1. CSE 444: Lecture 24 Query Execution Monday, March 7, 2005

2. Outline External Sorting Sort-based algorithms An example

3. The I/O Model of Computation In main memory: CPU time –Big O notation ! In databases time is dominated by I/O cost –Big O too, but for I/O’s –Often big O becomes a constant Consequence: need to redesign certain algorithms See sorting next

4. Sorting Problem: sort 1 GB of data with 1MB of RAM. Where we need this: –Data requested in sorted order (ORDER BY) –Needed for grouping operations –First step in sort-merge join algorithm –Duplicate removal –Bulk loading of B+-tree indexes.

5. 2-Way Merge-sort: Requires 3 Buffers in RAM Pass 1: Read a page, sort it, write it. Pass 2, 3, …, etc.: merge two runs, write them Main memory buffers INPUT 1 INPUT 2 OUTPUT Disk Runs of length L Runs of length 2L

6. Two-Way External Merge Sort Assume block size is B = 4Kb Step 1  runs of length L = 4Kb Step 2  runs of length L = 8Kb Step 3  runs of length L = 16Kb...... Step 9  runs of length L = 1MB... Step 19  runs of length L = 1GB (why ?) Need 19 iterations over the disk data to sort 1GB

7. Can We Do Better ? Hint: We have 1MB of main memory, but only used 12KB

8. Cost Model for Our Analysis B: Block size ( = 4KB) M: Size of main memory ( = 1MB) N: Number of records in the file R: Size of one record

9. External Merge-Sort Phase one: load M bytes in memory, sort –Result: runs of length M bytes ( 1MB ) M bytes of main memory Disk... M/R records

10. Phase Two Merge M/B – 1 runs into a new run (250 runs ) Result: runs of length M (M/B – 1) bytes (250MB) M bytes of main memory Disk... Input M/B Input 1 Input 2.. Output

11. Phase Three Merge M/B – 1 runs into a new run Result: runs of length M (M/B – 1) 2 records (625GB) M bytes of main memory Disk... Input M/B Input 1 Input 2.. Output

12. Cost of External Merge Sort Number of passes: How much data can we sort with 10MB RAM? –1 pass  10MB data –2 passes  25GB data (M/B = 2500) Can sort everything in 2 or 3 passes !

13. External Merge Sort The xsort tool in the XML toolkit sorts using this algorithm Can sort 1GB of XML data in about 8 minutes

14. Two-Pass Algorithms Based on Sorting Assumption: multi-way merge sort needs only two passes Assumption: B(R) <= M 2 Cost for sorting: 3B(R)

15. Two-Pass Algorithms Based on Sorting Duplicate elimination  (R) Trivial idea: sort first, then eliminate duplicates Step 1: sort chunks of size M, write –cost 2B(R) Step 2: merge M-1 runs, but include each tuple only once –cost B(R) Total cost: 3B(R), Assumption: B(R) <= M 2

16. Two-Pass Algorithms Based on Sorting Grouping:  a, sum(b) (R) Same as before: sort, then compute the sum(b) for each group of a’s Total cost: 3B(R) Assumption: B(R) <= M 2

17. Two-Pass Algorithms Based on Sorting R ∪ S x = first(R) y = first(S) While (_______________) do { case x y; } x = first(R) y = first(S) While (_______________) do { case x y; } Complete the program in class:

18. Two-Pass Algorithms Based on Sorting R ∩ S x = first(R) y = first(S) While (_______________) do { case x < y: case x=y: case x > y; } x = first(R) y = first(S) While (_______________) do { case x < y: case x=y: case x > y; } Complete the program in class:

19. Two-Pass Algorithms Based on Sorting R - S Complete the program in class: x = first(R) y = first(S) While (_______________) do { case x < y: case x=y: case x > y; } x = first(R) y = first(S) While (_______________) do { case x < y: case x=y: case x > y; }

20. Two-Pass Algorithms Based on Sorting Binary operations: R ∪ S, R ∩ S, R – S Idea: sort R, sort S, then do the right thing A closer look: –Step 1: split R into runs of size M, then split S into runs of size M. Cost: 2B(R) + 2B(S) –Step 2: merge M/2 runs from R; merge M/2 runs from S; ouput a tuple on a case by cases basis Total cost: 3B(R)+3B(S) Assumption: B(R)+B(S)<= M 2

