Q1: The select (o) operation and (∞) join operations are commutative. Give one scenario or case under which the right hand side of the following equivalence rule is more efficient that its left hand side, explain your answer: (1) Oc1(0c2(R)) = 0c2(0c1(R)) (2) Rab S = S 00a-b R (when nested block join algorithm is used) Where ∞ is an equi-join operation R and S are tables a is an attribute in R b is an attribute in S c1, and c2 are conditions
Q1: The select (o) operation and (∞) join operations are commutative. Give one scenario or case under which the right hand side of the following equivalence rule is more efficient that its left hand side, explain your answer: (1) Oc1(0c2(R)) = 0c2(0c1(R)) (2) Rab S = S 00a-b R (when nested block join algorithm is used) Where ∞ is an equi-join operation R and S are tables a is an attribute in R b is an attribute in S c1, and c2 are conditions
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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![Q1:
The select (o) operation and (c) join operations are commutative. Give one scenario or
case under which the right hand side of the following equivalence rule is more efficient
that its left hand side, explain your answer:
(1) Oc1(0c2(R)) = 0c2(0c1(R))
(2) Rab S = Scoa-b R (when nested block join algorithm is used)
Where
co is an equi-join operation
R and S are tables
a is an attribute in R
b is an attribute in S
c1, and c2 are conditions](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F942c8d0f-f1c4-4293-a355-ff4ab799bb7f%2F05c7fad6-24a0-45a2-b575-dcb02f7af82b%2Fktbialw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Q1:
The select (o) operation and (c) join operations are commutative. Give one scenario or
case under which the right hand side of the following equivalence rule is more efficient
that its left hand side, explain your answer:
(1) Oc1(0c2(R)) = 0c2(0c1(R))
(2) Rab S = Scoa-b R (when nested block join algorithm is used)
Where
co is an equi-join operation
R and S are tables
a is an attribute in R
b is an attribute in S
c1, and c2 are conditions
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