O

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

)nloglogn(logOn log2

1log n log

2

1

)nloglogn(O

Observation

Let two arrays X & Y such that

CY[i]X[i] i

3 4 6 5 5 4 5 6 4 5

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

0 13 2 2 1 2 3 1 2

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

+1 -1 -1-1 +1 +1 -1+1 +1

)j,i(RMQ)j,i(RMQ j,i YX

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 =

nlog2

1

1nlog2

1

n2

12

1nlog2

1

nlognO 2

)1(O),n(O

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.

)1(O),n(O

)1(O),n(O

Cartesian Tree

10 1634 26 7 19 10 1225 22

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

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 1634 26 7 19 10 1225 22

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

4

Cartesian Tree

10 1634 26 7 19 10 1225 22

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

4

6

Cartesian Tree

10 1634 26 7 19 10 1225 22

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

4

7

6

Cartesian Tree

10 1634 26 7 19 10 1225 22

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

4

7

6

9

Cartesian Tree

10 1634 26 7 19 10 1225 22

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

4

7

6

9

8

Cartesian Tree

10 1634 26 7 19 10 1225 22

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

4

5 7

6

9

8

Cartesian Tree

10 1634 26 7 19 10 1225 22

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

4

0

5 7

6

9

8

Cartesian Tree

10 1634 26 7 19 10 1225 22

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

4

0

2

1 3

5 7

6

9

8

Linear Cartesian Tree Construction

10 1634 26 7 19 10 1225 22

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

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 1634 26 7 19 10 1225 22

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

0

10

Linear Cartesian Tree Construction

10 1634 26 7 19 10 1225 22

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

0

1

10

25

Linear Cartesian Tree Construction

10 1634 26 7 19 10 1225 22

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

0

2

1

10

22

25

34

Linear Cartesian Tree Construction

10 1626 7 19 10 1225 22

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

0

2

1 33

73434

10

22

25 34

Linear Cartesian Tree Construction

10 1634 26 7 19 10 1225 22

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

4

0

2

1 3

10

22

25 34

7

Linear Cartesian Tree Construction

10 1634 26 7 19 10 1225 22

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

4

0

2

1 3

5 7

6

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)

)1(O),n(O

)1(O),n(O

Linear Cartesian Tree Construction

10 1634 26 7 19 10 1225 22

A[0] A[2]A[1] A[9]A[3] A[4] A[5] A[6] A[7] A[8]

4

0

2

1 3

5 7

6

9

8