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Huffman Codes and Asssociation Rules (II) Prof. Sin-Min Lee Department of Computer Science.

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Presentation on theme: "Huffman Codes and Asssociation Rules (II) Prof. Sin-Min Lee Department of Computer Science."— Presentation transcript:

1 Huffman Codes and Asssociation Rules (II) Prof. Sin-Min Lee Department of Computer Science

2 Huffman Code Example Given: ABCDE By using an increasing algorithm (changing from smallest to largest), it changes to: BCADE 12346

3 Huffman Code Example – Step 1 Because B and C are the lowest values, they can be appended. The new value is 3

4 Huffman Code Example – Step 2 Reorder the problem using the increasing algorithm again. This gives us: BCADE 3346

5 Huffman Code Example – Step 3 Doing another append will give:

6 Huffman Code Example – Step 4 From the initial BC A D E code we get: DEABC 46 6 DEBCA 46 6 DABCE 4 66 DBCAE 4 66

7 Huffman Code Example – Step 5 Taking derivates from the previous step, we get: DEBCA 466 EDBCA 6 10 DABCE 106 DEABC 46 6

8 Huffman Code Example – Step 6 Taking derivates from the previous step, we get: BCADE 646 EDBCA 6 10 EDABC ABCDE 64 6

9 Huffman Code Example – Step 7 After the previous step, we’re supposed to map a 1 to each right branch and a 0 to each left branch. The results of the codes are:

10 Example Items={milk, coke, pepsi, beer, juice}. Support = 3 baskets. B1 = {m, c, b}B2 = {m, p, j} B3 = {m, b} B4 = {c, j} B5 = {m, p, b}B6 = {m, c, b, j} B7 = {c, b, j}B8 = {b, c} Frequent itemsets: {m}, {c}, {b}, {j}, {m, b}, {c, b}, {j, c}.

11 Association Rules Association rule R : Itemset1 => Itemset2 –Itemset1, 2 are disjoint and Itemset2 is non- empty –meaning: if transaction includes Itemset1 then it also has Itemset2 Examples –A,B => E,C –A => B,C

12 Example B1 = {m, c, b}B2 = {m, p, j} B3 = {m, b}B4 = {c, j} B5 = {m, p, b}B6 = {m, c, b, j} B7 = {c, b, j}B8 = {b, c} An association rule: {m, b} → c. –Confidence = 2/4 = 50%. + _ +

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17 From Frequent Itemsets to Association Rules Q: Given frequent set {A,B,E}, what are possible association rules? –A => B, E –A, B => E –A, E => B –B => A, E –B, E => A –E => A, B –__ => A,B,E (empty rule), or true => A,B,E

18 Classification vs Association Rules Classification Rules Focus on one target field Specify class in all cases Measures: Accuracy Association Rules Many target fields Applicable in some cases Measures: Support, Confidence, Lift

19 Rule Support and Confidence Suppose R : I => J is an association rule –sup (R) = sup (I  J) is the support count support of itemset I  J (I or J) –conf (R) = sup(J) / sup(R) is the confidence of R fraction of transactions with I  J that have J Association rules with minimum support and count are sometimes called “strong” rules

20 Association Rules Example: Q: Given frequent set {A,B,E}, what association rules have minsup = 2 and minconf= 50% ? A, B => E : conf=2/4 = 50% A, E => B : conf=2/2 = 100% B, E => A : conf=2/2 = 100% E => A, B : conf=2/2 = 100% Don’t qualify A =>B, E : conf=2/6 =33%< 50% B => A, E : conf=2/7 = 28% < 50% __ => A,B,E : conf: 2/9 = 22% < 50%

21 Find Strong Association Rules A rule has the parameters minsup and minconf: –sup(R) >= minsup and conf (R) >= minconf Problem: –Find all association rules with given minsup and minconf First, find all frequent itemsets

22 Finding Frequent Itemsets Start by finding one-item sets (easy) Q: How? A: Simply count the frequencies of all items

23 Finding itemsets: next level Apriori algorithm (Agrawal & Srikant) Idea: use one-item sets to generate two-item sets, two-item sets to generate three-item sets, … –If (A B) is a frequent item set, then (A) and (B) have to be frequent item sets as well! –In general: if X is frequent k-item set, then all (k-1)- item subsets of X are also frequent  Compute k-item set by merging (k-1)-item sets

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25 Finding Association Rules A typical question: “find all association rules with support ≥ s and confidence ≥ c.” –Note: “support” of an association rule is the support of the set of items it mentions. Hard part: finding the high-support (frequent ) itemsets. –Checking the confidence of association rules involving those sets is relatively easy.

