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### 1/30

negative. So, can vector space

similarity be negative?

Project part A discussion

Relevance feedback; correlation analysis; LSI

So many ways things can go wrong…

Reasons that ideal effectiveness hard to achieve:

- Document representation loses information.
- Users’ inability to describe queries precisely.
- Similarity function used not be good enough.
- Importance/weight of a term in representing a document and query may be inaccurate
- Same term may have multiple meanings and different terms may have similar meanings.

Query expansion

Relevance feedback

LSI

Co-occurrence analysis

Improving Vector Space Ranking

- We will consider three techniques
- Relevance feedback—which tries to improve the query quality
- Correlation analysis, which looks at correlations between keywords (and thus effectively computes a thesaurus based on the word occurrence in the documents) to do query elaboration
- Principal Components Analysis (also called Latent Semantic Indexing) which subsumes correlation analysis and does dimensionality reduction.

Relevance Feedback for Vector Model

In the “ideal” case where we know the relevant

Documents a priori

Cr = Set of documents that are truly relevant to Q

N = Total number of documents

Rocchio Method

Qo is initial query. Q1 is the query after one iteration

Dr are the set of relevant docs

Dn are the set of irrelevant docs

Alpha =1; Beta=.75, Gamma=.25 typically.

Other variations possible,

but performance similar

1.0

D1

Q’

0.5

Q0

Q”

D2

0

0.5

1.0

Retrieval

Rocchio/Vector IllustrationQ0 = retrieval of information = (0.7,0.3)

D1 = information science = (0.2,0.8)

D2 = retrieval systems = (0.9,0.1)

Q’ = ½*Q0+ ½ * D1 = (0.45,0.55)

Q” = ½*Q0+ ½ * D2 = (0.80,0.20)

Example Rocchio Calculation

Relevant

docs

Non-rel doc

Original Query

Constants

Rocchio Calculation

Resulting feedback query

Rocchio Method

- Rocchio automatically
- re-weights terms
- adds in new terms (from relevant docs)
- have to be careful when using negative terms
- Rocchio is not a machine learning algorithm

- Most methods perform similarly
- results heavily dependent on test collection

- Machine learning methods are proving to work better than standard IR approaches like Rocchio

Rocchio is just one approach for relevance feedback…

- Relevance feedback—in the most general terms—involves learning.
- Given a set of known relevant and known irrelevant documents, learn relevance metric so as to predict whether a new doc is relevant or not
- Essentially a classification problem!
- Can use any classification learning technique (e.g. naïve bayes, neural nets, support vector m/c etc.)

- Viewed this way, Rocchio is just a simple classification method
- That summarizes positive examples by the positive centroid, negative examples by the negative centroid, and assumes the most compact description of the relevance metric is vector difference between the two centroids.

- Essentially a classification problem!

- Given a set of known relevant and known irrelevant documents, learn relevance metric so as to predict whether a new doc is relevant or not

Correlation/Co-occurrence analysis

Co-occurrence analysis:

- Terms that are related to terms in the original query may be added to the query.
- Two terms are related if they have high co-occurrence in documents.
Let n be the number of documents;

n1 and n2 be # documents containing terms t1 and t2,

m be the # documents having both t1 and t2

If t1 and t2 are independent

If t1 and t2 are correlated

Measure degree of correlation

>> if

Inversely correlated

Terms and Docs as mutually dependent vectors in

In addition to doc-doc similarity,

We can compute term-term distance

Document vector

Term vector

- If terms are independent, the
- T-T similarity matrix would
- be diagonal
- =If it is not diagonal, we can
- use the correlations to add
- related terms to the query
- =But can also ask the question
- “Are there independent
- dimensions which define the
- space where terms & docs are
- vectors ?”

