1. Introduction to Information Retrieval Information Retrieval and Data Mining (AT71.07) Comp. Sc. and Inf. Mgmt. Asian Institute of Technology Instructor: Prof. Sumanta Guha Slide Sources: Introduction to Information Retrieval book slides from Stanford University, adapted and supplemented Chapter 6: Scoring, term weighting, and the vector space model 1

2. Introduction to Information Retrieval Introduction to Information Retrieval CS276: Information Retrieval and Web Search Christopher Manning and Prabhakar Raghavan Lecture 6: Scoring, term weighting, and the vector space model

3. Introduction to Information Retrieval This lecture; IIR Sections 6.2-6.4.3  Ranked retrieval  Scoring documents  Term frequency  Collection statistics  Weighting schemes  Vector space scoring

4. Introduction to Information Retrieval Ranked retrieval  Thus far, our queries have all been Boolean.  Documents either match or don’t.  Good for expert users with precise understanding of their needs and the collection.  Also good for applications: Applications can easily consume 1000s of results.  Not good for the majority of users.  Most users incapable of writing Boolean queries (or they are, but they think it’s too much work).  Most users don’t want to wade through 1000s of results.  This is particularly true of web search. Ch. 6

5. Introduction to Information Retrieval Problem with Boolean search: feast or famine  Boolean queries often result in either too few (=0) or too many (1000s) results.  Query 1: “standard user dlink 650” → 200,000 hits  Query 2: “standard user dlink 650 no card found”: 0 hits  It takes a lot of skill to come up with a query that produces a manageable number of hits.  AND gives too few; OR gives too many Ch. 6

6. Introduction to Information Retrieval Ranked retrieval models  Rather than a set of documents satisfying a query expression, in ranked retrieval models, the system returns an ordering over the (top) documents in the collection with respect to a query  Free text queries: Rather than a query language of operators and expressions, the user’s query is just one or more words in a human language  In principle, there are two separate choices here, but in practice, ranked retrieval models have normally been associated with free text queries and vice versa 6

7. Introduction to Information Retrieval Feast or famine: not a problem in ranked retrieval  When a system produces a ranked result set, large result sets are not an issue  Indeed, the size of the result set is not an issue  We just show the top k ( ≈ 10) results  We don’t overwhelm the user  Premise: the ranking algorithm works Ch. 6

8. Introduction to Information Retrieval Scoring as the basis of ranked retrieval  We wish to return in order the documents most likely to be useful to the searcher  How can we rank-order the documents in the collection with respect to a query?  Assign a score – say in [0, 1] – to each document  This score measures how well document and query “match”. Ch. 6

9. Introduction to Information Retrieval Parametric and zone indexes  Consider query: “find docs authored by Shakespeare in 1601 containing the phrase alas poor Yorick ”  Fields can have well-defined set of values, e.g., numeric or character strings of fixed max length. 9 date field author field Parametric search interface to enter parameter values (=field values)

10. Introduction to Information Retrieval Parametric and zone indexes  Zone are similar to fields, except contents of a zone can be arbitrary free text. E.g., title zone, abstract zone, body zone, …  Indexes for fields and zones can be  Separate (drawback larger dictionary):  Combined (drawback larger postings, but dictionary is not enlarged – preferable to get a compact dictionary, e.g., to fit in main): 10 william.author11177244255 william.title11134244255 william.body4134213255 william4.body11.author.title134.title.body

11. Introduction to Information Retrieval Weighted zone scoring  Given a Boolean query q and doc d, weighted zone* scoring assigns each pair (q, d) a score between 0 and 1 by computing a linear combination of zone scores.  Suppose docs each have l zones.  Let g 1, g 2, …, g l be the respective zone weights, s.t., ∑ i=1..l g i = 1.  Let s 1, s 2, …, s l be a Boolean score for the respective zones, being 1/0 according as q occurs/does not occur in the i th zone of doc d.  Then, weighted zone score is ∑ i=1..l g i s i  E.g., for indexes of previous slide, if zone weights of author, title and body are, resp., g 1 = 0.2, g 2 = 0.3, g 3 = 0.5, then W EIGHTED Z ONE (“william”, 11) = 1*0.2 + 1*0.3 + 0*0.5 = 0.5 11 *Here and in the following, by zone we mean zone and fields.

