CSE 4412 Study Guide - Data Mining, Association Rule Learning, The Fp

• School: York University
• Department: Computer Science & Engineering
• Course: CSE 4412
• Professor: Aijun An

Document Summary

Solutions to assignment 1 (fall 2012: we prove this by contradiction. Let"s assume there is a frequent itemset of length k+1 which is missed by the join step. Let {i1, i2, , ik, ik+1} be that frequent itemset. Since {i1, i2, , ik, ik+1} is frequent, its subsets, {i1, i2, , ik-1, ik}and {i1, i2, , ik-1, ik+1}, must be frequent according to the. This means that {i1, i2, , ik-1, ik}and {i1, i2, , ik-1, ik+1} are in lk. Because they are in lk and they share the first k-1 items, they will be joined into {i1, i2, , ik, ik+1} when generating ck+1. This contradicts the assumption made at the beginning. (a) finding all frequent itemsets containing item i. min_sup_count=min_sup* (number of transactions) =30%*10=3. So 1-item frequent itemset as follow in the header table; Since we are only looking for frequent itemsets containing item i, i put i to be the last item in the head.