CSC 261/461 Database Systems Lecture 19
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1 CSC 261/461 Database Systems Lecture 19 Fall 2017
2 Announcements CIRC: CIRC is down!!! MongoDB and Spark (mini) projects are at stake. L Project 1 Milestone 4 is out Due date: Last date of class We will check your website after that date But, finish early Due Dates: Suggestions:
3 Due Dates 11/12 to 11/18 11/19 to 11/25 (Thanksgiving Week) 11/26 to 12/02 12/03 to 12/09: Term Paper Due: 12/08 12/10 to 12/13 (Last Class): Poster Session on: 12/11 Project 1 Milestone 4 is due on 12/13 MongoDB Spark Term Paper Poster Session Final: December 18, 2017 at 7:15 pm
4 Topics for Today Query Processing (Chapter 18) Query Optimization (Chapter 19) on Wednesday
5 QUERY PROCESSING
6 Steps in Query Processing Scanning Parsing Validation Query Tree Creation Query Optimization (Query planning) Code generation (to execute the plan) Running the query code
7 Steps in Query Processing
8 SQL Queries SQL Queries are decomposed into Query blocks: Select From Where Group By Having Translate Query blocks into Relational Algebraic expression Remember, SQL includes aggregate operators: MIN, MAX, SUM, COUNT etc. Part of the extended algebra Let s go back to Chapter 8 (Section 8.4.2)
9 Aggregate Functions and Grouping (Relational Algebra) Aggregate function: I < grouping attributes > I < function list > (R) Dno I COUNT Ssn, AVERAGE Salary (EMPLOYEE).
10 Semijoin ( ) R S = P A1,,An (R S) Where A 1,, A n are the attributes in R Example: Employee Dependents Students(sid,sname,gpa) People(ssn,pname,address) SQL: SELECT DISTINCT sid,sname,gpa FROM Students,People WHERE sname = pname; OR SELECT DISTINCT sid,sname,gpa FROM Students WHERE sname IN (SELECT pname FROM People); RA: Students People
11 EXTERNAL SORTING
12 External Merge Sort
13 Why are Sort Algorithms Important? Data requested from DB in sorted order is extremely common e.g., find students in increasing GPA order Why not just use quicksort in main memory?? What about if we need to sort 1TB of data with 1GB of RAM A classic problem in computer science!
14 So how do we sort big files? 1. Split into chunks small enough to sort in memory ( runs ) 2. Merge pairs (or groups) of runs using the external merge algorithm 3. Keep merging the resulting runs (each time = a pass ) until left with one sorted file!
15 2. EXTERNAL MERGE & SORT
16 Challenge: Merging Big Files with Small Memory How do we efficiently merge two sorted files when both are much larger than our main memory buffer?
17 External Merge Algorithm Input: 2 sorted lists of length M and N Output: 1 sorted list of length M + N Required: At least 3 Buffer Pages IOs: 2(M+N)
18 Key (Simple) Idea To find an element that is no larger than all elements in two lists, one only needs to compare minimum elements from each list. If: A : A < A > B : B < Then: Min(A :, B : ) A E Min(A :, B : ) B F for i=1.n and j=1.m
19 External Merge Algorithm Main Memory Input: Two sorted files F 1 F 2 1,5 2,22 7,11 20,31 23,24 25,30 Buffer Output: One merged sorted file Disk
20 External Merge Algorithm Main Memory Input: Two sorted files F 1 F 2 7,11 20,31 23,24 25,30 Buffer 1,5 2,22 Output: One merged sorted file Disk
21 External Merge Algorithm Main Memory Input: Two sorted files F 1 F 2 7,11 20,31 23,24 25,30 Buffer ,2 Output: One merged sorted file Disk
22 External Merge Algorithm Main Memory Input: Two sorted files F 1 F 2 7,11 20,31 23,24 25,30 Buffer 5 22 Output: One merged sorted file 1,2 Disk
23 External Merge Algorithm Main Memory Input: Two sorted files F 1 F 2 7,11 20,31 23,24 25,30 Buffer 22 5 Output: One merged sorted file 1,2 Disk This is all the algorithm sees Which file to load a page from next?
24 External Merge Algorithm Main Memory Input: Two sorted files F 1 F 2 7,11 20,31 23,24 25,30 Buffer 22 5 Output: One merged sorted file 1,2 Disk We know that F 2 only contains values 22 so we should load from F 1!