12. Introduction to Information Retrieval Weighted zone scoring Algorithm to compute weighted zone score from two postings lists given an AND query. Z ONE S CORE (q1, q2) // Score for query “q 1 AND q 2 ” // Boolean score 1 if both queries present in zone; 0, otherwise 1 float scores[N] = [0] 2 constant g[ℓ] 3 p 1 ← postings(q 1 ) 4 p 2 ← postings(q 2 ) 5 // scores[] is an array with a score entry for each document, initialized to zero. 6 //p 1 and p 2 are initialized to point to the beginning of their respective postings. 7 //Assume g[] is initialized to the respective zone weights. 8 while p 1 ≠ NIL AND p 2 ≠ NIL 9 do if docID(p 1 ) == docID(p 2 ) 10 then scores[docID(p 1 )] ← W EIGHTED Z ONE (p 1, p 2, g) 11 p 1 ← next(p 1 ) 12p 2 ← next(p 2 ) 13 else if docID(p 1 ) < docID(p 2 ) 14 then p 1 ← next(p 1 ) 15 else p 2 ← next(p 2 ) 16return scores 12 ∑ i=1..l g i s i, where s i is 1 if both queries present in zone i; 0, otherwise. Assume a combined dictionary.

13. Introduction to Information Retrieval Learning weights (simple machine learning) Assume only two zones title (T) and body (B) with zone weights g and 1 – g, respectively. Example DocID d Query s T s B Judgment (human expert) r (quantized judgment) φ 1 37 linux 1 1 Relevant 1 φ 2 37 penguin 0 1 Non-relevant 0 φ 3 238 system 0 1 Relevant 1 φ 4 238 penguin 0 0 Non-relevant 0 φ 5 1741 kernel 1 1 Relevant 1 φ 6 2094 driver 0 1 Relevant 1 φ 7 3191 driver 1 0 Non-relevant 0 Training examples s T s B Score 0 0 0 0 1 1 – g 1 0 g 1 1 1 Four possible combinations of s T and s B and the corresponding score(d, q) = g * s T (d, q) + (1 – g) * s B (d, q) 13

14. Introduction to Information Retrieval Learning weights (simple machine learning) Squared error of the scoring function with weight g on example φ is ε(g, φ) = ( r(d, q) – score(d, q) ) 2 Example d Query s T s B Score r ε ε assuming g = 0.4 φ 1 37 linux 1 1 1 1 0 0 φ 2 37 penguin 0 1 1 – g 0 (1 – g) 2 0.36 φ 3 238 system 0 1 1 – g 1 g 2 0.16 φ 4 238 penguin 0 0 0 0 0 0 φ 5 1741 kernel 1 1 1 1 0 0 φ 6 2094 driver 0 1 1 – g 1 g 2 0.16 φ 7 3191 driver 1 0 g 0 g 2 0.16 Tot_ ε Tot_ ε = 0.84 s T s B Score Score assuming g = 0.4 0 0 0 0 0 1 1 – g 0.6 1 0 g 0.4 1 1 1 1 14

15. Introduction to Information Retrieval Learning weights (simple machine learning)  Total error of a set of training examples Tot_ ε = ∑ j ε(g, φ j ) = ∑ j ( r(d j, q) – score(d j, q) ) 2  Goal is to choose g to minimize the total error.  Note: Our example has only two zones with weights g and 1 – g, respectively! Generally, there will be l zones with weights g 1, …, g l. Same principles! s T s B Score r No. ε 0 0 0 0 n 00n 0 0 0 0 1 n 00r 1 0 1 1 – g 0 n 01n (1 – g) 2 0 1 1 – g 1 n 01r g 2 1 0 g 0 n 10n g 2 1 0 g 1 n 10r (1 – g) 2 1 1 1 0 n 11n 1 1 1 1 1 n 11r 0 Total error Tot_ ε : (n 01r + n 10n )g 2 + (n 10r + n 01n )(1 – g) 2 + n 00r + n 11n 15

