User:Mark Schröder/sandbox

In computer science a range minimum query (RMQ) solves the problem of finding the minimal value in a sub-array of an array of comparable objects. Range minimum queries have several use cases in computer science such as the lowest common ancestor problem or the longest common prefix problem (LCP).

Definition
Given an array $$A[1 ... n]$$ of $$n$$ objects taken from a well-ordered set (such as numbers), the range minimum query $$\operatorname{RMQ}_A(l,r) = \arg\min A[k]$$ (with $$1 \le l \le r \le n$$) returns the position of the minimal element in the specified sub-array $$A[l ... r]$$.

For example, when $$A = [0,5,2,5,4,3,1,6,3]$$, then the answer to the range minimum query for the sub-array $$A[3 \dots 8] = [2,5,4,3,1,6]$$ is $$7$$, as $$A[7] = 1$$.

Naïve solution
In a typical setting, the array $A$ is static, i.e., elements are not inserted or deleted during a series of queries, and the queries to be answered on-line (i.e., the whole set of queries are not known in advance to the algorithm). In this case a suitable preprocessing the array into a data structure (often called a preprocessing scheme) ensures faster query answering. A naïve solution is to precompute all possible queries, i.e. the minimum of all sub-arrays of $A$, and store these in an array $B$ such that $B[i, j] = min(A[i…j])$; then a range min query can be solved in constant time by array lookup in $B$. There are $Θ(n²)$ possible queries for a length-$n$ array, and the answers to these can be computed in $Θ(n²)$ time by dynamic programming.

Solution using constant time and linearithmic space
As in the solution above, answering in constant time will be achieved by pre-computing results. However, the array in which the results are stored will have a size of $$\mathcal{O}(n \log n)$$. For every left boundary $$l$$, the results to the queries $$\min(A[l \dots l + 2^{(k - 1)}])$$ are stored, with $$ k $$ growing from $$ 0 $$ to $$ \lfloor \log n \rfloor $$. Therefore, the table stores at most $$ \mathcal{O}(n \log n) $$ entries. A query $$ \min(A[l \dots r]) $$ can now be answered by splitting it into two separate queries: one query is the pre-computed query with range from $$ l $$ to the highest boundary smaller than $$ r $$. The other is the query with the same range that has $$ r $$ as its right boundary. The overall result can be obtained in constant time because these two queries can be answered in constant time and the only thing left to do is to to choose the smaller of the two results.

Solution using logarithmic time and linear space
This solution answers RMQs in $$ \mathcal{O}(\log n) $$ time. Its data structures use $$ \mathcal{O}(n) $$ space and its data structures can also be used to answer queries in constant time. The array is first conceptually divided into blocks of size $$ s = \frac{\log n}{4} $$. Then the minimum for each block can be computed in $$ \mathcal{O}(n) $$ time overall and the minima are stored in a new array.

RMQs can now be answered in logarithmic time trivially by looking at the blocks containing the left query boundary, the right query boundary and all the blocks in between: For example, the minimum array for $$A = [0,5,2,5,4,3,1,6,3]$$ would be $$A' = [0,3,1]$$.
 * The two blocks containing the boundaries can be searched naïvely. Elements outside the boundary need not even be looked at. This can be done in logarithmic time.
 * The minima of all blocks that are fully contained in the range, and the two minima mentioned above, need to be searched to answer the query.
 * Because the array was divided into blocks of size $$ \frac{\log n}{4} $$, there are at most $$ \frac{4n}{\log n} $$ blocks that are fully contained in the query.
 * By using the linearithmic solution one can find the overall minimum among these blocks. This data structure has size $$ \mathcal{O}(\frac{n}{\log n} \log (\frac{n}{\log n})) = \mathcal{O}(n) $$.
 * Now, only three minima need to be compared.

Solution using constant time and linear space
Only the sub-queries inside the blocks that are not fully contained in the query need to be answered in constant time. There are at most two of those queries after going through the steps above, which means they must be answered in constant time. This is achieved by storing the Cartesian trees for all the blocks in the array. A few observations: For every such tree, the possible result for all queries need to be stored. This comes down to $$ s^2 $$ or \mathcal{O}(\log^2 n) entries. This means the overall size of the table is $$ \mathcal{O}(n) $$.
 * Blocks with the same Cartesian trees give the same result for all queries on that block
 * The number of different Cartesian trees of $$ s $$ nodes is $$ C_s $$, the $$ s $$th Catalan number
 * Therefore, the number of different Cartesian trees for the blocks is in the range of $$ 4^s $$

To look up results efficiently, the Cartesian tree (row) corresponding to a specific block must be addressable in constant time. The solution is to do a breadth-first-search through the tree and represent every inner node as a 0-bit and every leaf as a 1-bit in a bit-word. Leave nodes are added so that every already existing node has two children. This leads to a size of $$ \frac{\log n}{4} $$ for every tree. To enable random access in constant time to any tree, the trees not contained in the original array need to be included as well. An array with indices of $$ \frac{\log n}{4} $$ bits length has size $$ 2^{\frac{\log n}{4}} = \mathcal{O}(n) $$.



It is known that a $O(n)$-time preprocessing is sufficient to answer subsequent queries in $O(1)$ time. The space of the resulting scheme is actually very small, namely $O(n)$ bits.

Applications
RMQs are used as a tool for many tasks in exact and approximate string matching. Severeal applications can be found in Fischer and Heun (2007, S.3).

Computing the lowest common ancestor in a tree
RMQs can be used to solve the lowest common ancestor problem and are used as a tool for many tasks in exact and approximate string matching. The LCA query $$ LCA_S(v,w) $$ of a rooted tree $$ S=(V,E) $$ and two edges $$v, w \in V $$ returns the last common edge $$u$$, or $$v$$ (or $$w$$) if the edge is on the path from the root to the edge $$w$$ (or $$v$$). Gabow, Bentley, and Tarjan (1984)[C] showed the the LCA Problem can be reduced in linear time to the RMQ problem. It follows that, same as the RMQ problem, the LCA problem can be solved in constant time and linear space as shown in J. Fischer and Heun (2007).

Computing the longest common prefix in a string
In the context of text indexing RMQs can be used to find the LCP (Longest Common Prefix), where $$ LCP_T(i, j) $$ computes the LCP of a common prefix of the suffixes that start indexes $$ i $$ and $$ j $$ in $$ T $$. For this purpose the LCP-Array $$H$$ of the Suffix-Array $$A$$ is used to compute an inverse Suffix-Array $$ A^{-1} $$ for $$T$$. Based on these two datastructures the length of the LCP can by computed in constant time by the formula: $$ LCP(i, j) = RMQ_H(A^{-1}[i]+1, A^{-1}[j]) $$.