Association Clusters

- Let Mij be the term-document matrix
- For the full corpus (Global)
- For the docs in the set of initial results (local)
- (also sometimes, stems are used instead of terms)

- Correlation matrix C = MMT (term-doc Xdoc-term = term-term)

Un-normalized Association Matrix

Normalized Association Matrix

Nth-Association Cluster for a term tu is the set of terms tv such that

Suv are the n largest values among Su1, Su2,….Suk

Example

11 4 6

4 34 11

6 11 26

Correlation

Matrix

d1d2d3d4d5d6d7

K1 2 1 0 2 1 1 0

K2 0 0 1 0 2 2 5

K3 1 0 3 0 4 0 0

Normalized

Correlation

Matrix

1.0 0.097 0.193

0.097 1.0 0.224

0.193 0.224 1.0

1thAssoc Cluster for K2is K3

Scalar clusters

Even if terms u and v have low correlations,

they may be transitively correlated

(e.g. a term w has high correlation with u and v).

Consider the normalized association matrix S

The “association vector” of term u Au is (Su1,Su2…Suk)

To measure neighborhood-induced correlation between terms:

Take the cosine-theta between the association vectors of terms

u and v

Nth-scalar Cluster for a term tu is the set of terms tv such that

Suv are the n largest values among Su1, Su2,….Suk

Example

Normalized Correlation

Matrix

AK1

USER(43): (neighborhood normatrix)

0: (COSINE-METRIC (1.0 0.09756097 0.19354838) (1.0 0.09756097 0.19354838))

0: returned 1.0

0: (COSINE-METRIC (1.0 0.09756097 0.19354838) (0.09756097 1.0 0.2244898))

0: returned 0.22647195

0: (COSINE-METRIC (1.0 0.09756097 0.19354838) (0.19354838 0.2244898 1.0))

0: returned 0.38323623

0: (COSINE-METRIC (0.09756097 1.0 0.2244898) (1.0 0.09756097 0.19354838))

0: returned 0.22647195

0: (COSINE-METRIC (0.09756097 1.0 0.2244898) (0.09756097 1.0 0.2244898))

0: returned 1.0

0: (COSINE-METRIC (0.09756097 1.0 0.2244898) (0.19354838 0.2244898 1.0))

0: returned 0.43570948

0: (COSINE-METRIC (0.19354838 0.2244898 1.0) (1.0 0.09756097 0.19354838))

0: returned 0.38323623

0: (COSINE-METRIC (0.19354838 0.2244898 1.0) (0.09756097 1.0 0.2244898))

0: returned 0.43570948

0: (COSINE-METRIC (0.19354838 0.2244898 1.0) (0.19354838 0.2244898 1.0))

0: returned 1.0

Scalar (neighborhood)

Cluster Matrix

1.0 0.226 0.383

0.226 1.0 0.435

0.383 0.435 1.0

1thScalar Cluster for K2is still K3

On the database/statistics example

database is most related

to SQL and second most

related to index

Scalar Clusters

t1= database

t2=SQL

t3=index

t4=regression

t5=likelihood

t6=linear

1.0000 0.9604 0.8240 0.0847 0.1459 0.1136

0.9604 1.0000 0.9245 0.0388 0.1063 0.0660

0.8240 0.9245 1.0000 0.0465 0.1174 0.0655

0.0847 0.0388 0.0465 1.0000 0.8972 0.8459

0.1459 0.1063 0.1174 0.8972 1.0000 0.8946

0.1136 0.0660 0.0655 0.8459 0.8946 1.0000

Notice that index became much closer to database

3679 2391 1308 238 302 273

2391 1807 953 0 123 63

1308 953 536 32 87 27

238 0 32 3277 1584 1573

302 123 87 1584 972 887

273 63 27 1573 887 1423

Association Clusters

1.0000 0.7725 0.4499 0.0354 0.0694 0.0565

0.7725 1.0000 0.6856 0 0.0463 0.0199

0.4499 0.6856 1.0000 0.0085 0.0612 0.0140

0.0354 0 0.0085 1.0000 0.5944 0.5030

0.0694 0.0463 0.0612 0.5944 1.0000 0.5882

0.0565 0.0199 0.0140 0.5030 0.5882 1.0000

Metric Clusters

- Let r(ti,tj) be the minimum distance (in terms of number of separating words) between ti and tj in any single document (infinity if they never occur together in a document)
- Define cluster matrix Suv= 1/r(ti,tj)

average..

