Query optimization in database

C H A P T E R
13
Query Optimization
Practice Exercises
13.1 Show that the following equivalences hold. Explain how you can apply
them to improve the efficiency of certain queries:
a. E1 1␪ (E2 − E3) = (E1 1␪ E2 − E1 1␪ E3).
b. ␴␪( AGF (E)) = AGF (␴␪(E)), where ␪ uses only attributes from A.
c. ␴␪(E1 1 E2) = ␴␪(E1) 1 E2, where ␪ uses only attributes from E1.
Answer:
a. E1 1␪ (E2 − E3) = (E1 1␪ E2 − E1 1␪ E3).
Let us rename (E1 1␪ (E2−E3)) as R1, (E1 1␪ E2) as R2 and (E1 1␪ E3)
as R3. It is clear that if a tuple t belongs to R1, it will also belong to R2.
If a tuple t belongs to R3, t[E3’s attributes] will belong to E3, hence
t cannot belong to R1. From these two we can say that
∀t, t ∈ R1 ⇒ t ∈ (R2 − R3)
It is clear that if a tuple t belongs to R2 − R3, then t[R2’s attributes] ∈
E2 and t[R2’s attributes] ∈ E3. Therefore:
∀t, t ∈ (R2 − R3) ⇒ t ∈ R1
The above two equations imply the given equivalence.
This equivalence is helpful because evaluation of the right hand
side join will produce many tuples which will finally be removed
from the result. The left hand side expression can be evaluated more
efficiently.
b. ␴␪( AGF (E)) = AGF (␴␪(E)), where ␪ uses only attributes from A.
␪ uses only attributes from A. Therefore if any tuple t in the output
of AGF (E) is filtered out by the selection of the left hand side, all the
tuples in E whose value in A is equal to t[A] are filtered out by the
selection of the right hand side. Therefore:
1

2 Chapter 13 Query Optimization
∀t, t ∈ ␴␪( AGF (E)) ⇒ t ∈ AGF (␴␪(E))
Using similar reasoning, we can also conclude that
∀t, t ∈ AGF (␴␪(E)) ⇒ t ∈ ␴␪( AGF (E))
This equivalence is helpful because evaluation of the right hand
side avoids performing the aggregation on groups which are any-
way going to be removed from the result. Thus the right hand side
expression can be evaluated more efficiently than the left hand side
expression.
c. ␴␪(E1 1 E2) = ␴␪(E1) 1 E2 where ␪ uses only attributes from E1.
␪ uses only attributes from E1. Therefore if any tuple t in the output
of (E1 1 E2) is filtered out by the selection of the left hand side, all
the tuples in E1 whose value is equal to t[E1] are filtered out by the
selection of the right hand side. Therefore:
∀t, t ∈ ␴␪(E1 1 E2) ⇒ t ∈ ␴␪(E1) 1 E2
Using similar reasoning, we can also conclude that
∀t, t ∈ ␴␪(E1) 1 E2 ⇒ t ∈ ␴␪(E1 1 E2)
This equivalence is helpful because evaluation of the right hand side
avoids producing many output tuples which are anyway going to
be removed from the result. Thus the right hand side expression can
be evaluated more efficiently than the left hand side expression.
13.2 For each of the following pairs of expressions, give instances of relations
that show the expressions are not equivalent.
a. A(R − S) and A(R) − A(S).
b. ␴B<4( AGmax(B) as B(R)) and AGmax(B) as B(␴B<4(R)).
c. In the preceding expressions, if both occurrences of max were re-
placed by min would the expressions be equivalent?
d. (R 1 S) 1 T and R 1 (S 1 T)
In other words, the natural left outer join is not associative. (Hint:
Assume that the schemas of the three relations are R(a, b1), S(a, b2),
and T(a, b3), respectively.)
e. ␴␪(E1 1 E2) and E1 1 ␴␪(E2), where ␪ uses only attributes from E2.
Answer:
a. R = {(1, 2)}, S = {(1, 3)}
The result of the left hand side expression is {(1)}, whereas the result
of the right hand side expression is empty.