21. Two-Pass Algorithms Based on Sorting R |x| R.A =S.B S x = first(R) y = first(S) While (_______________) do { case x.A < y.B: case x.A=y.B: case x.A > y.B; } x = first(R) y = first(S) While (_______________) do { case x.A < y.B: case x.A=y.B: case x.A > y.B; } Complete the program in class: R(A,C) sorted on A S(B,D) sorted on B

22. Two-Pass Algorithms Based on Sorting Join R |x| S Start by sorting both R and S on the join attribute: –Cost: 4B(R)+4B(S) (because need to write to disk) Read both relations in sorted order, match tuples –Cost: B(R)+B(S) Difficulty: many tuples in R may match many in S –If at least one set of tuples fits in M, we are OK –Otherwise need nested loop, higher cost Total cost: 5B(R)+5B(S) Assumption: B(R) <= M 2, B(S) <= M 2

23. Two-Pass Algorithms Based on Sorting Join R |x| S If the number of tuples in R matching those in S is small (or vice versa) we can compute the join during the merge phase Total cost: 3B(R)+3B(S) Assumption: B(R) + B(S) <= M 2

24. Summary of External Join Algorithms Block Nested Loop Join: B(S) + B(R)*B(S)/M Partitioned Hash Join: 3B(R)+3B(S) Assuming min(B(R),B(S)) <= M 2 Merge Join 3B(R)+3B(S) Assuming B(R)+B(S) <= M 2 Index Join B(R) + T(R)B(S)/V(S,a) Assuming…

25. Example Product(pname, maker), Company(cname, city) How do we execute this query ? Select Product.pname From Product, Company Where Product.maker=Company.cname and Company.city = “Seattle” Select Product.pname From Product, Company Where Product.maker=Company.cname and Company.city = “Seattle”

26. Example Product(pname, maker), Company(cname, city) Assume: Clustered index: Product.pname, Company.cname Unclustered index: Product.maker, Company.city

27.  city=“Seattle” Product (pname,maker) Company (cname,city) maker=cname Logical Plan:

28.  city=“Seattle” Product (pname,maker) Company (cname,city) cname=maker Physical plan 1: Index-based selection Index-based join

29.  city=“Seattle” Product (pname,maker) Company (cname,city) maker=cname Physical plans 2a and 2b: Index- scan Merge-join Scan and sort (2a) index scan (2b) Which one is better ??

30.  city=“Seattle” Product (pname,maker) Company (cname,city) cname=maker Physical plan 1: Index-based selection Index-based join T(Company) / V(Company, city)  T(Product) / V(Product, maker) Total cost: T(Company) / V(Company, city)  T(Product) / V(Product, maker) Total cost: T(Company) / V(Company, city)  T(Product) / V(Product, maker)

31.  city=“Seattle” Product (pname,maker) Company (cname,city) maker=cname Physical plans 2a and 2b: Table- scan Merge-join Scan and sort (2a) index scan (2b) B(Company) 3B(Product) T(Product) No extra cost (why ?) Total cost: (2a): 3B(Product) + B(Company) (2b): T(Product) + B(Company) Total cost: (2a): 3B(Product) + B(Company) (2b): T(Product) + B(Company)

32. Plan 1: T(Company)/V(Company,city)  T(Product)/V(Product,maker) Plan 2a: B(Company) + 3B(Product) Plan 2b: B(Company) + T(Product) Which one is better ?? It depends on the data !!

33. Example Case 1: V(Company, city)  T(Company) Case 2: V(Company, city) << T(Company) T(Company) = 5,000 B(Company) = 500 M = 100 T(Product) = 100,000 B(Product) = 1,000 We may assume V(Product, maker)  T(Company) (why ?) T(Company) = 5,000 B(Company) = 500 M = 100 T(Product) = 100,000 B(Product) = 1,000 We may assume V(Product, maker)  T(Company) (why ?) V(Company,city) = 2,000 V(Company,city) = 20

34. Which Plan is Best ? Plan 1: T(Company)/V(Company,city)  T(Product)/V(Product,maker) Plan 2a: B(Company) + 3B(Product) Plan 2b: B(Company) + T(Product) Case 1: Case 2:

35. Lessons Need to consider several physical plan –even for one, simple logical plan No magic “best” plan: depends on the data In order to make the right choice –need to have statistics over the data –the B’s, the T’s, the V’s