Range Minimum and Lowest Common Ancestor Queriesjrad/presentation.pdf · Range Minimum Queries...

Range Minimum and Lowest Common AncestorQueries

Slides by Solon P. Pissis

November 15, 2019

RMQs and LCA queries

Range Minimum Queries problem

DefinitionGiven an array A[1 . . . n], preprocess A so that a minimum of anyfragment A[i . . . j ] can be computed efficiently:

RMQA(i , j) = arg min A[k]

with 1 ≤ i ≤ k ≤ j ≤ n.

i jk1 n

Answering any query in O(1) time is trivial if we allow O(n2) timeand space preprocessing.

with 1 ≤ i ≤ k ≤ j ≤ n.

i jk1 n

with 1 ≤ i ≤ k ≤ j ≤ n.

i jk1 n

with 1 ≤ i ≤ k ≤ j ≤ n.

i jk1 n

O(n log n) time and space and O(1)-time queries

Let us show the following lemma.

LemmaThe RMQ problem can be solved in O(1) time after O(n log n)time and space preprocessing.

We apply the well-known doubling technique for preprocessing.

For all k = 0, 1, . . . , log n construct:

Bk [i ] = min{A[i ],A[i + 1], . . . ,A[i + 2k − 1]}.

How? B0[i ] = A[i ] and Bk+1[i ] = min{Bk [i ],Bk [i + 2k ]}, for all i .

Bk [i ] = min{A[i ],A[i + 1], . . . ,A[i + 2k − 1]}.

B0[i ] = A[i ] and Bk+1[i ] = min{Bk [i ],Bk [i + 2k ]}, for all i .

Bk [i ] = min{A[i ],A[i + 1], . . . ,A[i + 2k − 1]}.

And now the querying part.

Two cases:

I Fragment of length 2k : Trivial, use B;

I Otherwise, we can cover any range with two power-of-2ranges.

Compute k = blog(j − i + 1)c.

i j1 n

Return min{Bk [i ],Bk [j − 2k + 1]} in O(1) time!

Two cases:

i j1 n

Two cases:

i j1 n

Two cases:

i j1 n

Two cases:

i j1 n

Two cases:

i j1 n

O(n) time and space and O(log n)-time queries

LemmaThe RMQ problem can be solved in O(log n) time after O(n) timeand space preprocessing.

(Segment tree.)

We apply the well-known micro-macro decomposition technique.

b = log n

Decompose array A into blocks of length b = log n.

(Segment tree.)

b = log n

(Segment tree.)

b = log n

(Segment tree.)

b = log n

(Segment tree.)

b = log n

(Segment tree.)

b = log n

(Segment tree.)

b = log n

Decompose array A into blocks of length b = log n.RMQs and LCA queries

b = log n

I Construct a new array A′:

A′[i ] = min{A[i · b + 1],A[i · b + 2], . . . ,A[(i + 1) · b]}.

I Build the previously described structure for A′[1 . . . nlog n ].

How much time and space do we need? O( nlog n log( n

log n )) = O(n).

b = log n

A′[i ] = min{A[i · b + 1],A[i · b + 2], . . . ,A[(i + 1) · b]}.

log n )) = O(n).

b = log n

A′[i ] = min{A[i · b + 1],A[i · b + 2], . . . ,A[(i + 1) · b]}.

log n )) = O(n).

b = log n

A′[i ] = min{A[i · b + 1],A[i · b + 2], . . . ,A[(i + 1) · b]}.

How much time and space do we need?

O( nlog n log( n

log n )) = O(n).

b = log n

A′[i ] = min{A[i · b + 1],A[i · b + 2], . . . ,A[(i + 1) · b]}.

log n )) = O(n).

Let i , j be a query (Notice: endpoints are inside blocks).

I We do not have the min in the prefix and suffix of every block.

I For each block, precompute it: O(n) time and space!

I For the rest: use the structure we have built for A′.

I Now any such query takes O(1) time.

Any problems? Yes! When i , j are fully contained in a single block.

Solution: Naıve search in block gives O(log n)-time queries!

Any problems?

Yes! When i , j are fully contained in a single block.

Any problems? Yes!

When i , j are fully contained in a single block.

O(n) time and space and O(1)-time queries

OK. Now it is time to show the breakthrough.

Theorem (Bender and Farach-Colton, LATIN 2000)

The RMQ problem can be solved in O(1) time after O(n) timeand space preprocessing.

I We only have to deal with the strictly-inside-a-block case.

I We will show how to do that for a very restricted case:

|A[i + 1]− A[i ]| = 1

I We will then explain why this restricted case is sufficient!

|A[i + 1]− A[i ]| = 1

Assume: |A[i + 1]− A[i ]| = 1.

Observation: If two arrays X [1 . . . k], Y [1 . . . k] differ by somefixed value at each position; i.e. there is a c such thatX [i ] = Y [i ] + c for every i , then all RMQ answers will be the samefor X and Y .

I Normalize blocks by subtracting the initial offset from everyelement.

I Choose b = 1/2 · log n (instead of b = log n).

I A normalized block is a vector of length b − 1.

I There are 2b−1 = Θ(√n) distinct vectors.

I Precompute and store all answers!

Assume: |A[i + 1]− A[i ]| = 1.

I Create Θ(√n) tables.

I In each table put all (log n/2)2 answers to all in-block queries.

I Θ(√n log2 n) extra preprocessing.

I Query time is O(1).

I Preprocessing time and space is O(n).

We still need to explain why this restricted case is sufficient!

Assume: |A[i + 1]− A[i ]| = 1.

Lowest Common Ancestor queries

The lowest common ancestor (LCA) of two nodes u and v is thedeepest node that is an ancestor of both u and v .

DefinitionGiven a rooted tree T , preprocess T so that the LCA of u and vcan be computed efficiently.

Linear-Time Reduction: RMQ to LCA

Given array A we can build the Cartesian tree T of A:

I The root is a minimum element.

I Removing this element splits the array into two.

I Recurse on the two subarrays.

Querying:

A = [8, 7, 3, 20, 2, 17, 5, 21, 11, 12]

17 113

Querying:

A = [8, 7, 3, 20, 2, 17, 5, 21, 11, 12]

17 113

Querying:

A = [8, 7, 3, 20, 2, 17, 5, 21, 11, 12]

17 113

Querying:

A = [8, 7, 3, 20, 2, 17, 5, 21, 11, 12]

17 113

Querying:

A = [8, 7, 3, 20, 2, 17, 5, 21, 11, 12]

17 113

Querying:

A = [8, 7, 3, 20, 2, 17, 5, 21, 11, 12]

17 113

Querying:

A = [8, 7, 3, 20, 2, 17, 5, 21, 11, 12]

17 113

Linear-Time Reduction: LCA to RMQ

Given a rooted tree T on n nodes we construct:

I E [1 . . . 2n − 1]: node labels traversed in an Euler Tour (DFS).I L[1 . . . 2n − 1]: the distance of E [i ] from the root.I R[i ] = min{j : E [j ] = i} (first occurrence of E [i ] in the tour).

E = [0, 4, 5, 4, 1, 2, 3, 2, 1, 4, 0]