1. Nested-Loop joins “one-and-a-half” pass method, since one relation will be read just once. Tuple-Based Nested-loop Join Algorithm: FOR each tuple s in S DO FOR each tuple r in R DO IF r and s join to make a tuple t THEN output t Improvement to Take Advantage of Disk I/O Model Instead of retrieving tuples of R, T(S) times, load memory with as many tuples of S as can fit, and match tuples of R against all S­tuples in memory.

2. Block-based nested loops Assume B(S) ≤ B(R), and B(S) > M Read M-1 blocks of S into main memory and compare to all of R, block by block FOR each chunk of M-1 blocks of S DO FOR each block b of R DO FOR each tuple t of b DO find the tuples of S in memory that join with t output the join of t with each of these tuples

3. Example B(R) = 1000, B(S) = 500, M = 101 Important Aside: 101 buffer blocks is not as unrealistic as it sounds. There may be many queries at the same time, competing for main­ memory buffers. Outer loop iterates 5 times at 100 I/O’s each At each iteration we read M-1 (i.e. 100) blocks of S and all of R (i.e. 1000) blocks. Total time: 5*(100 + 1000) = 5500 I/O’s Question: What if we reversed the roles of R and S? We would iterate 10 times, and in each we would read 100+500 blocks, for a total of 6000 I/O’s. Compare with one-pass join, if it could be done! We would need 1500 disk I/O’s if B(S)  M-1

4. Analysis of blocks nested loops Number of disk I/O’s: [B(S)/(M-1)]*(M-1 +B(R)) or B(S) + [B(S)B(R)/(M-1)] or approximately B(S)*B(R)/M

5. Two-pass algorithms based on sorting This special case of multi-pass algorithms is sufficient for most of the relation sizes. Main idea for unary operations on R Suppose B(R)  M (main memory size in blocks) First pass: Read M blocks of R into MM Sort the content of MM Write the sorted result (sublist/run) into M blocks on disk. Second pass: create final result

6. Duplicate elimination  using sorting In the second phase (merging) we don’t sort but copy each tuple just once. We can do that because the identical tuples will appear “at the same time,” i.e. they will be all the first ones at the buffers (for the sorted sublists). As usual, if one buffer gets empty we refill it.

7. Duplicate-Elimination using Sorting Example Assume M=3, each block holds 2 records and relation R consists of the following 17 tuples: 2, 5, 2, 1, 2, 2, 4, 5, 4, 3, 4, 2, 1, 5, 2, 1, 3 After the first pass the following sorted sub-lists are created: 1, 2, 2, 2, 2, 5 2, 3, 4, 4, 4, 5 1, 1, 2, 3, 5 In the second pass we dedicate a memory buffer to each sub-list.

8. Example (Cont’d)

11. Analysis of  (R) 2B(R) when creating sorted sublists B(R) to read each sublist in phase 2 Total: 3B(R) How large can R be? –There can be no more than M sublists since we need one buffer for each one.  So, B(R)/M ≤ M, (B(R)/M is the number of sublists) i.e. B(R) ≤ M 2 To compute  (R) we need at least sqrt(B(R)) blocks of MM.

12. Sort-based , , - Exampe: set union. Create sorted sublists of R and S Use input buffers for sorted sublists of R and S, one buffer per sublist Output each tuple once. We can do that since all the identical tuples appear “at the same time.” Analysis: 3(B(R) + B(S)) disk I/O’s Condition: B(R) + B(S) ≤ M 2 Similar algorithms for sort based intersection and difference (bag or set versions).

13. Join A problem for joins but not for the previous operators: The number of joining tuples from the two relations can exceed what fits in memory. First, we can try to maximize the number of available buffers for putting the joining tuples. How, we can do this? By minimizing the number of sorted sublists (since we need a buffer for each one of them).

14. Simple sort-based join For R(X,Y) S(Y,Z) with M buffers of memory: Sort R on Y, sort S on Y Merge phase Use 2 input buffers: 1 for R, 1 for S. Pick tuple t with smallest Y value in the buffer for R (or for S) If t doesn’t match with the first tuple in the buffer for S, then just remove t. Otherwise, read all the tuples from R with the same Y value as t and put them in the M-2 part of the memory. When the input buffer for R is exhausted fill it again and again. Then, read the tuples of S that match. For each one we produce the join of it with all the tuples of R in the M-2 part of the memory.

