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Chapter 12: Query Processing
· Overview
· Catalog Information for Cost Estimation
· Measures of Query Cost
· Selection Operation
· Sorting
· Join Operation
· Other Operations
· Evaluation of Expressions
· Transformation of Relational Expressions
· Choice of Evaluation Plans
Database Systems Concepts 12.1 Silberschatz, Korth and Sudarshan c 1997
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Basic Steps in Query Processing
1. Parsing and translation
2. Optimization
3. Evaluation
Query
Parser &
Translator
Relational Algebra
Expression
Optimizer
Execution PlanEvaluation EngineQuery
Output
Data Statistics
About Data
Database Systems Concepts 12.2 Silberschatz, Korth and Sudarshan c 1997
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Basic Steps in Query Processing (Cont.)
Parsing and translation
· translate the query into its internal form. This is then translated
into relational algebra.
· Parser checks syntax, verifies relations
Evaluation
· The query-execution engine takes a query-evaluation plan,
executes that plan, and returns the answers to the query.
Database Systems Concepts 12.3 Silberschatz, Korth and Sudarshan c 1997
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Basic Steps in Query Processing
Optimization ­ finding the cheapest evaluation plan for a query.
· Given relational algebra expression may have many equivalent
expressions
E.g. balance<2500(balance(account)) is equivalent to
balance(balance<2500(account))
· Any relational-algebra expression can be evaluated in many
ways. Annotated expression specifying detailed evaluation
strategy is called an evaluation-plan.
E.g. can use an index on balance to find accounts with balance
< 2500, or can perform complete relation scan and discard
accounts with balance  2500
· Amongst all equivalent expressions, try to choose the one with
cheapest possible evaluation-plan. Cost estimate of a plan
based on statistical information in the DBMS catalog.
Database Systems Concepts 12.4 Silberschatz, Korth and Sudarshan c 1997
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Catalog Information for Cost Estimation
· nr : number of tuples in relation r.
· br : number of blocks containing tuples of r.
· sr : size of a tuple of r in bytes.
· fr : blocking factor of r -- i.e., the number of tuples of r that fit
into one block.
· V(A, r): number of distinct values that appear in r for attribute
A; same as the size of A(r).
· SC(A, r): selection cardinality of attribute A of relation r;
average number of records that satisfy equality on A.
· If tuples of r are stored together physically in a file, then:
br =
nr
fr
Database Systems Concepts 12.5 Silberschatz, Korth and Sudarshan c 1997
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Catalog Information about Indices
· fi : average fan-out of internal nodes of index i, for
tree-structured indices such as B+-trees.
· HTi : number of levels in index i -- i.e., the height of i.
­ For a balanced tree index (such as a B+-tree) on attribute A
of relation r, HTi = logfi
(V(A, r) .
­ For a hash index, HTi is 1.
· LBi: number of lowest-level index blocks in i -- i.e., the number
of blocks at the leaf level of the index.
Database Systems Concepts 12.6 Silberschatz, Korth and Sudarshan c 1997
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Measures of Query Cost
· Many possible ways to estimate cost, for instance disk
accesses, CPU time, or even communication overhead in a
distributed or parallel system.
· Typically disk access is the predominant cost, and is also
relatively easy to estimate. Therefore number of block transfers
from disk is used as a measure of the actual cost of evaluation.
It is assumed that all transfers of blocks have the same cost.
· Costs of algorithms depend on the size of the buffer in main
memory, as having more memory reduces need for disk
access. Thus memory size should be a parameter while
estimating cost; often use worst case estimates.
· We refer to the cost estimate of algorithm A as EA. We do not
include cost of writing output to disk.
Database Systems Concepts 12.7 Silberschatz, Korth and Sudarshan c 1997
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Selection Operation
· File scan ­ search algorithms that locate and retrieve records
that fulfill a selection condition.
· Algorithm A1 (linear search). Scan each file block and test all
records to see whether they satisfy the selection condition.
­ Cost estimate (number of disk blocks scanned) EA1 = br
­ If selection is on a key attribute, EA1 = (br / 2) (stop on
finding record)
­ Linear search can be applied regardless of
 selection condition, or
 ordering of records in the file, or
 availability of indices
Database Systems Concepts 12.8 Silberschatz, Korth and Sudarshan c 1997
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Selection Operation (Cont.)
· A2 (binary search). Applicable if selection is an equality
comparison on the attribute on which file is ordered.
­ Assume that the blocks of a relation are stored contiguously
­ Cost estimate (number of disk blocks to be scanned):
EA2 = log2(br ) +
SC(A, r)
fr
- 1
 log2(br ) -- cost of locating the first tuple by a binary
search on the blocks
 SC(A, r) -- number of records that will satisfy the
selection
 SC(A, r)/fr -- number of blocks that these records will
occupy
­ Equality condition on a key attribute: SC(A, r) = 1; estimate
reduces to EA2 = log2(br )
Database Systems Concepts 12.9 Silberschatz, Korth and Sudarshan c 1997
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Statistical Information for Examples
· faccount = 20 (20 tuples of account fit in one block)
· V(branch-name, account) = 50 (50 branches)
· V(balance, account) = 500 (500 different balance values)
· naccount = 10000 (account has 10,000 tuples)
· Assume the following indices exist on account:
­ A primary, B+
-tree index for attribute branch-name
­ A secondary, B+
-tree index for attribute balance
Database Systems Concepts 12.10 Silberschatz, Korth and Sudarshan c 1997
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Selection Cost Estimate Example
branch-name="Perryridge"(account)
· Number of blocks is baccount = 500: 10, 000 tuples in the
relation; each block holds 20 tuples.