16. Introduction to Information Retrieval Learning weights (simple machine learning)  Want to minimize total error Tot_ ε = (n 01r + n 10n )g 2 + (n 10r + n 01n )(1 – g) 2 + n 00r + n 11n  Differentiating w.r.t. g: d(Tot_ ε)/dg = 2(n 01r + n 10n )g – 2(n 10r + n 01n )(1 – g)  Find minimum by solving: 2(n 01r + n 10n )g – 2(n 10r + n 01n )(1 – g) = 0 → (n 10r + n 10n + n 01r + n 01n )g = n 10r + n 01n → g = (n 10r + n 01n )/ (n 10r + n 10n + n 01r + n 01n ) 16

17. Introduction to Information Retrieval Exercises  Exercise 6.1: When using weighted scoring, is it necessary for all zones to use the same match function?  Exercise 6.2: If author, title and body zones have weights g1 = 0.2, g2 = 0.31 and g3 = 0.49, what are all the distinct score values a doc may get?  Exercise 6.3: Rewrite the algorithm in Fig. 6.4 to the case of more than two queries, viz., q 1 AND q 2 AND … AND q m.  Exercise 6.4: Write pseudocode for the function WeightedZone for the case of two postings lists in Fig. 6.4.  Exercise 6.5: Apply Eq. 6.6 to the sample training set in Fig. 6.5 to estimate the best value of g for this example.  Exercise 6.6: For the value of g estimated in Ex. 6.5, compute the weighted zone score of each (query, doc) example. How do these scores relate to the relevance judgments of Fig. 6.4 (quantized to 0/1)?  Exercise 6.7: Why does the expression for g in Eq. 6.6 not involve the training examples in which s T (d, q) and s B (d, q) have the same value? 17

18. Introduction to Information Retrieval Query-document matching scores  We need a way of assigning a score to a query/document pair  Let’s start with a one-term query  If the query term does not occur in the document: score should be 0  The more frequent the query term in the document, the higher the score (should be)  We will look at a number of alternatives for this. Ch. 6

19. Introduction to Information Retrieval Take 1: Jaccard coefficient  A commonly used measure of overlap of two sets A and B:  jaccard(A,B) = |A ∩ B| / |A ∪ B|  jaccard(A, A) = 1  jaccard(A, B) = 0 if A ∩ B = 0  jaccard(B, A) = jaccard(A, B)  A and B don’t have to be the same size.  Always assigns a number between 0 and 1.  Exercise: A = {1, 2, 3, 4}, B = {1, 2, 4}, C = {1, 2, 4, 5}. Calculate jaccard(A, B), jaccard(B, C), jaccard(A, C). Ch. 6

20. Introduction to Information Retrieval Jaccard coefficient: Scoring example  Exercise: What is the query-document match score that the Jaccard coefficient computes for each of the two documents below?  Query: ides of march  Document 1: caesar died in march  Document 2: the long march Ch. 6

21. Introduction to Information Retrieval Issues with Jaccard for scoring  It doesn’t consider term frequency (how many times a term occurs in a document)  Rare terms in a collection are more informative than frequent terms. Jaccard doesn’t consider this information  We need a more sophisticated way of normalizing for length Ch. 6

22. Introduction to Information Retrieval Recall (Lecture 1): Binary term- document incidence matrix Each document is represented by a binary vector ∈ {0,1} |V| ! Sec. 6.2

23. Introduction to Information Retrieval Term-document count matrices  Consider the number of occurrences of a term in a document*: Sec. 6.2 Each document is represented by a count vector ∈ ℕ |V| ! * Recall from Ch. 1 that term frequency can be stored with a document in the inverted index.

