Finding Similar Items Set Similarity Problem Find similar

Finding Similar Items

Set Similarity • Problem: Find similar sets. • Motivation: Many things can be modeled/represented as sets • Applications: – – Face Recognition Recommender Systems (Collaborative Filtering) Document Similarity/Clustering Similar (but infrequent) Items

Application: Face Recognition • We have a database of (say) 1 million face images. • We want to find the most similar images in the database. • Represent faces by (relatively) invariant values, e. g. , ratio of nose width to eye width. • Many-one problem : given a new face, see if it is close to any of the 1 million old faces. • Many-Many problem : which pairs of the 1 million faces are similar.

Simple Solution • Represent each face by a vector of 1000 values and score the comparisons. • Sort-of OK for many-one problem. • Out of the question for the many-many problem (106*1000/2 numerical comparisons). • We can do better!

Applications - Collaborative Filtering (1) • Represent a customer, e. g. , of Netflix, by the set of movies they rented. – Similar customers have a relatively large fraction of their choices in common. • Motivation: A customer can be pitched an item that a similar customer bought, but which they did not buy.

Applications - Collaborative Filtering (2) • Products are similar if they are bought by many of the same customers. – E. g. , movies of the same genre are typically rented by similar sets of Netflix customers. • Motivation: A customer can be pitched an item that is a similar to an item that he/she already bought. • …isn’t this mining frequent pairs… • Well, not exactly…

Applications: Similar Items Problem • Search for pairs of items that appear together a large fraction of the times that either appears, – even if neither item appears in very many baskets. – Such items are considered "similar" – E. g. caviar and crackers Modeling • Each item is a set: the set of baskets in which it appears. – Thus, the problem becomes: Find similar sets!

Applications: Similar Documents (1) • Given a body of documents, e. g. , Web pages, find pairs of docs that have a lot of text in common, e. g. : – Mirror sites, or approximate mirrors. – Plagiarism, including large quotations. – Repetitions of news articles at news sites. • How do you represent a document so it is easy to compare with others? – Special cases are easy, e. g. , identical documents, or one document contained verbatim in another. – General case, where many small pieces of one doc appear out of order in another, is hard.

Applications: Similar Documents (2) • Represent doc by its set of shingles (or k -grams). • A k-shingle (or k-gram) for a document is a sequence of k characters that appears in the document. Example. • k=2; doc = abcab. • Set of 2 -shingles = {ab, bc, ca}. • At that point, doc problem becomes finding similar sets.

The Jaccard Measure of Similarity • The similarity of sets S and T is the ratio of the sizes of the intersection and union of S and T. – Sim (C 1, C 2) = |S T|/|S T| = Jaccard similarity. • Disjoint sets have a similarity of 0, and the similarity of a set with itself is 1. • Another example: similarity of sets {1, 2, 3} and {1, 3, 4, 5} is – 2/5.

Solutions 1. Divide-Compute-Merge (DCM) uses external sorting, merging. 2. Locality-Sensitive Hashing (LSH) can be carried out in main memory, but admits some false negatives.

Divide-Compute-Merge • Originally designed for “shingles” and docs. – Applicable for other set similarity problems. • Based on the merge-sort paradigm.

DCM Steps doc 1: s 11, s 12, …, s 1 k doc 2: s 21, s 22, …, s 2 k … Invert s 11, doc 1 s 12, doc 1 … s 1 k, doc 1 s 21, doc 2 … sort on shingle. Id t 1, doc 11 t 1, doc 12 … t 2, doc 21 t 2, doc 22 … Invert and pair doc 11, doc 12, 1 doc 11, doc 12, 2 doc 11, doc 12, 1 doc 11, doc 13, 10 … … Merge doc 11, doc 13, 1 … sort on <doc. Id 1, doc. Id 2> doc 11, doc 12, 1 doc 11, doc 13, 1 … doc 21, doc 22, 1 …

DCM Summary 1. 2. 3. 4. 5. 6. Start with the pairs <shingle. Id, doc. Id>. Sort by shingle. Id. In a sequential scan, generate triplets <doc. Id 1, doc. Id 2, 1> for pairs of docs that share a shingle. Sort on <doc. Id 1, doc. Id 2>. Merge triplets with common doc. Ids to generate triplets of the form <doc. Id 1, doc. Id 2, count>. Output document pairs with count > threshold.

Some Optimizations • “Invert and Pair” is the most expensive step. • Speed it up by eliminating very common shingles. – “the”, “ 404 not found”, “<A HREF”, etc. • Also, eliminate exact-duplicate docs first.

Locality-Sensitive Hashing • First represent sets as bit vectors, and organize them as columns of a signature matrix M. • Big idea: hash columns of matrix M several times. • Arrange that (only) similar columns are likely to hash to the same bucket. • Candidate pairs are those that hash at least once to the same bucket.

LSH Roadmap Similar customers Similar products Documents Sets Technique: Shingling Technique: Minhashing Signatures Technique: Locality-Sensitive Hashing Buckets containing mostly similar items (sets)

Minhashing • Suppose that the elements of each set are chosen from a "universal" set of n elements e 0, el, . . . , en-1. • Pick a random permutation of the n elements. • Minhash value of a set S is the first element, in the permuted order, that is a member of S. Example • Suppose the universal set is {1, 2, 3, 4, 5} and the permutation we choose is (3, 5, 4, 2, 1). – Set {2, 3, 5} hashes to 3. – Set {1, 2, 5} hashes to 5. – Set {1, 2} hashes to 2.

