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Problem solving on Boolean Model and Vector Space Model

  • Difficulty Level : Expert
  • Last Updated : 30 May, 2021

Boolean Model: 

It is a simple retrieval model based on set theory and boolean algebra. Queries are designed as boolean expressions which have precise semantics. Retrieval strategy is based on binary decision criterion. Boolean model considers that index terms are present or absent in a document.

Problem Solving: 

Consider 5 documents with a vocabulary of 6 terms

  • document 1 = ‘ term1 term3 ‘
  • document 2 = ‘ term 2 term4 term6 ‘
  • document 3 = ‘ term1 term2 term3 term4 term5 ‘
  • document 4 = ‘ term1 term3 term6 ‘
  • document 5 = ‘ term3 term4 ‘

Our documents in boolean model



  term 1 term 2 term 3 term 4 term 5 term 6
document 1  1 0 1 0 0 0
document 2 0 1 0 1 0 1
document 3 1 1 1 1 1 0
document 4 1 0 1 0 0 1
document 5 0 0 1 1 0 0

Consider the query

Find the document consisting of term1 and term3 and not term2

term1 ∧ term3 ∧ ¬ term2

  term1  ¬term 2 term 3 term 4 term 5 term 6
document 1 1 1 1 0 0 0
document 2 0 0 0 1 0 1
document 3  1 0 1 1 1 0
document 4 1 1 1 0 0 1
document  5 0 1 1 1 0 0

  • document 1 : 1 ∧ 1∧ 1 = 1
  • document 2 : 0 ∧ 0 ∧ 0 = 0
  • document 3 : 1 ∧ 1 ∧ 0 = 0
  • document 4 : 1 ∧ 1 ∧ 1 = 1
  • document 5 : 0 ∧ 1 ∧ 1 = 0

Based on the above computation document1 and document4 are relevant to the given query

Vector Model:

The method of performing the operations and the formulas required for the computation is present in the previous document that is part 1. Consider the following collection of documents.

  • document1 = ‘one two ‘
  • document2 = ‘three two four ‘
  • document3 =’one two three ‘
  • document4 =’one two ‘

The formulas used



tf_i,_j = \frac {freq_i,_j}{max_l(freq_l,_j)}

idf_i = log\frac{N}{n_i}

w_i,_j = tf_i * log\frac{N}{n_i}

sim(dj,q) = \frac{\sum_{i=1}^t w_i,_j * w_i,_q}{\sqrt{\sum_{i=1}^t w^2_i,_j} * \sqrt{\sum_{i=1}^t w^2_i,_q}}

Some terms appear thrice, twice and sometimes only once in the document.The total number of documents N=4. Therefore, the IDF values of the terms are:

one --> log2(4/3) = 0.4147
two --> log2(4/4) = 0
three --> log2(4/2) = 1
four -->log2(4/1) = 2

Representation in boolean model

  one two three four
document1 1 1 0 0
document2 0 1 1 1
document3 1 1 1 0
document4 1 1 0 0



Calculation of term frequency

one --> 3/4 = 0.75
two --> 4/4 = 1
three --> 2/4 = 0.5
four --> 1/4 = 0.25

Calculation of weights ( tf * idf )

weight(one) --> 0.75 * 0.4147 = 0.3110
weight(two) --> 1 * 0 = 0
weight(three) --> 0.5 * 1 = 0.5
weight(four) --> 0.25 * 2 = 0.5

Representation of vector model in terms of weights

  one two three four
document1  0.3110 0 0 0
document2  0 0 0.5 0.5
document3  0.3110 0 0.5 0
document4  0.3110 0 0 0

QUERY: Document containing ‘ one three three ‘

Calculation of weights for query terms(term frequency)

  • weight(one) –> 1/3 = 0.333
  • weight(three) –> 2/3 = 0.667

Vector representation

  • Document    \vec{d}_j = \{0.3110, 0, 0.5, 0.5 \}
  • Query \vec{q} = \{0.333, 0, 0.667, 0 \}

Similarity calculation: the 



sim(d1,q) = \frac{0.3110 * 0.333 + 0 * 0 + 0 * 0.667 + 0 * 0}{\sqrt{ (0.3110^2 + 0^2 + 0^2 + 0^2) } *\sqrt {(0.333^2+ 0^2 + 0.667^2 + 0^2)}} = 0.4466\\ sim(d2,q) = \frac{0 * 0.333 + 0 * 0 + 0.5 * 0.667 + 0.5 * 0}{\sqrt{ (0^2 + 0^2 + 0.5^2 + 0.5^2) } *\sqrt {(0.333^2 + 0^2 + 0.667^2 + 0^2)} }= 0.4001 \\ sim(d3,q) = \frac{0.3110 * 0.333 + 0 * 0 + 0.5 * 0.667 + 0 * 0}{\sqrt{ (0.3110^2 + 0^2 + 0.5^2 + 0^2)} * \sqrt{(0.333^2 + 0^2 + 0.667^2 + 0^2)}} = 0.9086\\ sim(d4,q) = \frac{0.3110 * 0.333 + 0 * 0 + 0 * 0.667 + 0 * 0}{\sqrt {(0.3110^2 + 0^2 + 0^2 + 0^2)} * \sqrt{(0.333^2 + 0^2 + 0.667^2 + 0^2)}} = 0.4466\\

Ranking of the documents, ( for ranking we have followed the method in statistics for the case of allocating same rank to two different items) 

document1 2nd
document2 4th
document3 1st
document4 2nd

Since the similarity between document 3 is greater than the similarities between the other documents, 3rd document is more relevant to the query.

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