Question

2. The Apriori algorithm makes use of prior knowledge of subset support properties.

(a) Prove that all nonempty subsets of a frequent itemset must also be frequent.

(b) Prove that the support of any nonempty subset s′ of itemset s must be at least

as great as the support of s.

(c) Given frequent itemset l and subset s of l, prove that the confidence of the rule

“s′ ⇒(l−s′)” cannot be more than the confidence of“s⇒(l−s),” where s′ is a subset of s.

(d) A partitioning variation of Apriori subdivides the transactions of a database D

into n nonoverlapping partitions. Prove that any itemset that is frequent in D must be frequent in at least one partition of D.

Answer 1

2. a)  Prove that all nonempty subsets of a frequent itemset must also be frequent.

Actually we can prove this property in many ways, including by direct applying the proof of the question which is given in part (b).

Another approach to this problem is proved by contradiction. Let i be a frequent itemset and assume one of its nonempty subset X is infrequent. Since X is infrequent, we know that P(X) < min_sup, where P(X) is the support of itemset X and min_sup is a user-defined minimum support threshold. By definition of support, we know that if an itemset Y is merged with X, the resulting itemset X∪Y cannot occur more frequently than X. Therefore, since i is a superset of X, then i must be infrequent which contradicts with the known fact. It concludes that our assumption is wrong and thus all nonempty subsets of a frequent itemset must be frequent as well.

2.(b) Prove that the support of any nonempty subset s’ of itemset s must be at least as great as the support of s.

Let s be an itemset with support P(s) = x, and s’ be a nonempty proper subset of s. If s occurs x times in the dataset, then each of the items of s must occur at least x times. Therefore, since s’ is a non-empty proper subset of s, s’ must also occur at least x times,

i.e., P(s’) is at least as large as P(s)

Note: according to new rules of Chegg India if any question contains more than two subparts then it need not to answer all subparts.

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