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2$[IZ:ah2$umX ] i 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.
Cost 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 algorithm A as EA. We do not include cost of writing output to disk.2#w
1
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 indicesVT*a
c.
$
ES6 + 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):
log2(br) cost of locating the first tuple by a binary search on the blocks
SC(A, r) numbers 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) qZsZZ*Z]ZnsK5
{
L $Statistical Information for Examples%$$
% faccount = 20 (20 tuples of account fit in one block)
V(branchname, 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 branchname
A secondary, B+tree index for attribute balanceh
T"
&"
",
Selection Cost Estimate Example
sbranchname = Perryridge (account)
Number of blocks is baccount = 500: 10,000 tuples in the relation; each block holds 20 tuples.
Assume account is sorted on branchname.
V(branchname, 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) (ZZZZZ l%"4b
: Selections Using Indices
CIndex scan search algorithms that use an index; condition is on searchkey 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 searchkey attribute be A.
A5(equality on searchkey of secondary index).
Retrieve a single record if the searchkey is a candidate key
EA5 = HTi + 1
Retrieve multiple records (each may be on a different block) if the searchkey is not a candidate key. EA5 = HTi + SC(A, r)
tZ?ZZ}ZZ
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Cost Estimate Example (Indices) (
Consider the query is sbranchname = Perryridge (account), with the primary index on branchname.
Since V(branchname, 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.gZZ
3
6*

Selections Involving Comparisons! $
! Implement selections of the form sAv(r) or sA>v(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:
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:
where c is defined as before. (Linear file scan may be cheaper if
c is large!)$Z6ZZ8ZFXZ%Z6
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$Implementation of Complex Selections%$$
% The selectivity of a condition qi is the probability that a tuple in the relation r satisfies qi . If si is the number of satisfying tuples in r, qi s selectivity is given by si/nr.
Conjunction: sq1q2& qn(r) . The estimate for number of tuples in the result is:
Disjunction: sq1q2& qn(r) .Estimated number of tuples:
Negation: sq(r). Estimated number of tuples:
4
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'Lf * !Algorithms for Complex Selections"!$
" A8 (conjunctive selection using one index). Select a combination of qi and algorithms A1 through A7 that results in the least cost for . Test other conditions in memory buffer.
A9 (conjunctive selection using multiplekey index). Use appropriate composite (multiplekey) 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.SZP%m.@4.S
.Example of Cost Estimate for Complex Selection/.
/ Consider a selection on account with the following condition:
where branchname = Perryridge and balance = 1200
Consider using algorithm A8:
The branchname index is clustering, and if we use it the cost estimate is 12 block reads (as we saw before).
The balance index is nonclustering, and V(branch, 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 branchname index is preferable, even though its condition is less selective.
If both indices were nonclustering, it would be preferable to use the balance index.>8<PPB9 W
Example (cont.)(
VConsider using algorithm A10:
Use the index on balance to retrieve set S1 of pointers to records with balance = 1200.
Use index on branchname to retrieve set S2 of pointers to records with branchname = Perryridge .
S1 S2 = set of pointers to records with branchname = 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 the one tuple in 50 * 500 meets both conditions. Since naccout = 10000, conservatively overestimate that S1 S2 contains one pointer.
The total estimated cost of this strategy is five block reads.P
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Sorting
XWe 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 sortmerge is a good choice.*
[ 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 inmemory 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.ZvZZZ$Lf
H *Example: External Sorting Using SortMerge++
+ 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)(%FwD`
=
x A Join Operation(
Several different algorithms to implement joins
Nestedloop join
Block nestedloop join
Indexed nestedloop join
Mergejoin
Hashjoin
Choice based on cost estimate
Join size estimates required, particularly for cost estimates for outerlevel operations in a relationalalgebra expression.<0V0V
! Join Operation: Running Example (
Running example:
Depositor customer
Catalog information for join examples:
ncusmter = 10, 000.
fcustomer = 25, which implies that
bcustomer = 10000/25 = 400.
ndepositor = 50, which implies that
bdepositor = 5000/50 = 100.
V(customername, depositor) = 2500, which implies that, on average, each customer has two accounts.
Also assume that customername in depositor is a foreign key on customer.vZZ'Z8ZZ%ZZdZJZ'
xO !Estimation of the Size of Joins (
BThe Cartesian product r s contains nrns tuples; each tuple occupies sr + ss types.
