Database Systems
(資料庫系統)
December 6, 2004
Chapter 14
By Hao-hua Chu (朱浩華)
1
Announcement
• Next week reading: Chapter 15
• Assignment #9 will be out today.
– It is due in two weeks.
2
Evaluating Relational Operators
Chapter 14
3
Outline
• Continue from Chapter 12 (Overview of Query
Evaluation)
• Discuss more on the different evaluation algorithms for
relational operators:
–
–
–
–
–
Selection
Projection
Join
Set Operations
Aggregate
4
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
• 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.
5
Schema for Examples
Sailors (sid: integer, sname: string, rating: integer, age: real)
Reserves (sid: integer, bid: integer, day: dates, rname: string)
• Similar to old schema; rname added for variations.
– Rname is the name of the agent making the reservation on behalf
of a sailor.
• 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.
Again, we will look at cost measured in # of disk I/Os.
6
Selection Operation
SELECT
FROM
WHERE
*
Reserves R
R.rname=‘Joe’
• Several alternative algorithms for selection
• No algorithm is always better (disk I/O costs) than the others,
depending on the file organizations (sorted, clustered index, etc.).
• Consider four situations:
–
–
–
–
No Index, Unsorted Data
No Index, Sorted Data
B+ Tree Index
Hash Index
• Two types of selection conditions:
– Conjunctions (term_1 AND term_2)
– Disjunctions (term_1 OR term_2)
7
No Index for Selections
SELECT
FROM
WHERE
*
Reserves R
R.rname=‘Joe’
• Unsorted Data, no Index (on rname)
– Algorithm: scan the file
– Cost: 1000 page I/Os
• Sorted Data (on rname), no Index (on rname)
– Algorithm: binary search to find the first qualifying tuple, then
scan to find remaining qualifying tuples.
– Cost of binary search is O(log2 1000) ~ 10 page I/Os.
– Cost of scanning: number of pages containing qualifying tuples.
8
Using B+ Tree Index for
Selections
• Cost depends on (#qualifying tuples), and (whether the
index is clustered or not).
–
–
–
–
Cost of finding qualifying data entries (typically small) plus cost of
retrieving tuples (could be large without clustering).
In example, assuming uniform distribution of names, about 10% of
tuples qualify (100 pages, 10,000 tuples).
With clustered index, cost is little more than 100 page I/Os.
With unclustered index, cost can be up to 10,000 page I/Os!
•
•
Why? Each qualified tuple may be on a different page. This is worse
than the naïve file scan (1,000 page I/Os).
There is one possible refinement that can bound the cost <= cost of file
scan.
9
Refinement for Unclustered
Indexes
1. Find qualifying data entries.
2. Sort the rid’s of the data tuples to be retrieved (e.g., by their pageNos)
RID = <pageNO, slotNo>
3. Fetch rids in order. This ensures that each data page is looked at
just once.
• The cost is bounded by the number of page I/Os (no
worse than file scan).
10
Using Hash Index for Selections
• Good only for equality selection conditions.
– Does not apply to range selections.
• The cost of using the hash index is a sum of
– Retrieving the bucket (overflow) page in the index requires only
1~2 page I/Os.
– Retrieving all qualified tuples requires up to (# qualified tuples)
page I/Os, given that each qualified tuple can be on different
page.
• Can apply the refinement of sorting of rids on PageNo.
11
General Selections
• So far, we have considered simple selection condition in
the form of (R.rname=‘Joe’).
• A general selection condition is a boolean combination of
– (term_1 op term_2 … op term_n), where op can be AND | OR.
• Given a general selection condition:
(1) rewrite the general condition in conjunctive normal form (CNF):
(day < 8/9/02 AND rname = ‘Joe’) OR bid=5 OR sid=3
Can be transformed into
(day < 8/9/02 OR bid=5 OR sid=3) AND (rname=‘Joe’ OR bid=5 OR sid=3)
(2) Try to match conjunct with index (Chapter 12)
(3) Evaluating Selections with Conjunctions / with Disjunctions
12
Two Approaches in Evaluating
Selections with Conjunction
• 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.
– The selectivity of a conjunct is proportional to the reduction factor
of applying this conjunct to the tuples.
– 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;
day<8/9/94 must then be checked.