25 External Merge Algorithm Main Memory Input: Two sorted files F 1 20,31 Buffer 7,11 F 2 23,24 25, Output: One merged sorted file 1,2 Disk
26 External Merge Algorithm Main Memory Input: Two sorted files F 1 20,31 Buffer 11 F 2 23,24 25, ,7 Output: One merged sorted file 1,2 Disk
27 External Merge Algorithm Main Memory Input: Two sorted files F 1 20,31 Buffer 11 F 2 23,24 25,30 22 Output: One merged sorted file 1,2 5,7 Disk
28 External Merge Algorithm Main Memory Input: Two sorted files F 1 20,31 Buffer F 2 23,24 25,30 Output: One merged sorted file 1,2 5,7 And so on Disk
29 We can merge lists of arbitrary length with only 3 buffer pages. If lists of size M and N, then Cost: 2(M+N) IOs Each page is read once, written once
30 External Merge Sort Algorithm Example: 3 Buffer pages 6-page file F 1 44,10 Disk 33,12 55,31 Main Memory Buffer Orange file = unsorted F 2 18,22 27,24 3,1 1. Split into chunks small enough to sort in memory
31 EXTERNAL MERGE SORT (BEFORE MERGE) CSC 261, Spring 2017, UR
32 External Merge Sort Algorithm Example: 3 Buffer pages 6-page file F 1 44,10 Disk 33,12 55,31 Main Memory Buffer Orange file = unsorted F 2 18,22 27,24 3,1 1. Split into chunks small enough to sort in memory
33 External Merge Sort Algorithm Example: 3 Buffer pages 6-page file F 1 Disk Main Memory Buffer Orange file = unsorted 44,10 33,12 55,31 F 2 18,22 27,24 3,1 1. Split into chunks small enough to sort in memory
34 External Merge Sort Algorithm Example: 3 Buffer pages 6-page file F 1 Disk Main Memory Buffer Orange file = unsorted 10,12 31,33 44,55 F 2 18,22 27,24 3,1 1. Split into chunks small enough to sort in memory
35 External Merge Sort Algorithm Example: 3 Buffer pages 6-page file Each sorted file is a called a run F 1 F 2 10,12 18,22 Disk 31,33 44,55 27,24 3,1 Main Memory Buffer 1,3 18,22 24,27 And similarly for F 2 1. Split into chunks small enough to sort in memory
36 External Merge Sort Algorithm Example: 3 Buffer pages 6-page file F 1 10,12 Disk 31,33 44,55 Main Memory Buffer F 2 1,3 18,22 24,27 2. Now just run the external merge algorithm & we re done!
37 Calculating IO Cost For 3 buffer pages, 6 page file: 1. Split into two 3-page files and sort in memory = 1 R + 1 W for each file = 2*(3 + 3) = 12 IO operations 2. Merge each pair of sorted chunks using the external merge algorithm = 2*(3 + 3) = 12 IO operations 3. Total cost = 24 IO
38 Running External Merge Sort on Larger Files 10,12 45,38 Disk 31,33 44,55 18,43 24,27 Assume we still only have 3 buffer pages (Buffer not pictured) 10,12 31,33 47,55 41,3 18,22 23,20 42,46 31,33 39,55 1,3 18,23 24,27 10,12 48,33 44,40 16,31 18,22 24,27
39 Running External Merge Sort on Larger Files 10,12 Disk 31,33 44,55 1. Split into files small enough to sort in buffer Assume we still only have 3 buffer pages (Buffer not pictured) 45,38 18,43 24,27 10,12 31,33 47,55 41,3 18,22 23,20 42,46 31,33 39,55 1,3 18,23 24,27 10,12 48,33 44,40 16,31 18,22 24,27
40 Running External Merge Sort on Larger Files 10,12 18,24 Disk 31,33 44,55 27,38 43,45 1. Split into files small enough to sort in buffer and sort Assume we still only have 3 buffer pages (Buffer not pictured) 10,12 31,33 47,55 3,18 20,22 23,41 31,33 39,42 46,55 1,3 18,23 24,27 10,12 16,18 33,40 44,48 22,24 27,31 Call each of these sorted files a run
41 Running External Merge Sort on Larger Files 10,12 18,24 10,12 3,18 31,33 1,3 10,12 Disk 31,33 44,55 27,38 43,45 31,33 47,55 20,22 23,41 39,42 46,55 18,23 24,27 33,40 44,48 10,12 33,38 3,10 23,31 1,3 31,33 10,12 Disk 18,24 27,31 43,44 45,55 12,18 20,22 33,41 47,55 18,23 24,27 39,42 46,55 16,18 22,24 Assume we still only have 3 buffer pages (Buffer not pictured) 2. Now merge pairs of (sorted) files the resulting files will be sorted! 16,18 22,24 27,31 27,31 33,40 44,48