If R S = , the r 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 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 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 customer, customername in depositor is a foreign key of customer, hence, the result has exactly ndepositor tuples, which is 5000.F.PPP
, b$ 2 "'Estimation of the Size of Joins (Cont.)('$
( 2If R S = {A} is not a key for R or S.
If we assume that every tuple t in R produces tuples in R S, number of tuples in R S is estimated to be:
If the reverse is true, the estimate obtained will be:
The lower of these two estimates is probably the more accurate one. `)r
#'Estimation of the Size of Joins (Cont.)('$
( Compute the size estimates for depositor customer without using information about foreign keys:
V(customername, depositor) = 2500, and
V(customername, 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.V`nnnnn`f
j $NestedLoop Join
Compute the theta join, r q 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 q
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.!ZZ0P ,LQ D I %NestedLoop 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 fist entirely in memory, use that as the inner relation. This reduce 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 estimates will be 500 disk accesses.
Block nestedloops algorithm (next slide) is preferable.Pma?bq &Block NestedLoop Join
Variant of nestedloop 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
Worse 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)bqZ P P#PYPIPZq
%L ? 'Block NestedLoop Join (Cont.)(
Worst case estimate: br * bs + br block accesses. Best case: br + bs block accesses.
Improvements to nestedloop and block nested loop algorithms:
If equijoin attribute forms a key on inner relation, stop inner loop with first match
In block nestedloop, 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 availableZZOs"a P ( Indexed NestedLoop Join(
If an index is available on the inner loop s join attribute and join is an equijoin 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.
ZZ`ZFQN^bJ I { )!!Example of Index NestedLoop Join""$
" Compute depositor customer, with depositor as the outer relation.
Let customer have a primary B+tree index on the join attribute customername, 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 + 500 * 5 = 25, 100 disk accesses.
This cost is lower than the 40,100 accesses needed for a block nestedloop join.pgZ+ZQZd
+Q ? *"
MergeJoin
RFirst sort both relations on their join attribute ( if not already sorted on the join attributes).
Join step is similar to the merge stage of the sortmerge algorithm. Main difference is handling of duplicate values in join attribute every pair with same values on join attribute must be matched**?"
* +#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 equijoins and natural joins
If one relation is sorted, and the other has a secondary B+tree index on the join attribute, hybrid mergejoins 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.@}
#K
L J n ,$ HashJoin
Applicable for equijoins 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]).
Hso, Hs1, & , Hsmax denote partitions of s tuples, each initially empty. Each tuple ts s is put in partition Hsi, where i = h (ts[JoinAttrs]).:ZZ o
B
B
P + %HashJoin (Cont.)
Ur 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.V
,
V .&HashJoin (Cont.)
/'HashJoin algorithm
The hashjoin 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 inmemory 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 inmemory hash index. Output the concatenation of their attributes.
Relation s is called the build input and r is called the probe input.
1ZZXZFZZ1
E
O 0(HashJoin algorithm (Cont.)
HThe 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 1 GB or less with memory size of 2MB, with block size of 4KB.
Hashtable 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. BZZZ@
S1 >!!6] " $ 1)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.
Z,ZZ.[D b O 2*Example of Cost of HashJoin
customer 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).Vn&x? g6=
g 3+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 hashjoin, instead of 1500 with plain hashjoin.
Hybrid hashjoin most useful if .P
4,
Complex Joins
Join with a conjunctive condition:
r q1 q2& qn s
Compute the result of one of the simpler joins r qis
final result comprises those tuples in the intermediate result that satisfy the remaining conditions
q1 & qi 1 qi+1 & qn
Test these conditions as tuples in r qi s are generated.
Join with a disjunctive condition:
r q1 q2& qn s
Compute as the union of the records in individual join r qi s:
(r q1 s) (r q2 s) & (r qn s)
#?#E8#6l*#?
5Complex Joins (Cont.)
Join involving three relations: loan depositor customer
Strategy 1. Compute depositor customer, use result to compute loan (depositor customer)
Strategy 2. Compute loan depositor first, and then join the result with customer.
Strategy 3. Perform the pair of joins at once. Build an index on loan for loannumber, and on customer for customername.
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 specialpurpose operation that is more efficient than implementing two joins of two relationsoPPP<XKo6 & 6.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 sortmerge.
Hashing is similar duplicates will come into the same bucket.
