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Apriori on HotDogs data with rules
Short answer
Apriori with min_sup=2 yields frequent itemsets: all singles, pairs like {HotDogs,Buns}, and triple {HotDogs,Coke,Chips}. Strong rules (conf≥60%) include Buns→HotDogs (100%), Coke→Chips (100%), Chips,HotDogs→Coke (100%), and Coke→HotDogs (66.7%), sorted by confidence.
Given min_sup=33.34% and min_conf=60%, use apriori algorithm on following transaction data to determine frequent item sets. Also, indicate association rules generated, underline the strong ones and sort them by confidence.
| Transaction ID | Item set |
|---|---|
| TID1 | HotDogs, Buns, Ketchup |
| TID2 | HotDogs, Buns |
| TID3 | HotDogs, Coke, Chips |
| TID4 | Coke, Chips |
| TID5 | Chips, Ketchup |
| TID6 | HotDogs, Coke, Chips |
Assumption: min_sup = 33.34% of 6 transactions is taken as 2 transactions (2/6 = 33.3%). min_conf = 60%.
Transactions: TID1 {HotDogs, Buns, Ketchup}; TID2 {HotDogs, Buns}; TID3 {HotDogs, Coke, Chips}; TID4 {Coke, Chips}; TID5 {Chips, Ketchup}; TID6 {HotDogs, Coke, Chips}.
Frequent itemsets by Apriori
C₁ → L₁: HotDogs 4, Buns 2, Ketchup 2, Coke 3, Chips 4. All have count ≥ 2, so L₁ contains all five.
C₂ (10 pairs) counts:
| Pair | Count | Frequent? |
|---|---|---|
| HotDogs, Buns | 2 | Yes |
| HotDogs, Ketchup | 1 | No |
| HotDogs, Coke | 2 | Yes |
| HotDogs, Chips | 2 | Yes |
| Buns, Ketchup | 1 | No |
| Buns, Coke | 0 | No |
| Buns, Chips | 0 | No |
| Ketchup, Coke | 0 | No |
| Ketchup, Chips | 1 | No |
| Coke, Chips | 3 | Yes |
L₂ = {HotDogs-Buns:2, HotDogs-Coke:2, HotDogs-Chips:2, Coke-Chips:3}
C₃: HotDogs-Coke + HotDogs-Chips → {HotDogs, Coke, Chips}; its subsets Coke-Chips, HotDogs-Coke, HotDogs-Chips are all frequent. Another join, HotDogs-Buns + HotDogs-Coke → {HotDogs, Buns, Coke}, is pruned (Buns-Coke infrequent), likewise {HotDogs, Buns, Chips}. Count({HotDogs, Coke, Chips}) = 2 (TID3, TID6) → L₃ = {HotDogs, Coke, Chips}.
Frequent itemsets: 5 single items, the 4 pairs of L₂, and the triple {HotDogs, Coke, Chips}.
Association rules (sorted by confidence; strong rules in bold)
| Rule | Support count | Confidence | Strong? |
|---|---|---|---|
| Buns → HotDogs | 2 | 2/2 = 100.0% | Yes |
| Coke → Chips | 3 | 3/3 = 100.0% | Yes |
| Chips, HotDogs → Coke | 2 | 2/2 = 100.0% | Yes |
| Coke, HotDogs → Chips | 2 | 2/2 = 100.0% | Yes |
| Chips → Coke | 3 | 3/4 = 75.0% | Yes |
| Coke → Chips, HotDogs | 2 | 2/3 = 66.7% | Yes |
| Coke → HotDogs | 2 | 2/3 = 66.7% | Yes |
| Chips, Coke → HotDogs | 2 | 2/3 = 66.7% | Yes |
| Chips → Coke, HotDogs | 2 | 2/4 = 50.0% | No |
| Chips → HotDogs | 2 | 2/4 = 50.0% | No |
| HotDogs → Buns | 2 | 2/4 = 50.0% | No |
| HotDogs → Chips | 2 | 2/4 = 50.0% | No |
| HotDogs → Chips, Coke | 2 | 2/4 = 50.0% | No |
| HotDogs → Coke | 2 | 2/4 = 50.0% | No |
Answer: 8 strong rules (confidence ≥ 60%): Buns → HotDogs (100%), Coke → Chips (100%), Chips, HotDogs → Coke (100%), Coke, HotDogs → Chips (100%), Chips → Coke (75%), Chips, Coke → HotDogs (66.7%), Coke → HotDogs (66.7%), Coke → Chips, HotDogs (66.7%). (The table lists all 14; those below 60% are not strong.)
If the threshold is applied strictly (33.34% needs count > 2), only items HotDogs, Coke, Chips and the pair {Coke, Chips} stay frequent, giving Coke → Chips (100%) and Chips → Coke (75%).
✓ Answer checked by Claude Opus 5.5 (high effort)
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