1.1

CS220CS220Database SystemsDatabase Systems

Query Evaluation IQuery Evaluation I

Warning: There is enough detail in here and Warning: There is enough detail in here and the next set of slides for 3-4 classes. We’re the next set of slides for 3-4 classes. We’re

only looking at the high-level concepts.only looking at the high-level concepts.Slides courtesy G. Kollios, Boston University

1.2

IntroductionIntroduction

We’ve covered the basic underlying storage, buffering, and indexing technology. Now we can move on to query

processing. Some database operations are

EXPENSIVE Can greatly improve performance by

being “smart” e.g., can speed up 1,000,000x over

naïve approach Main weapons are:

1. clever implementation techniques for operators

2. exploiting “equivalencies” of relational operators

3. using statistics and cost models to choose among these.

Query Optimizationand Execution

Relational Operators

Files and Access Methods

Buffer Management

Disk Space Management

DB

SQL Query

1.3

Cost-based Query Sub-SystemCost-based Query Sub-System

Query Plan Evaluator

Query Optimizer

Plan Generator

Plan Cost Estimator

Usually there is aheuristics-basedrewriting step beforethe cost-based steps.

Statistics

Catalog Manager

Schema

Select *From Blah BWhere B.blah = blah

Queries

Query Parser

1.4

Query Processing OverviewQuery Processing Overview The query optimizer translates SQL to a special internal

“language” Query Plans

The query executor is an interpreter for query plans Think of query plans as “box-and-arrow”

dataflow diagrams

Each box implements a relational operator

Edges represent a flow of tuples (columns as specified)

For single-table queries, these diagrams arestraight-line graphs

HeapScan

Sort

Distinct

name, gpa

name, gpa

name, gpa

OptimizerSELECT DISTINCT name, gpa FROM Students

1.5

Query OptimizationQuery Optimization

A deep subject, focuses on multi-table queries We will only need a cookbook version for now.

Build the dataflow bottom up: Choose an Access Method (HeapScan or

IndexScan) Non-trivial, we’ll learn about this later!

Next apply any WHERE clause filters Next apply GROUP BY and aggregation

Can choose between sorting and hashing!

Next apply any HAVING clause filters Next Sort to help with ORDER BY and DISTINCT

In absence of ORDER BY, can do DISTINCT via hashing!

Distinct

HeapScan

Filter

HashAgg

Filter

Sort

1.6

IteratorsIterators

The relational operators are all subclasses of the class iterator:

class iterator { void init(); tuple next(); void close(); iterator inputs[];

// additional state goes here}

Note: Edges in the graph are specified by inputs (max 2, usually 1) Encapsulation: any iterator can be input to any other! When subclassing, different iterators will keep different kinds of

state information

iterator

1.7

Example: Example: ScanScan

init(): Set up internal state call init() on child – often a file open

next(): call next() on child until qualifying tuple found or EOF keep only those fields in “proj_list” return tuple (or EOF -- “End of File” -- if no tuples remain)

close(): call close() on child clean up internal state

Note: Scan also applies “selection” filters and “projections”

(without duplicate elimination)

class Scan extends iterator { void init(); tuple next(); void close(); iterator inputs[1]; bool_expr filter_expr; proj_attr_list proj_list;}

1.8

Example: Example: SortSort

init(): generate the sorted runs on disk Allocate runs[] array and fill in with disk pointers. Initialize numberOfRuns Allocate nextRID array and initialize to NULLs

next(): nextRID array tells us where we’re “up to” in each run find the next tuple to return based on nextRID array advance the corresponding nextRID entry return tuple (or EOF -- “End of File” -- if no tuples remain)

close(): deallocate the runs and nextRID arrays

class Sort extends iterator { void init(); tuple next(); void close(); iterator inputs[1]; int numberOfRuns; DiskBlock runs[]; RID nextRID[];}

1.9

Streaming through RAMStreaming through RAM

Simple case: “Map”. (assume many records per disk page) Goal: Compute f(x) for each record, write out the result Challenge: minimize RAM, call read/write rarely

Approach Read a chunk from INPUT to an Input Buffer Write f(x) for each item to an Output Buffer When Input Buffer is consumed, read another chunk When Output Buffer fills, write it to OUTPUT

Reads and Writes are not coordinated (i.e., not in lockstep) E.g., if f() is Compress(), you read many chunks per write. E.g., if f() is DeCompress(), you write many chunks per read.

f(x)RAM

InputBuffer

OutputBuffer

OUTPUTINPUT

1.10

RendezvousRendezvous

Streaming: one chunk at a time. Easy.