15. Example of sort join B(R) = 1000, B(S) = 500, M= 101 To sort R, we need 4*B(R) I/O’s, same for S. –Total disk I/O’s = 4*(B(R) + B(S)) Doing the join in the merge phase: –Total disk I/O’s = B(R) + B(S) Total disk I/O’s = 5*(B(R) + B(S)) = 7500 Memory Requirement: To be able to do the sort, should have B(R) ≤ M 2 and B(S) ≤ M 2 Recall: for nested-loop join, we needed 5500 disk I/O’s, but the memory requirement was quadratic (it is linear, here), i.e., nested-loop join is not good for joining relations that are much larger than MM.

16. Potential problem... R(X, Y) ----------- x 1 a x 2 a … x n a S(Y, Z) --------- a z 1 a z 2... a z m What if n+1 > M-1 and m+1 > M-1? If the tuples from R (or S) with the same value y of Y do not fit in M-1 buffers, then we use all M-1 buffers to do a nested-loop join on the tuples with Y-value y from both relations. Observe that we can “smoothly” continue with the nested loop join when we see that the R tuples with Y-value y do not fit in M-1 buffers.

17. Do we really need the fully sorted files? Can We Improve on Sort Join? R S Join? sorted runs

18. A more efficient sort-based join Suppose we are not worried about many common Y values Create Y-sorted sublists of R and S Bring first block of each sublist into a buffer (assuming we have at most M sublists) Find smallest Y-value from heads of buffers. Join with other tuples in heads of buffers, use other possible buffers, if there are “many” tuples with the same Y values. Disk I/O: 3*(B(R) + B(S)) Requirement: B(R) + B(S) ≤ M 2

19. Example of more efficient sort-join B(R) = 1000, B(S) = 500, M= 101 Total of 15 sorted sublists If too many tuples join on a value y, use the remaining 86 MM buffers for a one pass join on y Total cost: 3(1000 + 500) = 4500 disk I/O’s M 2 =10201 > B(R) + B(S), so the requirement is satisfied

20. Summary of sort-based algorithms OperatorsApprox. M requiredDisk I/O ,  Sqrt(B)3B , , - Sqrt(B(R) + B(S))3(B(R)+B(S))  Sqrt(max(B(R)+B(S)))5(B(R)+B(S))  Sqrt(max(B(R)+B(S)))3(B(R)+B(S))

21. Two-pass algorithms based on hashing Main idea: Let B(R) > M instead of sorted sublists, create partitions, based on hashing Second pass to create result from partitions

22. Creating partitions Here partitions are created based on all attributes of the relation except for grouping and join, where the partitions are based on the grouping and join-attributes respectively. Why bucketize? Tuples with “matching” values end up in the same bucket. Initialize M-1 buckets using M-1 empty buffers; FOR each block b of relation R DO read block b into the M-th buffer; IF the buffer for bucket h(t) has no room for t THEN copy the buffer to disk; initialize a new empty block in that buffer; ENDIF ; copy t to the buffer for bucket h(t); ENDFOR; FOR each bucket DO IF the buffer for this bucket is not empty THEN write the buffer to disk;

23. Hash : even / odd Simple Example S 5 4 12 3 13 8 11 14 R243589R243589 2 4 84 12 8 14 3 5 95 3 13 11 Even Odd Buckets RS

24. Hash-based duplicate elimination Pass 1: create partitions by hashing on all attributes Pass 2: for each partition, use the one-pass method for duplicate elimination Cost: 3B(R) disk I/O’s Requirement: B(R) ≤ M*(M-1) (B(R)/M is the approximate size of one bucket) i.e. the req. is approximately B(R) ≤ M 2

25. Hash-based grouping and aggregation Pass 1: create partitions by hashing on grouping attributes Pass 2: for each partition, use one-pass method. Cost: 3B(R), Requirement: B(R) ≤ M 2 If B(R) > M 2 Read blocks of partition one by one Create one slot in memory for each group-value Requirement: where L is the list of grouping attributes

