Solutions week 9
===============

Word Sense Disambiguation
-------------------------

1. Context window:

  - single most important chip in the computer
  - new Pentium III chip falls into the
  - will connect the chip to the computer

  - the ultimate chocolate chip cookie perfectly baked
  - which were potato chip cookies with chocolate

2. Vectors:

  - (1, 1, 0, 0, 0)
  - (0, 1, 0, 0, 0)
  - (1, 0, 0, 0, 0)

  - (0, 0, 1, 0, 1)
  - (0, 0, 1, 1, 1) 

3. P(s1) = 3/5, P(s2) = 2/5, P(v[0]=1|s1) = 2/3, P(v[3] = 1|s2) = 1

4. sense = argmax P(s) P(v|s)
   
   New data: (0,1,0,0,1)

  - s1: 3/5 * 1/3 * 2/3 * 1 * 1   * 0 = 0
  - s2: 2/5 * 1   * 0   * 0 * 1/2 * 1 = 0

  Need some smoothing (avoid zeroes). This way the first sense would
  be selected. 


Information Retrieval
---------------------

1. machine = {D1,D2,D3}
   technology = {D1}
   program = {D2}
   human = {D3}
   language = {D1,D2,D3}

   (machine OR human) AND NOT program = {{D1,D2,D3} JOIN {D3}} - {D2}
                                      =
                                      = {D1,D2,D3} - {D2} = {D1,D3}

2. D1 = (1log(3/3), 1log(3/1), 0        , 0        , 1log(3/3))
   D2 = (1log(3/3), 0        , 1log(3/1), 0        , 1log(3/3))
   D3 = (1log(3/3), 0,       , 0,       , 1log(3/1), 1log(3/3))

3. Since the tf.idf of "machine" and "language" are 0 no documents are
retrieved. To solve this, consider smoothing all 0 values to some
small values. For example, IDF can change from log(N/n) to
log((N+1)/n). Then the query is:

   Q = (1log(4/3), 0, 0, 0, 1log(4/3)

And the new D1 to D3:

   D1 = (1log(4/3), 1log(4/1), 0        , 0        , 1log(4/3))
   D2 = (1log(4/3), 0        , 1log(4/1), 0        , 1log(4/3))
   D3 = (1log(4/3), 0,       , 0,       , 1log(4/1), 1log(4/3))

The scores are computed after applying the formula.