Mid 2 revision and Asssociation Rules II

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Mid 2 revision and Asssociation Rules (II)
Lecture 16

• Prof. Sin-Min Lee
• Department of Computer Science

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Example Data Instance
STUDENT
COURSE
Takes
PROFESSOR
Teaches
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Natural Join and Intersection

• Natural join special case of join where ? is
  implicit attributes with same name must be
  equal
• STUDENT ? Takes STUDENT ?STUDENT.sid
  Takes.sid Takes
• Intersection as with set operations, derivable
  from difference

A ? B
B-A
A-B
A B
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Division

• A somewhat messy operation that can be expressed
  in terms of the operations we have already
  defined
• Used to express queries such as The fid's of
  faculty who have taught all subjects
• Paraphrased The fids of professors for which
  there does not exist a subject that they havent
  taught

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Division Using Our Existing Operators

• All possible teaching assignments Allpairs
• NotTaught, all (fid,subj) pairs for which
  professor fid has not taught subj
• Answer is all faculty not in NotTaught

?fid,subj (PROFESSORx ?subj(COURSE))
Allpairs - ?fid,subj(Teaches ? COURSE)
?fid(PROFESSOR) - ?fid(NotTaught)
?fid(PROFESSOR) - ?fid(
?fid,subj (PROFESSOR x ?subj(COURSE)) -
?fid,subj(Teaches ? COURSE))
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Division R1 R2

• Requirement schema(R1) ¾ schema(R2)
• Result schema schema(R1) schema(R2)
• Professors who have taught all courses
• What about Courses that have been taught by all
  faculty?

?fid (?fid,subj(Teaches ? COURSE) ?subj(COURSE))
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Division

• Goal Produce the tuples in one relation, r, that
  match all tuples in another relation, s
• r (A1, An, B1, Bm)
• s (B1 Bm)
• r/s, with attributes A1, An, is the set of all
  tuples ltagt such that for every tuple ltbgt in s,
  lta,bgt is in r
• Can be expressed in terms of projection, set
  difference, and cross-product

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Division (contd)
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Division - Example

• List the Ids of students who have passed all
  courses that were taught in spring 2000
• Numerator
• StudId and CrsCode for every course passed by
  every student
• ?StudId, CrsCode (?Grade? F (Transcript) )
• Denominator
• CrsCode of all courses taught in spring 2000
• ?CrsCode (?SemesterS2000 (Teaching) )
• Result is numerator/denominator

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Relational Calculus

• Important features
• Declarative formal query languages for relational
  model
• Based on the branch mathematical logic known as
  predicate calculus
• Two types of RC
• 1) tuple relational calculus
• 2) domain relational calculus
• A single statement can be used to perform a query

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Tuple Relational Calculus

• based on specifying a number of tuple variables
• a tuple variable refers to any tuple

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Generic Form

• t COND (t)
• where
• t is a tuple variable and
• COND(t) is Boolean expression involving t

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Simple example 1

• To find all employees whose salary is greater
  than 50,000
• t EMPLOYEE(t) and t.Salarygt5000
• where
• EMPLOYEE(t) specifies the range of tuple variable
  t
• The above operation selects all the attributes

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Simple example 2

• To find only the names of employees whose salary
  is greater than 50,000
• t.FNAME, t.NAME EMPLOYEE(t) and t.Salarygt5000
• The above is equivalent to
• SELECT T.FNAME, T.LNAME
• FROM EMPLOYEE T
• WHERE T.SALARY gt 5000

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Elements of a tuple calculus

• In general, we need to specify the following in a
  tuple calculus expression
• Range Relation (I.e, R(t)) FROM
• Selected combination WHERE
• Requested attributes SELECT

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More ExampleQ0

• Retrieve the birthrate and address of the
  employee(s) whose name is John B. Smith
• t.BDATE, t.ADDRESS EMPLOYEE(t) AND
  t.FNAMEJohn AND t.MINITB AND
  t.LNAMESmith

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Formal Specification of tuple Relational Calculus

