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

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

3. 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

4. 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

5. Complexity Notation
Preprocessing Time
Query Time

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

7. 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.

8. Example 1 4 5 6 7 9 2 3 8 E: L: R:

9. 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).

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

11. 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 :

12. 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 -

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

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

15. 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

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

17. 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]
...

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

19. Observation Let two arrays X & Y such that 3 4 5 6 5 4 5 6 5 4 1 2 3 21 2 3 2 1 2 3 2 1 +1 +1 +1 -1 -1 +1 +1 -1 -1

20.  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 =

21. 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.

22. 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.

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

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

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

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

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

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

29. 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

30. 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

31. Linear Cartesian Tree Construction10 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.

32. Linear Cartesian Tree Construction10 25 22 34 7 19 10 12 26 16 10

33. Linear Cartesian Tree Construction10 25 22 34 7 19 10 12 26 16 10 25 1

34. Linear Cartesian Tree Construction10 25 22 34 7 19 10 12 26 16 10 22 2 25 1

35. Linear Cartesian Tree Construction10 25 22 34 34 34 7 7 19 10 12 26 16 10 22 2 34 25 1 3 3

36. Linear Cartesian Tree Construction10 25 22 34 7 19 10 12 26 16 7 4 10 22 2 34 25 1 3

37. Linear Cartesian Tree Construction10 25 22 34 7 19 10 12 26 16 4 6 2 5 7 1 3 9 8

38. 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)

39. Linear Cartesian Tree Construction10 25 22 34 7 19 10 12 26 16 4 6 2 5 7 1 3 9 8