24. Introduction to Information Retrieval Bag of words model  Vector representation doesn’t consider the ordering of words in a document  John is quicker than Mary and Mary is quicker than John have the same vectors  This is called the bag of words model.  In a sense, this is a step back: The positional index was able to distinguish these two documents.  We will look at “recovering” positional information later in this course.  For now: bag of words model

25. Introduction to Information Retrieval Term frequency tf  The term frequency tf t,d of term t in document d is defined as the number of times that t occurs in d.  We want to use tf when computing query-document match scores. But how?  Raw term frequency is not what we want:  A document with 10 occurrences of the term is more relevant than a document with 1 occurrence of the term.  But not 10 times more relevant!  Relevance does not increase proportionally with term frequency.

26. Introduction to Information Retrieval Log-frequency weighting  The log frequency weight of term t in d is  0 → 0, 1 → 1, 2 → 1.3, 10 → 2, 1000 → 4, etc.  Score for a document-query pair: sum over terms t in both q and d:  score  The score is 0 if none of the query terms is present in the document. Sec. 6.2

27. Introduction to Information Retrieval Document frequency  Rare terms are more informative than frequent terms  Recall stop words  Consider a term in the query that is rare in the collection (e.g., capricious) vs. a term that is frequent (e.g., person)  A document containing this term “capricious” is very likely to be relevant to a query containing “capricious”  → We want a high weight for rare terms like capricious. Sec. 6.2.1

28. Introduction to Information Retrieval Document frequency, continued  Frequent terms are less informative than rare terms  Consider a query term that is frequent in the collection (e.g., high, increase, line)  A document containing such a term is more likely to be relevant than a document that doesn’t  But it’s not a sure indicator of relevance.  → For frequent terms, we want high positive weights for words like high, increase, and line  But lower weights than for rare terms.  We will use document frequency (df) to capture this. Sec. 6.2.1

29. Introduction to Information Retrieval idf weight  df t is the document frequency of t: the number of documents that contain t  df t is an inverse measure of the informativeness of t  df t  N  We define the idf (inverse document frequency) of t by  We use log (N/df t ) instead of N/df t to “dampen” the effect of idf. Will turn out the base of the log is immaterial. Sec. 6.2.1

30. Introduction to Information Retrieval idf example, suppose N = 1 million termdf t idf t calpurnia16 animal1004 sunday1,0003 fly10,0002 under100,0001 the1,000,0000 There is one idf value for each term t in a collection. Sec. 6.2.1

31. Introduction to Information Retrieval Effect of idf on ranking  Question: Does idf have an effect on ranking for one- term queries, like  iPhone  idf has no effect on ranking one term queries  idf affects the ranking of documents for queries with at least two terms  For the query capricious person, idf weighting makes occurrences of capricious count for much more in the final document ranking than occurrences of person. 31

32. Introduction to Information Retrieval Collection vs. Document frequency  The collection frequency of t is the number of occurrences of t in the collection, counting multiple occurrences.  Example:  Which word is a better search term (and should get a higher weight)? WordCollection frequencyDocument frequency insurance104403997 try104228760 Sec. 6.2.1

33. Introduction to Information Retrieval tf-idf weighting  The tf-idf weight of a term is the product of its tf weight and its idf weight.  Best known weighting scheme in information retrieval  Note: the “-” in tf-idf is a hyphen, not a minus sign!  Alternative names: tf.idf, tf x idf  Increases with the number of occurrences within a document  Increases with the rarity of the term in the collection Sec. 6.2.2

34. Introduction to Information Retrieval Final ranking of documents for a query 34 Sec. 6.2.2

35. Introduction to Information Retrieval Binary → count → weight matrix Each document is now represented by a real-valued vector of tf-idf weights ∈ R |V| Sec. 6.3