Nth-metric Cluster for a term tu is the set of terms tv such that

Suv are the n largest values among Su1, Su2,….Suk

r(ti,tj) is also useful

For proximity queries

And phrase queries

Beyond Correlation analysis: PCA/LSI

- Suppose I start with documents described in terms of just two key words, u and v, but then
- Add a bunch of new keywords (of the form 2u-3v; 4u-v etc), and give the new doc-term matrix to you. Will you be able to tell that the documents are really 2-dimensional (in that there are only two independent keywords)?
- Suppose, in the above, I also add a bit of noise to each of the new terms (i.e. 2u-3v+noise; 4u-v+noise etc). Can you now discover that the documents are really 2-D?
- Suppose further, I remove the original keywords, u and v, from the doc-term matrix, and give you only the new linearly dependent keywords. Can you now tell that the documents are 2-dimensional?
- Notice that in this last case, the true dimensions of the data are not even present in the representation! You have to re-discover the true dimensions as linear combinations of the given dimensions.
- Which means the current terms themselves are vectors in the original space..

- Notice that in this last case, the true dimensions of the data are not even present in the representation! You have to re-discover the true dimensions as linear combinations of the given dimensions.

added

PCA/LSI continued

- The fact that keywords in the documents are not actually independent, and that they have synonymy and polysemy among them, often manifests itself as if some malicious oracle mixed up the data as above.
- Need Dimensionality Reduction Techniques
- If the keyword dependence is only linear (as above), a general polynomial complexity technique called Principal Components Analysis is able to do this dimensionality reduction
- PCA applied to documents is called Latent Semantic Indexing
- If the dependence is nonlinear, you need non-linear dimensionality reduction techniques (such as neural networks); much costlier.

Better if one axis accounts for most data variation

What should we call the red axis? Size (“factor”)

Reduce Dimensions

What if we only consider “size”

We retain 1.75/2.00 x 100 (87.5%)

of the original variation.

Thus, by discarding the yellow axis

we lose only 12.5%

of the original information.

If you can do it for fish, why not to docs?

- We have documents as vectors in the space of terms
- We want to
- “Transform” the axes so that the new axes are
- “Orthonormal” (independent axes)
- Notice that the new fish axes are uncorrelated..

- Can be ordered in terms of the amount of variation in the documents they capture
- Pick top K dimensions (axes) in this ordering; and use these new K dimensions to do the vector-space similarity ranking

- “Orthonormal” (independent axes)

- “Transform” the axes so that the new axes are
- Why?
- Can reduce noise
- Can eliminate dependent variabales
- Can capture synonymy and polysemy

- How?
- SVD (Singular Value Decomposition)

Rank and Dimensionality

What we want to do: Given M of rank R, find a matrix M’ of rank R’ < R such that ||M-M’|| is the smallest

If you do a bit of calculus of variations, you will find that the solution is related to Eigen decomposition

More specifically, Singular Value Decomposition, of a matrix

Rank of a matrix M is defined as the size of the largest square sub-matrix of M which has a non-zero determinant.

The rank of a matrix M is also equal to the number of non-zero singular values it has

Rank of M is related to the true dimensionality of M. If you add a bunch of rows to M that are linear combinations of the existing rows of M, the rank of the new matrix will still be the same as the rank of M.

Distance between two equi-sized matrices M and M’; ||M-M’|| is defined as the sum of the squares of the differences between the corresponding entries (Sum (muv-m’uv)2)

Will be equal to zero when M = M’

Bunch of Facts about SVD

- Relation between SVD and Eigen value decomposition
- Eigen value decomp is defined only for square matrices
- Only square symmetric matrices have real-valued eigen values
- PCA (principle component analysis) is normally done on correlation matriceswhich are square symmetric (think of d-d or t-t matrices).