–
13
Intersection of Rids
• Second approach (if we have 2 or more matching indexes
that use Alternatives (2) or (3) for data entries):
–
–
–
–
•
Get sets of rids of data tuples using each matching index.
Then intersect these sets of rids.
Retrieve the tuples and apply any remaining terms.
(Intersection on rids, not on the tuples)
Example: day<8/9/94 AND bid=5 AND sid=3.
–
If we have a B+ tree index on day and an index on sid, both using
Alternative (2), we can retrieve rids of tuples satisfying day<8/9/94
using the first, rids of recs satisfying sid=3 using the second,
intersect, retrieve tuples and check bid=5.
14
Selections with Disjunction
• Consider the example #1:
(day < 8/9/02 OR rname=‘Joe’), index on rname
– We could check (rname=‘Joe’) using index, but evaluating (day <
8/9/02) requires a file scan, so might as well as checking for
condition during file scan.
– File scan is the most selective path.
• Consider example #2:
(day < 8/9/02 OR rname=‘Joe’) AND sid=3, indexes on rname and sid
– (sid = 3) is the most selective path, similar to approach #1.
• Consider example #3:
(day < 8/9/02 OR rname=‘Joe’), indexes on day and rname
– Union of Rids matching (day < 8/9/02) and matching
(rname=‘Joe’), similar to approach #2.
15
The Projection Operation
• A simple approach based on sorting:
SELECT DISTINCT
FROM
R.sid, R.bid
Reserves R
(1) Scan R and produce a set of tuples that contain only the projected
attributes
(2) External Merge-Sort this set of tuples.
(3) Scan the sorted result, comparing adjacent tuples to discard
duplicates.
•
What is the cost?
–
–
–
–
Step (1) requires reading1,000 page I/Os, and writing out
(assuming that projected attributes is ¼ size of a original tuple) 250
page I/Os.
Step (2) requires (assuming 20 buffer pages) 2 passes (1 + logB-1
250/B). The cost is 2*2*250 = 1000 page I/Os.
Step (3) requires 250 page I/Os.
Total cost is 2,500 page I/Os.
16
The Projection Operation
SELECT DISTINCT
FROM
R.sid, R.bid
Reserves R
• An improved approach on sorting:
–
–
–
(Combine step 1 with step 2): modify Pass 0 of external sort to
eliminate unwanted fields. Also use replacement sort to produce
runs of about 2B pages.
(Combine step 2 with step 3): Modify merging passes to eliminate
duplicates.
Cost: In Pass 0, read original relation (1000 page I/Os), write out
same number of smaller tuples (250 page I/Os). In the one
merging pass, read the runs (250 page I/Os). The total cost is 1500
page I/Os (compared to 2500 page I/Os in the simple approach).
17
Projection Based on Hashing
• The basic idea is that duplicate tuples will be hashed to the same
bucket. So you can look at tuples in each bucket and eliminate
duplicates.
• Partitioning phase: Read R using one input buffer. For each tuple,
discard unwanted fields, apply hash function h1 to choose one of B-1
output buffers.
–
Result is B-1 partitions (of tuples with no unwanted fields). 2 tuples from
different partitions guaranteed to be distinct.
• Duplicate elimination phase: For each partition, read it and build an inmemory hash table, using hash fn h2 (<> h1) on all fields, while
discarding duplicates.
– Why hashing 2nd time? Minimize collisions (two unique tuples hashing to
the same bucket).
• Cost: In partitioning phase, read 1000 pages I/Os and write out 250 page I/Os.
In elimination phase, read 250 page I/Os. Total is 1500.
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Discussion of Projection
• Sort-based approach is the standard, why?
– No need to worry if distribution of hash function is uniform.
– Result is sorted.
• If an index on the relation contains all projected attributes in
its search key, can do index-only scan.
–
Apply projection techniques to data entries (much smaller!)
19
Equality Joins With One Join Column
SELECT
FROM
WHERE
*
Reserves R1, Sailors S1
R1.sid=S1.sid
• R ⋈ S is very common, so must be carefully optimized.
• R x S is large; so, R x S followed by a selection is
inefficient.
• Assume: M tuples in R, pR tuples per page, N tuples in S,
pS tuples per page.
–
In our examples, R is Reserves and S is Sailors.
• We will consider more complex join conditions later.
• Cost metric: # of page I/Os.
