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===Majority===
Finding frequent elements in a given set of items is one of the most important tasks in data mining. Finding frequent elements might be a difficult task to achieve when most items have similar frequencies. Therefore, it might be more beneficial if some threshold of significance was used for detecting such items. One of the most famous algorithms for finding the majority of an array was proposed by Boyer and Moore <ref>{{cite book |last1=Boyer|first1=Robert S.|date=1991|chapter-url=http://dx.doi.org/10.1007/978-94-011-3488-0_5|pages=105–117|place=Dordrecht|publisher=Springer Netherlands|access-date=2021-12-18|last2=Moore|first2=J. Strother|title=Automated Reasoning |chapter=MJRTY—A Fast Majority Vote Algorithm |series=Automated Reasoning Series |volume=1 |doi=10.1007/978-94-011-3488-0_5 |isbn=978-94-010-5542-0 }}</ref> which is also known as the [[Boyer–Moore majority vote algorithm]]. Boyer and Moore proposed an algorithm to find the majority element of a string (if it has one) in <math>O(n)</math> time and using <math>O(1)</math> space. In the context of Boyer and
==== Two-dimensional arrays ====
Gagie et al.<ref>{{cite book |last1=Gagie|first1=Travis |date=2011|chapter-url=http://dx.doi.org/10.1007/978-3-642-24583-1_29 |pages=295–300|place=Berlin, Heidelberg|publisher=Springer Berlin Heidelberg|isbn=978-3-642-24582-4|access-date=2021-12-18|last2=He|first2=Meng|last3=Munro|first3=J. Ian|last4=Nicholson|first4=Patrick K.|title=String Processing and Information Retrieval |chapter=Finding Frequent Elements in Compressed 2D Arrays and Strings |series=Lecture Notes in Computer Science |volume=7024 |doi=10.1007/978-3-642-24583-1_29 }}</ref> proposed a data structure that supports range
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==== One-dimensional arrays ====
Chan et al.<ref name=":0">{{cite book |last1=Chan|first1=Timothy M. |date=2012|chapter-url=http://dx.doi.org/10.1007/978-3-642-31155-0_26 |pages=295–306|place=Berlin, Heidelberg|publisher=Springer Berlin Heidelberg|isbn=978-3-642-31154-3|access-date=2021-12-20|last2=Durocher|first2=Stephane|last3=Skala|first3=Matthew|last4=Wilkinson|first4=Bryan T.|title=Algorithm Theory – SWAT 2012 |chapter=Linear-Space Data Structures for Range Minority Query in Arrays |series=Lecture Notes in Computer Science |volume=7357 |doi=10.1007/978-3-642-31155-0_26 }}</ref> proposed a data structure that given a one-dimensional array<math>A</math>, a subrange <math>R</math> of <math>A</math> (specified at query time) and a threshold <math>\tau</math> (specified at query time), is able to return the list of all <math>\tau</math>-majorities in <math>O(1/\tau)</math> time requiring <math>O(n \log n)</math> words of space. To answer such queries, Chan et al.<ref name=":0" /> begin by noting that there exists a data structure capable of returning the ''top-k'' most frequent items in a range in <math>O(k)</math> time requiring <math>O(n)</math> words of space. For a one-dimensional array <math>A[0,..,n-1]</math>, let a one-sided top-k range query to be of form <math>A[0..i] \text { for } 0 \leq i \leq n-1</math>. For a maximal range of ranges <math>A[0..i] \text { through } A[0..j]</math> in which the frequency of a distinct element <math>e</math> in <math>A</math> remains unchanged (and equal to <math>f</math>), a horizontal line segment is constructed. The <math>x</math>-interval of this line segment corresponds to <math>[i,j]</math> and it has a <math>y</math>-value equal to <math>f</math>. Since adding each element to <math>A</math> changes the frequency of exactly one distinct element, the aforementioned process creates <math>O(n)</math> line segments. Moreover, for a vertical line <math>x=i</math> all
Chan et al.<ref name=":0" /> first construct a [[range tree]] in which each branching node stores one copy of the data structure described above for one-sided range top-k queries and each leaf represents an element from <math>A</math>. The top-k data structure at each node is constructed based on the values existing in the subtrees of that node and is meant to answer one-sided range top-k queries. Please note that for a one-dimensional array <math>A</math>, a range tree can be constructed by dividing <math>A</math> into two halves and recursing on both halves; therefore, each node of the resulting range tree represents a range. It can also be seen that this range tree requires <math>O(n \log n)</math> words of space, because there are <math>O(\log n)</math> levels and each level <math>\ell</math> has <math>2^{\ell}</math> nodes. Moreover, since at each level <math>\ell</math> of a range tree all nodes have a total of <math>n</math> elements of <math>A</math> at their subtrees and since there are <math>O(\log n)</math> levels, the space complexity of this range tree is <math>O(n \log n)</math>.
