1 IncSpan :Incremental Mining of Sequential Patterns in Large Database Hong Cheng, Xifeng Yan, Jiawei Han Proc Int. Conf. on Knowledge Discovery and Data Mining (KDD'04) Advisor ： Jia-Ling Koh Speaker ： Chun-Wei Hsieh 02/25/2005

2 Problem Databases are updated incrementally. (Customer shopping transaction sequences, Weather sequences and patient treatment sequences) Two kinds of database updates (1) INSERT :inserting new sequences (New customers) (2) APPEND: appending new itemsets/items to the existing sequences (newly purchased items for existing customers)

3 The property of updates : INSERT : If a sequence is infrequent in both and,it cannot be frequent in APPEND: Even if a sequence is infrequent in both and,it might be frequent in When the database is updated with a combination of INSERT and APPEND, we can treat INSERT as a special case of APPEND – treating the inserted sequences as appended transactions to an empty sequence in the original database.

4 Examples : Examples in INSERT and APPEND database

5 Preliminary Concepts An original sequence database An appended sequence database Min_sup: a minimum support threshold FS: the set of frequent sequential pattern Buffer ratio : SFS: the set of semi-frequent sequential pattern The problem of incremental sequential pattern mining is to mine the set of frequent subsequences FS ’ in based on FS instead of mining on from scratch.

6 Buffering Semi-frequent Patterns When the database is updated to, there are several possibilities: 1. A pattern which is frequent in is still frequent in 2. A pattern which is semi-frequent in becomes frequent in 3. A pattern which is semi-frequent in is still semi-frequent in 4. Appended database brings new items. 5. A pattern which is infrequent in becomes frequent in 6. A pattern which is infrequent in becomes semi-frequent in Case (1) – (3) are trivial cases

7 Case (4): Appended database brings new items. It does not appear in Property: An item which does not appear in and is brought by has no information in FS or SFS. Solution: Scan the database LDB for single items. Then use the new frequent item as prefix to construct projected database and discover frequent and semi-frequent sequences recursively.

8 LDB and ODB LDB is the set of sequences in DB ’ which are appended with items/itemsets. ODB is the set of sequences in DB which are appended with items/itemsets in DB ’. LDB ODB

9 Case (4):examples (c) Min_sup=3 u=0.6

10 Case (5): A pattern which is infrequent in becomes frequent in Property: If an infrequent sequence p ’ in becomes frequent in, all of its prefix subsequences must also be frequent in. Solution: Start from its frequent prefix p in FS and construct p-projected database, we will discover p ’. A sequence p ’ which changes from infrequent to frequent must have sup(p ’ ) > (1 - )*min_sup. If supLDB(p) < (1 - )*min_sup, we can safely prune search with prefix p.

11 Case (5):examples (a,c) Min_sup=3 u=0.6

12 Case (5):theorem For a frequent pattern p, if its support in LDB supLDB(p) < (1 - )*min_sup, then there is no sequence p’ having p as prefix changing from infrequent in to frequent in Proof : p’ was infrequent in, so sup (p’) < *min_sup (1) If supLDB(p) < (1 - )*min_sup, then supLDB(p’ ) supLDB(p) < (1 - )*min_sup Since supLDB(p’ ) = supODB(p’ ) + sup(p’ ). Then we have sup(p’ ) supLDB(p’ ) < (1 - )*min_sup.(2) Since sup (p’ ) = sup (p’) + sup(p’), combining (1)and (2), we have sup (p’) < min_sup. So p’ cannot be frequent in

13 Case (6): A pattern which is infrequent in becomes semi-frequent in Property: If an infrequent sequence p ’ becomes semifrequent in, all of its prefix subsequences must be either frequent or semi-frequent. Solution: Start from its prefix p in FS or SFS and construct p-projected database, we will discover p ’

14 Case (6):examples (be) Min_sup=3 u=0.6

15 IncSpan Step 1: Scan LDB for single items, as shown in case (4). Step 2: Check every pattern in FS and SFS in LDB to adjust the support of those patterns. Step 2.1: If a pattern becomes frequent, add it to FS ’. Then check whether it meets the projection condition. If so,use it as prefix to project database, as shown in case (5). Step 2.2: If a pattern is semi-frequent, add it to SFS ’.

16 Algorithm

17 Reverse Pattern Matching Since the appended items are always at the end part of the original sequence, reverse pattern matching would be more efficient than projection from the front If the last item of p is not supported by sa, we can prune searching. If the last item of p is supported by sa, we have to check whether s ’ supports p. If p is not supported by s ’, we can prune searching and keep sup(p) unchanged. Otherwise we have to check whether s supports p. If s supports p, keep sup(p) unchanged; otherwise, increase sup(p) by 1.

18 Shared Projection when we detect some subsequence that needs projecting database, we do not do the projection immediately. Instead we label it. After finishing checking and labeling all the sequences, we do the projection by traversing the sequential pattern tree. DB ’ |

19 Experiment (a) varying min sup(b) varying percentage of updated sequences

20 Experiment (a) varying buffer ratio(c) Memory Usage under varied min sup

21 Experiment (b) multiple increments of database (c) varying # of sequences (in 1000) in DB