# Problem solving on Boolean Model and Vector Space Model

**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

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 --> log_{2}(4/3) = 0.4147 two --> log_{2}(4/4) = 0 three --> log_{2}(4/2) = 1 four -->log_{2}(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
- Query

Similarity calculation: the

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.**