1. Indexing and Execution May 20 th, 2002

2. Indexing - Recap Primary file organization: heap, sorted, hashed. Primary vs. secondary Clustered vs. unclustered Dense vs. sparse Composite search key indexes. Not all combinations are possible. Next: B+-trees.

3. Tree-Based Indexes ``Find all students with gpa > 3.0’’ –If data is in sorted file, do binary search to find first such student, then scan to find others. –Cost of binary search can be quite high. Simple idea: Create an `index’ file. * Can do binary search on (smaller) index file! Page 1 Page 2 Page N Page 3 Data File k2 kN k1 Index File

4. Tree-Based Indexes (2) P 0 K 1 P 1 K 2 P 2 K m P m index entry 10*15*20*27*33*37* 40* 46* 51* 55* 63* 97* 2033 5163 40 Root

5. B+ Tree: The Most Widely Used Index Insert/delete at log F N cost; keep tree height- balanced. (F = fanout, N = # leaf pages) Minimum 50% occupancy (except for root). Each node contains d <= m <= 2d entries. The parameter d is called the order of the tree. Index Entries Data Entries Root

6. Example B+ Tree Search begins at root, and key comparisons direct it to a leaf. Search for 5*, 15*, all data entries >= 24*... 17 24 30 2* 3*5* 7*14*16* 19*20* 22*24*27* 29*33*34* 38* 39* 13

7. B+ Trees in Practice Typical order: 100. Typical fill-factor: 67%. –average fanout = 133 Typical capacities: –Height 4: 133 4 = 312,900,700 records –Height 3: 133 3 = 2,352,637 records Can often hold top levels in buffer pool: –Level 1 = 1 page = 8 Kbytes –Level 2 = 133 pages = 1 Mbyte –Level 3 = 17,689 pages = 133 MBytes

8. Inserting a Data Entry into a B+ Tree Find correct leaf L. Put data entry onto L. –If L has enough space, done! –Else, must split L (into L and a new node L2) Redistribute entries evenly, copy up middle key. Insert index entry pointing to L2 into parent of L. This can happen recursively –To split index node, redistribute entries evenly, but push up middle key. (Contrast with leaf splits.)

9. Insertion in a B+ Tree Insert (K, P) Find leaf where K belongs, insert If no overflow (2d keys or less), halt If overflow (2d+1 keys), split node, insert in parent: If leaf, keep K3 too in right node When root splits, new root has 1 key only K1K2K3K4K5 P0P1P2P3P4p5 K1K2 P0P1P2 K4K5 P3P4p5 (K3, ) to parent

10. Insertion in a B+ Tree 80 2060100120140 101518203040506065808590 101518203040506065808590 Insert K=19

11. Insertion in a B+ Tree 80 2060100120140 10151819203040506065808590 10151820304050606580859019 After insertion

12. Insertion in a B+ Tree 80 2060100120140 10151819203040506065808590 10151820304050606580859019 Now insert 25

13. Insertion in a B+ Tree 80 2060100120140 1015181920253040506065808590 10151820253040606580859019 After insertion 50

14. Insertion in a B+ Tree 80 2060100120140 1015181920253040506065808590 10151820253040606580859019 But now have to split ! 50

15. Insertion in a B+ Tree 80 203060100120140 1015181920256065808590 10151820253040606580859019 After the split 50 304050

16. Deleting a Data Entry from a B+ Tree Start at root, find leaf L where entry belongs. Remove the entry. –If L is at least half-full, done! –If L has only d-1 entries, Try to re-distribute, borrowing from sibling (adjacent node with same parent as L). If re-distribution fails, merge L and sibling. If merge occurred, must delete entry (pointing to L or sibling) from parent of L. Merge could propagate to root, decreasing height.

17. Deletion from a B+ Tree 80 203060100120140 1015181920256065808590 10151820253040606580859019 Delete 30 50 304050

18. Deletion from a B+ Tree 80 203060100120140 1015181920256065808590 101518202540606580859019 After deleting 30 50 4050 May change to 40, or not

19. Deletion from a B+ Tree 80 203060100120140 1015181920256065808590 101518202540606580859019 Now delete 25 50 4050

20. Deletion from a B+ Tree 80 203060100120140 10151819206065808590 1015182040606580859019 After deleting 25 Need to rebalance Rotate 50 4050

21. Deletion from a B+ Tree 80 193060100120140 10151819206065808590 1015182040606580859019 Now delete 40 50 4050

22. Deletion from a B+ Tree 80 193060100120140 10151819206065808590 10151820606580859019 After deleting 40 Rotation not possible Need to merge nodes 50

23. Deletion from a B+ Tree 80 1960100120140 1015181920506065808590 10151820606580859019 Final tree 50

24. Issues in Indexing Multi-dimensional indexing: –how do we index regions in space? –Document collections? –Multi-dimensional sales data –How do we support nearest neighbor queries? Indexing is still a hot and unsolved problem!