42 Running External Merge Sort on Larger Files 10,12 18,24 Disk 31,33 44,55 27,38 43,45 10,12 33,38 Disk 18,24 27,31 43,44 45,55 3,10 18,20 Disk 10,12 12,18 22,23 24,27 Assume we still only have 3 buffer pages (Buffer not pictured) 10,12 3,18 31,33 47,55 20,22 23,41 3,10 23,31 12,18 20,22 33,41 47,55 31,31 43,44 33,33 38,41 45,47 55,55 3. And repeat 31,33 39,42 46,55 1,3 18,23 24,27 1,3 10,12 16,18 1,3 10,12 16,18 18,23 24,27 33,40 44,48 22,24 27,31 31,33 10,12 27,31 39,42 46,55 16,18 22,24 33,40 44,48 18,22 27,31 40,42 23,24 24,27 31,33 33,39 44,46 48,55 Call each of these steps a pass
43 Running External Merge Sort on Larger Files Disk Disk Disk Disk 10,12 31,33 44,55 10,12 18,24 27,31 3,10 10,12 12,18 1,3 3,10 10,10 18,24 27,38 43,45 33,38 43,44 45,55 18,20 22,23 24,27 12,12 12,16 18,18 10,12 31,33 47,55 3,10 12,18 20,22 31,31 33,33 38,41 18,18 20,22 22,23 3,18 20,22 23,41 23,31 33,41 47,55 43,44 45,47 55,55 23,24 24,24 27,27 31,33 39,42 46,55 1,3 18,23 24,27 1,3 10,12 16,18 27,31 31,31 31,33 1,3 18,23 24,27 31,33 39,42 46,55 18,22 23,24 24,27 33,33 33,38 39,40 10,12 33,40 44,48 10,12 16,18 22,24 27,31 31,33 33,39 41,42 43,44 44,45 16,18 22,24 27,31 27,31 33,40 44,48 40,42 44,46 48,55 46,47 48,55 55,55 4. And repeat!
44 Simplified 3-page Buffer Version Assume for simplicity that we split an N-page file into N singlepage runs and sort these; then: First pass: Merge N/2 pairs of runs each of length 1 page Unsorted input file Split & sort Second pass: Merge N/4 pairs of runs each of length 2 pages In general, for N pages, we do log 2 N passes +1 for the initial split & sort Each pass involves reading in & writing out all the pages = 2N IO Sorted! Merge Merge à 2N*( log 2 N +1) total IO cost!
45 Using B+1 buffer pages to reduce # of passes Suppose we have B+1 buffer pages now; we can: 1. Increase length of initial runs. Sort B+1 at a time! At the beginning, we can split the N pages into runs of length B+1 and sort these in memory IO Cost: 2N( log < N + 1) Starting with runs of length 1 2N( log < N B ) Starting with runs of length B+1
46 Using B+1 buffer pages to reduce # of passes Suppose we have B+1 buffer pages now; we can: 2. Perform a B-way merge. On each pass, we can merge groups of B runs at a time (vs. merging pairs of runs)! IO Cost: 2N( log < N + 1) 2N( log < N B ) Starting with runs of length 1 Starting with runs of length B+1 2N( log V N B ) Performing B-way merges
47 Algorithm fro Select Operation Read Section 18.3 (18.3.1, , , ) Mostly covers searching: 1. Linear Search 2. Binary Search 3. Indexing 4. Hashing 5. B+ Tree (Skip bitmap index and functional index)
48 Algorithm for Join Operation The most time consuming operation
49 What you will learn about in this section 1. Nested Loop Join (NLJ) 2. Block Nested Loop Join (BNLJ) 3. Index Nested Loop Join (INLJ) 4. Sorted-Merge Join 5. Hash Join
50 RECAP: Joins
51 Joins: Example R S SELECT R.A,B,C,D FROM R, S WHERE R.A = S.A Example: Returns all pairs of tuples r R, s S such that r. A = s. A R S A B C A D A B C D
52 Joins: Example R S SELECT R.A,B,C,D FROM R, S WHERE R.A = S.A Example: Returns all pairs of tuples r R, s S such that r. A = s. A R S A B C A D A B C D
53 Joins: Example R S SELECT R.A,B,C,D FROM R, S WHERE R.A = S.A Example: Returns all pairs of tuples r R, s S such that r. A = s. A R S A B C A D A B C D
54 Joins: Example R S SELECT R.A,B,C,D FROM R, S WHERE R.A = S.A Example: Returns all pairs of tuples r R, s S such that r. A = s. A R S A B C A D A B C D
55 Joins: Example R S SELECT R.A,B,C,D FROM R, S WHERE R.A = S.A Example: Returns all pairs of tuples r R, s S such that r. A = s. A R S A B C A D A B C D
56 Semantically: A Subset of the Cross Product R S SELECT R.A,B,C,D FROM R, S WHERE R.A = S.A Example: Returns all pairs of tuples r R, s S such that r. A = s. A R S A B C D A B C A D Cross Produc t Filter by conditions (r.a = s.a) Can we actually implement a join in this way?