• A general format
• t1.A1, t2.A2,,tn.An COND ( t1 ,t2 ,, tn,
  tn1, tn2,,tnm)
• where
• t1,,tnm are tuple var
• Ai attribute?R(ti)
• COND (formula)
• Where COND corresponds to statement about the
  world, which can be True or False

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Elements of formula

• A formula is made of Predicate Calculus atoms
• an atom of the from R(ti)
• ti.A op tj.B op?, lt,gt,..
• F1 And F2 where F1 and F2 are formulas
• F1 OR F2
• Not (F1)
• F(?t) (F) or F (?t) (F)
• ? Y friends (Y, John)
• ?X likes(X, ICE_CREAM)

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Example Queries Using the Existential Quantifier

• Retrieve the name and address of all employees
  who work for the Research department
• t.FNAME, t.LNAME, t.ADDRESS EMPLOYEE(t) AND (?
  d) (DEPARTMENT (d) AND d.DNAMEResearch AND
  d.DNUMBERt.DNO)

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More Example

• For every project located in Stafford, retrieve
  the project number, the controlling department
  number, and the last name, birthrate, and address
  of the manger of that department.

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Cont.

• p.PNUMBER,p.DNUM,m.LNAME,m.BDATE,
  m.ADDRESSPROJECT(p) and EMPLOYEE(M) and
  P.PLOCATIONStafford and (? d) (DEPARTMENT(D)
  AND P.DNUMd.DNUMBER and d.MGRSSNm.SSN))

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Logical Equivalences

• There are two logical equivalences that will be
  heavily used
• p?q ? ?p ? q (Whenever p is true, q must also
  be true.)
• ?x. p(x) ? ??x. ?p(x) (p is true for all x)
• The second can be a lot easier to check!

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Normalization

• Review on Keys
• superkey a set of attributes which will uniquely
  identify each tuple in a relation
• candidate key a minimal superkey
• primary key a chosen candidate key
• secondary key all the rest of candiate keys
• prime attribute an attribute that is a part of a
  candidate key (key column)
• nonprime attribute a nonkey column

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Normalization

• Functional Dependency Type by Keys
• whole (candidate) key ? nonprime attribute
  full FD (no violation)
• partial key ? nonprime attribute partial FD
  (violation of 2NF)
• nonprime attribute ? nonprime attribute
  transitive FD (violation of 3NF)
• not a whole key ? prime attribute violation of
  BCNF

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Functional Dependencies

• Let R be a relation schema
• ? ? R and ? ? R
• The functional dependency
• ? ? ?holds on R iff for any legal relations
  r(R), whenever two tuples t1 and t2 of r have
  same values for ?, they have same values for ?.
  
• t1? t2 ? ? t1? t2 ?
• On this instance, A ? B does NOT hold, but B ? A
  does hold.

A B

• 4
• 1 5
• 3 7

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1. Closure

• Given a set of functional dependencies, F, its
  closure, F , is all FDs that are implied by FDs
  in F.
• e.g. If A ? B, and B ? C,
• then clearly A ? C

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Armstrongs Axioms

• We can find F by applying Armstrongs Axioms
• if ? ? ?, then ? ? ?
  (reflexivity)
• if ? ? ?, then ? ? ? ? ?
  (augmentation)
• if ? ? ?, and ? ? ?, then ? ? ? (transitivity)
• These rules are
• sound (generate only functional dependencies that
  actually hold) and
• complete (generate all functional dependencies
  that hold).

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Additional rules

• If ? ? ? and ? ? ?, then ? ? ? ? (union)
• If ? ? ? ?, then ? ? ? and ? ? ? (decomposition)
• If ? ? ? and ? ? ? ?, then ? ? ? ?
  (pseudotransitivity)
• The above rules can be inferred from Armstrongs
  axioms.