· Assume account is sorted on branch-name.
­ V(branch-name, account) is 50
­ 10000/ 50 = 200 tuples of the account relation pertain to
Perryridge branch
­ 200/ 20 = 10 blocks for these tuples
­ A binary search to find the first record would take
log2(500) = 9 block accesses
· Total cost of binary search is 9 + 10 - 1 = 18 block accesses
(versus 500 for linear scan)
Database Systems Concepts 12.11 Silberschatz, Korth and Sudarshan c 1997
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Selections Using Indices
· Index scan ­ search algorithms that use an index; condition is
on search-key of index.
· A3 (primary index on candidate key, equality). Retrieve a
single record that satisfies the corresponding equality
condition. EA3 = HTi + 1
· A4 (primary index on nonkey, equality) Retrieve multiple
records. Let the search-key attribute be A.
EA4 = HTi + SC(A,r)
fr
· A5 (equality on search-key of secondary index).
­ Retrieve a single record if the search-key is a candidate key
EA5 = HTi + 1
­ Retrieve multiple records (each may be on a different block)
if the search-key is not a candidate key. EA5 = HTi +SC(A, r)
Database Systems Concepts 12.12 Silberschatz, Korth and Sudarshan c 1997
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Cost Estimate Example (Indices)
Consider the query is branch-name="Perryridge"(account), with the
primary index on branch-name.
· Since V(branch-name, account) = 50, we expect that
10000/50 = 200 tuples of the account relation pertain to the
Perryridge branch.
· Since the index is a clustering index, 200/20 = 10 block reads
are required to read the account tuples
· Several index blocks must also be read. If B+
-tree index stores
20 pointers per node, then the B+
-tree index must have
between 3 and 5 leaf nodes and the entire tree has a depth of
2. Therefore, 2 index blocks must be read.
· This strategy requires 12 total block reads.
Database Systems Concepts 12.13 Silberschatz, Korth and Sudarshan c 1997
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Selections Involving Comparisons
Implement selections of the form Av (r) or Av (r) by using a
linear file scan or binary search, or by using indices in the following
ways:
· A6 (primary index, comparison). The cost estimate is:
EA6 = HTi +
c
fr
where c is the estimated number of tuples satisfying the
condition. In absence of statistical information c is assumed to
be nr / 2.
· A7 (secondary index, comparison). The cost estimate is:
EA7 = HTi +
LBi  c
nr
+ c
where c is defined as before. (Linear file scan may be cheaper
if c is large!)
Database Systems Concepts 12.14 Silberschatz, Korth and Sudarshan c 1997
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Implementation of Complex Selections
· The selectivity of a condition i is the probability that a tuple in
the relation r satisfies i . If si is the number of satisfying tuples
in r, i 's selectivity is given by si /nr .
· Conjunction: 12...n (r). The estimate for number of
tuples in the result is:
nr 
s1  s2  . . .  sn
nn
r
· Disjunction: 12...n (r). Estimated number of tuples:
nr  1 - (1 -
s1
nr
)  (1 -
s2
nr
)  . . .  (1 -
sn
nr
)
· Negation: (r). Estimated number of tuples:
nr - size((r))
Database Systems Concepts 12.15 Silberschatz, Korth and Sudarshan c 1997
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Algorithms for Complex Selections
· A8 (conjunctive selection using one index). Select a
combination of i and algorithms A1 through A7 that results in
the least cost for i
(r). Test other conditions in memory buffer.
· A9 (conjunctive selection using multiple-key index). Use
appropriate composite (multiple-key) index if available.
· A10 (conjunctive selection by intersection of identifiers).
Requires indices with record pointers. Use corresponding
index for each condition, and take intersection of all the
obtained sets of record pointers. Then read file. If some
conditions did not have appropriate indices, apply test in
memory.
· A11 (disjunctive selection by union of identifiers). Applicable if
all conditions have available indices. Otherwise use linear
scan.
Database Systems Concepts 12.16 Silberschatz, Korth and Sudarshan c 1997
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Example of Cost Estimate for Complex Selection
· Consider a selection on account with the following condition:
where branch-name = "Perryridge" and balance = 1200
· Consider using algorithm A8:
­ The branch-name index is clustering, and if we use it the
cost estimate is 12 block reads ( as we saw before).
­ The balance index is non-clustering, and
V(balance, account) = 500, so the selection would retrieve
10, 000/ 500 = 20 accounts. Adding the index block reads,
gives a cost estimate of 22 block reads.
­ Thus using branch-name index is preferable, even though
its condition is less selective.
­ If both indices were non-clustering, it would be preferable to
use the balance index.
Database Systems Concepts 12.17 Silberschatz, Korth and Sudarshan c 1997
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Example (cont.)
· Consider using algorithm A10:
­ Use the index on balance to retrieve set S1 of pointers to
records with balance = 1200.
­ Use index on branch-name to retrieve set S2 of pointers to
records with branch-name = "Perryridge".
­ S1  S2 = set of pointers to records with branch-name =
"Perryridge" and balance = 1200.
­ The number of pointers retrieved (20 and 200) fit into a
single leaf page; we read four index blocks to retrieve the
two sets of pointers and compute their intersection.
­ Estimate that one tuple in 50  500 meets both conditions.
Since naccount = 10000, conservatively overestimate that
S1  S2 contains one pointer.
­ The total estimated cost of this strategy is five block reads.