Projection is implemented by performing projection on each tuple followed by duplicate elimination.TA,d,,
Z
7/Other Operation (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 mergejoin after sorting, or variant of hashjoin.jMkBZ
80Other Operations (Cont.)
NE.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 inmemory 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.`$$L
q
$^y
91Other Operations (Cont.)
JOuter join can be computed either as
A join followed by addition of nullpadded nonparticipating tuples.
By modifying the join algorithms.
Example:
In r s, non participating tuples are those in r  PR(r s)
Modify mergejoin to compute r s: During merging, for every tuples tr from r that do not match any tuple in s, output tr padded with nulls.
Right outerjoin and full outerjoin can be computed similarly.%g Dn@%g 7N1T :2Evaluation of Expressions
:Materialization: evaluate one operation at a time, starting at the lowestlevel. Use intermediate results materialized into temporary relations to evaluate nextlevel operations.
E.g., in figure below, compute and store sbalance<2500(account); then compute and store its join with customer, and finally compute the projection on customername.
Pcustomername
sbalance<2500 customer
accountXZnZZ3Zo
;3!Evaluation of Expressions (Cont.)"!$
" DPipelining: 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 sbalance<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, hashjoin.
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.dP
<4(Transformation of Relational Expressions)(
) zGeneration of queryevaluation 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.rKZZZ=7
{ =5Equivalence 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.
Pcustomername
Pcustomername
T/ >6Equivalence Rules
1. Conjunctive selection operations can be deconstructed into a sequence of individual selections.
sq1 q2 (E) = sq1 ( q2 (E))
2. Selection operations are commutative.
sq1 ( sq2 (E))= sq2 (sq1 (E))
3. Only the last in a sequence of projection operations is needed, the others can be omitted.
PL1(PL2(& (PLn(E))& )) = PL1(E)
4. Selections can be combined with Cartesian products and theta joins.
(a) sq (E1 E2) = E1 q E2
(b) sq1 (E1 q2E2) = E1 q1 q2 E22cZ Z)Z"ZZc)aP6 ' ?7Equivalence Rules (Cont.)
5. Thetajoin operations (and natural joins) are commutative.
E1 q E2 = E2 q E1
6. (a) Natural join operations are associative:
(E1 E2) E3 = E1 (E2 E3)
(b) Theta joins are associative in the following manner:
(E1 q1 E2) q2 q3 E3 = E1 q1 q3 (E2 q2 E3)
where q2 involves attributes from only E2 and E3.~>1
>
@8Equivalence Rules (Cont.)
67. The selection operation distributes over the theta join operation under the following two conditions:
(a) When all the attributes in q0 involve only the attributes of one of the expressions (E1) being joined.
sq0 (E1 q E2) = (sq0 (E1)) q E2
(b) When q1 involves only the attributes of E1 and q2 involves only the attributes of E2.
sq1 q2 (E1 q E2) = (sq1 (E1)) q (sq2 ( E2)) Z8
!
"
n  A9Equivalence Rules (Cont.)
8. The projection operation distributes over the theta join operation as follows:
(a) if q involves only attributes from L1 L2:
PL1 L2 (E1 q E2) = (PL1(E1)) q (PL2(E2))
(b) Consider a join E1 q 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 q ,
but are not in L1 L2, and let L4 be attributes of E2 that
are involved in join condition q , but are not in L1 L2.
PL1 L2 (E1 q E2) = PL1 L2((PL1 L3 (E1)) q (PL2 L4 (E2)))5xVZIxZ
*"ID
W B: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.:
sp(E1  E2) = sp(E1)  sp(E2)
For difference and intersection, we also have:
sp(E1  E2) = sp(E1)  E2
12. The projection operation distributes over the union operation.
PL(E1 E2) = (PL(E1)) PL(E2))Zf
^
5
F
C;Selection Operation Example
Query: Find the names of all customers who have an account at some branch located in Brooklyn.
Pcustomername(sbranchcity = Brooklyn
(branch (account depositor)))
Transformation using rule 7a.