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Example

• R (A, B, C, G, H, I)F A ? B A ? C CG
  ? H CG ? I B ? H
• Some members of F
• A ? H
• by transitivity from A ? B and B ? H
• AG ? I
• by augmenting A ? C with G, to get AG ? CG
  and then transitivity with CG ? I
• CG ? HI
• by augmenting CG ? I to infer CG ? CGI,
• and augmenting of CG ? H to infer CGI ? HI,
• and then transitivity

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2. Closure of an attribute set

• Given a set of attributes A and a set of FDs F,
  closure of A under F is the set of all attributes
  implied by A
• In other words, the largest B such that
• A ? B
• Redefining super keys
• The closure of a super key is the entire
  relation schema
• Redefining candidate keys
• 1. It is a super key
• 2. No subset of it is a super key

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Computing the closure for A

• Simple algorithm
• 1. Start with B A.
• 2. Go over all functional dependencies, ? ? ? ,
  in F
• 3. If ? ? B, then
• Add ? to B
• 4. Repeat till B changes

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Example

• R (A, B, C, G, H, I)F A ? B A ? C CG
  ? H CG ? I B ? H
• (AG) ?
• 1. result AG
• 2. result ABCG (A ? C and A ? B)
• 3. result ABCGH (CG ? H and CG ? AGBC)
• 4. result ABCGHI (CG ? I and CG ? AGBCH
• Is (AG) a candidate key ?
• 1. It is a super key.
• 2. (A) BC, (G) G.
• YES.

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Support

• Simplest question find sets of items that appear
  frequently in the baskets.
• Support for itemset I the number of baskets
  containing all items in I.
• Given a support threshold s, sets of items that
  appear in gt s baskets are called frequent
  itemsets.

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Example

• Itemsmilk, coke, pepsi, beer, juice.
• Support 3 baskets.
• B1 m, c, b B2 m, p, j B3 m, b
• B4 c, j B5 m, p, b B6 m,
  c, b, j
• B7 c, b, j B8 b, c
• Frequent itemsets m, c, b, j, m,
  b, c, b, j, c.

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Association Rules

• Association rule R Itemset1 gt Itemset2
• Itemset1, 2 are disjoint and Itemset2 is
  non-empty
• meaning if transaction includes Itemset1 then
  it also has Itemset2
• Examples
• A,B gt E,C
• A gt B,C

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Example

• B1 m, c, b B2 m, p, j
• B3 m, b B4 c, j
• B5 m, p, b B6 m, c, b, j
• B7 c, b, j B8 b, c
• An association rule m, b ? c.
• Confidence 2/4 50.

_ _

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From Frequent Itemsets to Association Rules

• Q Given frequent set A,B,E, what are possible
  association rules?
• A gt B, E
• A, B gt E
• A, E gt B
• B gt A, E
• B, E gt A
• E gt A, B
• __ gt A,B,E (empty rule), or true gt A,B,E

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Classification vs Association Rules

• Classification Rules
• Focus on one target field
• Specify class in all cases
• Measures Accuracy

• Association Rules
• Many target fields
• Applicable in some cases
• Measures Support, Confidence, Lift

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• Apriori Algorithm (1997)
• Let I a,b,c, be a set of all items in the
  domain
• Let T S S ? I be a bag of all transaction
  records of item sets
• Let support(S) ? A A ? T ? S ? A
• Let L1 a a ? I ? support(a) ?
  minSupport
• ?k (k gt 1 ? Lk-1 ? ?) Let
• Lk Si ? Sj (Si ? Lk-1) ? (Sj ? Lk-1) ?
• ( Si Sj 1 ) ? (
  Sj Si 1) ?
• ( ?S ((S ? Si ? Sj) ?
  (S k-1)) ? S ? Lk-1 ) ?
• ( support(Si ? Sj) ?
  minSupport )
• The set of all frequent item sets is given by
• L ?Lk
• and the set of all association rules is
  given by
• R A ? C A ? ?(Lk) ? (C Lk A) ? (A ?
  ?) ? (C ? ?)
• support(Lk) /
  support(A) ? minConfidence

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Rule Support and Confidence

• Suppose R I gt J is an association rule
• sup (R) sup (I ? J) is the support count
• support of itemset I ? J (I or J)
• conf (R) sup(J) / sup(R) is the confidence of R
• fraction of transactions with I ? J that have J
• Association rules with minimum support and count
  are sometimes called strong rules

