1. Association Analysis (2)

2. Example TIDList of item ID’s T1I1, I2, I5 T2I2, I4 T3I2, I3 T4I1, I2, I4 T5I1, I3 T6I2, I3 T7I1, I3 T8I1, I2, I3, I5 T9I1, I2, I3 Min_sup_count = 2 Itemset {I5} {I4} {I3} {I2} {I1} C1 Sup. count Itemset 2{I5} 2{I4} 6{I3} 7{I2} 6{I1} F1

3. Generate C2 from F1  F1 TIDList of item ID’s T1I1, I2, I5 T2I2, I4 T3I2, I3 T4I1, I2, I4 T5I1, I3 T6I2, I3 T7I1, I3 T8I1, I2, I3, I5 T9I1, I2, I3 Min_sup_count = 2 ItemsetSup. count {I1}6 {I2}7 {I3}6 {I4}2 {I5}2 F1 {I4,I5} {I3,I5} {I3,I4} {I2,I5} {I2,I4} Itemset {I2,I3} {I1,I5} {I1,I4} {I1,I3} {I1,I2} C2 ItemsetSup. C {I1,I2}4 {I1,I3}4 {I1,I4}1 {I1,I5}2 {I2,I3}4 {I2,I4}2 {I2,I5}2 {I3,I4}0 {I3,I5}1 {I4,I5}0

4. Generate C3 from F2  F2 TIDList of item ID’s T1I1, I2, I5 T2I2, I4 T3I2, I3 T4I1, I2, I4 T5I1, I3 T6I2, I3 T7I1, I3 T8I1, I2, I3, I5 T9I1, I2, I3 Min_sup_count = 2 ItemsetSup. C {I1,I2}4 {I1,I3}4 {I1,I5}2 {I2,I3}4 {I2,I4}2 {I2,I5}2 F2 Itemset {I1,I2,I3} {I1,I2,I5} {I1,I3,I5} {I2,I3,I4} {I2,I3,I5} {I2,I4,I5} Prune Itemset {I1,I2,I3} {I1,I2,I5} {I1,I3,I5} {I2,I3,I4} {I2,I3,I5} {I2,I4,I5} ItemsetSup. C {I1,I2,I3}2 {I1,I2,I5}2 F3

5. Generate C4 from F3  F3 TIDList of item ID’s T1I1, I2, I5 T2I2, I4 T3I2, I3 T4I1, I2, I4 T5I1, I3 T6I2, I3 T7I1, I3 T8I1, I2, I3, I5 T9I1, I2, I3 Min_sup_count = 2 ItemsetSup. C {I1,I2,I3,I5}2 C4 {I1,I2,I3,I5} is pruned because {I2,I3,I5} is infrequent ItemsetSup. C {I1,I2,I3}2 {I1,I2,I5}2 F3

6. Candidate support counting Scan the database of transactions to determine the support of each candidate itemset Brute force: Match each transaction against every candidate. –Too many comparisons! Better method: Store the candidate itemsets in a hash structure – A transaction will be tested for match only against candidates contained in a few buckets TID Items 1 Bread, Milk 2 Bread, Diaper, Beer, Eggs 3 Milk, Diaper, Beer, Coke 4 Bread, Milk, Diaper, Beer 5 Bread, Milk, Diaper, Coke N Transactions Hash Structure k Buckets

7. Generate Hash Tree 2 3 4 5 6 7 1 4 5 1 3 6 1 2 4 4 5 7 1 2 5 4 5 8 1 5 9 3 4 5 3 5 6 3 5 7 6 8 9 3 6 7 3 6 8 1,4,7 2,5,8 3,6,9 Hash function You need: A hash function (e.g. p mod 3) Max leaf size: max number of itemsets stored in a leaf node (if number of candidate itemsets exceeds max leaf size, split the node) Suppose you have 15 candidate itemsets of length 3 and leaf size is 3: {1 4 5}, {1 2 4}, {4 5 7}, {1 2 5}, {4 5 8}, {1 5 9}, {1 3 6}, {2 3 4}, {5 6 7}, {3 4 5}, {3 5 6}, {3 5 7}, {6 8 9}, {3 6 7}, {3 6 8}