Pcustomername
((sbranchcity = Brooklyn (branch))
(account depositor))
Performing the selection as early as possible reduces the size of the relation to be joined.`TF]`
]
D<"Selection Operation Example(Cont.)#"$
# Query: Find the names of all customers with an account at a Brooklyn branch whose account balance is over $1000.
Pcustomername(sbranchcity = Brooklyn balance > 1000
(branch (account depositor))
Transformation using join associativity (Rule 6a):
Pcustomername(sbranchcity = Brooklyn balance > 1000
(branch account)) depositor)
Second form provides an opportunity to apply the Perform selections early rule, resulting in the subexpression
sbranchcity = Brooklyn (branch) sbalance > 1000 (account)
Thus a sequence of transformations can be useful4qb3fqF1q
3
q
1
T E=Projection Operation Example
Pcustomername((sbranchcity = Brooklyn (branch)
account) depositor)
When we compute
(sbranchcity = Brooklyn (branch) account)
We obtain a relation whose schema is:
(branchname, branchcity, assets, accountnumber, balance)
Push projections using equivalence rules 8a and 8b; eliminate unneeded attributes from intermediate results to get:
Pcustomername ((Paccountnumber (
sbranchcity = Brooklyn (branch)) account)) depositor)
x\ZZ9bZuZnZZZ
&<u
F>Join Ordering Example
For all relations r1, r2 and r3,
(r1 r2) r3 = r1 (r2 r3)
If r2 r3 is quite large and r1 r2 is small, we choose
(r1 r2) r3
so that we compute and store a smaller temporary relation.!+=n<
=
G?Join Ordering Example (Cont.)
Consider the expression
Pcustomername((sbranchcity = Brooklyn (branch))
account) depositor)
Could compute account depositor first, and join result with
sbranchcity = Brooklyn (branch)
but account 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
sbranchcity = Brooklyn (branch) account
first.`An2>ZZ2xF
A
H@Evaluation Plan
An evaluation plan defines exactly what algorithm is used for each operation, and how the execution of the operations is coordinated.
IAChoice of Evaluation Plans
EMust 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.
Mergejoin may be costlier than hashjoin, but may provide a sorted output which reduces the cost for an outer level aggregation.
Nestedloop 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 costbased fashion.
2. Use heuristics to choose a plan.hWnGWn
F JBCost Based OptimizationConsider finding the best joinorder for r1 r2 & rn.
There are (2(n1))!/(n )! 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 leastcost 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.
*
]&KCCostBased Optimization (Cont.) (In leftdeep join trees, the righthandside input for each join is a relation, not the result of an intermediate join.
If only leftdeep join trees are considered, cost of finding best join order becomes O(2n).
2
LD#Dynamic Programming in Optimization$$$0To find best leftdeep join tree for a set of n relations:
Consider n alternatives with one relation as righthandside input and the other relations as lefthandside input.
Using (recursively computed and stored) leastcost join order for each alternative on lefthandside, 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 of S of n relations, consider all possible plans of the form: S1 (S  S1) where S1 is any nonempty 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
;ZZ1Z&ZZ;1[
MEInteresting Orders in CostBased Optimization.. Consider the expression ( r1 r2 r3) r4 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 r2 r3 sorted on the attributes common with r4 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 r2 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.
{n
R
Lb! V[NFHeuristic OptimizationhCostbased optimization is expensive, even with dynamic programming.
Systems may use heuristics to reduce the number of choices that must be made in a costbased fashion.
Heuristic optimization transforms the querytree 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 costbased optimization.
L:ZZcZ:bOG'Steps in Typical Heuristic Optimization(($Deconstruct conjunctive selections into a sequence of single selection operations (Equiv. Rule 1).
Move selection operations down the query tree for the earliest possible execution (Equiv. Rules 2, 7a, 7b, 11).
Execute first those selection and join operations that will produce the smallest relations (Equiv. rule 6).
Replace Cartesian product operations that are followed by a selection condition by join operations (Equiv. Rule 4a).
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).
Identify those subtrees whose operations can be pipelined, and execute them using pipelining.
,PPHStructure of Query Optimizers~The System R optimizer considers only leftdeep 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.
~QI%Structure of Query Optimizers (Cont.)&&$8Some query optimizers integrate heuristic selection and the generation of alternative access plans.
System R and Starburst use a hierarchical procedure based on the nestedblock concept of SQL: heuristic rewriting followed by costbased joinorder optimization.
The Oracle7 optimizer supports a heuristic based on available access paths.
Even with the use of heuristics, costbased query optimization impose a substantial overhead.
This expense is usually more than offset by savings at queryexecution time, particularly by reducing the number of slow disk accesses.
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