25. Indexing Exercise #1 The Purchase table: – date, buyer, seller, store, product Reports are generated once a day. What’s the best file-organization/indexing strategy?

26. Indexing Exercise #2 Airline database: Reservations table -- –flight#, seat#, date#, occupied, customer-id

27. Indexing Exercise #3 Web log application: load all the logs every night into a database. Generate reports every day (for curious professors).

28. Query Execution Query compiler Execution engine Index/record mgr. Buffer manager Storage manager storage User/ Application Query update Query execution plan Record, index requests Page commands Read/write pages

29. Query Execution Plans Purchase Person Buyer=name City=‘seattle’ phone>’5430000’ buyer (Simple Nested Loops) SELECT S.sname FROM Purchase P, Person Q WHERE P.buyer=Q.name AND Q.city=‘seattle’ AND Q.phone > ‘5430000’  Query Plan: logical tree implementation choice at every node scheduling of operations. (Table scan)(Index scan) Some operators are from relational algebra, and others (e.g., scan, group) are not.

30. The Leaves of the Plan: Scans Table scan: iterate through the records of the relation. Index scan: go to the index, from there get the records in the file (when would this be better?) Sorted scan: produce the relation in order. Implementation depends on relation size.

31. How do we combine Operations? The iterator model. Each operation is implemented by 3 functions: –Open: sets up the data structures and performs initializations –GetNext: returns the the next tuple of the result. –Close: ends the operations. Cleans up the data structures. Enables pipelining! Contrast with data-driven materialize model. Sometimes it’s the same (e.g., sorted scan).

32. Implementing Relational Operations We will consider how to implement: –Selection ( ) Selects a subset of rows from relation. –Projection ( ) Deletes unwanted columns from relation. –Join ( ) Allows us to combine two relations. –Set-difference Tuples in reln. 1, but not in reln. 2. –Union Tuples in reln. 1 and in reln. 2. –Aggregation ( SUM, MIN, etc.) and GROUP BY

33. Schema for Examples Purchase: –Each tuple is 40 bytes long, 100 tuples per page, 1000 pages (i.e., 100,000 tuples, 4MB for the entire relation). Person: –Each tuple is 50 bytes long, 80 tuples per page, 500 pages (i.e, 40,000 tuples, 2MB for the entire relation). Purchase ( buyer :string, seller : string, product : integer), Person ( name :string, city :string, phone : integer)

34. Simple Selections Of the form With no index, unsorted: Must essentially scan the whole relation; cost is M (#pages in R). With an index on selection attribute: Use index to find qualifying data entries, then retrieve corresponding data records. (Hash index useful only for equality selections.) Result size estimation: (Size of R) * reduction factor. More on this later. SELECT * FROM Person R WHERE R.phone < ‘543%’

35. Using an Index for Selections Cost depends on #qualifying tuples, and clustering. –Cost of finding qualifying data entries (typically small) plus cost of retrieving records. –In example, assuming uniform distribution of phones, about 54% of tuples qualify (500 pages, 50000 tuples). With a clustered index, cost is little more than 500 I/Os; if unclustered, up to 50000 I/Os! Important refinement for unclustered indexes: 1. Find sort the rid’s of the qualifying data entries. 2. Fetch rids in order. This ensures that each data page is looked at just once (though # of such pages likely to be higher than with clustering).

36. Two Approaches to General Selections First approach: Find the most selective access path, retrieve tuples using it, and apply any remaining terms that don’t match the index: –Most selective access path: An index or file scan that we estimate will require the fewest page I/Os. –Consider city=“seattle AND phone<“543%” : A hash index on city can be used; then, phone<“543%” must be checked for each retrieved tuple. Similarly, a b-tree index on phone could be used; city=“seattle” must then be checked.

37. Intersection of Rids Second approach –Get sets of rids of data records using each matching index. –Then intersect these sets of rids. –Retrieve the records and apply any remaining terms.

