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 ? V20000.
• Vi ? Vj0.
• 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
  2t minterms 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)
• ..
• m 2t-1(0, 0, , 1).
• where each vector mi is associated with the
  respective minterm mini .
• For mi .mj0 for all

5
The new vector kki is defined as
1.1
1.2
6
(No Transcript)
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
  product0)
• 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(c1,1V1c1,2V2c1,3V3c1,4V4c1,5V5c1,6V6)/C
• where c1,1w1,1w1,2w1,8 251 (D1, D2, and D8
  has minterm min1), c1,2w1,3w1,9 123(D3 and
  D9 has minterm min2), c1,3w1,4w1,6w1,120000
  (D4, D6 and D12 has minterm min3.),
  c1,4w1,5w1,1000. c1,5w1,70. c1,6w1,111.
• C(c1,1 2c1,2 2c1,3 2c1,4 2c1,5 2c1,6 2)0.5

10
Generalized Vector Space Model

• k2(c2,1V1c2,2V2c2,3V3c2,4V4c2,5V5c2,6V6)/C
• where c2,1w2,1w2,2w2,8 111 (D1, D2, and D8
  has minterm m1), c2,2w2,3w2,9 112(D3 and D9
  has minterm m2), c2,3w2,4w2,6w2,120000
  (D4, D6 and D12 has minterm m3.),
  c2,4w2,5w2,10123. c2,5w2,70. c2,6w2,110.
• C(c2,1 2c2,2 2c2,3 2c2,4 2c2,5 2c2,6 2)0.5

11
Generalized Vector Space Model

• k3(c3,1V1c3,2V2c3,3V3c3,4V4c3,5V5c3,6V6)/C
• where c3,1w3,1w3,2w3,8 0 (D1, D2, and D8 has
  minterm m1), c3,2w3,3w3,9 112(D3 and D9 has
  minterm m2), c3,3w3,4w3,6w2,122125 (D4, D6
  and D12 has minterm m3.), c3,4w3,5w3,10123.
  c3,5w3,71. c3,6w3,112.
• C(c3,1 2c3,2 2c3,3 2c3,4 2c3,5 2c3,6 2)0.5

12
Generalized Vector Space Model

• k4(c4,1V1c4,2V2c4,3V3c4,4V4c4,5V5c4,6V6)/C
• where c4,1w4,1w4,2w4,8 0 (D1, D2, and D8 has
  minterm m1), c4,2w4,3w4,9 112(D3 and D9 has
  minterm m2), c4,3w4,4w4,6w4,122114 (D4, D6
  and D12 has minterm m3.), c4,4w4,5w4,10224.
  c4,5w4,70. c4,6w4,110.
• C(c4,1 2c4,2 2c4,3 2c4,4 2c4,5 2c4,6 2)0.5
• Kis 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 qKx 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 Wu703
• 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 logic considering the space
composed of two terms kx and ky only.

• ky

• ky

• kx

• kx

17
Extended Boolean Model

• For query qKx 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 qKx 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

• qork1 ?p k2 ?p ?p Km
• qandk1 ?p k2 ?p ?p km

20
Properties

• The p norm as defined above enjoys a couple of
  interesting properties as follows. First, when
  p1 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.