Query processing: phrase queries and positional indexes Prof. Paolo Ferragina, Algoritmi per "Information Retrieval" Query processing: phrase queries and positional indexes Paolo Ferragina Dipartimento di Informatica Università di Pisa

Sec. 2.4 Phrase queries Want to be able to answer queries such as “stanford university” – as a phrase Thus the sentence “I went at Stanford my university” is not a match.

Solution #1: 2-word indexes Sec. 2.4.1 Solution #1: 2-word indexes For example the text “Friends, Romans, Countrymen” would generate the biwords friends romans romans countrymen Each of these 2-words is now an entry in the dictionary Two-word phrase query-processing is immediate.

Can have false positives! Sec. 2.4.1 Longer phrase queries Longer phrases are processed by reducing them to bi-word queries in AND stanford university palo alto can be broken into the Boolean query on biwords, such as stanford university AND university palo AND palo alto Need the docs to verify + They are combined with other solutions Can have false positives! Index blows up

Solution #2: Positional indexes Sec. 2.4.2 Solution #2: Positional indexes In the postings, store for each term and document the position(s) in which that term occurs: <term, number of docs containing term; doc1: position1, position2 … ; doc2: position1, position2 … ; etc.>

Processing a phrase query Sec. 2.4.2 Processing a phrase query “to be or not to be”. to: 2:1,17,74,222,551; 4:8,16,190,429,433; 7:13,23,191; ... be: 1:17,19; 4:17,191,291,430,434; 5:14,19,101; ... Same general method for proximity searches

Sec. 7.2.2 Query term proximity Free text queries: just a set of terms typed into the query box – common on the web Users prefer docs in which query terms occur within close proximity of each other Would like scoring function to take this into account – how?

Positional index size You can compress position values/offsets Sec. 2.4.2 Positional index size You can compress position values/offsets Nevertheless, a positional index expands postings storage by a factor 2-4 in English Nevertheless, a positional index is now commonly used because of the power and usefulness of phrase and proximity queries … whether used explicitly or implicitly in a ranking retrieval system.

Sec. 2.4.3 Combination schemes 2-Word + Positional index is a profitable combination 2-word is particularly useful for particular phrases (“Michael Jackson”, “Britney Spears”) More complicated mixing strategies do exist!

Soft-AND E.g. query rising interest rates Sec. 7.2.3 Soft-AND E.g. query rising interest rates Run the query as a phrase query If <K docs contain the phrase rising interest rates, run the two phrase queries rising interest and interest rates If we still have <K docs, run the “vector space query” rising interest rates (…see next…) “Rank” the matching docs (…see next…)

Prof. Paolo Ferragina, Algoritmi per "Information Retrieval" Zone indexes Paolo Ferragina Dipartimento di Informatica Università di Pisa

Parametric and zone indexes Sec. 6.1 Parametric and zone indexes Thus far, a doc has been a term sequence But documents have multiple parts: Author Title Date of publication Language Format etc. These are the metadata about a document

Sec. 6.1 Zone A zone is a region of the doc that can contain an arbitrary amount of text e.g., Title Abstract References … Build inverted indexes on fields AND zones to permit querying E.g., “find docs with merchant in the title zone and matching the query gentle rain”

Sec. 6.1 Example zone indexes Encode zones in dictionary vs. postings.

Caching for faster query Prof. Paolo Ferragina, Algoritmi per "Information Retrieval" Caching for faster query Two opposite approaches: Cache the query results (exploits query locality) Cache pages of posting lists (exploits term locality)

Tiered indexes for faster query Sec. 7.2.1 Tiered indexes for faster query Break postings up into a hierarchy of lists Most important … Least important Inverted index thus broken up into tiers of decreasing importance At query time use top tier unless it fails to yield K docs If so drop to lower tiers

Sec. 7.2.1 Example tiered index

Query processing: optimizations Prof. Paolo Ferragina, Algoritmi per "Information Retrieval" Query processing: optimizations

Skip pointers (at indexing time) Sec. 2.3 Skip pointers (at indexing time) 41 128 128 2 4 8 41 48 64 11 31 1 2 3 8 11 17 21 31 How do we deploy them ? Where do we place them ?

Sec. 2.3 Using skips 41 128 2 4 8 41 48 64 128 11 31 But the skip successor of 11 on the lower list is 31, so we can skip ahead past the intervening postings. 1 2 3 8 11 17 21 31 Suppose we’ve stepped through the lists until we process 8 on each list. We match it and advance. We then have 41 and 11 on the lower. 11 is smaller.

Placing skips Tradeoff: Sec. 2.3 Placing skips Tradeoff: More skips  shorter spans  more likely to skip. But lots of comparisons to skip pointers. Fewer skips  longer spans  few successful skips. Less pointer comparisons.

Placing skips Simple heuristic for postings of length L Sec. 2.3 Placing skips Simple heuristic for postings of length L use L evenly-spaced skip pointers. This ignores the distribution of query terms. Easy if the index is relatively static. This definitely useful for in-memory index The I/O cost of loading a bigger list can outweigh the gains!

