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05-09-2014, 04:59 PM
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APRIORI Algorithm SEMINAR REPORT

APRIORI Algorithm




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The Apriori Algorithm: Basics



The Apriori Algorithm is an influential algorithm for
mining frequent itemsets for boolean association rules.
Key Concepts :
• Frequent Itemsets: The sets of item which has minimum
support (denoted by Li for ith-Itemset).
• Apriori Property: Any subset of frequent itemset must be
frequent.
• Join Operation: To find L
k , a set of candidate k-itemsets
is generated by joining Lk-1 with itself.


The Apriori Algorithm in a
Nutshell



• Find the frequent itemsets: the sets of items that have
minimum support
– A subset of a frequent itemset must also be a
frequent itemset
• i.e., if {AB} is a frequent itemset, both {
A} and {
B}
should be a frequent itemset
– Iteratively find frequent itemsets with cardinality
from 1 to k (k-itemset)
• Use the frequent itemsets to generate association rules.

Generating 2-itemset Frequent Pattern


• 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 haven’t used Apriori Property yet.Step 3: Generating 3-itemset Frequent Pattern
Itemset
{I1, I2, I3}
{I1, I2, I5}
Itemset Sup.
Count
{I1, I2, I3} 2
{I1, I2, I5} 2
Itemset Sup
Count
{I1, I2, I3} 2
{I1, I2, I5} 2
C3 C3 L3
Scan D for
count of
each
candidate
C


Generating 3-itemset Frequent Pattern[/b]



• 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.



Methods to Improve Apriori’s Efficiency



• Hash-based itemset counting: A
k-itemset whose corresponding
hashing bucket count is below the threshold cannot be frequent.
• Transaction reduction: A transaction that does not contain any
frequent k-itemset is useless in subsequent scans.
• Partitioning: Any itemset that is potentially frequent in DB must be
frequent in at least one of the partitions of DB.
• Sampling: mining on a subset of given data, lower support threshold
+ a method to determine the completeness.
• Dynamic itemset counting: add new candidate itemsets only when
all of their subsets are estimated to be frequent.



Why Frequent Pattern Growth Fast ?


• Performance study shows
– FP-growth is an order of magnitude faster than Apriori,
and is also faster than tree-projection
• Reasoning
– No candidate generation, no candidate test
– Use compact data structure
– Eliminate repeated database scan
– Basic operation is counting and FP-tree building
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