Generalized Vector Space Model

1. Generalized Vector Space Model • Definition Let ki be a vector associated with the index term ki . Independence of index terms in the vector model implies that the set of vectors {k1 ,k2 ,…,kt}is linearly independent and forms a basis for the subspace of interest. The dimension of this space is the number t of index terms in the collection.

2. An example for independent • V1=(1, 0, 0), V2=(0, 1, 0), V3=(0, 0, 1). • V1 V2=0+0+0=0. • Vi Vj=0. • Each element represents a keywords. • Different keywords are treated as totally different items. This is not reasonable since sometimes they are related.

3. Definition Given the set {k1 ,k2 ,…,kt} of index terms in a collection, as before, let wi,j be the weight associated with the term-document pair [ki ,dj]. If the wi,j weights are all binary, then all possible patterns of term co-occurrence (inside documents) can be represented by a set of 2tminterms given by min1 =(0,0,…,0), min2 =(1,0,…,0),…, min2t =(1,1,…,1). • Let gi (minj ) return the weight {0,1} of the index term ki in the minterm mini. (gi(dj) is defined similarly.)

4. Definition Let us define the following set of vectors (containing 2t elements) • m1=(0, 0, …, 1) m2=(0, 0, …, 1, 0) ….. m2t-1=(0, 0, …, 1). where each vector mi is associated with the respective minterm mini. For mi .•mj=0 for all

5. The new vector kki is defined as: 1.1 1.2

7. An example for Generalized Vector Space Model • Suppose that the system has 12 documents and 4 keywords. • D1=(2, 1, 0, 0), D2=(5, 1, 0, 0), D3=(1, 1, 1, 1), • D4=(0, 0, 2, 2), D5=(0, 1, 1, 2), D6=(0, 0, 1, 1), • D7=(0, 0, 1, 0), D8=(1, 1, 0, 0), D9=(2, 1, 1, 1), • D10=(0, 2, 2, 2). D11=(1, 0, 2, 0), D12=(0,0, 2,1). • Minterms: 6 minterms are used as independent vectors to form a base. • min1=(1, 1, 0, 0), min2=(1, 1, 1, 1), min3=(0, 0, 1, 1), min4=(0, 1, 1, 1), min5=(0, 0,1, 0), min6=(1, 0, 1, 0).

8. Generalized Vector Space Model • Independent vectors: v1= (1, 0, 0, 0, 0, 0), v2=(0, 1, 0, 0, 0, 0), v3=(0, 0, 1, 0, 0, 0), v4=(0, 0, 0, 1, 0, 0), v5=(0, 0, 0, 0, 1, 0), v6=(0, 0, 0, 0, 0, 1). • Vi represents minterm mini. • Each pair of Vi and Vj is orthogonal. (dot product=0) • The four keywords k1, k2, k3, and k4 are represent by a combination of the independent vectors.

9. Generalized Vector Space Model • The four keywords k1, k2, k3, and k4 are represent by a combination of the independent vectors. k1=(c1,1V1+c1,2V2+c1,3V3+c1,4V4+c1,5V5+c1,6V6)/C where c1,1=w1,1+w1,2+w1,8 =2+5+1 (D1, D2, and D8 has minterm min1), c1,2=w1,3+w1,9 =1+2=3(D3 and D9 has minterm min2), c1,3=w1,4+w1,6+w1,12=0+0+0=0 (D4, D6 and D12 has minterm min3.), c1,4=w1,5+w1,10=0+0. c1,5=w1,7=0. c1,6=w1,11=1. C=(c1,12+c1,22+c1,32+c1,42+c1,52+c1,62)0.5

