User:Mark Schröder/sandbox

A Constructing the corresponding cartesian tree to solve a range minimum query.
Range minimum query reduced to the lowest common ancestor problem.

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

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Given an array   of   objects taken from a well-ordered set (such as numbers), the range minimum query   (with  ) returns the position of the minimal element in the specified sub-array  .

For example, when  , then the answer to the range minimum query for the sub-array   is  , as  .

Algorithms

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Naïve solution

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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[ij]); 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.[1]

Solution using constant time and linearithmic space

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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  . For every left boundary  , the results to the queries   are stored, with   growing from   to  . Therefore, the table stores at most   entries. A query   can now be answered by splitting it into two separate queries: one query is the pre-computed query with range from   to the highest boundary smaller than  . The other is the query with the same range that has   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.

r\l 0 1 2 3
1 1 1 1 1
2 2 3 3 3
3 3 3 3 7
4 4 5 6 7
5 5 6 7 7
6 6 7 7 7
7 7 7 7 7
8 8 7 7 7
9 9 7 7 7
Result table for  :

Solution using logarithmic time and linear space

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This solution answers RMQs in   time. Its data structures use   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  . Then the minimum for each block can be computed in   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:

  • 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  , there are at most   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  .
  • Now, only three minima need to be compared.

For example, the minimum array for   would be  .

Solution using constant time and linear space

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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:

  • Blocks with the same Cartesian trees give the same result for all queries on that block
  • The number of different Cartesian trees of   nodes is  , the  th Catalan number
  • Therefore, the number of different Cartesian trees for the blocks is in the range of  

For every such tree, the possible result for all queries need to be stored. This comes down to   or </math> \mathcal{O}(\log^2 n) </math> entries. This means the overall size of the table is  .

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   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   bits length has size  .

 
Example of Cartesian trees for  . Notice that the first and third tree have the same layout, so there are exactly two pre-computed sets of queries in the table on the left.
Index 1 2 3
1 2 3 1 2 3 1 2 3
0 /
23 (Bitword 0010111) 1 2 3 / 2 3 / / 3
39 (Bitword 0100111) 1 1 1 / 2 3 / / 3
127 /
Pre-computed results for the three Cartesian block trees of  :

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.[2]

Applications

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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)[3].

Computing the lowest common ancestor in a tree

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RMQs can be used to solve the lowest common ancestor problem[1][4] and are used as a tool for many tasks in exact and approximate string matching. The LCA query   of a rooted tree   and two edges   returns the last common edge  , or   (or  ) if the edge is on the path from the root to the edge   (or  ). 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)[3].

Computing the longest common prefix in a string

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In the context of text indexing RMQs can be used to find the LCP (Longest Common Prefix), where   computes the LCP of a common prefix of the suffixes that start indexes   and   in  . For this purpose the LCP-Array   of the Suffix-Array   is used to compute an inverse Suffix-Array   for  . Based on these two datastructures the length of the LCP can by computed in constant time by the formula:  [5].

See also

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References

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  • Berkman, Omer; Vishkin, Uzi (1993). "Recursive Star-Tree Parallel Data Structure". SIAM Journal on Computing. 22 (2): 221–242. doi:10.1137/0222017.
  1. ^ a b Bender, Michael A.; Farach-Colton, Martín; Pemmasani, Giridhar; Skiena, Steven; Sumazin, Pavel (2005). "Lowest common ancestors in trees and directed acyclic graphs" (PDF). Journal of Algorithms. 57 (2): 75–94. doi:10.1016/j.jalgor.2005.08.001.
  2. ^ Fischer, Johannes; Heun, Volker (2007). A New Succinct Representation of RMQ-Information and Improvements in the Enhanced Suffix Array. Proceedings of the International Symposium on Combinatorics, Algorithms, Probabilistic and Experimental Methodologies. LNCS. Vol. 4614. Springer. pp. 459–470. doi:10.1007/978-3-540-74450-4_41.
  3. ^ a b Fischer, J. and V. Heun (2007). "A new succinct representation of RMQ-information and improvements in the enhanced suffix array". Combinatorics, Algorithms, Probabilistic and Experimental Methodologies: 459–470. doi:10.1007/978-3-540-74450-4_41.
  4. ^ Bender, Michael; Farach-Colton, Martín (2000). The LCA Problem Revisited. LATIN 2000: Theoretical Informatics. LNCS. Vol. 1776. Springer. pp. 88–94. doi:10.1007/10719839_9.
  5. ^ Fischer, J. and V. Heun (2006). "Theoretical and practical improvements on the RMQ-problem, with applications to LCA and LCE". Combinatorial Pattern Matching: 36–48. doi:10.1007/11780441_5.
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Category:Search algorithms