57 Notes We write R S to mean join R and S by returning all tuple pairs where all shared attributes are equal We write R S on A to mean join R and S by returning all tuple pairs where attribute(s) A are equal For simplicity, we ll consider joins on two tables and with equality constraints ( equijoins ) However joins can merge > 2 tables, and some algorithms do support non-equality constraints!
58 Nested Loop Joins
59 Notes We are again considering IO aware algorithms: care about disk IO Given a relation R, let: T(R) = # of tuples in R P(R) = # of pages in R Recall that we read / write entire pages with disk IO Note also that we omit ceilings in calculations good exercise to put back in!
60 Nested Loop Join (NLJ) Compute R S on A: for r in R: for s in S: if r[a] == s[a]: yield (r,s)
![Nested Loop Join (NLJ) Cost: Compute R S on A: for r in R: for s in S: if r[a] == s[a]: yield (r,s) P(R) 1.](/docs-images/96/127260808/images/61-7.jpg "Loop over the tuples in R Note that our IO cost is based on the number of pages loaded, not the number of")
61 Nested Loop Join (NLJ) Cost: Compute R S on A: for r in R: for s in S: if r[a] == s[a]: yield (r,s) P(R) 1. Loop over the tuples in R Note that our IO cost is based on the number of pages loaded, not the number of tuples!
62 Nested Loop Join (NLJ) Cost: Compute R S on A: for r in R: for s in S: if r[a] == s[a]: yield (r,s) P(R) + T(R)*P(S) 1. Loop over the tuples in R 2. For every tuple in R, loop over all the tuples in S Have to read all of S from disk for every tuple in R!
63 Nested Loop Join (NLJ) Cost: Compute R S on A: for r in R: for s in S: if r[a] == s[a]: yield (r,s) P(R) + T(R)*P(S) 1. Loop over the tuples in R 2. For every tuple in R, loop over all the tuples in S 3. Check against join conditions Note that NLJ can handle things other than equality constraints just check in the if statement!
64 Nested Loop Join (NLJ) Cost: Compute R S on A: for r in R: for s in S: if r[a] == s[a]: yield (r,s) P(R) + T(R)*P(S) + OUT 1. Loop over the tuples in R 2. For every tuple in R, loop over all the tuples in S 3. Check against join conditions 4. Write out (to page, then when page full, to disk)
65 Nested Loop Join (NLJ) Cost: Compute R S on A: for r in R: for s in S: if r[a] == s[a]: yield (r,s) P(R) + T(R)*P(S) + OUT What if R ( outer ) and S ( inner ) switched? P(S) + T(S)*P(R) + OUT Outer vs. inner selection makes a huge difference- DBMS needs to know which relation is smaller!
66 Block Nested Loop Join (BNLJ)
67 Block Nested Loop Join (BNLJ) Given 3 pages of memory Compute R S on A: for each page pr of R: for page ps of S: for each tuple r in pr: for each tuple s in ps: if r[a] == s[a]: yield (r,s) Cost: P(R) 1. Load in 1 page of R at a time (leaving 1 page each free for S & output) Note: There could be some speedup here due to the fact that we re reading in multiple pages sequentially however we ll ignore this here!
68 Block Nested Loop Join (BNLJ) Cost: Given 3 pages of memory Compute R S on A: for each page pr of R: for page ps of S: for each tuple r in pr: for each tuple s in ps: if r[a] == s[a]: yield (r,s) P R + P R. P(S) 1. Load in 1 page of R at a time (leaving 1 page each free for S & output) 2. For each page segment of R, load each page of S Note: Faster to iterate over the smaller relation first!
69 Block Nested Loop Join (BNLJ) Cost: Given 3 pages of memory Compute R S on A: for each page pr of R: for page ps of S: for each tuple r in pr: for each tuple s in ps: if r[a] == s[a]: yield (r,s) P R + P R. P(S) 1. Load in 1 page of R at a time (leaving 1 page each free for S & output) 2. For each page segment of R, load each page of S 3. Check against the join conditions BNLJ can also handle non-equality constraints
![Block Nested Loop Join (BNLJ) Cost: Given 3 pages of memory Compute R S on A: for each page pr of R: for page ps of S: for each tuple r in pr: for each tuple s in ps: if r[a] == s[a]: yield](/docs-images/96/127260808/images/70-5.jpg "(r,s) P R + P R. P(S) 1. Load 1 page of R at a time (leaving 1 page each free for S & output) 2. For each page segment of R, load each page of S 3. Check against the join conditions 4.")