But some algorithms need certain items to be co-resident in memorynot guaranteed to appear in the same input chunk

Time-space Rendezvous in the same place (RAM) at the same time

There may be many combos of such items

1.11

B-2

Divide and ConquerDivide and Conquer

Out-of-core algorithms orchestrate rendezvous.

Typical RAM Allocation: Assume B pages worth of RAM available

Use 1 page of RAM to read into

Use 1 page of RAM to write into

B-2 pages of RAM as workspace

IN OUT

OUTPUTINPUT

1.12

Divide and ConquerDivide and Conquer

Phase 1 “streamwise” divide into N/(B-2) megachunks

output (write) to disk one megachunk at a time

B-2

IN OUT

OUTPUTINPUT

1.13

Divide and ConquerDivide and Conquer

Phase 2Now megachunks will be the input

process each megachunk individually.

B-2

IN OUT

OUTPUTINPUT

1.14

Better: Double BufferingBetter: Double Buffering Main thread runs f(x) on one pair of I/O buffers

2nd “I/O thread” fills/drains simultaneously fills and drains the other pair of I/O bufs

Main thread ready for a new buf? All set!

Why better??????

Usable in any of the following discussionAssuming you have RAM buffers to spare!

For simplicity we won’t use this trick in our examples

f(x)RAM

InputBuffers

OutputBuffers

OUTPUTINPUT

1.15

Sometimes you need orderSometimes you need order

Rendezvous Eliminating duplicates

Summarizing groups of items

Sorted Data output must be ordered

e.g., return results in decreasing order of relevance

Upcoming fundamentals: Sort-merge join algorithm involves sorting (rendezvous)

First step in bulk loading B+ tree index (ordering)

Problem: sort 100GB of data with 1GB of RAM. why not virtual memory?

Relational Model: Unordered collections of data items (i.e.,

sets).

Relational Model: Unordered collections of data items (i.e.,

sets).

1.16

Sorting: 2-WaySorting: 2-Way

sortRAM

I/OBuffer

OUTPUTINPUT

• Pass 0: – read a page, sort it, write it.– only one buffer page is used– a repeated “batch job”

1.17

Sorting: 2-Way (cont.)Sorting: 2-Way (cont.)

Pass 1, 2, 3, …, etc. (merge): requires 3 buffer pages

note: this has nothing to do with double buffering!

merge pairs of runs into runs twice as long

a streaming algorithm, as in the previous slide!

INPUT 1

INPUT 2

OUTPUT

RAM

Merge

1.18

Two-Way External Merge SortTwo-Way External Merge Sort

Sort subfiles and Merge

How many passes?

N pages in the file => the number of passes =

Total I/O cost? (reads + writes)

Each pass we read + write each page in file. So total cost is:

Input file

1-page runs

2-page runs

4-page runs

8-page runs

PASS 0

PASS 1

PASS 2

PASS 3

9

3,4 6,2 9,4 8,7 5,6 3,1 2

3,4 5,62,6 4,9 7,8 1,3 2

2,34,6

4,7

8,91,35,6 2

2,3

4,46,7

8,9

1,23,56

1,22,3

3,4

4,56,6

7,8

1.19

Pass 0 – Create Sorted Runs

General External Merge SortGeneral External Merge Sort

More than 3 buffer pages. How can we utilize them?

To sort a file with N pages using B buffer pages: Pass 0: use B buffer pages. Produce sorted runs of

B pages each.

INPUT 1

INPUT B

Disk

INPUT 2

. . .

RAM

sort

1.20

Merging Runs

General External Merge SortGeneral External Merge Sort

Pass 1, 2, …, etc.: merge B-1 runs.

Creates runs of (B-1) * size of runs from previous pass.

INPUT 1

INPUT B-1

OUTPUT

Disk

INPUT 2

. . .

