Reducing the collection of itemsets: alternative representations and combinatorial problemsToo many frequent itemsets • If {a1, …, a100} is a frequent itemset, then there are 1.27*1030 frequent sub-patterns! • There should be some more condensed way to describe the data 1210010021001100100Frequent itemsets maybe too many to be helpful • If there are many and large frequent itemsets enumerating all of them is costly. • We may be interested in finding the boundary frequent patterns. • Question: Is there a good definition of such boundary?all items empty set Frequent itemsets Non-frequent itemsets borderBorders of frequent itemsets • Itemset X is more specific than itemset Y if X superset of Y (notation: Y<X). Also, Y is more general than X (notation: X>Y) • The Border: Let S be a collection of frequent itemsets and P the lattice of itemsets. The border Bd(S) of S consists of all itemsets X such that all more general itemsets than X are in S and no pattern more specific than X is in S. then with allfor and , then with allfor )(SWWXPWSYXYPYPXSBdPositive and negative border • Border • Positive border: Itemsets in the border that are also frequent (belong in S) • Negative border: Itemsets in the border that are not frequent (do not belong in S) then with allfor and , then with allfor )(SWWXPWSYXYPYPXSBd then with allfor )( SYYXPYSXSBd then with allfor \)( SYXYPYSPXSBd Examples with borders • Consider a set of items from the alphabet: {A,B,C,D,E} and the collection of frequent sets S = {{A},{B},{C},{E},{A,B},{A,C},{A,E},{C,E},{A,C,E}} • The negative border of collection S is Bd-(S) = {{D},{B,C},{B,E}} • The positive border of collection S is Bd+(S) = {{A,B},{A,C,E}}Descriptive power of the borders • Claim: A collection of frequent sets S can be fully described using only the positive border (Bd+(S)) or only the negative border (Bd-(S)).Maximal patterns Frequent patterns without proper frequent super patternMaximal Frequent Itemset nullAB AC AD AE BC BD BE CD CE DEA B C D EABC ABD ABE ACD ACE ADE BCD BCE BDE CDEABCD ABCE ABDE ACDE BCDEABCDEBorder Infrequent Itemsets Maximal Itemsets An itemset is maximal frequent if none of its immediate supersets is frequentMaximal patterns • The set of maximal patterns is the same as the positive border • Descriptive power of maximal patterns: – Knowing the set of all maximal patterns allows us to reconstruct the set of all frequent itemsets!! – We can only reconstruct the set not the actual frequenciesClosed patterns • An itemset is closed if none of its immediate supersets has the same support as the itemset TID Items1 {A,B}2 {B,C,D}3 {A,B,C,D}4 {A,B,D}5 {A,B,C,D}Itemset Support{A} 4{B} 5{C} 3{D} 4{A,B} 4{A,C} 2{A,D} 3{B,C} 3{B,D} 4{C,D} 3Itemset Support{A,B,C} 2{A,B,D} 3{A,C,D} 2{B,C,D} 3{A,B,C,D} 2Maximal vs Closed Itemsets TID Items1 ABC2 ABCD3 BCE4 ACDE5 DEnullAB AC AD AE BC BD BE CD CE DEA B C D EABC ABD ABE ACD ACE ADE BCD BCE BDE CDEABCD ABCE ABDE ACDE BCDEABCDE124 1231234245 34512 124 24412323243445122244 423 424Transaction Ids Not supported by any transactionsMaximal vs Closed Frequent Itemsets nullAB AC AD AE BC BD BE CD CE DEA B C D EABC ABD ABE ACD ACE ADE BCD BCE BDE CDEABCD ABCE ABDE ACDE BCDEABCDE124 1231234245 34512 124 24412323243445122244 423 424Minimum support = 2 # Closed = 9 # Maximal = 4 Closed and maximal Closed but not maximalWhy are closed patterns interesting? • s({A,B}) = s(A), i.e., conf({A}{B}) = 1 • We can infer that for every itemset X , s(A U {X}) = s({A,B} U X) • No need to count the frequencies of sets X U {A,B} from the database! • If there are lots of rules with confidence 1, then a significant amount of work can be saved – Very useful if there are strong correlations between the items and when the transactions in the database are similarWhy closed patterns are interesting? • Closed patterns and their frequencies alone are sufficient representation for all the frequencies of all frequent patterns • Proof: Assume a frequent itemset X: – X is closed s(X) is known – X is not closed s(X) = max {s(Y) | Y is closed and X subset of Y}Maximal vs Closed sets • Knowing all maximal patterns (and their frequencies) allows us to reconstruct the set of frequent patterns • Knowing all closed patterns and their frequencies allows us to reconstruct the set of all frequent patterns and their frequencies FrequentItemsetsClosedFrequentItemsetsMaximalFrequentItemsetsA more algorithmic approach to reducing the collection of frequent itemsetsPrototype problems: Covering problems • Setting: – Universe of N elements U = {U1,…,UN} – A set of n sets S = {s1,…,sn} – Find a collection C of sets in S (C subset of S) such that UcєCc contains many elements from U • Example: – U: set of documents in a collection – si: set of documents that contain term ti – Find a collection of terms that cover most of the documentsPrototype covering problems • Set cover problem: Find a small collection C of sets from S such that all elements in the universe U are covered by some set in C • Best collection problem: find a collection C of k sets from S such that the collection covers as many elements from the universe U as possible • Both problems are NP-hard • Simple approximation algorithms with provable properties are available and very useful in practiceSet-cover problem • Universe of N elements U = {U1,…,UN} • A set of n sets S = {s1,…,sn} such that Uisi =U • Question: Find the smallest number of sets from S to form collection C (C subset of S) such that UcєCc=U • The set-cover problem is NP-hard (what does this mean?)Trivial algorithm • Try all subcollections of S • Select the smallest one that covers all the elements in U • The running time of the trivial algorithm is O(2|S||U|) • This is way too slowGreedy algorithm for set cover • Select first the largest-cardinality set s from S • Remove the elements from s from U • Recompute the sizes of the remaining sets in S • Go back to the first stepAs an algorithm • X = U • C = {} • while X is not empty do – For all sєS let as=|s intersection X| – Let s be such that as is maximal – C = C U {s} – X = X\ sHow can this go wrong? • No global consideration of how good or bad a selected set is going to beHow good is the greedy algorithm? • Consider a
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