Database Systems Concepts 12.18 Silberschatz, Korth and Sudarshan c 1997
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Sorting
· We may build an index on the relation, and then use the index
to read the relation in sorted order. May lead to one disk block
access for each tuple.
· For relations that fit in memory, techniques like quicksort can
be used. For relations that don't fit in memory, external
sort-merge is a good choice.
Database Systems Concepts 12.19 Silberschatz, Korth and Sudarshan c 1997
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External Sort­Merge
Let M denote memory size (in pages).
1. Create sorted runs as follows. Let i be 0 initially. Repeatedly
do the following till the end of the relation:
(a) Read M blocks of relation into memory
(b) Sort the in-memory blocks
(c) Write sorted data to run Ri ; increment i.
2. Merge the runs; suppose for now that i < M. In a single merge
step, use i blocks of memory to buffer input runs, and 1 block
to buffer output. Repeatedly do the following until all input
buffer pages are empty:
(a) Select the first record in sort order from each of the buffers
(b) Write the record to the output
(c) Delete the record from the buffer page; if the buffer page is
empty, read the next block (if any) of the run into the buffer.
Database Systems Concepts 12.20 Silberschatz, Korth and Sudarshan c 1997
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Example: External Sorting Using Sort­Merge
g 24
a 19
d 31
c 33
b 14
e 16
r 16
d 21
m 3
p 2
d 7
a 14
a 14
a 19
b 14
c 33
d 7
d 21
d 31
e 16
g 24
m 3
p 2
r 16
a 19
b 14
c 33
d 31
e 16
g 24
a 14
d 7
d 21
m 3
p 2
r 16
a 19
d 31
g 24
b 14
c 33
e 16
d 21
m 3
r 16
a 14
d 7
p 2
Initial
Relation
Create
Runs
Merge
Pass-1
Merge
Pass-2
Runs Runs
Sorted
Output
Database Systems Concepts 12.21 Silberschatz, Korth and Sudarshan c 1997
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External Sort­Merge (Cont.)
· If i  M, several merge passes are required.
­ In each pass, contiguous groups of M - 1 runs are merged.
­ A pass reduces the number of runs by a factor of M - 1,
and creates runs longer by the same factor.
­ Repeated passes are performed till all runs have been
merged into one.
· Cost analysis:
­ Disk accesses for initial run creation as well as in each pass
is 2br (except for final pass, which doesn't write out results)
­ Total number of merge passes required: logM-1(br /M) .
Thus total number of disk accesses for external sorting:
br (2 logM-1(br /M) + 1)
Database Systems Concepts 12.22 Silberschatz, Korth and Sudarshan c 1997
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Join Operation
· Several different algorithms to implement joins
­ Nested-loop join
­ Block nested-loop join
­ Indexed nested-loop join
­ Merge-join
­ Hash-join
· Choice based on cost estimate
· Join size estimates required, particularly for cost estimates for
outer-level operations in a relational-algebra expression.
Database Systems Concepts 12.23 Silberschatz, Korth and Sudarshan c 1997
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Join Operation: Running Example
Running example:
depositor 1 customer
Catalog information for join examples:
· ncustomer = 10, 000.
· fcustomer = 25, which implies that
bcustomer = 10000/ 25 = 400.
· ndepositor = 5000.
· fdepositor = 50, which implies that
bdepositor = 5000/ 50 = 100.
· V(customer-name, depositor) = 2500, which implies that, on
average, each customer has two accounts.
Also assume that customer-name in depositor is a foreign key on
customer.
Database Systems Concepts 12.24 Silberschatz, Korth and Sudarshan c 1997
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Estimation of the Size of Joins
· The Cartesian product r × s contains nr ns tuples; each tuple
occupies sr + ss bytes.
· If R  S = , then r 1 s is the same as r × s.
· If R  S is a key for R, then a tuple of s will join with at most
one tuple from r; therefore, the number of tuples in r 1 s is no
greater than the number of tuples in s.
If R  S in S is a foreign key in S referencing R, then the
number of tuples in r 1 s is exactly the same as the number of
tuples in s.
The case for R  S being a foreign key referencing S is
symmetric.
· In the example query depositor 1 customer, customer-name
in depositor is a foreign key of customer; hence, the result has
exactly ndepositor tuples, which is 5000.
Database Systems Concepts 12.25 Silberschatz, Korth and Sudarshan c 1997
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Estimation of the Size of Joins (Cont.)
· If R  S = {A} is not a key for R or S.
If we assume that every tuple t in R produces tuples in R 1 S,
number of tuples in R 1 S is estimated to be:
nr  ns
V(A, s)
If the reverse is true, the estimate obtained will be:
nr  ns
V(A, r)
The lower of these two estimates is probably the more
accurate one.
Database Systems Concepts 12.26 Silberschatz, Korth and Sudarshan c 1997
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Estimation of the Size of Joins (Cont.)
· Compute the size estimates for depositor 1 customer without
using information about foreign keys:
­ V(customer-name, depositor) = 2500, and
V(customer-name, customer) = 10000
­ The two estimates are 5000  10000/ 2500 = 20, 000 and
5000  10000/ 10000 = 5000
­ We choose the lower estimate, which, in this case, is the
same as our earlier computation using foreign keys.
Database Systems Concepts 12.27 Silberschatz, Korth and Sudarshan c 1997
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Nested-Loop Join
· Compute the theta join, r 1 s
for each tuple tr in r do begin
for each tuple ts in s do begin
test pair (tr , ts) to see if they satisfy the join condition 
if they do, add tr  ts to the result.
end
end
· r is called the outer relation and s the inner relation of the join.