26 Naïve Algorithm A simple way to find frequent pairs is: –Read file once, counting in main memory the occurrences of each pair. Expand each basket of n items into its n (n -1)/2 pairs. Fails if #items-squared exceeds main memory.

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28 C1C1 L1L1 C2C2 L2L2 C3C3 Filter Construct First pass Second pass

29 Fast Algorithms for Mining Association Rules, by Rakesh Agrawal and Ramakrishan Sikant, IBM Almaden Research Center [Agrawal, Srikant 94]

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39 ItemsTID Set-of- itemsets TID { {1},{3},{4} }100 { {2},{3},{5} }200 { {1},{2},{3},{5} }300 { {2},{5} }400 SupportItemset 2{1} 3{2} 3{3} 3{5} itemset {1 2} {1 3} {1 5} {2 3} {2 5} {3 5} Set-of-itemsetsTID { {1 3} }100 { {2 3},{2 5} {3 5} }200 { {1 2},{1 3},{1 5}, {2 3}, {2 5}, {3 5} } 300 { {2 5} }400 SupportItemset 2{1 3} 3{2 3} 3{2 5} 2{3 5} itemset {2 3 5} Set-of-itemsetsTID { {2 3 5} }200 { {2 3 5} }300 SupportItemset 2{2 3 5} Database C^ 1 L2L2 C2C2 C^ 2 C^ 3 L1L1 L3L3 C3C3

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43 Dynamic Programming Approach Dynamic Programming Approach l Want proof of principle of optimality and overlapping subproblems l Principle of Optimality F The optimal solution to L k includes the optimal solution of L k-1 F Proof by contradiction l Overlapping Subproblems F Lemma of every subset of a frequent item set is a frequent item set F Proof by contradiction

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56 The Apriori Algorithm: Example Consider a database, D, consisting of 9 transactions. Suppose min. support count required is 2 (i.e. min_sup = 2/9 = 22 % ) Let minimum confidence required is 70%. We have to first find out the frequent itemset using Apriori algorithm. Then, Association rules will be generated using min. support & min. confidence. TIDList of Items T100I1, I2, I5 T100I2, I4 T100I2, I3 T100I1, I2, I4 T100I1, I3 T100I2, I3 T100I1, I3 T100I1, I2,I3, I5 T100I1, I2, I3

57 Step 1 : Generating 1-itemset Frequent Pattern ItemsetSup.Count {I1}6 {I2}7 {I3}6 {I4}2 {I5}2 ItemsetSup.Count {I1}6 {I2}7 {I3}6 {I4}2 {I5}2 In the first iteration of the algorithm, each item is a member of the set of candidate. The set of frequent 1-itemsets, L 1, consists of the candidate 1-itemsets satisfying minimum support. Scan D for count of each candidate Compare candidate support count with minimum support count C1C1 L1L1

58 Step 2 : Generating 2-itemset Frequent Pattern Itemset {I1, I2} {I1, I3} {I1, I4} {I1, I5} {I2, I3} {I2, I4} {I2, I5} {I3, I4} {I3, I5} {I4, I5} ItemsetSup. Count {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 Itemse t Sup Count {I1, I2}4 {I1, I3}4 {I1, I5}2 {I2, I3}4 {I2, I4}2 {I2, I5}2 Generate C 2 candidates from L 1 C2C2 C2C2 L2L2 Scan D for count of each candidate Compare candidate support count with minimum support count

59 Step 2 : Generating 2-itemset Frequent Pattern [Cont.] To discover the set of frequent 2-itemsets, L 2, the algorithm uses L 1 Join L 1 to generate a candidate set of 2-itemsets, C 2. Next, the transactions in D are scanned and the support count for each candidate itemset in C 2 is accumulated (as shown in the middle table). The set of frequent 2-itemsets, L 2, is then determined, consisting of those candidate 2-itemsets in C 2 having minimum support. Note: We haven’t used Apriori Property yet.