Exercises 3
b. R = {(1, 2), (1, 5)}
The left hand side expression has an empty result, whereas the right
hand side one has the result {(1, 2)}.
c. Yes, on replacing the max by the min, the expressions will become
equivalent. Any tuple that the selection in the rhs eliminates would
not pass the selection on the lhs if it were the minimum value, and
would be eliminated anyway if it were not the minimum value.
d. R = {(1, 2)}, S = {(2, 3)}, T = {(1, 4)}. The left hand expression
gives {(1, 2, null, 4)} whereas the the right hand expression gives
{(1, 2, 3, null)}.
e. Let R be of the schema (A, B) and S of (A, C). Let R = {(1, 2)}, S =
{(2, 3)} and let ␪ be the expression C = 1. The left side expres-
sion’s result is empty, whereas the right side expression results in
{(1, 2, null)}.
13.3 SQL allows relations with duplicates (Chapter 3).
a. Define versions of the basic relational-algebra operations ␴, , ×,
1, −, ∪, and ∩ that work on relations with duplicates, in a way
consistent with SQL.
b. Check which of the equivalence rules 1 through 7.b hold for the
multiset version of the relational-algebra defined in part a.
Answer:
a. We define the multiset versions of the relational-algebra operators
here. Given multiset relations r1 and r2,
i. ␴
Let there be c1 copies of tuple t1 in r1. If t1 satisfies the selection ␴␪,
then there are c1 copies of t1 in ␴␪(r1), otherwise there are none.
ii.
For each copy of tuple t1 in r1, there is a copy of tuple A(t1) in
A(r1), where A(t1) denotes the projection of the single tuple t1.
iii. ×
If there are c1 copies of tuple t1 in r1 and c2 copies of tuple t2 in
r2, then there are c1 ∗ c2 copies of the tuple t1.t2 in r1 × r2.
iv. 1
The output will be the same as a cross product followed by a
selection.
v. −
If there are c1 copies of tuple t in r1 and c2 copies of t in r2, then
there will be c1 − c2 copies of t in r1 − r2, provided that c1 − c2 is
positive.
vi. ∪

there will be c1 + c2 copies of t in r1 ∪ r2.
vii. ∩
there will be min(c1, c2) copies of t in r1 ∩ r2.
b. All the equivalence rules 1 through 7.b of section 13.2.1 hold for the
multiset version of the relational-algebra defined in the first part.
There exist equivalence rules which hold for the ordinary relational-
algebra, but do not hold for the multiset version. For example con-
sider the rule :-
A∩ B = A∪ B − (A− B) − (B − A)
This is clearly valid in plain relational-algebra. Consider a multiset
instance in which a tuple t occurs 4 times in A and 3 times in B. t
will occur 3 times in the output of the left hand side expression, but
6 times in the output of the right hand side expression. The reason
for this rule to not hold in the multiset version is the asymmetry in
the semantics of multiset union and intersection.
13.4 Consider the relations r1(A, B, C), r2(C, D, E), and r3(E, F), with primary
keys A, C, and E, respectively. Assume that r1 has 1000 tuples, r2 has 1500
tuples, and r3 has 750 tuples. Estimate the size of r1 1 r2 1 r3, and give
an efficient strategy for computing the join.
Answer:
• The relation resulting from the join of r1, r2, and r3 will be the same
no matter which way we join them, due to the associative and com-
mutative properties of joins. So we will consider the size based on the
strategy of ((r1 1 r2) 1 r3). Joining r1 with r2 will yield a relation
of at most 1000 tuples, since C is a key for r2. Likewise, joining that
result with r3 will yield a relation of at most 1000 tuples because E is
a key for r3. Therefore the final relation will have at most 1000 tuples.
• An efficient strategy for computing this join would be to create an
index on attribute C for relation r2 and on E for r3. Then for each tuple
in r1, we do the following:
a. Use the index for r2 to look up at most one tuple which matches
the C value of r1.
b. Use the created index on E to look up in r3 at most one tuple which
matches the unique value for E in r2.
13.5 Consider the relations r1(A, B, C), r2(C, D, E), and r3(E, F) of Practice
Exercise 13.4. Assume that there are no primary keys, except the entire
schema. Let V(C,r1) be 900, V(C,r2) be 1100, V(E,r2) be 50, and V(E,r3)
be 100. Assume that r1 has 1000 tuples, r2 has 1500 tuples, and r3 has 750