8. Generate Hash Tree 1,4,7 2,5,8 3,6,9 Hash function Suppose you have 15 candidate itemsets of length 3 and leaf size is 3: {1 4 5}, {1 2 4}, {4 5 7}, {1 2 5}, {4 5 8}, {1 5 9}, {1 3 6}, {2 3 4}, {5 6 7}, {3 4 5}, {3 5 6}, {3 5 7}, {6 8 9}, {3 6 7}, {3 6 8} 2 3 4 5 6 7 1 4 5 1 3 6 1 2 4 4 5 7 1 2 5 4 5 8 1 5 9 3 5 6 3 5 7 6 8 9 3 4 5 3 6 7 3 6 8 Split nodes with more than 3 candidates using the second item

9. Generate Hash Tree 1,4,7 2,5,8 3,6,9 Hash function Suppose you have 15 candidate itemsets of length 3 and leaf size is 3: {1 4 5}, {1 2 4}, {4 5 7}, {1 2 5}, {4 5 8}, {1 5 9}, {1 3 6}, {2 3 4}, {5 6 7}, {3 4 5}, {3 5 6}, {3 5 7}, {6 8 9}, {3 6 7}, {3 6 8} 2 3 4 5 6 7 3 5 6 3 5 7 6 8 9 3 4 5 3 6 7 3 6 8 1 2 4 4 5 7 1 2 5 4 5 8 1 5 9 1 4 5 1 3 6 Now split nodes using the third item

10. Generate Hash Tree 1,4,7 2,5,8 3,6,9 Hash function Suppose you have 15 candidate itemsets of length 3 and leaf size is 3: {1 4 5}, {1 2 4}, {4 5 7}, {1 2 5}, {4 5 8}, {1 5 9}, {1 3 6}, {2 3 4}, {5 6 7}, {3 4 5}, {3 5 6}, {3 5 7}, {6 8 9}, {3 6 7}, {3 6 8} 2 3 4 5 6 7 3 5 6 3 5 7 6 8 9 3 4 5 3 6 7 3 6 8 1 4 5 1 3 6 1 2 4 4 5 7 1 2 5 4 5 8 1 5 9 Now, split this similarly.

11. Subset Operation Given a (lexicographically ordered) transaction t, say {1,2,3,5,6} how can we enumerate the possible subsets of size 3?

12. Subset Operation Using Hash Tree 1 5 9 1 4 5 1 3 6 3 4 5 3 6 7 3 6 8 3 5 6 3 5 7 6 8 9 2 3 4 5 6 7 1 2 4 4 5 7 1 2 5 4 5 8 1 2 3 5 6 1 +2 3 5 6 3 5 62 + 5 63 + 1,4,7 2,5,8 3,6,9 Hash Function transaction

13. Subset Operation Using Hash Tree 1 5 9 1 4 5 1 3 6 3 4 5 3 6 7 3 6 8 3 5 6 3 5 7 6 8 9 2 3 4 5 6 7 1 2 4 4 5 7 1 2 5 4 5 8 1,4,7 2,5,8 3,6,9 Hash Function 1 2 3 5 6 3 5 61 2 + 5 61 3 + 61 5 + 3 5 62 + 5 63 + 1 +2 3 5 6 transaction

14. Subset Operation Using Hash Tree 1 5 9 1 4 5 1 3 6 3 4 5 3 6 7 3 6 8 3 5 6 3 5 7 6 8 9 2 3 4 5 6 7 1 2 4 4 5 7 1 2 5 4 5 8 1,4,7 2,5,8 3,6,9 Hash Function 1 2 3 5 6 3 5 61 2 + 5 61 3 + 61 5 + 3 5 62 + 5 63 + 1 +2 3 5 6 transaction Match transaction against 7 out of 15 candidates

15. Rule Generation An association rule can be extracted by partitioning a frequent itemset Y into two non­empty subsets, X and Y -X, such that X  Y-X satisfies the confidence threshold. Each frequent k-itemset, Y, can produce up to 2 k -2 association rules –ignoring rules that have empty antecedents or consequents. Example Let Y = {1, 2, 3} be a frequent itemset. Six candidate association rules can be generated from Y: {1, 2}  {3}, {1, 3}  {2}, {2, 3}  {1}, {1}  {2, 3}, {2}  {1, 3}, {3}  {1, 2}. Computing the confidence of an association rule does not require additional scans of the transactions. Consider {1, 2}  {3}. The confidence is  ({1, 2, 3}) /  ({1, 2}) Because {1, 2, 3} is frequent, the anti­ monotone property of support ensures that {1, 2} must be frequent, too, and we know the supports of frequent itemsets.