10. Generalized Vector Space Model k2=(c2,1V1+c2,2V2+c2,3V3+c2,4V4+c2,5V5+c2,6V6)/C where c2,1=w2,1+w2,2+w2,8 =1+1++1 (D1, D2, and D8 has minterm m1), c2,2=w2,3+w2,9 =1+1=2(D3 and D9 has minterm m2), c2,3=w2,4+w2,6+w2,12=0+0+0=0 (D4, D6 and D12 has minterm m3.), c2,4=w2,5+w2,10=1+2=3. c2,5=w2,7=0. c2,6=w2,11=0. C=(c2,12+c2,22+c2,32+c2,42+c2,52+c2,62)0.5

11. Generalized Vector Space Model k3=(c3,1V1+c3,2V2+c3,3V3+c3,4V4+c3,5V5+c3,6V6)/C where c3,1=w3,1+w3,2+w3,8 =0 (D1, D2, and D8 has minterm m1), c3,2=w3,3+w3,9 =1+1=2(D3 and D9 has minterm m2), c3,3=w3,4+w3,6+w2,12=2+1+2=5 (D4, D6 and D12 has minterm m3.), c3,4=w3,5+w3,10=1+2=3. c3,5=w3,7=1. c3,6=w3,11=2. C=(c3,12+c3,22+c3,32+c3,42+c3,52+c3,62)0.5

12. Generalized Vector Space Model k4=(c4,1V1+c4,2V2+c4,3V3+c4,4V4+c4,5V5+c4,6V6)/C where c4,1=w4,1+w4,2+w4,8 =0 (D1, D2, and D8 has minterm m1), c4,2=w4,3+w4,9 =1+1=2(D3 and D9 has minterm m2), c4,3=w4,4+w4,6+w4,12=2+1+1=4 (D4, D6 and D12 has minterm m3.), c4,4=w4,5+w4,10=2+2=4. c4,5=w4,7=0. c4,6=w4,11=0. C=(c4,12+c4,22+c4,32+c4,42+c4,52+c4,62)0.5 Ki’s are converted from a vector of length 4 into a vector of length 6.

13. Extended Boolean Model: • Disadvantages of “Boolean Model” : • No term weight is used • Counterexample: query q=Kx AND Ky. Documents containing just one term, e,g, Kx is considered as irrelevant as another document containing none of these terms. • No term weight is used • The size of the output might be too large or too small

14. Extended Boolean Model: • The Extended Boolean model was introduced in 1983 by Salton, Fox, and Wu[703] • The idea is to make use of term weight as vector space model. • Strategy: Combine Boolean query with vector space model. • Why not just use Vector Space Model? • Advantages: It is easy for user to provide query.

15. Extended Boolean Model: • Each document is represented by a vector (similar to vector space model.) • Remember the formula. • Query is in terms of Boolean formula. • How to rank the documents?

16. Fig. Extended Boolean logicconsidering the space composed of two terms kx and ky only. • ky • ky • kx • kx

17. Extended Boolean Model: • For query q=Kx or Ky, (0,0) is the point we try to avoid. Thus, we can use to rank the documents • The bigger the better.

18. Extended Boolean Model: • For query q=Kx and Ky, (1,1) is the most desirable point. • We use to rank the documents. • The bigger the better.

19. Extend the idea to m terms • qor=k1 p k2 p … p Km • qand=k1 p k2 p … p km

20. Properties: • The p norm as defined above enjoys a couple of interesting properties as follows. First, when p=1 it can be verified that • Second, when p= it can be verified that • Sim(qor,dj)=max(xi) • Sim(qand,dj)=min(xi)

21. Example: • For instance, consider the query q=(k1 k2)  k3. The similarity sim(q,dj) between a document dj and this query is then computed as • Any boolean can be expressed as a numeral formula.

22. Exercise: 1. Give the numeral formula for extended Boolean model of the query q=(k1 or k2 or k3)and (not k4 or k5). (assume that there are 5 terms in total.) 2. Assume that the document is represented by the vector (0.8, 0.1, 0.0, 0.0, 1.0). What is sim(q, d) for extended Boolean model? Also try to do more exercise for other Boolean formulas.