RAM

Merge

1.21

Cost of External Merge SortCost of External Merge Sort

Number of passes:

Cost = 2N * (# of passes)

E.g., with 5 buffer pages, to sort 108 page file:Pass 0: = 22 sorted runs of 5 pages

each (last run is only 3 pages)

Pass 1: = 6 sorted runs of 20 pages each (last run is only 8 pages)

Pass 2: 2 sorted runs, 80 pages and 28 pages

Pass 3: Sorted file of 108 pages

Formula check: 1+┌log4 22┐= 1+3 4 passes √

1.22

# of Passes of External Sort# of Passes of External Sort

( I/O cost is 2N times number of passes)

1.23

Memory Requirement for External Memory Requirement for External SortingSorting

How big of a table can we sort in two passes?

Each “sorted run” after Phase 0 is of size B

Can merge up to B-1 sorted runs in Phase 1

Answer: B(B-1).

Sort N pages of data in about space

1.24

Alternative: HashingAlternative: Hashing

Idea:Many times we don’t require order

E.g.: removing duplicates

E.g.: forming groups

Often just need to rendezvous matches

Hashing does thisAnd may be cheaper than sorting! (Hmmm…!)

But how to do it out-of-core??

1.25

DivideDivide

Streaming Partition (divide): Use a hash f’n hp to stream records to disk partitionsAll matches rendezvous in the same partition.

Streaming alg to create partitions on disk: “Spill” partitions to disk via output buffers

1.26

Divide & ConquerDivide & Conquer

Streaming Partition (divide): Use a hash function hp to stream records to disk-based partitionsAll matches rendezvous in the same partition.

Streaming alg to create partitions on disk: “Spill” partitions to disk via output buffers

ReHash (conquer): Read partitions into RAM-based hash table one at a time, using hash function hr

Then go through each bucket of this hash table to achieve rendezvous in RAM

Note: Two different hash functions

hp is coarser-grained than hr

1.27

Two PhasesTwo Phases

Partition:

B main memory buffers DiskDisk

Original Relation OUTPUT

2INPUT

1

hashfunction

hp B-1

Partitions

1

2

B-1

. . .

1.28

Two PhasesTwo Phases

Partition:

Rehash:

PartitionsHash table for partition

Ri (k <= B pages)

B main memory buffersDisk

Result

hashfnhr

B main memory buffers DiskDisk

Original Relation OUTPUT

2INPUT

1

hashfunction

hp B-1

Partitions

1

2

B-1

. . .

1.29

Cost of External HashingCost of External Hashing

cost = 4*N IO’s

1.30

Memory RequirementMemory Requirement

How big of a table can we hash in two passes?B-1 “partitions” result from Phase 0

Each should be no more than B pages in size

Answer: B(B-1). We can hash a table of size N pages in about space

Note: assumes hash function distributes records evenly!

Have a bigger table? Recursive partitioning!

1.31

How does this compare with How does this compare with external sortingexternal sorting??

1.32

So which is better ??So which is better ??

Simplest analysis: Same memory requirement for 2 passes

Same I/O cost

But we can dig a bit deeper…

Sorting pros: Great if input already sorted (or almost sorted) w/heapsort

Great if need output to be sorted anyway

Not sensitive to “data skew” or “bad” hash functions

Hashing pros: For duplicate elimination, scales with # of values

Not # of items! We’ll see this again.

Can exploit extra memory to reduce # IOs (stay tuned…)

1.33

Summing Up 1Summing Up 1

Unordered collection model

Read in chunks to avoid fixed I/O costs

Patterns for Big Data Streaming

Divide & Conquer

also Parallelism (but we didn’t cover this here)

1.34

Summary Part 2Summary Part 2

Sort/Hash Duality Sorting is Conquer & Merge

Hashing is Divide & Conquer

Sorting is overkill for rendezvous But sometimes a win anyhow

Sorting sensitive to internal sort alg Quicksort vs. HeapSort

In practice, QuickSort tends to be used

Don’t forget double buffering (with threads)

1.35

CS220CS220Database SystemsDatabase Systems

Query Evaluation IIQuery Evaluation II

1.36

Cost-based Query Sub-SystemCost-based Query Sub-System

Query Plan Evaluator

Query Optimizer

Plan Generator

Plan Cost Estimator

Statistics

Catalog Manager

Schema

Select *From Blah BWhere B.blah = blah

Queries

Query Parser

1.37

Review - Relational OperationsReview - 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.

Also: Aggregation (SUM, MIN, etc.) and GROUP BY Since each op returns a relation, ops can be composed! After we

cover the operations, we will discuss how to optimize queries formed by composing them.