36. Introduction to Information Retrieval  Exercise 6.8: Why is the idf of a term always finite?  Exercise 6.9: What is the idf of a term that occurs in every document? Compare this with the use of stop word lists.  Exercise 6.10: Consider the table of term frequencies for 3 docs: term df t idf t Doc1 Doc2 Doc3 car 18,165 1.65 27 4 24 auto 6,723 2.08 3 33 0 insurance 19,241 1.62 0 33 29 best 25,235 1.50 14 0 17 Compute the tf-idf weights for the terms car, auto, insurance and best for each doc using the given idf values.  Exercise 6.11: Can the tf-idf weight of a term in a doc exceed 1?  Exercise 6.12: How does the base of the logarithm in affect the score calculation by ? How does it affect the relative scores of two docs on a given query?  Exercise 6.13: If the logarithm in is computed base 2, suggest a simple approximation of the idf of a term. 36

37. Introduction to Information Retrieval Documents as vectors  So we have a |V|-dimensional vector space  Terms are axes of the space  Documents are points or vectors in this space  Very high-dimensional: tens of millions of dimensions when you apply this to a web search engine  These are very sparse vectors – most entries are zero. Sec. 6.3

38. Introduction to Information Retrieval Queries as vectors  Key idea 1: Do the same for queries: represent them as vectors in the space  Key idea 2: Rank documents according to their proximity to the query in this space  proximity = similarity of vectors  proximity ≈ inverse of distance  Recall: We do this because we want to get away from the you’re-either-in-or-out Boolean model.  Instead: rank more relevant documents higher than less relevant documents Sec. 6.3

39. Introduction to Information Retrieval Formalizing vector space proximity  First cut: distance between two points  ( = distance between the end points of the two vectors)  Euclidean distance?  Euclidean distance is a bad idea... ... because Euclidean distance is large for vectors of different lengths. Sec. 6.3

40. Introduction to Information Retrieval Why distance is a bad idea The Euclidean distance between q and d 2 is large even though the distribution of terms in the query q and the distribution of terms in the document d 2 are very similar. Sec. 6.3

41. Introduction to Information Retrieval Use angle instead of distance  Thought experiment: take a document d and append it to itself. Call this document d′.  “Semantically” d and d′ have the same content  The Euclidean distance between the two documents can be quite large  The angle between the two documents is 0, corresponding to maximal similarity.  Key idea: Rank documents according to angle with query. Sec. 6.3

42. Introduction to Information Retrieval From angles to cosines  The following two notions are equivalent.  Rank documents in decreasing order of the angle between query and document  Rank documents in increasing order of cosine(query,document)  Cosine is a monotonically decreasing function for the interval [0 o, 180 o ] Sec. 6.3

43. Introduction to Information Retrieval From angles to cosines  But how – and why – should we be computing cosines? Sec. 6.3

44. Introduction to Information Retrieval Length normalization  A vector can be (length-) normalized by dividing each of its components by its length – for this we use the L 2 norm:  Dividing a vector by its L 2 norm makes it a unit (length) vector (on surface of unit hypersphere)  Effect on the two documents d and d′ (d appended to itself) from earlier slide: they have identical vectors after length-normalization.  Long and short documents now have comparable weights Sec. 6.3

45. Introduction to Information Retrieval cosine(query,document) Dot product Unit vectors q i is the tf-idf weight of term i in the query d i is the tf-idf weight of term i in the document cos(q,d) is the cosine similarity of q and d … or, equivalently, the cosine of the angle between q and d. Sec. 6.3

46. Introduction to Information Retrieval Cosine for length-normalized vectors  For length-normalized vectors, cosine similarity is simply the dot product (or scalar product): for q, d length-normalized. 46

47. Introduction to Information Retrieval Cosine similarity illustrated 47

48. Introduction to Information Retrieval Cosine similarity amongst 3 documents termSaSPaPWH affection1155820 jealous10711 gossip206 wuthering0038 How similar are the novels SaS: Sense and Sensibility PaP: Pride and Prejudice, and WH: Wuthering Heights? Term frequencies (counts) Sec. 6.3 Note: To simplify this example, we don’t do idf weighting.

