Frequent itemsets: {1}, {2}, {3}, {5}, {1,3}, {1,5}, {2,3}, {2,5}, {3,5}, {1,3,5}, {2,3,5}. Strong rules (confidence ≥ 60%) are listed in Step 4.
Transactions: T10 {1,3,4}, T20 {2,3,5}, T30 {1,2,3,5}, T40 {2,5}, T50 {1,3,5}. Minimum support count = 2.
Step 1: C1 → L1
{1}: 3, {2}: 3, {3}: 4, {4}: 1, {5}: 4. Remove {4} (1 < 2). L1 = {1}, {2}, {3}, {5}.
Step 2: C2 → L2
| Itemset | {1,2} | {1,3} | {1,5} | {2,3} | {2,5} | {3,5} |
|---|
| Count | 1 | 3 | 2 | 2 | 3 | 3 |
L2 = {1,3}, {1,5}, {2,3}, {2,5}, {3,5} ({1,2} removed).
Step 3: C3 → L3
Joining L2 gives {1,3,5} and {2,3,5} (any candidate containing {1,2} is pruned). Counts: {1,3,5} in T30, T50 = 2; {2,3,5} in T20, T30 = 2. Both frequent. C4 = {1,2,3,5} is pruned because {1,2} is not frequent, so the algorithm stops.
Step 4: Strong rules (confidence = support(A∪B)/support(A))
From 2-itemsets:
- 1 → 3: 3/3 = 100%; 3 → 1: 3/4 = 75%
- 1 → 5: 2/3 = 66.7%; (5 → 1: 2/4 = 50%, rejected)
- 2 → 3: 2/3 = 66.7%; (3 → 2: 50%, rejected)
- 2 → 5: 3/3 = 100%; 5 → 2: 3/4 = 75%
- 3 → 5: 3/4 = 75%; 5 → 3: 3/4 = 75%
From {1,3,5} (support 2):
- 1,5 → 3: 2/2 = 100%; 1,3 → 5: 2/3 = 66.7%; 3,5 → 1: 2/3 = 66.7%; 1 → 3,5: 2/3 = 66.7%
- Rejected: 3 → 1,5 and 5 → 1,3 (2/4 = 50%)
From {2,3,5} (support 2):
- 2,3 → 5: 2/2 = 100%; 2,5 → 3: 2/3 = 66.7%; 3,5 → 2: 2/3 = 66.7%; 2 → 3,5: 2/3 = 66.7%
- Rejected: 3 → 2,5 and 5 → 2,3 (50%)
All rules listed without "rejected" have confidence ≥ 60% and are strong. (If your syllabus asks only for rules from the largest itemsets, give the eight rules from {1,3,5} and {2,3,5}.)