· Requires no indices and can be used with any kind of join
condition.
· Expensive since it examines every pair of tuples in the two
relations. If the smaller relation fits entirely in main memory,
use that relation as the inner relation.
Database Systems Concepts 12.28 Silberschatz, Korth and Sudarshan c 1997
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Nested-Loop Join (Cont.)
· In the worst case, if there is enough memory only to hold one
block of each relation, the estimated cost is nr  bs + br disk
accesses.
· If the smaller relation fits entirely in memory, use that as the
inner relation. This reduces the cost estimate to br + bs disk
accesses.
· Assuming the worst case memory availability scenario, cost
estimate will be 5000  400 + 100 = 2, 000, 100 disk accesses
with depositor as outer relation, and
10000  100 + 400 = 1, 000, 400 disk accesses with customer
as the outer relation.
· If the smaller relation (depositor) fits entirely in memory, the
cost estimate will be 500 disk accesses.
· Block nested-loops algorithm (next slide) is preferable.
Database Systems Concepts 12.29 Silberschatz, Korth and Sudarshan c 1997
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Block Nested-Loop Join
· Variant of nested-loop join in which every block of inner relation
is paired with every block of outer relation.
for each block Br of r do begin
for each block Bs of s do begin
for each tuple tr in Br do begin
for each tuple ts in Bs do begin
test pair (tr , ts) for satisfying the join condition
if they do, add tr  ts to the result.
end
end
end
end
· Worst case: each block in the inner relation s is read only once
for each block in the outer relation (instead of once for each
tuple in the outer relation)
Database Systems Concepts 12.30 Silberschatz, Korth and Sudarshan c 1997
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Block Nested-Loop Join (Cont.)
· Worst case estimate: br  bs + br block accesses. Best case:
br + bs block accesses.
· Improvements to nested-loop and block nested loop
algorithms:
­ If equi-join attribute forms a key on inner relation, stop inner
loop with first match
­ In block nested-loop, use M - 2 disk blocks as blocking unit
for outer relation, where M = memory size in blocks; use
remaining two blocks to buffer inner relation and output.
Reduces number of scans of inner relation greatly.
­ Scan inner loop forward and backward alternately, to make
use of blocks remaining in buffer (with LRU replacement)
­ Use index on inner relation if available
Database Systems Concepts 12.31 Silberschatz, Korth and Sudarshan c 1997
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Indexed Nested-Loop Join
· If an index is available on the inner loop's join attribute and join
is an equi-join or natural join, more efficient index lookups can
replace file scans.
· Can construct an index just to compute a join.
· For each tuple tr in the outer relation r, use the index to look up
tuples in s that satisfy the join condition with tuple tr .
· Worst case: buffer has space for only one page of r and one
page of the index.
­ br disk accesses are needed to read relation r, and, for
each tuple in r, we perform an index lookup on s.
­ Cost of the join: br + nr  c, where c is the cost of a single
selection on s using the join condition.
· If indices are available on both r and s, use the one with fewer
tuples as the outer relation.
Database Systems Concepts 12.32 Silberschatz, Korth and Sudarshan c 1997
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Example of Index Nested-Loop Join
· Compute depositor 1 customer, with depositor as the outer
relation.
· Let customer have a primary B+
-tree index on the join attribute
customer-name, which contains 20 entries in each index node.
· Since customer has 10,000 tuples, the height of the tree is 4,
and one more access is needed to find the actual data.
· Since ndepositor is 5000, the total cost is
100 + 5000  5 = 25, 100 disk accesses.
· This cost is lower than the 40, 100 accesses needed for a block
nested-loop join.
Database Systems Concepts 12.33 Silberschatz, Korth and Sudarshan c 1997
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Merge­Join
1. First sort both relations on their join attribute (if not already
sorted on the join attributes).
2. Join step is similar to the merge stage of the sort-merge
algorithm. Main difference is handling of duplicate values in
join attribute -- every pair with same value on join attribute
must be matched
a 3
b 1
d 8
d 13
f 7
m 5
q 6
a A
b G
c L
d N
m B
a1 a2 a1 a3
pr ps
r s
Database Systems Concepts 12.34 Silberschatz, Korth and Sudarshan c 1997
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Merge­Join (Cont.)
· Each tuple needs to be read only once, and as a result, each
block is also read only once. Thus number of block accesses is
br + bs, plus the cost of sorting if relations are unsorted.
· Can be used only for equi-joins and natural joins
· If one relation is sorted, and the other has a secondary B+
-tree
index on the join attribute, hybrid merge-joins are possible.
The sorted relation is merged with the leaf entries of the
B+
-tree. The result is sorted on the addresses of the unsorted
relation's tuples, and then the addresses can be replaced by
the actual tuples efficiently.
Database Systems Concepts 12.35 Silberschatz, Korth and Sudarshan c 1997
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Hash­Join
· Applicable for equi-joins and natural joins.
· A hash function h is used to partition tuples of both relations
into sets that have the same hash value on the join attributes,
as follows:
­ h maps JoinAttrs values to {0, 1, . . . , max}, where JoinAttrs
denotes the common attributes of r and s used in the
natural join.
­ Hr0
, Hr1
, . . . , Hrmax denote partitions of r tuples, each initially
empty. Each tuple tr  r is put in partition Hri
, where
i = h(tr [JoinAttrs]).