49. Introduction to Information Retrieval 3 documents example contd. Log frequency weighting termSaSPaPWH affection3.062.762.30 jealous2.001.852.04 gossip1.3001.78 wuthering002.58 After length normalization termSaSPaPWH affection0.7890.8320.524 jealous0.5150.5550.465 gossip0.33500.405 wuthering000.588 cos(SaS,PaP) ≈ 0.789 × 0.832 + 0.515 × 0.555 + 0.335 × 0.0 + 0.0 × 0.0 ≈ 0.94 cos(SaS,WH) ≈ 0.79 cos(PaP,WH) ≈ 0.69 Why do we have cos(SaS,PaP) > cos(SAS,WH)? SaS and PaP were both written by Jane Austen, WH by Emily Brontë. Sec. 6.3

50. Introduction to Information Retrieval Computing cosines Suppose the terms in the query q are t1, t2, …, tn. E.g., if q = “cat dog cat rat”, then n =3 and t1 = cat, t2 = dog, t3 = rat. Suppose the document is d. Now, q = (0, …, 0, w t1,q, 0, …, 0, w t2,q, …, 0, …, 0, w tn,q, 0, …, 0) d = (x, …, x, w t1,d, x, …, x, w t2,d, …, x, …, x, w tn,d, x, …, x) where 0’s are terms not in q and x’s are corresponding values in d. Now, q d = w t1,q x w t1,d + w t2,q x w t2,d + … + w tn,q x w tn,d and cos(q, d) = w t1,q x w t1,d + w t2,q x w t2,d + … + w tn,q x w tn,d |q| x |d| where |q| = Length(q) = (w 2 t1,q + w 2 t2,q + …+ w 2 tn,q ) and |d| = Length(d) = (Σ t runs over all terms in d w 2 t,d ) 50

51. Introduction to Information Retrieval Computing cosines example E.g., if q = “cat dog cat rat”, then n =3 and t1 = cat, t2 = dog, t3 = rat. Suppose the document is d = “cat cat cat dog goat dog horse”. Then, cos(q, d) = w cat,q x w cat,d + w dog,q x w dog,d + w rat,q x w rat,d |q| x |d| where |q| = Length(q) = (w 2 cat,q + w 2 dog,q + w 2 rat,q ) and |d| = Length(d) = (w 2 cat,d + w 2 dog,d + w 2 goat,d + w 2 horse,d ) 51

52. Introduction to Information Retrieval Computing cosine scores Sec. 6.3 Can traverse posting lists one term at time – which is called term-at-a-time scoring. Or can traverse them concurrently as in the INTERSECT algorithm of Ch. 1 – which is called document-at-a-time scoring No need to store these per doc per posting list. Can be computed on-the-fly from the df t value at the head of the postings list and the tf t,d value in the doc. Priority queue=heap! No need to divide by Length[q] as it is the same for all docs! Pre-calculated off-line.

53. Introduction to Information Retrieval tf-idf weighting has many variants Why is the base of the log in idf immaterial? Sec. 6.4

54. Introduction to Information Retrieval Weighting may differ in queries vs documents  Many search engines allow for different weightings for queries vs. documents  SMART Notation: denotes the combination in use in an engine, with the notation ddd.qqq, using the acronyms from the previous table  A very standard weighting scheme is: lnc.ltc  Document: logarithmic tf (l first character), no idf (n second character), cosine normalization…  Query: logarithmic tf (l first character), idf (t second character), cosine normalization … A bad idea? Sec. 6.4