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Association Rules Example

• Q Given frequent set A,B,E, what association
  rules have minsup 2 and minconf 50 ?
• A, B gt E conf2/4 50
• A, E gt B conf2/2 100
• B, E gt A conf2/2 100
• E gt A, B conf2/2 100
• Dont qualify
• A gtB, E conf2/6 33lt 50
• B gt A, E conf2/7 28 lt 50
• __ gt A,B,E conf 2/9 22 lt 50

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Find Strong Association Rules

• A rule has the parameters minsup and minconf
• sup(R) gt minsup and conf (R) gt minconf
• Problem
• Find all association rules with given minsup and
  minconf
• First, find all frequent itemsets

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Finding Frequent Itemsets

• Start by finding one-item sets (easy)
• Q How?
• A Simply count the frequencies of all items

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Finding itemsets next level

• Apriori algorithm (Agrawal Srikant)
• Idea use one-item sets to generate two-item
  sets, two-item sets to generate three-item sets,
  
• If (A B) is a frequent item set, then (A) and (B)
  have to be frequent item sets as well!
• In general if X is frequent k-item set, then all
  (k-1)-item subsets of X are also frequent
• Compute k-item set by merging (k-1)-item sets

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Finding Association Rules

• A typical question find all association rules
  with support s and confidence c.
• Note support of an association rule is the
  support of the set of items it mentions.
• Hard part finding the high-support (frequent )
  itemsets.
• Checking the confidence of association rules
  involving those sets is relatively easy.

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Naïve Algorithm

• A simple way to find frequent pairs is
• Read file once, counting in main memory the
  occurrences of each pair.
• Expand each basket of n items into its n (n
  -1)/2 pairs.
• Fails if items-squared exceeds main memory.

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Filter
Filter
Construct
Construct
C1
L1
C2
L2
C3
First pass
Second pass
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Agrawal, Srikant 94
Fast Algorithms for Mining Association Rules, by
Rakesh Agrawal and Ramakrishan Sikant, IBM
Almaden Research Center
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C1
L1
Database
C2
L2
C2
C3
L3
C3
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• Dynamic Programming Approach
• Want proof of principle of optimality and
  overlapping subproblems
• Principle of Optimality
• The optimal solution to Lk includes the
  optimal solution of Lk-1
• Proof by contradiction
• Overlapping Subproblems
• Lemma of every subset of a frequent item set is a
  frequent item set
• Proof by contradiction

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The Apriori Algorithm Example

• Consider a database, D , consisting of 9
  transactions.
• Suppose min. support count required is 2 (i.e.
  min_sup 2/9 22 )
• Let minimum confidence required is 70.
• We have to first find out the frequent itemset
  using Apriori algorithm.
• Then, Association rules will be generated using
  min. support min. confidence.

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Step 1 Generating 1-itemset Frequent Pattern
Compare candidate support count with minimum
support count
Scan D for count of each candidate
C1
L1

• In the first iteration of the algorithm, each
  item is a member of the set of candidate.
• The set of frequent 1-itemsets, L1 , consists of
  the candidate 1-itemsets satisfying minimum
  support.

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Step 2 Generating 2-itemset Frequent Pattern
Generate C2 candidates from L1
Compare candidate support count with minimum
support count
Scan D for count of each candidate
L2
C2
C2
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Step 2 Generating 2-itemset Frequent Pattern
Cont.

• To discover the set of frequent 2-itemsets, L2 ,
  the algorithm uses L1 Join L1 to generate a
  candidate set of 2-itemsets, C2.
• Next, the transactions in D are scanned and the
  support count for each candidate itemset in C2 is
  accumulated (as shown in the middle table).
• The set of frequent 2-itemsets, L2 , is then
  determined, consisting of those candidate
  2-itemsets in C2 having minimum support.
• Note We havent used Apriori Property yet.