­ Hs0
, Hs1
, ..., Hsmax denote partitions of s tuples, each initially
empty. Each tuple ts  s is put in partition Hsi
, where
i = h(ts[JoinAttrs]).
Database Systems Concepts 12.36 Silberschatz, Korth and Sudarshan c 1997
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Hash­Join (Cont.)
· r tuples in Hri
need only to be compared with s tuples in Hsi
;
they do not need to be compared with s tuples in any other
partition, since:
­ An r tuple and an s tuple that satisfy the join condition will
have the same value for the join attributes.
­ If that value is hashed to some value i, the r tuple has to be
in Hri
and the s tuple in Hsi
.
Database Systems Concepts 12.37 Silberschatz, Korth and Sudarshan c 1997
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Hash­Join (Cont.)
0
1
2
3
4
0
1
2
3
4
r s
.
.
.
.
.
.
.
.
Partitions
of r
Partitions
of s
Database Systems Concepts 12.38 Silberschatz, Korth and Sudarshan c 1997
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Hash­Join algorithm
The hash-join of r and s is computed as follows.
1. Partition the relations s using hashing function h. When
partitioning a relation, one block of memory is reserved as the
output buffer for each partition.
2. Partition r similarly.
3. For each i:
(a) Load Hsi
into memory and build an in-memory hash index
on it using the join attribute. This hash index uses a
different hash function than the earlier one h.
(b) Read the tuples in Hri
from disk one by one. For each tuple
tr locate each matching tuple ts in Hsi
using the in-memory
hash index. Output the concatenation of their attributes.
Relation s is called the build input and r is called the probe input.
Database Systems Concepts 12.39 Silberschatz, Korth and Sudarshan c 1997
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Hash­Join algorithm (Cont.)
· The value max and the hash function h is chosen such that
each Hsi
should fit in memory.
· Recursive partitioning required if number of partitions max is
greater than number of pages M of memory.
­ Instead of partitioning max ways, partition s M - 1 ways;
­ Further partition the M - 1 partitions using a different hash
function
­ Use same partitioning method on r
­ Rarely required: e.g., recursive partitioning not needed for
relations of 1GB or less with memory size of 2MB, with
block size of 4KB.
· Hash-table overflow occurs in partition Hsi
if Hsi
does not fit in
memory. Can resolve by further partitioning Hsi
using different
hash function. Hri
must be similarly partitioned.
Database Systems Concepts 12.40 Silberschatz, Korth and Sudarshan c 1997
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Cost of Hash­Join
· If recursive partitioning is not required: 3(br + bs) + 2  max
· If recursive partitioning is required, number of passes required
for partitioning s is logM-1(bs) - 1 . This is because each final
partition of s should fit in memory.
· The number of partitions of probe relation r is the same as that
for build relation s; the number of passes for partitioning of r is
also the same as for s. Therefore it is best to choose the
smaller relation as the build relation.
· Total cost estimate is:
2(br + bs) logM-1(bs) - 1 + br + bs
· If the entire build input can be kept in main memory, max can
be set to 0 and the algorithm does not partition the relations
into temporary files. Cost estimate goes down to br + bs.
Database Systems Concepts 12.41 Silberschatz, Korth and Sudarshan c 1997
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Example of Cost of Hash­Join
customer 1 depositor
· Assume that memory size is 20 blocks.
· bdepositor = 100 and bcustomer = 400.
· depositor is to be used as build input. Partition it into five
partitions, each of size 20 blocks. This partitioning can be done
in one pass.
· Similarly, partition customer into five partitions, each of size
80. This is also done in one pass.
· Therefore total cost: 3(100 + 400) = 1500 block transfers
(Ignores cost of writing partially filled blocks).
Database Systems Concepts 12.42 Silberschatz, Korth and Sudarshan c 1997
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Hybrid Hash­Join
· Useful when memory sizes are relatively large, and the build
input is bigger than memory.
· With a memory size of 25 blocks, depositor can be partitioned
into five partitions, each of size 20 blocks.
· Keep the first of the partitions of the build relation in memory. It
occupies 20 blocks; one block is used for input, and one block
each is used for buffering the other four partitions.
· customer is similarly partitioned into five partitions each of size
80; the first is used right away for probing, instead of being
written out and read back in.
· Ignoring the cost of writing partially filled blocks, the cost is
3(80 + 320) + 20 + 80 = 1300 block transfers with hybrid
hash-join, instead of 1500 with plain hash-join.
· Hybrid hash-join most useful if M >>

bs.
Database Systems Concepts 12.43 Silberschatz, Korth and Sudarshan c 1997
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Complex Joins
· Join with a conjunctive condition:
r 112...n s
­ Compute the result of one of the simpler joins r 1i
s
­ final result comprises those tuples in the intermediate result
that satisfy the remaining conditions
1  . . .  i-1  i+1  . . .  n
­ Test these conditions as tuples in r 1i
s are generated.
· Join with a disjunctive condition:
r 112...n s
Compute as the union of the records in individual joins r 1i
s:
(r 11
s)  (r 12
s)  . . .  (r 1n s)
Database Systems Concepts 12.44 Silberschatz, Korth and Sudarshan c 1997
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Complex Joins (Cont.)
· Join involving three relations: loan 1 depositor 1 customer
· Strategy 1. Compute depositor 1 customer; use result to
compute loan 1 (depositor 1 customer)
· Strategy 2. Compute loan 1 depositor first, and then join the
result with customer.
· Strategy 3. Perform the pair of joins at once. Build an index on
loan for loan-number, and on customer for customer-name.