26. Hash-based set union Pass 1: create partitions R 1,…,R M-1 of R, and S 1,…,S M-1 of S (with the same hash function) Pass 2: for each pair R i, S i compute R i  S i using the one- pass method. Cost: 3(B(R) + B(S)) Requirement? min(B(R),B(S)) ≤ M 2 Similar algorithms for intersection and difference (set and bag versions)

27. Partition hash-join Pass 1: create partitions R 1,..,R M-1 of R, and S 1,..,S M-1 of S, based on the join attributes (the same hash function for both R and S) Pass 2: for each pair R i, S i compute R i  S i using the one-pass method. Cost: 3(B(R) + B(S)) Requirement: min(B(R),B(S)) ≤ M 2

28. B(R) = 1000 blocks B(S) = 500 blocks Memory available = 101 blocks R  S on common attribute C Use 100 buckets –Read R –Hash –Write buckets Example... 10 blocks 100 R  Same for S

29. Read one R bucket Build memory hash table Read corresponding S bucket block by block. R S... R Memory... Cost “Bucketize:” –Read + write R –Read + write S Join –Read R –Read S Total cost = 3*[1000+500] = 4500

30. Saving some disk I/O’s (I) If we have more memory than we need to hold one block per bucket, then we can use several buffers for each bucket, and write them out as a group saving in seek time and rotational latency. Also, we can read the buckets in group in the second pass and saving in seek time and rotational latency. Well, these techniques don’t save disk I/O’s, but make them faster. What about saving some I/O’s?

31. Suppose that to join R with S we decide to create k buckets where k is much smaller than M. When we hash S we can keep m of the k buckets in memory, while keeping only one block for each of the other k-m buckets. We can do so provided –m*(B(S)/k) +(k-m)  M –B(S)/k is the approximate size of a bucket of S. Now, when we read the tuples of R, to hash them into buckets, we keep in memory: The m buckets of S that were never written out to disk, and One block for each of the k-m buckets of R whose corresponding buckets of S were written to disk. Saving some disk I/O’s (II)

32. If a tuple t of R hashes to one of the first m buckets, then we immediately join it with all the tuples of the corresponding S-bucket. If a tuple t of R hashes to a bucket whose corresponding S-bucket is on disk, then t is sent to the main memory buffer for that bucket, and eventually migrates to disk, as for a two pass, hash-based join. In the second pass, we join the corresponding buckets of R and S as usual (but only m-k). The savings in I/O’s is equal to two for every block of the S-buckets that remain in memory, and their corresponding R-buckets. Since m/k of the buckets are in memory we save 2(m/k)(B(S)+B(R)). Saving some disk I/O’s (III)

33. All but k-m of the memory buffers can be used to hold tuples of S, and the more of these tuples, the fewer the disk I/O’s. Thus, we want to minimize k, the number of buckets. We do so by making each bucket about as big as can fit in memory, i.e. the buckets are of (approximately) M size, and therefore k=B(S)/M. If that is the case, then there is room for one bucket in memory, i.e. m=1. –We have to make the bucket actually M-k blocks, but we are talking here approximately, when k<<M. So, we have that the savings in I/O’s are: 2(M/B(S))*(B(R) + B(S)) And, the total cost is: (3 - 2(M/B(S)))*(B(R) + B(S)) Saving some disk I/O’s (IV) How can we choose m and k?

34. Summary of hash-based methods OperatorsApprox. M required Disk I/O ,  Sqrt(B)3B , , - Sqrt(B(S))3(B(R)+B(S))  Sqrt(B(S))3(B(R)+B(S))  Sqrt(B(S))(3-2M/B(S))(B(R)+B(S))

35. Sort vs. Hash based algorithms Hash-based algorithms have a size requirement that depends only on the smaller of the two arguments rather than on the sum of the argument sizes, as for sort-based algorithms. Sort-based algorithms allow us to produce the result in sorted order and take advantage of that sort later. The result can be used in another sort-based algorithm later. Hash-based algorithms depend on the buckets being of nearly equal size. Well, what about a join with a very few values for the join attribute…