1.   a. = 12 + 22 + 32 + 42 + 52 + 62 + 72 + 82 + 92 + 102 + 112  = 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121 = 506
b. = -1 + -2 + -3 + -4 + -5 + -6 + -7 + -8 = -36
c. = 0*3 + 1*4 + 2*5 + 3*6 = 0 + 4 + 10 + 18 = 32

2. a. fk >= nk always.  In order for a document to be included in nk it must have at least one instance Tk.  However, it might also contain 2 instances of Tk.  If so, the document would be counted once in nk, but twice in fk.   Therefore, for n times the term occurs in the document, nk would be incremented once, while fk would be incremented n times the term occurs.  So nk can be the same as fk if all documents contain 1 instance of a term, but nk will not be larger than fk.  
	
  b.   = 12 * log(100/5) = 15.612
c.	= 1 * log(100/5) = .301

3. a.
Document	“information” weight	“retrieval” weight	
D1	20 * log (1000/120) =  18.416	3 * log (1000/70) = 3.465	
D2	1 * log (1000/120) = .9208	15 * log (1000/70) = 17.324	
D3	12 * log (1000/120) = 11.050	10 * log (1000/70) = 11.550	

3. b.   log(1000/120) = .9208           log(1000/70) = 1.1550
          (log(1000/70))2  = 1.335           (log(1000/120))2 = .8479

 wq info =                              1 * log(1000/120)
		((12 * (log (1000/120)) 2)  + (12 * (log (1000/70)) 2))-2	

	       =               1 * .9208
                         (1 * .8479)) –2+ ((1 * 1.335)  = .6234
	
wq ret =                        1 * log(1000/70)
	      ((12 * (log (1000/120)) 2)  + (12 * (log (1000/70)) 2))-2	

              = 	               1 * 1.1550
                         (1 * .8479)) –2 + ((1 * 1.335)      = .7815 

wD1info =                   20 * log(1000/120)
		((202 * (log (1000/120)) 2) + (32 * (log (1000/70)) 2))-2
	      
         =                 20* .9208
                         ((202 * .8479) + (32  * 1.335)) –2   = .9827
		
wD1ret =                                  3 * log(1000/70)
		((202 * (log (1000/120)) 2) + (32 * (log (1000/70)) 2))-2	        

        =                    3 * 1.1550
                         ((202 * .8479) + (32  * 1.335)) –2  =.1849

wD2 info =                     1* log(1000/120)
		((12 * (log (1000/120)) 2) + (152 * (log (1000/70)) 2))-2

        =                         1* .9208
                         ((12 * .8479) + (152  * 1.335)) –2 = .0500

wD2 ret =                            15 * log(1000/70)
		((12 * (log (1000/120)) 2) + (152 * (log (1000/70)) 2))-2

     =                 15 * 1.1550
                     ((12 * .8479) + (152  * 1.335)) –2  = .9379

wD3 info = 12* log(1000/120)
		((122 * (log (1000/120)) 2) + (102 * (log (1000/70)) 2))-2

        =   12* .9208
                         ((122 * .8479) + (102  * 1.335)) –2  = .6911

wD3 ret = 10 * log(1000/70)
		((122 * (log (1000/120)) 2) + (102 * (log (1000/70)) 2))-2	     

     =   10 * 1.1550
                     ((122 * .8479) + (102  * 1.335)) –2   = .7224

          Sim(Q, D1) = (.6234 * .9827) + (.7815  * .1849) = .7571
          Sim(Q, D2) = (.6234 * .0500) + (.7815  * .9379) = .7641
          Sim(Q, D3) = (.6234 * .6911) + (.7815  * .7224) = .9954

Document 3 is closest and most similar to the query.  Document 3 has a ratio of information frequency: retrieval frequency very close to 1:1.  The query’s ratio for these terms is 1:1.  Therefore, in this document, the weight of the terms is the closest to the weight of the terms in the query.  The other two documents have differing ratios of frequency of terms.  The amount that this ratio differs from the 1:1 ratio of the query is reflected in these results.

c.

a.	Yes, I expected results as shown on this graph.  Document 3 is the closest to the query, while the other two documents show their differential by lying farther away from the query.

b.	This graph is the same as the graph in part a.  The relationship of the normalized weights to the raw frequencies is about the same.  This works because by normalizing the weights we consider the subset of documents that contain that term.  This number is relatively small in number of documents; therefore the frequency of one document is fairly significant.  The normalization does not do much, except to bring the scale of the graph down.  Normalizing compares any results on the same scale.  So if the graph had been very different and either term had occurred more times in the collection, it would have allowed us a way to view the relationship of weights (because the occurrences of each term in the collection were small).

d.    If tf for the term frequencies had been used, then we would not weight the terms.  Both information and retrieval would have been given the same weight.  Therefore, Document 1 would contain 23 “hits”, Document 2 “16” hits, and Document 3 - 22 “hits”, since the number of hits would now equal the frequencies of both terms.  Document 1 would show the most similar results to the query since it contained the most number of terms hits.  The idf takes into account the fact that information is a more common word and occurs in120 documents, compared to retrieval, which occurs in 70 documents.  Retrieval is given a greater weight than information, because it’s not so frequent.  Taking into account these weights, the relative frequencies of terms to the number of documents they occur in, the document is weighted accordingly.  Otherwise, the document would only be a summation of the raw frequencies.  This doesn’t reflect the query, since one word could just occur very often in the collection and thereby throwing off the results.                                                                                                                                                                                                  
          
        If information had occurred in 500 documents, information would have had even less value.   So the number of times a “retrieval” occurs would be given a more relevant value.  Document 2 would be more similar to the query and Document 1 and Document 3 would be less similar.  Documents 1 and 3 would loose some of their weight since much of their weight depends on information and Document 2 would be more weighted, since little depends on information and much depends on retrieval, which would hold more value.