to Improvements APriori
ParkChenYu Algorithm Multistage Algorithm Approximate Algorithms Compacting Results
1
PCY Algorithm
x Hashbased improvement to APriori. x During Pass 1 of Apriori, most memory is idle. x Use that memory to keep counts of buckets into which pairs of items are hashed. x Gives extra condition that candidate pairs must satisfy on Pass 2.
2
Just the count, not the pairs themselves.
Picture of PCY
Item counts Hash table
Frequent items
Bitmap Counts of candidate pairs Pass 2 3
Pass 1
PCY Algorithm Before Pass 1 Organize Main Memory
x Space to count each item.
One (typically) 4byte integer per item.
x Use the rest of the space for as many integers, representing buckets, as we can.
4
PCY Algorithm Pass 1
FOR (each basket) { FOR (each item) add 1 to item's count; FOR (each pair of items) { hash the pair to a bucket; add 1 to the count for that bucket } }
5
Observations About Buckets
1. If a bucket contains a frequent pair, then the bucket is surely frequent.
1. Even without any frequent pair, a bucket can be frequent.
Again, nothing in the bucket can be eliminated.
6
We cannot use the hash table to eliminate any member of this bucket.
Observations (2)
3. But in the best case, the count for a bucket is less than the support s.
Now, all pairs that hash to this bucket can be eliminated as candidates, even if the pair consists of two frequent items.
7
PCY Algorithm Between Passes
x Replace the buckets by a bitvector:
1 means the bucket count exceeds the support s (a frequent bucket ); 0 means it did not.
x 4byte integers are replaced by bits, so the bitvector requires 1/32 of memory. x Also, decide which items are frequent and list them for the second pass.
8
PCY Algorithm Pass 2
x Count all pairs {i, j } that meet the conditions:
x Notice all these conditions are necessary for the pair to have a chance of being frequent.
9
1. Both i and j are frequent items. 2. The pair {i, j }, hashes to a bucket number whose bit in the bit vector is 1.
Memory Details
x Hash table requires buckets of 24 bytes.
x On second pass, a table of (item, item, count) triples is essential.
Number of buckets thus almost 1/41/2 of the number of bytes of main memory.
Thus, hash table must eliminate 2/3 of the candidate pairs to beat apriori.
10
Multistage Algorithm
x Key idea: After Pass 1 of PCY, rehash only those pairs that qualify for Pass 2 of PCY. x On middle pass, fewer pairs contribute to buckets, so fewer false positives frequent buckets with no frequent pair.
11
Multistage Picture
Item counts Freq. items Freq. items
First hash table
Bitmap 1 Second hash table
Bitmap 1 Bitmap 2 Counts of candidate pairs Pass 3 12
Pass 1
Pass 2
Multistage Pass 3
x Count only those pairs {i, j } that satisfy:
1. Both i and j are frequent items. 2. Using the first hash function, the pair hashes to a bucket whose bit in the first bitvector is 1. 3. Using the second hash function, the pair hashes to a bucket whose bit in the second bitvector is 1.
13
Important Points
1. The two hash functions have to be independent. 2. We need to check both hashes on the third pass.
If not, we would wind up counting pairs of frequent items that hashed first to an infrequent bucket but happened to hash second to a frequent bucket.
14
Multihash
x Key idea: use several independent hash tables on the first pass. x Risk: halving the number of buckets doubles the average count. We have to be sure most buckets will still not reach count s. x If so, we can get a benefit like multistage, but in only 2 passes.
15
Multihash Picture
Item counts Freq. items
First hash table Second hash table Pass 1
Bitmap 1 Bitmap 2 Counts of candidate pairs Pass 2 16
Extensions
x Either multistage or multihash can use more than two hash functions. x In multistage, there is a point of diminishing returns, since the bitvectors eventually consume all of main memory. x For multihash, the bitvectors occupy exactly what one PCY bitmap does, but too many hash functions makes all counts > s.
17
All (Or Most) Frequent Itemsets < In 2 Passes
x Simple algorithm. x SON (Savasere, Omiecinski, and Navathe). x Toivonen.
18
Simple Algorithm (1)
x Take a random sample of the market baskets. x Run apriori or one of its improvements (for sets of all sizes, not just pairs) in main memory, so you don't pay for disk I/O each time you increase the size of itemsets.
Be sure you leave enough space for counts.
19
MainMemory Picture
Copy of sample baskets Space for counts
20
Simple Algorithm (2)
x Use as your support threshold a suitable, scaledback number.
E.g., if your sample is 1/100 of the baskets, use s /100 as your support threshold instead of s .
21
Simple Algorithm Option
x Optionally, verify that your guesses are truly frequent in the entire data set by a second pass. x But you don't catch sets frequent in the whole but not in the sample.
But requires more space.
Smaller threshold, e.g., s /125, helps catch more truly frequent itemsets.
22
SON Algorithm (1)
x Repeatedly read small subsets of the baskets into main memory and perform the first pass of the simple algorithm on each subset. x An itemset becomes a candidate if it is found to be frequent in any one or more subsets of the baskets.
23
SON Algorithm (2)
x On a second pass, count all the candidate itemsets and determine which are frequent in the entire set. x Key "monotonicity" idea: an itemset cannot be frequent in the entire set of baskets unless it is frequent in at least one subset.
24
SON Algorithm Distributed Version
x This idea lends itself to distributed data mining. x If baskets are distributed among many nodes, compute frequent itemsets at each node, then distribute the candidates from each node. x Finally, accumulate the counts of all candidates.
25
Toivonen's Algorithm (1)
x Start as in the simple algorithm, but lower the threshold slightly for the sample.
Example: if the sample is 1% of the baskets, use s /125 as the support threshold rather than s /100. Goal is to avoid missing any itemset that is frequent in the full set of baskets.
26
Toivonen's Algorithm (2)
x Add to the itemsets that are frequent in the sample the negative border of these itemsets. x An itemset is in the negative border if it is not deemed frequent in the sample, but all its immediate subsets are.
27
Example: Negative Border
x ABCD is in the negative border if and only if it is not frequent, but all of ABC, BCD, ACD, and ABD are.
28
Picture of Negative Border
Negative Border ... tripletons doubletons singletons Frequent Itemsets
29
Toivonen's Algorithm (3)
x In a second pass, count all candidate frequent itemsets from the first pass, and also count their negative border. x If no itemset from the negative border turns out to be frequent, then the candidates found to be frequent in the whole data are exactly the frequent itemsets.
30
Toivonen's Algorithm (4)
x What if we find that something in the negative border is actually frequent? x We must start over again! x Try to choose the support threshold so the probability of failure is low, while the number of itemsets checked on the second pass fits in mainmemory.
31
Theorem:
x If there is an itemset that is frequent in the whole, but not frequent in the sample, then there is a member of the negative border for the sample that is frequent in the whole.
32
Proof:
x Suppose not; i.e., there is an itemset S frequent in the whole but
Not frequent in the sample, and Not present in the sample's negative border.
x Let T be a smallest subset of S that is not frequent in the sample. x T is frequent in the whole (S is frequent, monotonicity). x T is in the negative border (else not "smallest").
33
Compacting the Output
1. Maximal Frequent itemsets : no immediate superset is frequent. 2. Closed itemsets : no immediate superset has the same count (> 0).
Stores not only frequent information, but exact counts.
34
Example: Maximal/Closed
Count Maximal (s=3) Closed A 4 No No B 5 No Yes C 3 No No AB 4 Yes Yes AC 2 No No BC 3 Yes Yes ABC 2 No Yes
35