70 Block Nested Loop Join (BNLJ) Cost: Given 3 pages of memory Compute R S on A: for each page pr of R: for page ps of S: for each tuple r in pr: for each tuple s in ps: if r[a] == s[a]: yield (r,s) P R + P R. P(S) 1. Load 1 page of R at a time (leaving 1 page each free for S & output) 2. For each page segment of R, load each page of S 3. Check against the join conditions 4. Write out
![Block Nested Loop Join (BNLJ) (B+1 pages of Memory) Given B+1 pages of memory Compute R S on A: for each B-1 pages pr of R: for page ps of S: for each tuple r in pr: for each tuple s in ps: if r[a]](/docs-images/96/127260808/images/71-3.jpg "== s[a]: yield (r,s) Cost: P(R) 1.")
71 Block Nested Loop Join (BNLJ) (B+1 pages of Memory) Given B+1 pages of memory Compute R S on A: for each B-1 pages pr of R: for page ps of S: for each tuple r in pr: for each tuple s in ps: if r[a] == s[a]: yield (r,s) Cost: P(R) 1. Load in B-1 pages of R at a time (leaving 1 page each free for S & output) Note: There could be some speedup here due to the fact that we re reading in multiple pages sequentially however we ll ignore this here!
72 Block Nested Loop Join (BNLJ) Given B+1 pages of memory Compute R S on A: for each B-1 pages pr of R: for page ps of S: for each tuple r in pr: for each tuple s in ps: if r[a] == s[a]: yield (r,s) Cost: P R + P R B 1 P(S) 1. Load in B-1 pages of R at a time (leaving 1 page each free for S & output) 2. For each (B-1)-page segment of R, load each page of S Note: Faster to iterate over the smaller relation first!
73 Block Nested Loop Join (BNLJ) Compute R S on A: ps: for each B-1 pages pr of R: for page ps of S: for each tuple r in pr: for each tuple s in if r[a] == s[a]: yield (r,s) Cost: Given B+1 pages of memory P R + P R B 1 P(S) 1. Load in B-1 pages of R at a time (leaving 1 page each free for S & output) 2. For each (B-1)-page segment of R, load each page of S 3. Check against the join conditions BNLJ can also handle non-equality constraints
![Block Nested Loop Join (BNLJ) Cost: Given B+1 pages of memory Compute R S on A: for each B-1 pages pr of R: for page ps of S: for each tuple r in pr: for each tuple s in ps: if r[a] == s[a]: yield](/docs-images/96/127260808/images/74-2.jpg "(r,s) P R + b c Vd: P(S) + OUT 1. Load in B-1 pages of R at a time (leaving 1 page each free for S & output) 2. For each (B-1)-page segment of R, load each page of S 3.")
74 Block Nested Loop Join (BNLJ) Cost: Given B+1 pages of memory Compute R S on A: for each B-1 pages pr of R: for page ps of S: for each tuple r in pr: for each tuple s in ps: if r[a] == s[a]: yield (r,s) P R + b c Vd: P(S) + OUT 1. Load in B-1 pages of R at a time (leaving 1 page each free for S & output) 2. For each (B-1)-page segment of R, load each page of S 3. Check against the join conditions 4. Write out
75 BNLJ vs. NLJ: Benefits of IO Aware In BNLJ, by loading larger chunks of R, we minimize the number of full disk reads of S We only read all of S from disk for every (B-1)-page segment of R! Still the full cross-product, but more done only in memory NLJ P(R) + T(R)*P(S) + OUT BNLJ P R + b c Vd: P(S) + OUT BNLJ is faster by roughly (Vd:)e(c) b(c)
76 BNLJ vs. NLJ: Benefits of IO Aware Example: R: 500 pages S: 1000 pages 100 tuples / page We have 12 pages of memory (B = 11) Ignoring OUT here NLJ: Cost = ,000*1000 = 50 Million IOs ~= 140 hours BNLJ: Cost = fgg :ggg :g = 50 Thousand IOs ~= 0.14 hours A very real difference from a small change in the algorithm!
77 Acknowledgement Some of the slides in this presentation are taken from the slides provided by the authors. Many of these slides are taken from cs145 course offered by Stanford University.
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