1.38

Selection (filter) OperatorsSelection (filter) Operators

1.39

Schema for ExamplesSchema for Examples

Similar to old schema; rname added for variations. Reserves:

Each tuple is 40 bytes long, 100 tuples per page, 1000 pages.

Sailors: Each tuple is 50 bytes long, 80 tuples per page, 500 pages.

Sailors (sid: integer, sname: string, rating: integer, age: real)Reserves (sid: integer, bid: integer, day: date, rname: string)

1.40

Simple SelectionsSimple Selections

Of the form Question: how best to perform? Depends on:

what indexes/access paths are available

what is the expected size of the result (in terms of number of tuples and/or number of pages)

Size of result approximated as

size of R * reduction factor “reduction factor” is usually called selectivity.

estimate of reduction factors is based on statistics – we will discuss shortly.

SELECT *

FROM Reserves R

WHERE R.rname < ‘C%’

1.41

Alternatives for Simple Alternatives for Simple SelectionsSelections

With no index, unsorted: Must essentially scan the whole relation

cost is M (#pages in R). For “reserves” = 1000 I/Os.

With no index, sorted: cost of binary search + number of pages containing results.

For reserves = 10 I/Os + selectivity*#pages

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.)

1.42

Using an Index for SelectionsUsing an Index for Selections

Cost depends on #qualifying tuples, and clustering. Cost:

finding qualifying data entries (typically small)

plus cost of retrieving records (could be large w/o clustering).

In example “reserves” relation, if 10% of tuples qualify (result size estimate: 100 pages, 10000 tuples).

With a clustered index, cost is little more than 100 I/Os;

if unclustered, could be more than 10000 I/Os! unless…

1.43

Selections using Index (cont)Selections using Index (cont)

Important refinement for unclustered indexes:

1. Find qualifying data entries.

2. Sort the rid’s of the data records to be retrieved.

3. 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).

Index entries

Data entries

direct search for

(Index File)

(Data file)

Data Records

data entries

Data entries

Data Records

CLUSTEREDUNCLUSTERED

1.44

General Selection ConditionsGeneral Selection Conditions

Such selection conditions are first converted to conjunctive normal form (CNF): (day<8/9/94 OR bid=5 OR sid=3 ) AND

(rname=‘Paul’ OR bid=5 OR sid=3)

We only discuss the case with no ORs (a conjunction of terms of the form attr op value).

A B-tree index matches (a conjunction of) terms that involve only attributes in a prefix of the search key. Index on <a, b, c> matches a=5 AND b= 3, but not b=3.

For Hash index, must have all attributes in search key

(day<8/9/94 AND rname=‘Paul’) OR bid=5 OR sid=3

1.45

Composite Search KeysComposite Search Keys

Search on a combination of fields.

Equality query: Every field value is equal to a constant value. E.g. wrt <age,sal> index: age=12 and sal =20

Range query: Some field value is not a constant. E.g.: age > 11; or age=12 and sal > 10

Data entries in index sorted by search key to support range queries.

Lexicographic order

Like the dictionary, but on fields, not letters!

sue 13 20

bob

cal

joe 12

10

20

8011

12

name age sal

<sal, age>

<age, sal> <age>

<sal>

12,20

12,10

11,80

13,20

20,12

10,12

20,13

80,11

11

12

12

13

10

20

20

80

Data recordssorted by name

Data entries in indexsorted by <sal,age>

Data entriessorted by <sal>

Examples of composite keyindexes using lexicographic order.

1.46

Two Approaches to General Two Approaches to General SelectionsSelections

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.

Terms that match this index reduce the number of tuples retrieved; other terms are used to discard some retrieved tuples, but do not affect number of tuples/pages fetched.

1.47

Most Selective Index - ExampleMost Selective Index - Example

Consider day < 8/9/94 AND bid=5 AND sid=3. A B+ tree index on day can be used;

then, bid=5 and sid=3 must be checked for each retrieved tuple.

Similarly, a hash index on <bid, sid> could be used; Then, day<8/9/94 must be checked.

How about a B+tree on <rname,day>? How about a B+tree on <day, rname>? How about a Hash index on <day, rname>?

1.48

Intersection of RidsIntersection of Rids

Second approach: if we have 2 or more matching indexes (w/Alternatives (2) or (3) for data entries): 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.