­ For each tuple t in depositor, look up the corresponding
tuples in customer and the corresponding tuples in loan.
­ Each tuple of deposit is examined exactly once.
· Strategy 3 combines two operations into one special-purpose
operation that is more efficient than implementing two joins of
two relations.
Database Systems Concepts 12.45 Silberschatz, Korth and Sudarshan c 1997
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Other Operations
· Duplicate elimination can be implemented via hashing or
sorting.
­ On sorting duplicates will come adjacent to each other, and
all but one of a set of duplicates can be deleted.
Optimization: duplicates can be deleted during run
generation as well as at intermediate merge steps in
external sort-merge.
­ Hashing is similar ­ duplicates will come into the same
bucket.
· Projection is implemented by performing projection on each
tuple followed by duplicate elimination.
Database Systems Concepts 12.46 Silberschatz, Korth and Sudarshan c 1997
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Other Operations (Cont.)
· Aggregation can be implemented in a manner similar to
duplicate elimination.
­ Sorting or hashing can be used to bring tuples in the same
group together, and then the aggregate functions can be
applied on each group.
­ Optimization: combine tuples in the same group during run
generation and intermediate merges, by computing partial
aggregate values.
· Set operations (,  and -): can either use variant of
merge-join after sorting, or variant of hash-join.
Database Systems Concepts 12.47 Silberschatz, Korth and Sudarshan c 1997
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Other Operations (Cont.)
· E.g., Set operations using hashing:
1. Partition both relations using the same hash function,
thereby creating Hr0
, . . . , Hrmax , and Hs0
, . . . , Hsmax .
2. Process each partition i as follows. Using a different
hashing function, build an in-memory hash index on Hri
after it is brought into memory.
3. ­ r  s: Add tuples in Hsi
to the hash index if they are not
already in it. Then add the tuples in the hash index to the
result.
­ r  s: output tuples in Hsi
to the result if they are already
there in the hash index.
­ r - s: for each tuple in Hsi
, if it is there in the hash index,
delete it from the index. Add remaining tuples in the hash
index to the result.
Database Systems Concepts 12.48 Silberschatz, Korth and Sudarshan c 1997
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Other Operations (Cont.)
· Outer join can be computed either as
­ A join followed by addition of null-padded non-participating
tuples.
­ by modifying the join algorithms.
· Example:
­ In r ­
­1 s, non participating tuples are those in r - R(r 1 s)
­ Modify merge-join to compute r ­
­1 s: During merging, for
every tuple tr from r that do not match any tuple in s, output
tr padded with nulls.
­ Right outer-join and full outer-join can be computed
similarly.
Database Systems Concepts 12.49 Silberschatz, Korth and Sudarshan c 1997
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Evaluation of Expressions
· Materialization: evaluate one operation at a time, starting at
the lowest-level. Use intermediate results materialized into
temporary relations to evaluate next-level operations.
· E.g., in figure below, compute and store balance<2500(account);
then compute and store its join with customer, and finally
compute the projection on customer-name.
 customer-name
 balance < 2500
account
customer
Database Systems Concepts 12.50 Silberschatz, Korth and Sudarshan c 1997
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Evaluation of Expressions (Cont.)
· Pipelining: evaluate several operations simultaneously,
passing the results of one operation on to the next.
· E.g., in expression in previous slide, don't store result of
balance<2500(account) ­ instead, pass tuples directly to the join.
Similarly, don't store result of join, pass tuples directly to
projection.
· Much cheaper than materialization: no need to store a
temporary relation to disk.
· Pipelining may not always be possible -- e.g., sort, hash-join.
· For pipelining to be effective, use evaluation algorithms that
generate output tuples even as tuples are received for inputs to
the operation.
· Pipelines can be executed in two ways: demand driven and
producer driven.
Database Systems Concepts 12.51 Silberschatz, Korth and Sudarshan c 1997
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Transformation of Relational Expressions
· Generation of query-evaluation plans for an expression
involves two steps:
1. generating logically equivalent expressions
2. annotating resultant expressions to get alternative query
plans
· Use equivalence rules to transform an expression into an
equivalent one.
· Based on estimated cost, the cheapest plan is selected. The
process is called cost based optimization.
Database Systems Concepts 12.52 Silberschatz, Korth and Sudarshan c 1997
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Equivalence of Expressions
Relations generated by two equivalent expressions have the same
set of attributes and contain the same set of tuples, although their
attributes may be ordered differently.
 customer-name
branch
depositor
 branch-city=Brooklyn
accountdepositoraccount
 branch-city=Brooklyn
(a) Initial Expression Tree (b) Transformed Expression Tree
branch
 customer-name
Equivalent expressions
Database Systems Concepts 12.53 Silberschatz, Korth and Sudarshan c 1997
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Equivalence Rules
1. Conjunctive selection operations can be deconstructed into a
sequence of individual selections.
12
(E) = 1
(2
(E))
2. Selection operations are commutative.
1
(2
(E)) = 2
(1
(E))
3. Only the last in a sequence of projection operations is needed,
the others can be omitted.
L1
(L2
(. . . (Ln
(E)) . . .)) = L1
(E)
4. Selections can be combined with Cartesian products and theta
joins.
(a) (E1 × E2) = E1 1 E2
(b) 1
(E1 12
E2) = E1 112
E2
Database Systems Concepts 12.54 Silberschatz, Korth and Sudarshan c 1997
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Equivalence Rules (Cont.)