38. Implementing Projection Two parts: (1) remove unwanted attributes, (2) remove duplicates from the result. Refinements to duplicate removal: –If an index on a relation contains all wanted attributes, then we can do an index-only scan. –If the index contains a subset of the wanted attributes, you can remove duplicates locally. SELECT DISTINCT R.name, R.phone FROM Person R

39. Equality Joins With One Join Column R S is a common operation. The cross product is too large. Hence, performing R S and then a selection is too inefficient. Assume: M pages in R, p R tuples per page, N pages in S, p S tuples per page. –In our examples, R is Person and S is Purchase. Cost metric: # of I/Os. We will ignore output costs. SELECT * FROM Person R, Purchase S WHERE R.name=S.buyer

40. Simple Nested Loops Join For each tuple in the outer relation R, we scan the entire inner relation S. –Cost: M + (p R * M) * N = 1000 + 100*1000*500 I/Os: 140 hours! Page-oriented Nested Loops join: For each page of R, get each page of S, and write out matching pairs of tuples, where r is in R-page and S is in S-page. –Cost: M + M*N = 1000 + 1000*500 (1.4 hours) For each tuple r in R do for each tuple s in S do if r i == s j then add to result

41. Index Nested Loops Join If there is an index on the join column of one relation (say S), can make it the inner. –Cost: M + ( (M*p R ) * cost of finding matching S tuples) For each R tuple, cost of probing S index is about 1.2 for hash index, 2-4 for B+ tree. Cost of then finding S tuples depends on clustering. –Clustered index: 1 I/O (typical), unclustered: up to 1 I/O per matching S tuple. foreach tuple r in R do foreach tuple s in S where r i == s j do add to result

42. Examples of Index Nested Loops Hash-index on name of Person (as inner): –Scan Purchase: 1000 page I/Os, 100*1000 tuples. –For each Person tuple: 1.2 I/Os to get data entry in index, plus 1 I/O to get (the exactly one) matching Person tuple. Total: 220,000 I/Os. (36 minutes) Hash-index on buyer of Purchase (as inner): –Scan Person: 500 page I/Os, 80*500 tuples. –For each Person tuple: 1.2 I/Os to find index page with data entries, plus cost of retrieving matching Purchase tuples. Assuming uniform distribution, 2.5 purchases per buyer (100,000 / 40,000). Cost of retrieving them is 1 or 2.5 I/Os depending on clustering.

43. Block Nested Loops Join Use one page as an input buffer for scanning the inner S, one page as the output buffer, and use all remaining pages to hold ``block’’ of outer R. –For each matching tuple r in R-block, s in S-page, add to result. Then read next R-block, scan S, etc.... R & S Hash table for block of R (k < B-1 pages) Input buffer for S Output buffer... Join Result

44. Sort-Merge Join (R S) Sort R and S on the join column, then scan them to do a ``merge’’ on the join column. –Advance scan of R until current R-tuple >= current S tuple, then advance scan of S until current S-tuple >= current R tuple; do this until current R tuple = current S tuple. –At this point, all R tuples with same value and all S tuples with same value match; output for all pairs of such tuples. –Then resume scanning R and S. i=j

45. Cost of Sort-Merge Join R is scanned once; each S group is scanned once per matching R tuple. Cost: M log M + N log N + (M+N) –The cost of scanning, M+N, could be M*N (unlikely!) With 35, 100 or 300 buffer pages, both Person and Purchase can be sorted in 2 passes; total: 7500. (75 seconds).

46. Hash-Join Partition both relations using hash fn h: R tuples in partition i will only match S tuples in partition i. v Read in a partition of R, hash it using h2 (<> h!). Scan matching partition of S, search for matches. Partitions of R & S Input buffer for Si Hash table for partition Ri (k < B-1 pages) B main memory buffers Disk Output buffer Disk Join Result hash fn h2 B main memory buffers Disk Original Relation OUTPUT 2 INPUT 1 hash function h B-1 Partitions 1 2 B-1...

47. Cost of Hash-Join In partitioning phase, read+write both relations; 2(M+N). In matching phase, read both relations; M+N I/Os. In our running example, this is a total of 4500 I/Os. (45 seconds!) Sort-Merge Join vs. Hash Join: –Given a minimum amount of memory both have a cost of 3(M+N) I/Os. Hash Join superior on this count if relation sizes differ greatly. Also, Hash Join shown to be highly parallelizable. –Sort-Merge less sensitive to data skew; result is sorted.

48. How are we doing?

49. Double Pipelined Join (Tukwila) Hash Join 8Partially pipelined: no output until inner read 8Asymmetric (inner vs. outer) — optimization requires source behavior knowledge Double Pipelined Hash Join 4Outputs data immediately 4Symmetric — requires less source knowledge to optimize