Consider day<8/9/94 AND bid=5 AND sid=3. With a B+ tree index on day and an index on sid, we can retrieve rids of records satisfying day<8/9/94 using the first, rids of recs satisfying sid=3 using the second, intersect, retrieve records and check bid=5.

Note: commercial systems use various tricks to do this:

bit maps, bloom filters, index joins

1.49

Join OperatorsJoin Operators

1.50

Join OperatorsJoin Operators

Joins are a very common query operation. Joins can be very expensive:

Consider an inner join of R and S each with 1M records. Q: How many tuples in the answer?

(cross product in worst case, 0 in the best(?))

Many join algorithms have been developed

Can have very different join costs.

1.51

Equality Joins With One Join Equality Joins With One Join ColumnColumn

In algebra: R ⋈ S. Common! Must be carefully optimized. R S is large; so, R S followed by a selection is inefficient.

Assume: M = 1000 pages in R, pR =100 tuples per page.

N = 500 pages in S, pS = 80 tuples per page. In our examples, R is Reserves and S is Sailors.

Cost metric : # of I/Os. We will ignore output costs. We will consider more complex join conditions later.

SELECT *

FROM Reserves R1, Sailors S1

WHERE R1.sid=S1.sid

1.52

Simple Nested Loops JoinSimple Nested Loops Join

For each tuple in the outer relation R, we scan the entire inner relation S.

How much does this Cost?

(pR * M) * N + M = 100,000*500 + 1000 I/Os.

At 10ms/IO, Total: ???

What if smaller relation (S) was outer?

(ps * N) *M + N = 40,000*1000 + 500 I/Os.

What assumptions are being made here?

foreach tuple r in R doforeach tuple s in S do

if ri == sj then add <r, s> to result

Q: What is cost if one relation can fit entirely in memory?

1.53

Page-Oriented Nested Loops JoinPage-Oriented Nested Loops Join

For each page of R, get each page of S, and write out matching pairs of tuples <r, s>, where r is in R-page and S is in S-page.

What is the cost of this approach?

M*N + M= 1000*500 + 1000 If smaller relation (S) is outer, cost = 500*1000 + 500

foreach page bR in R do foreach page bS in S do foreach tuple r in bR do

foreach tuple s in bSdoif ri == sj then add <r, s> to result

1.54

Block Nested Loops JoinBlock Nested Loops Join Page-oriented NL doesn’t exploit extra buffers. Alternative approach: 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 <r, s> to result. Then read next R-block, scan S, etc.

. . .

. . .

R & Sblock of R tuples(k < B-1 pages)

Input buffer for S Output buffer

. . .

Join Result

1.55

Examples of Block Nested LoopsExamples of Block Nested Loops Cost:

Scan of outer + #outer blocks * scan of inner #outer blocks = ceiling(#pages of outer/blocksize)

With Reserves (R) as outer, and 100 pages/Block: Cost of scanning R is 1000 I/Os; a total of 10 blocks.

Per block of R, we scan Sailors (S); 10*500 I/Os.

If space for just 90 pages of R, we would scan S 12 times.

With 100-page block of Sailors as outer: Cost of scanning S is 500 I/Os; a total of 5 blocks.

Per block of S, we scan Reserves; 5*1000 I/Os.

1.56

Index Nested Loops JoinIndex Nested Loops Join

If there is an index on the join column of one relation (say S), can make it the inner and exploit the index. Cost: M + ( (M*pR) * 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 (assuming Alt. (2) or (3) for data entries) depends on clustering.

Clustered index: 1 I/O per page of matching S tuples. Unclustered: up to 1 I/O per matching S tuple.

foreach tuple r in R doforeach tuple s in S where ri == sj do

add <r, s> to result

1.57

Hash-index (Alt. 2) on sid of Sailors (as inner): Scan Reserves: 1000 page I/Os, 100*1000 tuples.

For each Reserves tuple: 1.2 I/Os to get data entry in index, plus 1 I/O to get (the exactly one) matching Sailors tuple. Total:

Hash-index (Alt. 2) on sid of Reserves (as inner): Scan Sailors: 500 page I/Os, 80*500 tuples.

For each Sailors tuple: 1.2 I/Os to find index page with data entries, plus cost of retrieving matching Reserves tuples. Assuming uniform distribution, 2.5 reservations per sailor (100,000 / 40,000). Cost of retrieving them is 1 or 2.5 I/Os depending on whether the index is clustered.