Minhash signatures • Compute signatures for the sets by using m permutations. – Typically, m would be about 100. • Signature of a set S is the list of the minhash values of S, for each of the m permutations, in order. Example • Universal set is {1, 2, 3, 4, 5}, m = 3, and the permutations are: – 1= (1, 2, 3, 4, 5), – 2= (5, 4, 3, 2, 1), – 3= (3, 5, 1, 4, 2). • Signature of S = {2, 3, 4} is – (2, 4, 3).

Minhashing and Jaccard Distance Surprising relationship If we choose a permutation at random, the probability that it will produce the same minhash values for two sets is the same as the Jaccard similarity of those sets. Why? (see the next slides) • Thus, if we have the signatures of two sets S and T, we can estimate the Jaccard similarity of S and T by the fraction of corresponding minhash values for the two sets that agree.

Four Types of Rows • Given columns C 1 and C 2, each representing a set, rows may be classified as: a b c d C 1 1 1 0 0 C 2 1 0 • Also, a = # rows of type a , etc. • Note Sim (C 1, C 2) = a /(a +b +c ).

Minhashing–put differently • Imagine the rows permuted randomly. • Define “hash” function h (C ) = the number of the first (in the permuted order) row in which column C has 1. • Use several independent hash functions to create a signature.

Minhashing and Jaccard Distance (2) • The probability (over all permutations of the rows) that h (C 1) = h (C 2) is the same as Sim (C 1, C 2). • Both are a /(a +b +c )! • Why? – Look down columns C 1 and C 2 until we see a 1. – If it’s a type-a row, then h (C 1) = h (C 2). If a type-b or type -c row, then not. – Type-d row doesn’t play any role for either of them.

Implementing Minhashing • Generating a permutation of all the universe of objects is infeasible. • Rather, simulate the choice of a random permutation by picking a random hash function h. – Pretend that the permutation that h represents places element e in position h(e). • Several elements might wind up in the same position. • As long as number of buckets is large, we can break ties as we like, and the simulated permutations will be sufficiently random that the relationship between signatures and similarity still holds.

Algorithm for minhashing • To compute the minhash value for a set S = {a 1, a 2, . . . , an} using a hash function h, we can execute: V =infinity; FOR i : = 1 TO n DO IF h(ai) < V THEN V = h(ai); • As a result, V will be set to the hash value of the element of S that has the smallest hash value.

Algorithm for set signature • If we have m hash functions h 1, h 2, . . . , hm, then we can compute m minhash values in parallel, as we process each member of S. FOR j : = 1 TO m DO Vj : = infinity; FOR i : = 1 TO n DO FOR j : = 1 TO m DO IF hj(ai) < Vj THEN Vj = hj(ai)

Example S = {1, 3, 4} T = {2, 3, 5} h(x) = x mod 5 g(x) = 2 x+1 mod 5 h(1) = 1 h(3) = 3 h(4) = 4 g(1) = 3 g(3) = 2 g(4) = 4 h(2) = 2 h(3) = 3 h(5) = 0 g(2) = 0 g(3) = 2 g(5) = 1 sig(S) = 1, 3 sig(T) = 5, 2

Locality-Sensitive Hashing of Signatures • Organize signatures of the various sets as a matrix M, with a column for each set's signature and a row for each hash function. • Big idea: hash columns of signature matrix M several times. • Arrange that (only) similar columns are likely to hash to the same bucket. • Candidate pairs are those that hash at least once to the same bucket.

Partition Into Bands r rows per band b bands Matrix M

Partition Into Bands • For each band, hash its portion of each column to a hash table with k buckets. • Candidate column pairs are those that hash to the same bucket for at least one band. Buckets Matrix M r rows b bands

Analysis • Probability that the signatures agree on one row is s (Jaccard similarity) • Probability that they agree on all r rows of a given band is s^r. • Probability that they do not agree on all the rows of a band is 1 - s^r • Probability that for none of the b bands do they agree in all rows of that band is (1 - s^r)^b • Probability that the signatures will agree in all rows of at least one band is 1 - (1 - s^r)^b • This function is the probability that the signatures will be compared for similarity.

Example • • Suppose 100, 000 columns. Signatures of 100 integers. Therefore, signatures take 40 Mb. But 5, 000, 000 pairs of signatures take a while to compare. • Choose 20 bands of 5 integers/band.

Suppose C 1, C 2 are 80% Similar • Probability C 1, C 2 agree on one particular band: – (0. 8)5 = 0. 328. • Probability C 1, C 2 do not agree on any of the 20 bands: – (1 -0. 328)20 =. 00035. – i. e. , we miss about 1/3000 th of the 80%-similar column pairs. • The chance that we do find this pair of signatures together in at least one bucket is 1 - 0. 00035, or 0. 99965.

Suppose C 1, C 2 Only 40% Similar • Probability C 1, C 2 agree on one particular band: – (0. 4)5 = 0. 01. • Probability C 1, C 2 do not agree on any of the 20 bands: – (1 -0. 01)^20 . 80 – i. e. , we miss a lot. . . • The chance that we do find this pair of signatures together in at least one bucket is 1 - 0. 80, or 0. 20 (i. e. only 20%).

Analysis of LSH – What We Want Probability = 1 if s > t Probability of sharing a bucket No chance if s < t t Similarity s of two columns

What One Row Gives You Remember: probability of equal hash-values = similarity Probability of sharing a bucket t Similarity s of two columns

What b Bands of r Rows Gives You At least one band identical t ~ (1/b)1/r Probability of sharing a bucket t Similarity s of two columns No bands identical 1 - (1 - s r )b Some row All rows of a band unequal are equal

LSH Summary • Tune to get almost all pairs with similar signatures, but eliminate most pairs that do not have similar signatures. • Check in main memory that candidate pairs really do have similar signatures. • Optional: In another pass through data, check that the remaining candidate pairs really are similar columns.