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Step 3 Generating 3-itemset Frequent Pattern
Compare candidate support count with min support
count
Scan D for count of each candidate
Scan D for count of each candidate
C3
L3
C3

• The generation of the set of candidate
  3-itemsets, C3 , involves use of the Apriori
  Property.
• In order to find C3, we compute L2 Join L2.
• C3 L2 Join L2 I1, I2, I3, I1, I2, I5,
  I1, I3, I5, I2, I3, I4, I2, I3, I5, I2,
  I4, I5.
• Now, Join step is complete and Prune step will
  be used to reduce the size of C3. Prune step
  helps to avoid heavy computation due to large Ck.

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Step 3 Generating 3-itemset Frequent Pattern
Cont.

• Based on the Apriori property that all subsets of
  a frequent itemset must also be frequent, we can
  determine that four latter candidates cannot
  possibly be frequent. How ?
• For example , lets take I1, I2, I3. The 2-item
  subsets of it are I1, I2, I1, I3 I2, I3.
  Since all 2-item subsets of I1, I2, I3 are
  members of L2, We will keep I1, I2, I3 in C3.
• Lets take another example of I2, I3, I5 which
  shows how the pruning is performed. The 2-item
  subsets are I2, I3, I2, I5 I3,I5.
• BUT, I3, I5 is not a member of L2 and hence it
  is not frequent violating Apriori Property. Thus
  We will have to remove I2, I3, I5 from C3.
• Therefore, C3 I1, I2, I3, I1, I2, I5
  after checking for all members of result of Join
  operation for Pruning.
• Now, the transactions in D are scanned in order
  to determine L3, consisting of those candidates
  3-itemsets in C3 having minimum support.

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Step 4 Generating 4-itemset Frequent Pattern

• The algorithm uses L3 Join L3 to generate a
  candidate set of 4-itemsets, C4. Although the
  join results in I1, I2, I3, I5, this itemset
  is pruned since its subset I2, I3, I5 is not
  frequent.
• Thus, C4 f , and algorithm terminates, having
  found all of the frequent items. This completes
  our Apriori Algorithm.
• Whats Next ?
• These frequent itemsets will be used to generate
  strong association rules ( where strong
  association rules satisfy both minimum support
  minimum confidence).

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Step 5 Generating Association Rules from
Frequent Itemsets

• Procedure
• For each frequent itemset l, generate all
  nonempty subsets of l.
• For every nonempty subset s of l, output the rule
  s ? (l-s) if
• support_count(l) / support_count(s) gt min_conf
  where min_conf is minimum confidence threshold.
• Back To Example
• We had L I1, I2, I3, I4, I5,
  I1,I2, I1,I3, I1,I5, I2,I3, I2,I4,
  I2,I5, I1,I2,I3, I1,I2,I5.
• Lets take l I1,I2,I5.
• Its all nonempty subsets are I1,I2, I1,I5,
  I2,I5, I1, I2, I5.

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Step 5 Generating Association Rules from
Frequent Itemsets Cont.

• Let minimum confidence threshold is , say 70.
• The resulting association rules are shown below,
  each listed with its confidence.
• R1 I1 I2 ? I5
• Confidence scI1,I2,I5/scI1,I2 2/4 50
• R1 is Rejected.
• R2 I1 I5 ? I2
• Confidence scI1,I2,I5/scI1,I5 2/2 100
• R2 is Selected.
• R3 I2 I5 ? I1
• Confidence scI1,I2,I5/scI2,I5 2/2 100
• R3 is Selected.

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Step 5 Generating Association Rules from
Frequent Itemsets Cont.

• R4 I1 ? I2 I5
• Confidence scI1,I2,I5/scI1 2/6 33
• R4 is Rejected.
• R5 I2 ? I1 I5
• Confidence scI1,I2,I5/I2 2/7 29
• R5 is Rejected.
• R6 I5 ? I1 I2
• Confidence scI1,I2,I5/ I5 2/2 100
• R6 is Selected.
• In this way, We have found three strong
  association rules.

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Example
Simple algorithm
ABCDE
Large itemset
ACDE?B
ABCE?D
Rules with minsup
CDE?AB
BCE?AD
ABE?CD
ADE?BC
ACD?BE
ACE?BD
ACE?BD
ABC?ED
ABCDE
Fast algorithm
ACDE?B
ABCE?D
ACE?BD
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