Exercises 5
tuples. Estimate the size of r1 1 r2 1 r3 and give an efficient strategy for
computing the join.
Answer: The estimated size of the relation can be determined by cal-
culating the average number of tuples which would be joined with each
tuple of the second relation. In this case, for each tuple in r1, 1500/V(C,r2)
= 15/11 tuples (on the average) of r2 would join with it. The intermedi-
ate relation would have 15000/11 tuples. This relation is joined with r3
to yield a result of approximately 10,227 tuples (15000/11 × 750/100 =
10227). A good strategy should join r1 and r2 first, since the intermediate
relation is about the same size as r1 or r2. Then r3 is joined to this result.
13.6 Suppose that a B+
-tree index on building is available on relation department,
and that no other index is available. What would be the best way to handle
the following selections that involve negation?
a. ␴¬(building <“Watson”)(department)
b. ␴¬(building =“Watson”)(department)
c. ␴¬(building <“Watson” ∨ budget <50000)(department)
Answer:
a. Use the index to locate the first tuple whose building field has value
“Watson”. From this tuple, follow the pointer chains till the end,
retrieving all the tuples.
b. For this query, the index serves no purpose. We can scan the file
sequentially and select all tuples whose building field is anything
other than “Watson”.
c. This query is equivalent to the query:
␴building ≥’Watson’ ∧ budget <5000)(department).
Using the building index, we can retrieve all tuples with building
value greater than or equal to “Watson” by following the pointer
chains from the first “Watson” tuple. We also apply the additional
criteria of budget < 5000 on every tuple.
13.7 Consider the query:
select *
from r, s
where upper(r.A) = upper(s.A);
where “upper” is a function that returns its input argument with all low-
ercase letters replaced by the corresponding uppercase letters.
a. Find out what plan is generated for this query on the database system
you use.

b. Some database systems would use a (block) nested-loop join for this
query, which can be very inefficient. Briefly explain how hash-join
or merge-join can be used for this query.
Answer:
a. First create relations r and s, and add some tuples to the two re-
lations, before finding the plan chosen; or use existing relations in
place of r and s. Compare the chosen plan with the plan chosen for a
query directly equating r.A = s.B. Check the estimated statistics too.
Some databases may give the same plan, but with vastly different
statistics.
(On PostgreSQL, we found that the optimizer used the merge join
plan described in the answer to the next part of this question.)
b. To use hash join, hashing should be done after applying the upper()
function to r.Aand s.A. Similarly, for merge join, the relations should
be sorted on the result of applying the upper() function on r.A and
s.A. The hash or merge join algorithms can then be used unchanged.
13.8 Give conditions under which the following expressions are equivalent
A,BGagg(C)(E1 1 E2) and (AGagg(C)(E1)) 1 E2
where agg denotes any aggregation operation. How can the above condi-
tions be relaxed if agg is one of min or max?
Answer: The above expressions are equivalent provided E2 contains only
attributes Aand B, with Aas the primary key (so there are no duplicates).
It is OK if E2 does not contain some A values that exist in the result of
E1, since such values will get filtered out in either expression. However, if
there are duplicate values in E2.A, the aggregate results in the two cases
would be different.
If the aggregate function is min or max, duplicate A values do not have
any effect. However, there should be no duplicates on (A, B); the first
expression removes such duplicates, while the second does not.
13.9 Consider the issue of interesting orders in optimization. Suppose you are
given a query that computes the natural join of a set of relations S. Given
a subset S1 of S, what are the interesting orders of S1?
Answer: The interesting orders are all orders on subsets of attributes that
can potentially participate in join conditions in further joins. Thus, let T
be the set of all attributes of S1 that also occur in any relation in S − S1.
Then every ordering of every subset of T is an interesting order.
13.10 Show that, with n relations, there are (2(n − 1))!/(n − 1)! different join
orders. Hint: A complete binary tree is one where every internal node has
exactly two children. Use the fact that the number of different complete