Sec. 2.3 You can solve it by Shortest Path Placing skips, contd What if it is known a distribution of accesses pk to the k-th element of the inverted list? w(i,j) = sumk=i..j pk [prob access an item in pos i..j] L^0(i,j) = average cost of accessing an item in the sublist from i to j = sumk=i..j pk * (k-i+1) L^1(1,n) = average cost with one single skip 1 (first skip cmp) + (avg cost access the two lists) minu>1 w(1,u-1) * L^0(1,u-1) + w(u,n) * L^1(u,n) L^0(i,j) can be tabulated in O(n^2) time Computing L^1(i,n) takes O(n), given L^1(j,n), for j>i Computing the total L^1(1,n) takes O(n^2) time

Sec. 2.3 Placing skips, contd What if it is also fixed the maximum number of skip-pointers that can be allocated? Same as before but we add the parameter p L^1_p(1,n) = 1 + min_{u>1} w(1,u-1) * L^0(1,u-1) + w(u,n) * L^1_{p-1}(u,n) L^1_0(i,j) = L^0(i,j), i.e. no pointers left, so scan L^1_i(j,n) takes O(n) time [min calculation] if are available the values for L^{i-1}(h,n) with h > j So L^p(1,n) takes O(n^2) time for a fixed p

Auto-completion Search Prof. Paolo Ferragina, Algoritmi per "Information Retrieval" Auto-completion Search

How it works What’s the dictionary ?

What’s the ranking/scoring of the answers ? Top-1 P = sy s 1 y 8,1 z 2 2 omo stile aibelyite zyg 7 5 5 czecin 8 2 1 4 1 etic ygy ial 6 2 4 3 What’s the ranking/scoring of the answers ?

How to compute the top-1 in O(1) time ? Top-1: How to speed-up P = sy s 8,1 1 1 y z 7 1 2 2 omo stile aibelyite zyg 2 3 5 7 5 4 5 czecin 8 2 1 4 1 etic ygy ial 6 2 4 3 How to compute the top-1 in O(1) time ?

Top-2 P = sy How to compute the top-2 in O(1) time ? Top-k in O(1) time, but k× space P = sy s 1 1,7 y z 7,6 1,4 2 2 omo stile aibelyite zyg 2 3 5 7 5 4,2 5 czecin 8 2 1 4 1 etic ygy ial 6 2 4 3 How to compute the top-2 in O(1) time ?

Top-k: How to squeeze ? P = sy 2 3 5 8 2 1 4 P = sy s 1 y z 2 2 omo stile aibelyite zyg 2 3 5 7 5 5 czecin 8 2 1 4 1 etic ygy ial 6 2 4 3 Prefixed by P, proceed D&C Score 8 1 2 1 3 4 2 5 3 6 5 7 String

Time: O(k) time, and space Top-k: How to squeeze ? Prefixed by P, proceed D&C Score 8 1 2 1 3 4 2 5 3 6 5 7 String L R RMQ-query in O(1) time and O(n) space Let H be a max-heap of size k, keep also min[H] and max[H] Initialize H with k pairs <-, NULL> Given the range <L,R> (here <1,4>) Compute max-score in Array[L,R] (pos. M, value m) If m ≤ min[H], skip; else: Insert <m,string> in H; If size(H)>k then remove min[H]; Recurse on <L,M-1> and <M+1,R>, if not empty. Time: O(k) time, and space Depth-first visit of the possibilities, it might find bad results first

H = {<8,4> e <5,7>} Example for Top-2 Consider this other array Score 4 1 2 1 3 8 4 2 5 3 6 5 7 String L R Range : operations [1,7]: H  <8,4>; recurse on [1,3] and [5,7] [1,3]: H={<8,4>}  <4,1>; recurse on [1,0] and [2,3] [5,7]: H={<8,4>,<4,1>}  <5,7>; delete <4,1> from H, recurse on [5,6] and [8,7] [2,3]: H={<8,4>,<5,7>}  <2,2>; since min[H]=5, not insert in H [5,6]: H ={<8,4>,<5,7>}  <3,6>; since min[H]=5, not insert in H H = {<8,4> e <5,7>}

Time: still O(k) time, and space A smarter approach Prefixed by P, proceed D&C Score 8 1 2 1 3 4 2 5 3 6 5 7 String L R Let H be a max-heap, including items <val, string, [low,high]> Compute max-score in Array[L,R] (pos. M, value m) i=0; insert <m, string[M], L, R> in H While (i<k) do Extract <x, string[X], Lx, Rx> from H, where x is max-value in H Return String[X] as one of the top-k strings Compute max-score in Array[Lx,X-1] (pos. M’, value m’) insert <m’, string[M’], Lx, X-1> Compute max-score in Array[X+1,Rx] (pos. M’’, value m’’) insert <m’’, string[M’’], X+1, Rx> i++; Time: still O(k) time, and space