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Undo. Why are you converting LaTeX math to ersatz hacked template-math? That is the wrong direction, needed only for certain very special circumstances that do not apply here. |
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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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