Exercises 7
binary trees with n leaf nodes is:
1
n
2(n − 1)
(n − 1)
If you wish, you can derive the formula for the number of complete binary
trees with n nodes from the formula for the number of binary trees with
n nodes. The number of binary trees with n nodes is:
1
n + 1
2n
n
This number is known as the Catalan number, and its derivation can be
found in any standard textbook on data structures or algorithms.
Answer: Each join order is a complete binary tree (every non-leaf node
has exactly two children) with the relations as the leaves. The number
of different complete binary trees with n leaf nodes is 1
n
2(n−1)
(n−1)
. This is
because there is a bijection between the number of complete binary trees
with n leaves and number of binary trees with n − 1 nodes. Any complete
binary tree with n leaves has n − 1 internal nodes. Removing all the leaf
nodes, we get a binary tree with n−1 nodes. Conversely, given any binary
tree with n − 1 nodes, it can be converted to a complete binary tree by
adding n leaves in a unique way. The number of binary trees with n − 1
nodes is given by 1
n
2(n−1)
(n−1)
, known as the Catalan number. Multiplying this
by n! for the number of permutations of the n leaves, we get the desired
result.
13.11 Show that the lowest-cost join order can be computed in time O(3n
).
Assume that you can store and look up information about a set of relations
(such as the optimal join order for the set, and the cost of that join order)
in constant time. (If you find this exercise difficult, at least show the looser
time bound of O(22n
).)
Answer: Consider the dynamic programming algorithm given in Sec-
tion 13.4. For each subset having k + 1 relations, the optimal join order
can be computed in time 2k+1
. That is because for one particular pair of
subsets A and B, we need constant time and there are at most 2k+1
− 2
different subsets that Acan denote. Thus, over all the n
k+1
subsets of size
k + 1, this cost is n
k+1
2k+1
. Summing over all k from 1 to n − 1 gives the
binomial expansion of ((1 + x)n
− x) with x = 2. Thus the total cost is less
than 3n
.
13.12 Show that, if only left-deep join trees are considered, as in the System R
optimizer, the time taken to find the most efficient join order is around
n2n
. Assume that there is only one interesting sort order.
Answer: The derivation of time taken is similar to the general case, except
that instead of considering 2k+1
− 2 subsets of size less than or equal to

k for A, we only need to consider k + 1 subsets of size exactly equal to
k. That is because the right hand operand of the topmost join has to be a
single relation. Therefore the total cost for ﬁnding the best join order for
all subsets of size k + 1 is n
k+1 (k + 1), which is equal to n n−1
k . Summing
over all k from 1 to n − 1 using the binomial expansion of (1 + x)n−1
with
x = 1, gives a total cost of less than n2n−1
.
13.13 Consider the bank database of Figure 13.9, where the primary keys are un-
derlined. Construct the following SQL queries for this relational database.
a. Write a nested query on the relation account to ﬁnd, for each branch
with name starting with B, all accounts with the maximum balance
at the branch.
b. Rewrite the preceding query, without using a nested subquery; in
other words, decorrelate the query.
c. Give a procedure (similar to that described in Section 13.4.4) for
decorrelating such queries.
Answer:
a. The nested query is as follows:
select S.acount number
from account S
where S.branch name like ’B%’ and
S.balance =
(select max(T.balance)
from account T
where T.branch name = S.branch name)
b. The decorrelated query is as follows:
create table t1 as
select branch name, max(balance)
from account
group by branch name
select account number
from account, t1
where account.branch name like ’B%’ and
account.branch name = t1.branch name and
account.balance = t1.balance
c. In general, consider the queries of the form:

Exercises 9
select · · ·
from L1
where P1 and
A1 op
(select f(A2)
from L2
where P2)
where, f is some aggregate function on attributes A2, and op is some
boolean binary operator. It can be rewritten as
create table t1 as
select f(A2),V
from L2
where P1
2
group by V
select · · ·
from L1, t1
where P1 and P2
2 and
A1 op t1.A2
where P1
2 contains predicates in P2 without selections involving
correlation variables, and P2
2 introduces the selections involving the
correlation variables. V contains all the attributes that are used in
the selections involving correlation variables in the nested query.
13.14 The set version of the semijoin operator ⋉ is deﬁned as follows:
r ⋉␪ s = R(r 1␪ s)
where R is the set of attributes in the schema of r. The multiset version of
the semijoin operation returns the same set of tuples, but each tuple has
exactly as many copies as it had in r.
Consider the nested query we saw in Section 13.4.4 which ﬁnds the names
of all instructors who taught a course in 2007. Write the query in relational
algebra using the multiset semjoin operation, ensuring that the number
of duplicates of each name is the same as in the SQL query. (The semijoin
operation is widely used for decorrelation of nested queries.)
Answer: The query can be written as follows:
instructor ⋉instructor.I D=teaches.I D (␴year=2007(teaches))

Query optimization in database

Query optimization in database

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