--- title: "Apriori on 8 transactions, 25% support" subject: "Data Mining" chapter: 4 key: "semester_7/data_mining:apriori-b65-transactions" source: https://www.bctengineer.com/q/semester_7/data_mining/apriori-b65-transactions site: BCT Engineering --- # Apriori on 8 transactions, 25% support Data Mining · Chapter 4: Association Analysis - Asked in: 2065 Baisakh - Video: [3. Apriori Algorithm | Association Rule Mining | Frequent Item Sets Solved Example by Mahesh Huddar](https://www.youtube.com/watch?v=wpCeFC8-z-k) — Mahesh Huddar - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/data_mining/apriori-b65-transactions - Whole chapter: https://www.bctengineer.com/md/semester_7/data_mining/chapter-4 ## Question Write the Apriori Algorithm. Use this algorithm to find the frequent item set and generate few rules associated with the transaction. (Take minimum support=25%, minimum confidence% [value not printed in the paper]) | Transaction ID | Items | | --- | --- | | 1 | A, C, D | | 2 | C, D, E | | 3 | A, D, E | | 4 | B, D, E | | 5 | A, C, D, E | | 6 | A, D, E | | 7 | A, B, C | | 8 | C, E | ## Short answer With min support 2 (25%) and assumed min confidence 60%, frequent itemsets: A,B,C,D,E; AC,AD,AE,CD,CE,DE; ACD,ADE,CDE. Strong rules include A,E→D (100%), D→E (83.3%), E→D (83.3%), A→D (80%), A,D→E (75%), D→A (66.7%). ## Answer #### Apriori algorithm ```text L1 = frequent items (scan database) for k = 2 while L(k-1) is not empty: Ck = join(L(k-1)) and prune candidates with an infrequent subset scan database, count candidates Lk = candidates with support >= min_support output all Lk; then generate rules with confidence >= min_conf ``` Principle: every subset of a frequent itemset must be frequent. #### Solution 8 transactions, minimum support 25% → 0.25 × 8 = **2 transactions**. The minimum confidence is not given, so **60% is assumed**. **C₁ → L₁**: A 5, B 2, C 5, D 6, E 6. All items have count ≥ 2, so L₁ = {A, B, C, D, E}. **C₂** counts: AB 1, AC 3, AD 4, AE 3, BC 1, BD 1, BE 1, CD 3, CE 3, DE 5. **L₂** = {AC:3, AD:4, AE:3, CD:3, CE:3, DE:5} (AB, BC, BD, BE removed, so B appears in no larger itemset). **C₃** by joining L₂: ACD, ACE, ADE, CDE (others like ADC are the same). All 2-subsets are frequent (no pruning). Counts: ACD 2, ACE 1, ADE 3, CDE 2. **L₃** = {ACD:2, ADE:3, CDE:2}. **C₄**: ACDE would need ACE, which is infrequent, so it is pruned. Stop. **Frequent itemsets**: A, B, C, D, E; AC, AD, AE, CD, CE, DE; ACD, ADE, CDE. #### Rules from {A, D, E} (support 3) | Rule | Confidence | Strong (≥ 60%)? | | --- | --- | --- | | A, E → D | 3/3 = 100% | Yes | | A, D → E | 3/4 = 75% | Yes | | D, E → A | 3/5 = 60% | Yes | | A → D, E | 3/5 = 60% | Yes | | D → A, E | 3/6 = 50% | No | | E → A, D | 3/6 = 50% | No | #### Rules from some 2-itemsets D → E 5/6 = 83.3%; E → D 5/6 = 83.3%; A → D 4/5 = 80%; D → A 4/6 = 66.7%; A → C 3/5 = 60%. #### Rules from {A, C, D} and {C, D, E} (support 2) A, C → D 2/3 = 66.7%; C, D → A 66.7%; C, D → E 66.7%; C, E → D 66.7%; the others (A, D → C 50%, D, E → C 40%, etc.) are weak. **Answer: frequent itemsets are listed above (largest {A, D, E}, {A, C, D}, {C, D, E}). Strong rules (confidence ≥ 60%) include A,E → D (100%), D → E (83.3%), E → D (83.3%), A → D (80%), A,D → E (75%), D → A (66.7%).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Minimum support 25% of 8 transactions is: A. 2 transactions B. 3 transactions C. 4 transactions D. 1 transaction Answer: A. 0.25 x 8 = 2. (Key point: Min count) 2. Why is ACDE pruned? A. ACE is infrequent B. ACD is infrequent C. Its count is 0 D. It has too many items Answer: A. A subset ACE (count 1) is infrequent. (Key point: Pruning ACDE) 3. Confidence of A,E -> D (support of AED 3, AE 3)? A. 100% B. 75% C. 60% D. 50% Answer: A. 3/3. (Key point: Rule confidence) 4. Why does B not appear in larger itemsets? A. B has support 0 B. All pairs with B are infrequent C. B appears in every transaction D. B is the only item Answer: B. AB, BC, BD, BE each have count 1. 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