55. Introduction to Information Retrieval tf-idf example: lnc.ltc TermQueryDocumentPro d tf- raw tf-wtdfidfwtn’liz e tf-rawtf-wtwtn’liz e auto0050002.3001110.520 best11500001.3 0.3400000 car11100002.0 0.52111 0.27 insurance1110003.0 0.7821.3 0.680.53 Document: car insurance auto insurance Query: best car insurance N, the number of docs = 1,000,000 Score = 0+0+0.27+0.53 = 0.8 Doc length = Sec. 6.4

56. Introduction to Information Retrieval  Exercise 6.14: If we were to stem jealous and jealousy to a common stem before setting up the vector space, detail how the definitions of tf and idf should be modified.  Exercise 6.15: Recall the tf-idf weights computed in Exercise 6.10. Compute the Euclidean normalized document vectors for each of the docs, where each has four components, one for each of the four terms.  Exercise 6.16: Verify that the sum of the squares of the components of each of the document vectors in Exercise 6.15 is 1 (to within rounding error). Why?  Exercise 6.17: With term weights as computed in Exercise 6.15, rank the three documents by computed score for the query car insurance for each of the following cases of term weight in the query:  The weight of the term is 1 if present in the query, 0 otherwise.  Euclidean normalized idf.  Exercise 6.18: One measure of the similarity of two vectors x and y is the Euclidean (or L 2 ) distance between them: |x – y| = sqrt( ∑ i=1..m (x i – y i ) 2 ) Given a query q and docs d 1, d 2, …, we may rank the docs d i in order of increasing distance from q. Show that if q and the d i are all normalized to unit vectors, then the rank ordering produced by Euclidean distance is identical to that produced by cosine similarity.

57. Introduction to Information Retrieval  Exercise 6.19: Compare the vector space similarity between the query “digital cameras” and the document “digital cameras and video cameras” by filling out the empty columns in the table below. query document word tf wf df idf q i =wf-idf tf wf d i =normalized wf q i · d i digital 10,000 video 100,000 cameras 50,000 Assume N = 10,000,000, logarithmic term weighting (wf columns) for query and doc, idf weighting for the query only and cosine normalization for the doc only. Treat and as a stop word. Enter term counts in the tf columns. What is the final similarity score? 57

58. Introduction to Information Retrieval  Exercise 6.20: Show that for the query affection, the relative ordering of the scores of the three docs in the table below is the reverse of the ordering for the query jealous gossip. term SaS PaP WH affection 0.996 0.993 0.847 jealous 0.087 0.120 0.466 gossip 0.017 0 0.254  Exercise 6.21: In turning a query into a unit vector in the table above, we assigned equal weights to each of the query terms. What other principled approaches are plausible?  Exercise 6.22: Consider the case of a query term that is not in the set of M indexed terms.; thus, our standard construction of the query vector results in V(q) not being in the vector space created from the collection. How would one adapt the vector space construction to handle this case?  Exercise 6.23: Refer to the tf and idf values for the four terms and three docs in Exercise 6.10. Compute the two top-scoring docs on the query best car insurance for each of the following weighting schemes (i) nnn.atc (ii) ntc.atc.  Exercise 6.24: Suppose the word coyote does not occur in the collection used in Exercises 6.10 and 6.23. How would one compute ntc.atc scores for the query coyote insurance? 58

59. Introduction to Information Retrieval Summary – vector space ranking  Represent the query as a weighted tf-idf vector  Represent each document as a weighted tf-idf vector  Compute the cosine similarity score for the query vector and each document vector  Rank documents with respect to the query by score  Return the top K (e.g., K = 10) to the user

60. Introduction to Information Retrieval Resources for today’s lecture  IIR 6.2 – 6.4.3  http://www.miislita.com/information-retrieval- tutorial/cosine-similarity-tutorial.html http://www.miislita.com/information-retrieval- tutorial/cosine-similarity-tutorial.html  Term weighting and cosine similarity tutorial for SEO folk! Ch. 6