20
Two Classes of Algorithms to
Implement Join Operation
• Algorithms in class 1 require enumerating all tuples in
the cross-product and discard tuples that do not meet
the join condition.
– Simple Nested Loops Join
– Blocked Nested Loops Join
• Algorithms in class 2 avoid enumerating the crossproduct.
– Index Nested Loops Join
– Sort-Merge Join
– Hash Join
21
Simple Nested Loops Join
foreach tuple r in R do
foreach tuple s in S do
if ri == sj then add <r, s> to result
• For each tuple in the outer relation R, scan the inner relation S
– Scan outer table R once, but inner table S pR * M times!
– Cost: M + pR * M * N = 1000 + 100*1000*500 I/Os => very huge.
– Want to select the smaller table as the inner table
• Simple improvement Page-oriented Nested Loops join
– For each page of R, get each page of S, and write out matching pairs of
tuples (scan S total of M times) => require 3 buffer pages
– Cost: M + M*N = 1000 + 1000*500 => still huge.
– If smaller relation (S) is outer, cost = 500 + 500*1000
• Given buffer pages, what improvement can you make?
22
Block Nested Loops Join
• Use one page as an input buffer for scanning the inner S.
• Use one page as the output buffer.
• Use remaining (B-2) pages to hold ``block’’ of outer R.
–
–
•
For each matching tuple r in R-block & s in S-page, add <r, s> to result.
Continue with the next R-block.
How many times is the inner table (S) scanned?
–
Ceiling ( M / (B-2) )
foreach block of B-2 pages of R do
foreach page of S do
for all matching tuples r in R-block and
s in S-page, add <r, s> to result
23
Block Nested Loops Join: Efficient
Matching Pairs
• Large B -> slow to match S-tuple with tuples in R-block (R-block has B-2
pages).
• Solution: build in-memory hash table for R-block.
R&S
Hash table for block of R
Join Result
...
...
...
Input buffer for S
Output buffer
24
Examples of Block Nested Loops
• Cost: Scan of outer + #outer blocks * scan of inner
• With Reserves (R) as outer, and 102 buffer pages:
–
–
–
Cost of scanning R is 1000 I/Os; a total of 10 R-blocks.
For each R-block, S is scanned once; 10*500 I/Os.
Total cost = 1000 + 10 * 500 = 6000 page I/Os => huge improvement over
page-oriented nested loops join.
• With 100-page block of Sailors as outer:
–
–
–
•
Cost of scanning S is 500 I/Os; a total of 5 blocks.
For each S-block, R is scanned once; 5*1000 I/Os.
Total cost = 500 + 5*1000 = 5500 page I/Os
For blocked access (block I/Os are more efficient), it may be best to
divide buffers evenly between R and S.
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Index Nested Loops Join
foreach tuple r in R do
foreach tuple s in S where ri == sj do
add <r, s> to result
• Block/simple nested loops joins do not use existing (or create) index.
• 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 (typical) for all matching S tuples
Unclustered index: up to 1 I/O per matching S tuple.
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Examples of Index Nested Loops
•
Hash-index (Alt. 2) on sid of Sailors (as inner):
–
–
–
•
Hash-index (Alt. 2) on sid of Reserves (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 + 1 I/O to get (the
exactly one) matching Sailors tuple.
Total: 1000 + 100,000 * 2.2 = 221,000 I/Os.
Scan Sailors: 500 page I/Os, 80*500 tuples.
For each Sailors tuple: 1.2 I/Os to find index page with data entries + 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.
Total (Clustered): 500 + 40,000 * 2.2 = 88,500 I/Os.
Given choices, put the relation with higher # tuples as inner loop.
Index Nested Loop performs better than simple nested loop.
How does it compare to block nested loop?
–
Depend on the # matching tuples
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Sort-Merge Join
• Sort R and S on the join column, then scan them to do a
``merge’’ (on join col.), and output result tuples.
–
–
–
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 in Ri (current R group) and
all S tuples with same value in Sj (current S group) match; output <r,
s> for all pairs of such tuples.
Then resume scanning R and S.