5. Theta-join operations (and natural joins) are commutative.
E1 1 E2 = E2 1 E1
6. (a) Natural join operations are associative:
(E1 1 E2) 1 E3 = E1 1 (E2 1 E3)
(b) Theta joins are associative in the following manner:
(E1 11
E2) 123
E3 = E1 113
(E2 12
E3)
where 2 involves attributes from only E2 and E3.
Database Systems Concepts 12.55 Silberschatz, Korth and Sudarshan c 1997
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Equivalence Rules (Cont.)
7. The selection operation distributes over the theta join operation
under the following two conditions:
(a) When all the attributes in 0 involve only the attributes of
one of the expressions (E1) being joined.
0
(E1 1 E2) = (0
(E1)) 1 E2
(b) When 1 involves only the attributes of E1 and 2 involves
only the attributes of E2.
12
(E1 1 E2) = (1
(E1)) 1 (2
(E2))
Database Systems Concepts 12.56 Silberschatz, Korth and Sudarshan c 1997
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Equivalence Rules (Cont.)
8. The projection operation distributes over the theta join
operation as follows:
(a) if  involves only attributes from L1  L2:
L1L2
(E1 1 E2) = (L1
(E1)) 1 (L2
(E2))
(b) Consider a join E1 1 E2. Let L1 and L2 be sets of attributes
from E1 and E2, respectively. Let L3 be attributes of E1 that
are involved in join condition , but are not in L1  L2, and
let L4 be attributes of E2 that are involved in join condition ,
but are not in L1  L2.
L1L2
(E1 1 E2) = L1L2
((L1L3
(E1)) 1 (L2L4
(E2)))
Database Systems Concepts 12.57 Silberschatz, Korth and Sudarshan c 1997
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Equivalence Rules (Cont.)
9. The set operations union and intersection are commutative
(set difference is not commutative).
E1  E2 = E2  E1
E1  E2 = E2  E1
10. Set union and intersection are associative.
11. The selection operation distributes over ,  and -. E.g.:
P(E1 - E2) = P(E1) - P(E2)
For difference and intersection, we also have:
P(E1 - E2) = P(E1) - E2
12. The projection operation distributes over the union operation.
L(E1  E2) = (L(E1))  (L(E2))
Database Systems Concepts 12.58 Silberschatz, Korth and Sudarshan c 1997
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Selection Operation Example
· Query: Find the names of all customers who have an account
at some branch located in Brooklyn.
customer-name (branch-city = "Brooklyn"
(branch 1 (account 1 depositor)))
· Transformation using rule 7a.
customer-name
((branch-city = "Brooklyn" (branch))
1 (account 1 depositor))
· Performing the selection as early as possible reduces the size
of the relation to be joined.
Database Systems Concepts 12.59 Silberschatz, Korth and Sudarshan c 1997
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Selection Operation Example (Cont.)
· Query: Find the names of all customers with an account at a
Brooklyn branch whose account balance is over $1000.
customer-name (branch-city = "Brooklyn"  balance >1000
(branch 1 (account 1 depositor)))
· Transformation using join associativity (Rule 6a):
customer-name ((branch-city = "Brooklyn" balance>1000
(branch 1 account)) 1 depositor)
· Second form provides an opportunity to apply the "perform
selections early" rule, resulting in the subexpression
branch-city = "Brooklyn" (branch) 1 balance>1000 (account)
· Thus a sequence of transformations can be useful
Database Systems Concepts 12.60 Silberschatz, Korth and Sudarshan c 1997
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Projection Operation Example
customer-name((branch-city = "Brooklyn" (branch)
1 account) 1 depositor)
· When we compute
(branch-city = "Brooklyn" (branch) 1 account)
we obtain a relation whose schema is:
(branch-name, branch-city, assets, account-number, balance)
· Push projections using equivalence rules 8a and 8b; eliminate
unneeded attributes from intermediate results to get:
customer-name((account-number (
(branch-city = "Brooklyn" (branch)) 1 account)) 1 depositor)
Database Systems Concepts 12.61 Silberschatz, Korth and Sudarshan c 1997
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Join Ordering Example
· For all relations r1, r2, and r3,
(r1 1 r2) 1 r3 = r1 1 (r2 1 r3)
· If r2 1 r3 is quite large and r1 1 r2 is small, we choose
(r1 1 r2) 1 r3
so that we compute and store a smaller temporary relation.
Database Systems Concepts 12.62 Silberschatz, Korth and Sudarshan c 1997
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Join Ordering Example (Cont.)
· Consider the expression
customer-name ((branch-city = "Brooklyn" (branch))
1 account 1 depositor)
· Could compute account 1 depositor first, and join result with
branch-city = "Brooklyn" (branch)
but account 1 depositor is likely to be a large relation.
· Since it is more likely that only a small fraction of the bank's
customers have accounts in branches located in Brooklyn, it is
better to compute
branch-city = "Brooklyn" (branch) 1 account
first.
Database Systems Concepts 12.63 Silberschatz, Korth and Sudarshan c 1997
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Evaluation Plan
An evaluation plan defines exactly what algorithm is used for each
operation, and how the execution of the operations is coordinated.
 branch-city=Brooklyn
branch
 customer-name (sort to remove duplicates)
account
depositor
 balance < 1000
(use index 1) (use linear scan)
(hash-join)
(merge-join)
Pipeline Pipeline
An evaluation plan
Database Systems Concepts 12.64 Silberschatz, Korth and Sudarshan c 1997
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Choice of Evaluation Plans
· Must consider the interaction of evaluation techniques when
choosing evaluation plans: choosing the cheapest algorithm
for each operation independently may not yield the best overall
algorithm. E.g.