16. Confidence-Based Prunning I Theorem. If a rule X  Y – X does not satisfy the confidence threshold, then any rule X ’  Y – X ’, where X ’ is a subset of X, cannot satisfy the confidence threshold as well. Proof. Consider the following two rules: X ’  Y – X ’ and X  Y – X, where X ’  X. The confidence of the rules are  (Y ) /  (X ’) and  (Y ) /  (X), respectively. Since X ’ is a subset of X,  (X ’)   (X). Therefore, the former rule cannot have a higher confidence than the latter rule.

17. Confidence-Based Prunning II Observe that: X ’  X implies that Y – X ’  Y – X Y X X’

18. Confidence-Based Prunning III Initially, all the high­confidence rules that have only one item in the rule consequent are extracted. These rules are then used to generate new candidate rules. For example, if –{acd}  {b} and {abd}  {c} are high­confidence rules, then the candidate rule {ad}  {bc} is generated by merging the consequents of both rules.

19. Confidence-Based Prunning IV {Bread,Milk}  {Diaper} (confidence = 3/3) threshold=50% {Bread,Diaper}  {Milk} (confidence = 3/3) {Diaper,Milk}  {Bread} (confidence = 3/3) Items (1-itemsets) Pairs (2-itemsets) Triplets (3-itemsets)

20. Confidence-Based Prunning V Merge: {Bread,Milk}  {Diaper} {Bread,Diaper}  {Milk} {Bread}  {Diaper,Milk} (confidence = 3/4) …

21. Compact Representation of Frequent Itemsets Some itemsets are redundant because they have identical support as their supersets Number of frequent itemsets Need a compact representation

22. Maximal Frequent Itemsets Border Infrequent Itemsets Maximal Itemsets An itemset is maximal frequent if none of its immediate supersets is frequent Maximal frequent itemsets form the smallest set of itemsets from which all frequent itemsets can be derived.

23. Maximal Frequent Itemsets Despite providing a compact representation, maximal frequent itemsets do not contain the support information of their subsets. –For example, the support of the maximal frequent itemsets {a, c, e}, {a, d}, and {b,c,d,e} do not provide any hint about the support of their subsets. An additional pass over the data set is therefore needed to determine the support counts of the non­maximal frequent itemsets. It might be desirable to have a minimal representation of frequent itemsets that preserves the support information. –Such representation is the set of the closed frequent itemsets.

24. Closed Itemset An itemset is closed if none of its immediate supersets has the same support as the itemset. –Put another way, an itemset X is not closed if at least one of its immediate supersets has the same support count as X. An itemset is a closed frequent itemset if it is closed and its support is greater than or equal to minsup.

25. Lattice Transaction Ids Not supported by any transactions

26. Maximal vs. Closed Itemsets Minimum support = 2 # Closed = 9 # Maximal = 4 Closed and maximal Closed but not maximal

27. Maximal vs Closed Itemsets All maximal frequent itemsets are closed because none of the maximal frequent itemsets can have the same support count as their immediate supersets.

28. Deriving Frequent Itemsets From Closed Frequent Itemsets Consider {a, d}. –It is frequent because {a, b, d} is. –Since it isn't closed, its support count must be identical to one of its immediate supersets. –The key is to determine which superset among {a, b, d}, {a, c, d}, or {a, d, e} has exactly the same support count as {a, d}. The Apriori principle states that: –Any transaction that contains the superset of {a, d} must also contain {a, d}. –However, any transaction that contains {a, d} does not have to contain the supersets of {a, d}. –So, the support for {a, d} must be equal to the largest support among its supersets. –Since {a, c, d} has a larger support than both {a, b, d} and {a, d, e}, the support for {a, d} must be identical to the support for {a, c, d}.

29. Algorithm Let C denote the set of closed frequent itemsets Let k max denote the maximum length of closed frequent itemsets F kmax ={f | f  C, | f | = k max } {Find all frequent itemsets of size k max } for k = k max – 1 downto 1 do Set F k to be all sub-itemsets of length k from the frequent itemsets in F k+1 plus the closed frequent itemsets of size k. for each f  F k do if f  C then f.support = max{f’.support | f’  F k+1, f  f’} end if end for

30. Example C = {ABC:3, ACD:4, CE:6, DE:7} k max =3 F3 = {ABC:3, ACD:4} F2 = {AB:3, AC:4, BC:3, AD:4, CD:4, CE:6, DE:7} F1 = {A:4, B:3, C:6, D:7, E:7}

31. Computing Frequent Closed Itemsets How? Use the Apriori Algorithm. After computing, say F k and F k+1, check whether there is some itemset in F k which has a support equal to the support of one of its supersets in F k+1. Purge all such itemsets from F k.