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for <math>0\leq i \leq \log(depth(x))</math> where <math>\operatorname{par}(x)</math> returns the label of the direct parent of node <math>x</math>. Put another way, for each marked node, the set of all paths with a power of two length (plus one for the node itself) towards the root is stored. Moreover, for each <math>P_i(x)</math>, the set of all majority ''candidates'' <math>C_i(x)</math> are stored. More specifically, <math>C_i(x)</math> contains the set of all <math>(\tau/2)</math>-majorities in <math>P_i(x)</math> or labels that appear more than <math>(\tau/2).(2^i+1)</math> times in <math>P_i(x)</math>. It is easy to see that the set of candidates <math>C_i(x)</math> can have at most <math>2/\tau</math> distinct labels for each <math>i</math>. Gagie et al.<ref name=":2"/> then note that the set of all <math>\tau</math>-majorities in the path from any marked node <math>x</math> to one of its ancestors <math>z</math> is included in some <math>C_i(x)</math> (Lemma 2 in <ref name=":2"/>) since the length of <math>P_i(x)</math> is equal to <math>(2^i+1)</math> thus there exists a <math>P_i(x)</math> for <math>0\leq i \leq \log(depth(x))</math> whose length is between <math>d_{xz} \text{ and } 2 d_{xz}</math> where <math>d_{xz}</math> is the distance between x and z. The existence of such <math>P_i(x)</math> implies that a <math>\tau</math>-majority in the path from <math>x</math> to <math>z</math> must be a <math>(\tau/2)</math>-majority in <math>P_i(x)</math>, and thus must appear in <math>C_i(x)</math>. It is easy to see that this data structure require <math>O(n \log n)</math> words of space, because as mentioned above in the construction phase <math>O(\tau n)</math> nodes are marked and for each marked node some candidate sets are stored. By definition, for each marked node <math>O(\log n)</math> of such sets are stores, each of which contains <math>O(1/\tau)</math> candidates. Therefore, this data structure requires <math>O(\log n \times (1/\tau) \times \tau n)=O(n \log n)</math> words of space. Please note that each node <math>x</math> also stores <math>count(x)</math> which is equal to the number of instances of <math>label(x)</math> on the path from <math>x</math> to the root of <math>T</math>, this does not increase the space complexity since it only adds a constant number of words per node.
Each query between two nodes <math>u</math> and <math>v</math> can be answered by using the decomposability property (as explained above) of range <math>\tau</math>-majority queries and by breaking the query path between <math>u</math> and <math>v</math> into four subpaths. Let <math>z</math> be the lowest common ancestor of <math>u</math> and <math>v</math>, with <math>x</math> and <math>y</math> being the nearest marked ancestors of <math>u</math> and <math>v</math> respectively. The path from <math>u</math> to <math>v</math> is decomposed into the paths from <math>u</math> and <math>v</math> to <math>x</math> and <math>y</math> respectively (the size of these paths are smaller than <math>2\lceil 1 / \tau\rceil</math> by definition, all of which are considered as candidates), and the paths from <math>x</math> and <math>y</math> to <math>z</math> (by finding the suitable <math>C_i(x)</math> as explained above and considering all of its labels as candidates). Please note that, boundary nodes have to be handled accordingly so that all of these subpaths are disjoint and from all of them a set of <math>O(1/\tau)</math> candidates is derived. Each of these candidates is then verified using a combination of the <math>labelanc (x, \ell)</math> query which returns the lowest ancestor of node <math>x</math> that has label <math>\ell</math> and the <math>count(x)</math> fields of each node. On a <math>w</math>-bit RAM and an alphabet of size <math>\sigma</math>, the <math>labelanc (x, \ell)</math> query can be answered in <math>O\left(\log \log _{w} \sigma\right) </math> time whilst having linear space requirements.<ref>{{Cite journal|last1=He|first1=Meng|last2=Munro|first2=J. Ian|last3=Zhou|first3=Gelin|date=2014-07-08|title=A Framework for Succinct Labeled Ordinal Trees over Large Alphabets|url=http://dx.doi.org/10.1007/s00453-014-9894-4|journal=Algorithmica|volume=70|issue=4|pages=696–717|doi=10.1007/s00453-014-9894-4|s2cid=253977813 |issn=0178-4617|url-access=subscription}}</ref> Therefore, verifying each of the <math>O(1/\tau)</math> candidates in <math>O\left(\log \log _{w} \sigma\right) </math> time results in <math>O\left((1/\tau)\log \log _{w} \sigma\right) </math> total query time for returning the set of all <math>\tau </math>-majorities on the path from <math>u </math> to <math>v </math>.
==Related problems==
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