- SVD is defined for all matrices
- Given a matrix dt, we consider the eigen decomposion of the correlation matrices d-d (dt*dt’) and tt (dt’*dt). SVD is
- (a) the eigen vectors of d-d (2) positive square roots of eigen values of dd or tt (3)eigen vectors of tt
- Both dd and tt are symmetric (they are correlation matrices)

- They both will have the same eigen values

- (a) the eigen vectors of d-d (2) positive square roots of eigen values of dd or tt (3)eigen vectors of tt
- Unless M is symmetric, MMT and MTM are different
- So, in general their eigen vectors will be different (although their eigen values are same)

- Given a matrix dt, we consider the eigen decomposion of the correlation matrices d-d (dt*dt’) and tt (dt’*dt). SVD is
- Since SVD is defined in terms of the eigen values and vectors of the “Correlation matrices” of a matrix, the eigen values will always be real valued (even if the matrix M is not symmetric).
- In general, the SVD decomposition of a matrix M equals its eigen decomposition only if M is both square and symmetric

- Eigen value decomp is defined only for square matrices

Rank and Dimensionality: 2

Suppose we did SVD on a doc-term matrix d-t, and took the top-k eigen values and reconstructed the matrix d-tk. We know

d-tk has rank k (since we zeroed out all the other eigen values when we reconstructed d-tk)

There is no k-rank matrix M such that ||d-t –M|| < ||d-t – d-tk||

In other words d-tk is the best rank-k (dimension-k) approximation to d-t!

This is the guarantee given by SVD!

Rank of a matrix M is defined as the size of the largest square sub-matrix of M which has a non-zero determinant.

The rank of a matrix M is also equal to the number of non-zero singular values it has

Rank of M is related to the true dimensionality of M. If you add a bunch of rows to M that are linear combinations of the existing rows of M, the rank of the new matrix will still be the same as the rank of M.

Distance between two equi-sized matrices M and M’; ||M-M’|| is defined as the sum of the squares of the differences between the corresponding entries (Sum (muv-m’uv)2)

Will be equal to zero when M = M’

Note that because the LSI dimensions are uncorrelated,

finding the best k LSI dimensions is the same as sorting the dimensions in terms of their individual varianc (i.e., corresponding singualr values), and picking top-k

Overview of Latent Semantic Indexing

Eigen Slide

factor-factor

(+ve sqrt of eigen values of

d-t*d-t’or d-t’*d-t; both same)

Doc-factor

(eigen vectors of d-t*d-t’)

(term-factor)T

(eigen vectors of d-t’*d-t)

Term

Term

dt

df

dfk

dtk

tft

doc

ff

tfkt

ffk

Þ

doc

fxt

dxt

dxf

fxf

dxk

kxk

kxt

dxt

Reduce Dimensionality:

Throw out low-order

rows and columns

Recreate Matrix:

Multiply to produce

approximate term-

document matrix.

dtk is a k-rank matrix

That is closest to dt

Singular Value Decomposition

Convert doc-term

matrix into 3matrices

D-F, F-F, T-F

Where DF*FF*TF’ gives the

Original matrix back

d-f*f-f

t1= database

t2=SQL

t3=index

t4=regression

t5=likelihood

t6=linear

F-F

D-F

6 singular values

(positive sqrt of

eigen values of

dd or tt)

T-F

Eigen vectors of dd (dt*dt’)

(Principal document

directions)

Eigen vectors of tt (dt’*dt)

(Principal term directions)

t2=SQL

t3=index

t4=regression

t5=likelihood

t6=linear

For the database/regression

example

Suppose D1 is a new

Doc containing “database”

50 times and D2 contains

“SQL” 50 times

26 18 10 0 1 1

25 17 9 1 2 1

15 11 6 -1 0 0

7 5 3 0 0 0

44 31 17 0 2 1

2 0 0 19 10 11

1 -1 0 24 12 14

2 0 0 16 8 9

2 -1 0 39 20 22

4 2 1 20 11 12

Reconstruction (rounded)

With 2 LSI dimensions

Rank=2

Variance loss: 7.5%

Rank=6

24 20 10 -1 2 3

32 10 7 1 0 1

12 15 7 -1 1 0

6 6 3 0 1 0

43 32 17 0 3 0

2 0 0 17 10 15

0 0 1 32 13 0

3 -1 0 20 8 1

0 1 0 37 21 26

7 -1 -1 15 10 22

Rank=4

Variance loss: 1.4%

Reconstruction (rounded)