­ merge-join may be costlier than hash-join, but may provide
a sorted output which reduces the cost for an outer level
aggregation.
­ nested-loop join may provide opportunity for pipelining
· Practical query optimizers incorporate elements of the
following two broad approaches:
1. Search all the plans and choose the best plan in a
cost-based fashion.
2. Use heuristics to choose a plan.
Database Systems Concepts 12.65 Silberschatz, Korth and Sudarshan c 1997
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Cost-Based Optimization
· Consider finding the best join-order for r1 1 r2 1 . . . rn.
· There are (2(n - 1))!/ (n - 1)! different join orders for above
expression. With n = 7, the number is 665280, with n = 10, the
number is greater than 176 billion!
· No need to generate all the join orders. Using dynamic
programming, the least-cost join order for any subset of
{r1, r2, . . . , rn} is computed only once and stored for future use.
· This reduces time complexity to around O(3n
). With n = 10,
this number is 59000.
Database Systems Concepts 12.66 Silberschatz, Korth and Sudarshan c 1997
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Cost-Based Optimization (Cont.)
· In left-deep join trees, the right-hand-side input for each join
is a relation, not the result of an intermediate join.
· If only left-deep join trees are considered, cost of finding best
join order becomes O(2n
).
r4 r5r3
r1 r2
r5
r4
r3
r2r1
(a) Left-deep Join Tree (b) Non-left-deep Join Tree
Database Systems Concepts 12.67 Silberschatz, Korth and Sudarshan c 1997
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Dynamic Programming in Optimization
· To find best left-deep join tree for a set of n relations:
­ Consider n alternatives with one relation as right-hand-side
input and the other relations as left-hand-side input.
­ Using (recursively computed and stored) least-cost join
order for each alternative on left-hand-side, choose the
cheapest of the n alternatives.
· To find best join tree for a set of n relations:
­ To find best plan for a set S of n relations, consider all
possible plans of the form: S1 1 (S - S1) where S1 is any
non-empty subset of S.
­ As before, use recursively computed and stored costs for
subsets of S to find the cost of each plan. Choose the
cheapest of the 2n
- 1 alternatives.
Database Systems Concepts 12.68 Silberschatz, Korth and Sudarshan c 1997
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Interesting Orders in Cost-Based Optimization
· Consider the expression (r1 1 r2 1 r3) 1 r4 1 r5
· An interesting sort order is a particular sort order of tuples
that could be useful for a later operation.
­ Generating the result of r1 1 r2 1 r3 sorted on the attributes
common with r4 or r5 may be useful, but generating it sorted
on the attributes common to only r1 and r2 is not useful.
­ Using merge­join to compute r1 1 r2 1 r3 may be costlier,
but may provide an output sorted in an interesting order.
· Not sufficient to find the best join order for each subset of the
set of n given relations; must find the best join order for each
subset, for each interesting sort order of the join result for that
subset. Simple extension of earlier dynamic programming
algorithms.
Database Systems Concepts 12.69 Silberschatz, Korth and Sudarshan c 1997
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Heuristic Optimization
· Cost-based optimization is expensive, even with dynamic
programming.
· Systems may use heuristics to reduce the number of choices
that must be made in a cost-based fashion.
· Heuristic optimization transforms the query-tree by using a set
of rules that typically (but not in all cases) improve execution
performance:
­ Perform selection early (reduces the number of tuples)
­ Perform projection early (reduces the number of attributes)
­ Perform most restrictive selection and join operations
before other similar operations.
· Some systems use only heuristics, others combine heuristics
with partial cost-based optimization.
Database Systems Concepts 12.70 Silberschatz, Korth and Sudarshan c 1997
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Steps in Typical Heuristic Optimization
1. Deconstruct conjunctive selections into a sequence of single
selection operations (Equiv. rule 1).
2. Move selection operations down the query tree for the earliest
possible execution (Equiv. rules 2, 7a, 7b, 11).
3. Execute first those selection and join operations that will
produce the smallest relations (Equiv. rule 6).
4. Replace Cartesian product operations that are followed by a
selection condition by join operations (Equiv. rule 4a).
5. Deconstruct and move as far down the tree as possible lists of
projection attributes, creating new projections where needed
(Equiv. rules 3, 8a, 8b, 12).
6. Identify those subtrees whose operations can be pipelined,
and execute them using pipelining.
Database Systems Concepts 12.71 Silberschatz, Korth and Sudarshan c 1997
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Structure of Query Optimizers
· The System R optimizer considers only left-deep join orders.
This reduces optimization complexity and generates plans
amenable to pipelined evaluation.
System R also uses heuristics to push selections and
projections down the query tree.
· For scans using secondary indices, the Sybase optimizer takes
into account the probability that the page containing the tuple
is in the buffer.
Database Systems Concepts 12.72 Silberschatz, Korth and Sudarshan c 1997
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Structure of Query Optimizers (Cont.)
· Some query optimizers integrate heuristic selection and the
generation of alternative access plans.
­ System R and Starburst use a hierarchical procedure based
on the nested-block concept of SQL: heuristic rewriting
followed by cost-based join-order optimization.
­ The Oracle7 optimizer supports a heuristic based on
available access paths.
· Even with the use of heuristics, cost-based query optimization
imposes a substantial overhead.
This expense is usually more than offset by savings at
query-execution time, particularly by reducing the number of
slow disk accesses.
Database Systems Concepts 12.73 Silberschatz, Korth and Sudarshan c 1997