Totals:

Examples of Index Nested Examples of Index Nested LoopsLoops

1.58

Sort-Merge Join (R S)Sort-Merge Join (R S)

Sort R and S on the join column, then scan them to do a ``merge’’ (on join col.), and output result tuples.

Particularly useful if one or both inputs are already sorted on join attribute(s)

output is required to be sorted on join attributes(s)

“Merge” phase can require some back tracking if duplicate values appear in join column

R is scanned once; each S group is scanned once per matching R tuple. (Multiple scans of an S group will probably find needed pages in buffer.)

1.59

Example of Sort-Merge JoinExample of Sort-Merge Join

Cost: Sort S +Sort R + (M+N) The cost of merging: usually M+N,

worst case is M*N (but very unlikely!)

With 35, 100 or 300 buffer pages, both Reserves and Sailors can be sorted in 2 passes; total join cost: 7500.

(BNL cost: 2500 to 15000 I/Os)

1.60

Refinement of Sort-Merge JoinRefinement of Sort-Merge Join

We can combine the merging phases in the sorting of R and S with the merging required for the join. Pass 0 as before, but apply to both R then S before merge.

If B > , where L is the size of the larger relation, using the sorting refinement that produces runs of length 2B in Pass 0, #runs of each relation is < B/2.

In “Merge” phase: Allocate 1 page per run of each relation, and `merge’ while checking the join condition

Cost: read+write each relation in Pass 0 + read each relation in (only) merging pass (+ writing of result tuples).

In example, cost goes down from 7500 to 4500 I/Os.

In practice, the I/O cost of sort-merge join, like the cost of external sorting, is linear.

1.61

Impact of BufferingImpact of Buffering

If several operations are executing concurrently, estimating the number of available buffer pages is guesswork.

Repeated access patterns interact with buffer replacement policy. e.g., Inner relation is scanned repeatedly in Simple Nested Loop

Join. With enough buffer pages to hold inner, replacement policy does not matter. Otherwise, MRU is best, LRU is worst (sequential flooding).

Does replacement policy matter for Block Nested Loops?

What about Index Nested Loops? Sort-Merge Join?

1.62

Hash-JoinHash-Join

Partition both relations on the join attributes using hash function h.

R tuples in partition Ri

will only match S tuples in partition Si.

For i= 1 to #partitions {

Read in partition Ri

and hash it using h2 (not h).

Scan partition Si and

probe hash table for matches.

}

Partitionsof R & S

Input bufferfor Si

Hash table for partitionRi (k < B-1 pages)

B main memory buffersDisk

Output buffer

Disk

Join Result

hashfnh2

h2

B main memory buffers DiskDisk

Original Relation OUTPUT

2INPUT

1

hashfunction

h B-1

Partitions

1

2

B-1

. . .

1.63

Observations on Hash-JoinObservations on Hash-Join

#partitions k < B, and B-1 > size of largest partition to be held in memory. Assuming uniformly sized partitions, and maximizing k, we get:

k= B-1, and M/(B-1) < B-2, i.e., B must be >

Since we build an in-memory hash table to speed up the matching of tuples in the second phase, a little more memory is needed.

If the hash function does not partition uniformly, one or more R partitions may not fit in memory. Can apply hash-join technique recursively to do the join of this R-partition with corresponding S-partition.

1.64

Cost of Hash-JoinCost of Hash-Join

In partitioning phase, read+write both relns; 2(M+N). In matching phase, read both relns; M+N I/Os.

In our running example, this is a total of 4500 I/Os.

Sort-Merge Join vs. Hash Join: Given a minimum amount of memory (what is this, for each?) both

have a cost of 3(M+N) I/Os. Hash Join superior if relation sizes differ greatly (e.g., if one reln fits in memory). Also, Hash Join shown to be highly parallelizable.

Sort-Merge less sensitive to data skew; result is sorted.

1.65

Set OperationsSet Operations Intersection and cross-product special cases of join. Union (Distinct) and Except similar; we’ll do union.

Sorting based approach to union: Sort both relations (on combination of all attributes).

Scan sorted relations and merge them.

Alternative: Merge runs from Pass 0 for both relations.

Hash based approach to union: Partition R and S using hash function h.

For each S-partition, build in-memory hash table (using h2), scan corr. R-partition and add tuples to table while discarding duplicates.