60 Step 3 : Generating 3-itemset Frequent Pattern Itemset {I1, I2, I3} {I1, I2, I5} ItemsetSup. Count {I1, I2, I3}2 {I1, I2, I5}2 ItemsetSup Count {I1, I2, I3}2 {I1, I2, I5}2 C3C3 C3C3 L3L3 Scan D for count of each candidate Compare candidate support count with min support count Scan D for count of each candidate The generation of the set of candidate 3-itemsets, C 3, involves use of the Apriori Property. In order to find C 3, we compute L 2 Join L 2. C 3 = L2 Join L2 = {{I1, I2, I3}, {I1, I2, I5}, {I1, I3, I5}, {I2, I3, I4}, {I2, I3, I5}, {I2, I4, I5}}. Now, Join step is complete and Prune step will be used to reduce the size of C 3. Prune step helps to avoid heavy computation due to large C k.

61 Step 3 : Generating 3-itemset Frequent Pattern [Cont.] Based on the Apriori property that all subsets of a frequent itemset must also be frequent, we can determine that four latter candidates cannot possibly be frequent. How ? For example, lets take {I1, I2, I3}. The 2-item subsets of it are {I1, I2}, {I1, I3} & {I2, I3}. Since all 2-item subsets of {I1, I2, I3} are members of L 2, We will keep {I1, I2, I3} in C 3. Lets take another example of {I2, I3, I5} which shows how the pruning is performed. The 2-item subsets are {I2, I3}, {I2, I5} & {I3,I5}. BUT, {I3, I5} is not a member of L 2 and hence it is not frequent violating Apriori Property. Thus We will have to remove {I2, I3, I5} from C 3. Therefore, C 3 = {{I1, I2, I3}, {I1, I2, I5}} after checking for all members of result of Join operation for Pruning. Now, the transactions in D are scanned in order to determine L 3, consisting of those candidates 3-itemsets in C 3 having minimum support.

62 Step 4 : Generating 4-itemset Frequent Pattern The algorithm uses L 3 Join L 3 to generate a candidate set of 4-itemsets, C 4. Although the join results in {{I1, I2, I3, I5}}, this itemset is pruned since its subset {{I2, I3, I5}} is not frequent. Thus, C 4 = φ, and algorithm terminates, having found all of the frequent items. This completes our Apriori Algorithm. What’s Next ? These frequent itemsets will be used to generate strong association rules ( where strong association rules satisfy both minimum support & minimum confidence).

63 Step 5: Generating Association Rules from Frequent Itemsets Procedure: For each frequent itemset “l”, generate all nonempty subsets of l. For every nonempty subset s of l, output the rule “s  (l-s)” if support_count(l) / support_count(s) >= min_conf where min_conf is minimum confidence threshold. Back To Example: We had L = {{I1}, {I2}, {I3}, {I4}, {I5}, {I1,I2}, {I1,I3}, {I1,I5}, {I2,I3}, {I2,I4}, {I2,I5}, {I1,I2,I3}, {I1,I2,I5}}. –Lets take l = {I1,I2,I5}. –Its all nonempty subsets are {I1,I2}, {I1,I5}, {I2,I5}, {I1}, {I2}, {I5}.

64 Step 5: Generating Association Rules from Frequent Itemsets [Cont.] Let minimum confidence threshold is, say 70%. The resulting association rules are shown below, each listed with its confidence. –R1: I1 ^ I2  I5 Confidence = sc{I1,I2,I5}/sc{I1,I2} = 2/4 = 50% R1 is Rejected. –R2: I1 ^ I5  I2 Confidence = sc{I1,I2,I5}/sc{I1,I5} = 2/2 = 100% R2 is Selected. –R3: I2 ^ I5  I1 Confidence = sc{I1,I2,I5}/sc{I2,I5} = 2/2 = 100% R3 is Selected.

65 Step 5: Generating Association Rules from Frequent Itemsets [Cont.] – R4: I1  I2 ^ I5 Confidence = sc{I1,I2,I5}/sc{I1} = 2/6 = 33% R4 is Rejected. – R5: I2  I1 ^ I5 Confidence = sc{I1,I2,I5}/{I2} = 2/7 = 29% R5 is Rejected. – R6: I5  I1 ^ I2 Confidence = sc{I1,I2,I5}/ {I5} = 2/2 = 100% R6 is Selected. In this way, We have found three strong association rules.

66 ABCDE ACDE  B ABCE  D ACD  BE ADE  BC CDE  AB ACE  BD BCE  AD ACE  BD ABE  CD ABC  ED Large itemset Rules with minsup Simple algorithm: Fast algorithm: ACE  BD ABCDE ACDE  B ABCE  D Example

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