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IT444: Web Intelligence

IT444: Web Intelligence. Revision A priori and HITS algorithm. Association Rules. Apriori Algorithm. Pass 1 Generate the candidate itemsets in C 1 Save the frequent itemsets in L 1 Pass k Generate the candidate itemsets in C k from the frequent itemsets in L k -1

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IT444: Web Intelligence

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  1. IT444: Web Intelligence Revision Apriori and HITS algorithm

  2. Association Rules Apriori Algorithm

  3. Pass 1 • Generate the candidate itemsets in C1 • Save the frequent itemsets in L1 Pass k • Generate the candidate itemsets in Ck from the frequent itemsets in Lk-1 • Join Lk-1p with Lk-1q, as follows: insert intoCkselectp.item1, p.item2, . . . , p.itemk-1, q.itemk-1fromLk-1p, Lk-1q wherep.item1 = q.item1, . . . p.itemk-2 = q.itemk-2, p.itemk-1 < q.itemk-1 • Generate all (k-1)-subsets from the candidate itemsets in Ck • Prune all candidate itemsets from Ck where some (k-1)-subset of the candidate itemset is not in the frequent itemsetLk-1 • Scan the transaction database to determine the support for each candidate itemset in Ck • Save the frequent itemsets in Lk

  4. Example • Assume the user-specified minimum support is 40%, then generate all frequent itemsets. • Given: The transaction database shown below:

  5. Pass-1 C1 L1

  6. Pass-2 C2 C2 Before computing support, check for pruning. Nothing pruned since all subsets of these itemsets are frequent

  7. C2 L2 After saving only the frequent itemsets

  8. Pass-3 C3 • To create C3 only look at items that have the same first item (in pass k, the first k - 2 items must match)

  9. Pruning (k-1)-subset of the candidate itemset is not in the frequent itemsetLk-1 In pass-3: • Find all subsets of 2 items from the C3, and check if they are in the frequent itemset L2.

  10. C3 after pruning Pruning eliminates ABE since BE is not frequent

  11. Scan transactions in the database and compute support L3

  12. Pass-4 • First k - 2 = 2 items must match in pass k = 4

  13. Pruning • Pruning: For ABCD we check whether ABC, ABD, ACD, BCD are frequent. They are in all cases, so we do not prune ABCD. • For ACDE we check whether ACD, ACE, ADE, CDE are frequent. Yes, in all cases, so we do not prune ACDE • Both are frequent L4

  14. Pass-5 • For pass 5 we can't form any candidates because there aren't two frequent 4-itemsets beginning with the same 3 items.

  15. Association Rules • {A, B, C} • Non-empty sets: • {A}{B}{C} {AB}{AC} {BC} • Assume min confidence 70% • Compute confidence for each rule

  16. Rules • R1: A, BC • Confidence= support {A B C}/support {A B} = 0.6/ 0.6= 1 => 100% Compute confidence in R2 R2: A, CB

  17. HITS algorithm

  18. Example-1 • Apply the HITS algorithm on the following web graph: 1 2 3

  19. Initialize HUB and AUTH values HUB=1 AUTH=1 HUB=1 AUTH=1 1 2 HUB=1 AUTH=1 3

  20. Normalization Normalized HUB (1)= HUB(1)/ SQRT [HUB(1)2+HUB(2)2+HUB(3)2] Normalized AUTH (1)= AUTH(1)/ SQRT [AUTH(1)2+AUTH(2)2+AUTH(3)2] We do this for all pages in the graph.

  21. Normalized values • HUB (1)=0.58, AUTH (1)=0.58 • HUB (2)=0.58, AUTH (2)=0.58 • HUB (3)=0.58, AUTH (3)=0.58

  22. Compute new HUB and AUTH valuesNode (1) • HUB (1)= AUTH(2)+AUTH(3)= = 0.58 + 0.58 = 1.16 • AUTH (1)= =0 Authority of nodes pointed to by node (1) Hub value of nodes pointing to node (1)

  23. Node (2) • HUB (2)= =0 • AUTH (2)= = HUB (1)= 0.58 Authority of nodes pointed to by node (2) Hub value of nodes pointing to node (2)

  24. Node (3) • HUB (3)= =0 • AUTH (3)= = HUB (1)= 0.58 Authority of nodes pointed to by node (3) Hub value of nodes pointing to node (3)

  25. After Normalization • HUB (1)= 1.16/SQRT [(1.16)2+02+02] =1.16/SQRT (1.3456) =1.16/1.16=1 • AUTH (1)= 0 • HUB(2)=0, AUTH(2)=0.71 • HUB(3)=0, AUTH(3)=0.71

  26. Recalculating HUB and AUTH • HUB (1)= AUTH(2)+AUTH(3)= = 0.71 + 0.71 = 1.42 • AUTH (1)= 0 Normalizing: Hub(1)= 1.42/SQRT [(1.42)2+02+02] HUB(1)=1.42/SQRT(2.0164) = 1.42/1.42= 1

  27. Recalculations • HUB (2)= 0 • AUTH(2)=0.71 • HUB (3)=0 • AUTH(3)=0.71 • Because the values are unchanged, we stop here. • Page 1 is clearly the hub, and pages 1, and 2 share the honor of being authorities.

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