With 4 LSI dimensions

LSI Ranking… Do the vector-space similarity in the LSI space

- Given a query
- Either add query also as a document in the D-T matrix and do the svd OR
- Convert query vector (separately) to the LSI space
- DFq*FF=q*TF

- this is the weighted query document in LSI space

- Reduce dimensionality as needed

Using LSI

Can be used on the entire corpus

First compute the SVD of the entire corpus

Store first k columns of the df*ff matrix [df*ff]k

Keep the tf matrix handy

When a new query q comes, take the k columns of q*tf

Compute the vector similarity between [q*tf]k and all rows of [df*ff]k, rank the documents and return

Can be used as a way of clustering the results returned by normal vector space ranking

Take some top 50 or 100 of the documents returned by some ranking (e.g. vector ranking)

Do LSI on these documents

Take the first k columns of the resulting [df*ff] matrix

Each row in this matrix is the representation of the original documents in the reduced space.

Cluster the documents in this reduced space (We will talk about clustering later)

MANJARA did this

We will need fast SVD computation algorithms for this. MANJARA folks developed approximate algorithms for SVD

Added based on class discussion

SVD Computation complexity

- For an mxn matrix SVD computation is
- O( km2n+k’n3) complexity
- k=4 and k’=22 for best algorithms

- Approximate algorithms that exploit the sparsity of M are available (and being developed)

- O( km2n+k’n3) complexity

Summary:What LSI can do

- LSI analysis effectively does
- Dimensionality reduction
- Noise reduction
- Exploitation of redundant data
- Correlation analysis and Query expansion (with related words)

- Any one of the individual effects can be achieved with simpler techniques (see scalar clustering etc). But LSI does all of them together.

LSI (dimensionality reduction) vs. Feature Selection

- Before reducing dimensions, LSI first finds a new basis (coordinate axes) and then selects a subset of them
- Good because the original axes may be too correlated to find top-k subspaces containing most variance
- Bad because the new dimensions may not have any significance to the user
- What are the two dimensions of the database example?
- Something like 0.44*database+0.33*sql..

- What are the two dimensions of the database example?

- An alternative is to select a subset of the original features themselves
- Advantage is that the selected features are readily understandable by the users (to the extent they understood the original features).
- Disadvantage is that as we saw in the Fish example, all the original dimensions may have about the same variance, while a (linear) combination of them might capture much more variation.
- Another disadvantage is that since original features, unlike LSI features, may be correlated, finding the best subset of k features is not the same as sorting individual features in terms of the variance they capture and taking the top-K (as we could do with LSI)

LSI as a special case of LDA

- Dimensionality reduction (or feature selection) is typically done in the context of specific classification tasks
- We want to pick dimensions (or features) that maximally differentiate across classes, while having minimal variance within any given class

- When doing dimensionality reduction w.r.t a classification task, we need to focus on dimensions that
- Increase variance across classes
- and reduce variance within each class

- Doing this is called LDA (linear discriminant analysis)
- LSI—as given—is insensitive to any particular classification task and only focuses on data variance
- LSI is a special case of LDA where each point defines its own class
- Interestingly, LDA is also related to eigen values.

In the example above, the

Red line corresponds to the

Dimension with most data variance

However, the green line corresponds

To the axis that does a better job of

Capturing the class variance (assuming

That the two different blobs correspond

To the different classes)

LSI vs. Nonlinear dimensionality reduction

- LSI only captures linear correlations
- It cannot capture non-linear dependencies between original dimensions
- E.g. if the data points are all falling on a simple manifold (e.g. a circle in the example below),
- Then, the features are “non-linearly” correlated (here X2+Y2=c)
- LSI analysis can’t reduce dimensionality here

- One idea is to use techniques such as neural nets or manifold learning techniques
- Another—simpler—idea is to consider first blowing up the dimensionality of the data by introducing new axes that are nonlinear combinations of existing ones (e.g. X2, Y2, sqrt(xy) etc.)
- We can now capture linear correlations across these nonlinear dimensions by doing LSI in this enlarged space, and map the k important dimensions found back to original space.
- So, in order to reduce dimensions, we first increase them (talk about crazy!)

- A way of doing this implicitly is kernel trick..

Advanced; Optional

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