1.66

General Join ConditionsGeneral Join Conditions

Equalities over several attributes

(e.g., R.sid=S.sid AND R.rname=S.sname): For Index NL, build index on <sid, sname> (if S is inner); or use

existing indexes on sid or sname.

For Sort-Merge and Hash Join, sort/partition on combination of the two join columns.

Inequality conditions (e.g., R.rname < S.sname): For Index NL, need (clustered!) B+ tree index.

Range probes on inner; # matches likely to be much higher than for equality joins.

Hash Join, Sort Merge Join not applicable!

Block NL quite likely to be the best join method here.

1.67

ReviewReview Implementation of Relational Operations as Iterators

Focus largely on External algorithms (sorting/hashing)

Choices depend on indexes, memory, stats,…

Joins Blocked nested loops:

simple, exploits extra memory

Indexed nested loops:

best if 1 rel small and one indexed

Sort/Merge Join

good with small amount of memory, bad with duplicates

Hash Join

fast (if enough memory), bad with skewed data

Relatively easy to parallelize

1.68

Aggregation OperatorsAggregation Operators

1.69

Schema for ExamplesSchema for Examples

Similar to old schema; rname added for variations. Reserves:

Each tuple is 40 bytes long, 100 tuples per page, 1000 pages. So, M = 1000, pR = 100.

Sailors: Each tuple is 50 bytes long, 80 tuples per page, 500 pages.

So, N = 500, pS = 80.

Sailors (sid: integer, sname: string, rating: integer, age: real)Reserves (sid: integer, bid: integer, day: dates, rname: string)

1.70

Aggregate Operations (AVG, MIN, Aggregate Operations (AVG, MIN, etc.)etc.)

Without grouping: In general, requires scanning the relation.

Given a tree index whose search key includes all attributes in the SELECT or WHERE clauses, can do index-only scan.

With grouping: Sort on group-by attributes, then scan relation and compute

aggregate for each group. (Better: combine sorting and aggregate computation.)

Similar approach based on hashing on group-by attributes.

Given a tree index whose search key includes all attributes in SELECT, WHERE and GROUP BY clauses, can do index-only scan; if group-by attributes form prefix of search key, can retrieve data entries/tuples in group-by order.

1.71

Sort GROUP BY: Naïve SolutionSort GROUP BY: Naïve Solution

The Sort iterator naturally permutes its input so that all tuples are output in sequence

The Aggregate iterator keeps running info (“transition values” or “transVals”) on agg functions in the SELECT list, per group: Example transVals:

For COUNT, it keeps count-so-far

For SUM, it keeps sum-so-far

For AVERAGE it keeps sum-so-far and count-so-far As soon as the Aggregate iterator sees a tuple from a new

group:

1. It produces an output for the old group based on the agg function

E.g. for AVERAGE it returns (sum-so-far/count-so-far)

2. It resets its running info.

3. It updates the running info with the new tuple’s info

Sort

Aggregate

1.72

Sort GROUP BY: Naïve SolutionSort GROUP BY: Naïve Solution

Sort

Aggregate

AB

CD

AB

A

ABCDABA

A

<A, 1><B,1>

AAB

<B,2>

B

<A, 2><A, 3>

A, 3

C

B, 2C, 1D, 1

1.73

Hash GROUP BY: Naïve SolutionHash GROUP BY: Naïve Solution(similar to the Sort GROUPBY)(similar to the Sort GROUPBY)

The Hash iterator permutes its input so that all tuples are output in groups.

The Aggregate iterator keeps running info (“transition values” or “transVals”) on agg functions in the SELECT list, per group E.g., for COUNT, it keeps count-so-far For SUM, it keeps sum-so-far For AVERAGE it keeps sum-so-far and count-so-far

When the Aggregate iterator sees a tuple from a new group:

1. It produces an output for the old group based on the agg functionE.g. for AVERAGE it returns (sum-so-far/count-so-far)

2. It resets its running info.

3. It updates the running info with the new tuple’s info

Hash

Aggregate

1.74

ExternaExternall

HashinHashingg

Partition:

Each group will

be in a single

disk-based partition file. But those files have many groups inter-mixed.

Rehash:

For Each Partition i:

hash i into an in-memory hash table

Return results until

records exhuasted then i++

B main memory buffers DiskDisk

Original Relation OUTPUT

2INPUT

1

hashfunction

hp B-1

Partitions

1

2

B-1

. . .