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Example of Sort-Merge Join (ScanMerge Two Sorted Relations)
Gs
1
2
5
sid
22
28
31
44
58
sname rating age
dustin
7
45.0
yuppy
9
35.0
lubber
8
55.5
guppy
5
35.0
rusty
10 35.0
Tr
1
3
4
6
sid
28
28
31
31
31
58
Ts
2a
3a
2b
3b
bid
103
103
101
102
101
103
day
12/4/96
11/3/96
10/10/96
10/12/96
10/11/96
11/12/96
5a
6a
5b
6b
rname
guppy
yuppy
dustin
lubber
lubber
dustin
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Cost of Merge-Sort Join
• R is scanned once; each S group is scanned once per
matching R tuple. (Multiple scans of an S group are
likely to find needed pages in buffer.)
• Cost: 2 M log M + 2 N log N + (M+N)
–
–
–
–
The cost of merge-sorting two relations is 2 M log M + 2 N log N.
With 35, 100 or 300 buffer pages, both Reserves and Sailors can
be sorted in 2 passes
The cost of scan-merge two sorted relations is M+N.
Total join cost: 2*2*1000 + 2*2*500 + 1000 + 500 = 7500 page
I/Os.
30
Refinement of Sort-Merge Join
• We can combine the merging phases in the sorting of R
and S with the merging required for the join.
– We need to have buffer size B > L , where L is the size of the
–
–
–
larger relation. Why? It means that #runs after Pass 0 for both
relations will be < B (#runs is 2(L/2B) = L/B < B), then you can do
the next thing …
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.
R
sort-merge
S
scan-merge
31
Hash Join
• Hash both relations (R & S) on the join attribute using the
same hash function h.
– Create two hash tables called Partition R & Partition S
• Tuples in Partition Ri only match tuples in Partition Si !
– Similar idea to sort merge join
• For each bucket i, locate matching tupes in Partition Ri and
Partition Si
– For efficient matching pairs, (1) hash tuples in Partition Ri using
another hash function h2, and (2) for each tuple in Partition Si , use
h2 to find matching R tuples.
32
Hash-Join
• Partition both relations
using hash fn h: R
tuples in partition i will
only match S tuples in
partition i.
• Read in a partition of R,
hash it using h2 (<> h!).
Scan matching partition
of S, search for matches.
Original
Relation
OUTPUT
1
Partitions
1
2
INPUT
2
hash
function
...
h
B-1
B-1
Disk
B main memory buffers
Partitions
of R & S
Disk
Join Result
hash
fn
Hash table for partition
Ri
h2
h2
Input buffer
for Si
Disk
Output
buffer
B main memory buffers
Disk33
Observations on Hash-Join
• #partitions k < B-1 (need one buffer page for reading), and B-2
> 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 >
M
• If we build an in-memory hash table to speed up the matching
of tuples, 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.
34
Cost 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:
–
–
–
–
Same amount of buffer pages.
Same cost of 3(M+N) I/Os.
Hash Join shown to be highly parallelizable.
Sort-Merge less sensitive to data skew; result is sorted.
35
General Join Conditions
• So far, we have only discussed single equality join condition.
• How about equalities over several attributes? (e.g., R.sid=S.sid AND
R.rname=S.sname):
–
–
For Index Nested Loops join, build index on <sid, sname> on R (if R 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.
• How about inequality conditions? (e.g., R.rname < S.sname):
–
–
–
For Index Nested Loops join, need (clustered!) B+ tree index.
Hash Join not applicable.
How about Block Nested Loops join? Is it applicable?
36
Set Operations
• Intersection and cross-product are special cases of join.
– Intersection is equality on all fields in join condition. None for crossproduct.
• 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.
• 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. Rpartition and add tuples to table while discarding duplicates.
37
Aggregate Operations (AVG, MIN,
AVG(S.age)
etc.) SELECT
FROM Sailors S
• Without grouping:
–
–
In general, requires scanning the relation and maintain some
running information (sum of age & number of sailors).
Given index whose search key includes all attributes in the SELECT
or WHERE clauses, can do index-only scan.
• With grouping:
–
–
–
GROUP-BY S.rating
Sort on group-by attributes (rating), then scan relation and compute
aggregate for each group.
Similar approach based on hashing on rating.
Given 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.
38
Impact of Buffering
• If several operations are executing concurrently, they share
buffer pool; therefore it is difficult to estimate the number of
available buffer pages.
• 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, replacement policy is critical. MRU is best, LRU
is worst (sequential flooding).
Does replacement policy matter for Block Nested Loops? (No, why?)
39