PartitionsHash table for partition

Ri (k < B pages)

B main memory buffersDisk

Result

fn INPUT

hashfunction

h2p

1.75

We Can Do Better!We Can Do Better!

Put summarization into the hashing process During the ReHash phase, don’t store tuples, store pairs of

the form <GroupVals, TransVals> When we want to insert a new tuple into the hash table

If we find a matching GroupVals, just update the TransVals appropriately

Else insert a new <GroupVals,TransVals> pair

What’s the benefit? Q: How many pairs will we have to maintain in the rehash

phase? A: Number of distinct values of GroupVals columns

Not the number of tuples!! Also probably “narrower” than the tuples

HashAgg

1.76

We Can Do Even Better Than We Can Do Even Better Than That: That:

Hybrid HashingHybrid Hashing What if the set of <GroupVals,TransVals> pairs fits in

memory? It would be a waste to spill all the tuples to disk and

read them all back back again! Recall <G,T> pairs may fit even if there are tons of

tuples! Idea: keep <G,T> pairs for a smaller 1st partition in

memory during phase 1! Output its stuff

at the end of Phase 1.

Q: how do wechoose the number of buffers (k) to allocate to this special partition?

1

B main memory buffers DiskDisk

Original Relation OUTPUT

3

INPUT

2

hh B-k

Partitions

2

3

B-k

. . .hr

k-buffer hashtable of <G,T> pairs

1.77

A Hash Function for Hybrid A Hash Function for Hybrid HashingHashing

Assume we like the hash-partition function hp

Define hh operationally as follows:

hh(x) = 1 if x maps to a <G,T> already in the in-memory hashtable

hh(x) = 1 if in-memory hashtable is not yet full (add new <G,T>)

hh(x) = hp(x) otherwise

This ensures that:

Bucket 1 fits in kpages of memory

If the entire set ofdistinct hashtableentries is smaller than k, we do no spilling!

1

B main memory buffers DiskDisk

Original Relation OUTPUT

3

INPUT

2

hh B-k

Partitions

2

3

. . .hr

k-buffer hashtable

1.78

Projection Projection (DupElim)(DupElim)

Issue is removing duplicates. Basic approach is to use sorting

1. Scan R, extract only the needed attrs (why do this 1st?)

2. Sort the resulting set

3. Remove adjacent duplicates

Cost: Reserves with size ratio 0.25 = 250 pages. With 20 buffer pages can sort in 2 passes, so1000 +250 + 2 * 2 * 250 + 250 = 2500 I/Os

Can improve by modifying external sort algorithm:

Modify Pass 0 of external sort to eliminate unwanted fields.

Modify merging passes to eliminate duplicates.

Cost: for above case: read 1000 pages, write out 250 in runs of 40 pages, merge runs = 1000 + 250 +250 = 1500.

SELECT DISTINCT R.sid, R.bid

FROM Reserves R

1.79

DupElim Based on HashingDupElim Based on Hashing Just like our discussion of GROUP BY and aggregation from before!

But the aggregation function is missing

SELECT DISTINCT R.sid, R.bid FROM Reserves R SELECT R.sid, R.bid FROM Reserves R GROUP BY R.sid, R.bid

Cost for Hashing? Without “hybrid” assuming partitions fit in memory (i.e. #bufs >= square root of the

#of pages of projected tuples) read 1000 pages and write out partitions of projected tuples (250

pages) Do dup elim on each partition (total 250 page reads) Total : 1500 I/Os.

With “hybrid hash”: subtract the I/O costs of 1st partition

1.80

DupElim & IndexesDupElim & Indexes

If an index on the relation contains all wanted attributes in its search key, can do index-only scan. Apply projection techniques to data entries (much smaller!)

If an ordered (i.e., tree) index contains all wanted attributes as prefix of search key, can do even better: Retrieve data entries in order (index-only scan), discard unwanted

fields, compare adjacent tuples to check for duplicates.

Same tricks apply to GROUP BY/Aggregation

1.81

Summary of Query EvaluationSummary of Query Evaluation Queries are composed of a few basic operators;

The implementation of these operators can be carefully tuned (and it is important to do this!).

Operators are “plug-and-play” due to the Iterator model.

Many alternative implementation techniques for each operator; no universally superior technique for most.

Must consider alternatives for each operation in a query and choose best one based on statistics, etc.

This is part of the broader task of Query Optimization, which we will cover next!