Database System Concepts
7th Edition
ISBN: 9780078022159
Author: Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher: McGraw-Hill Education
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Chapter 7, Problem 43E
Program Plan Intro
Materialized view:
- Materialized view is a
database object that contains the result of a query. - It is mainly used for increasing application performance.
- It is used for replicating data and to cache expensive queries in a data warehouse environment.
</PROGRAM-PLAN-INNTRO>
Constraints on materialized views enforcing functional dependency:
- The constraints on materialized view helps for increasing execution speed.
- Materialized view is refreshed by the type of materialized view.
- The constraints under materialized view should be normalized.
- That means each dimension should be contained in one table.
- Also, the joins between tables must be normalized or partially normalized.
- Hence, there is a guarantee that each child side row joins with exactly one parent side row.
- Hence, constraints on materialized view can be used for enforcing functional dependency.
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Given the database schema R(A, B, C), and a relation r on the schema R, write an SQL query to test whether the functional dependency B → C holds on relation r. Also write an SQL assertion that enforces the functional dependency. Assume that no null values are present. (Although part of the SQL standard, such assertions are not supported by any database implementation currently.)
Give an SQL schema definition for the employee database of Figure 3.19.Choose an appropriate domain for each attribute and an appropriate primarykey for each relation schema. Include any foreign-key constraints that might beappropriate.
Apply the ER-to-Relational Mapping Algorithm to get the relational model
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Chapter 7 Solutions
Database System Concepts
Ch. 7 - Prob. 1PECh. 7 - Prob. 2PECh. 7 - Explain how functional dependencies can be used to...Ch. 7 - Prob. 4PECh. 7 - Prob. 5PECh. 7 - Prob. 6PECh. 7 - Prob. 7PECh. 7 - Prob. 8PECh. 7 - Prob. 9PECh. 7 - Prob. 10PE
Ch. 7 - Prob. 11PECh. 7 - Prob. 12PECh. 7 - Prob. 13PECh. 7 - Prob. 14PECh. 7 - Prob. 15PECh. 7 - Prob. 16PECh. 7 - Prob. 17PECh. 7 - Prob. 18PECh. 7 - Prob. 19PECh. 7 - Prob. 20PECh. 7 - Prob. 21ECh. 7 - Prob. 22ECh. 7 -
Explain what is meant by repetition of...Ch. 7 -
Why are certain functional dependencies called...Ch. 7 - Prob. 25ECh. 7 - Prob. 26ECh. 7 - Prob. 27ECh. 7 - Prob. 28ECh. 7 - Prob. 29ECh. 7 - Prob. 30ECh. 7 - Prob. 32ECh. 7 - Prob. 33ECh. 7 - Prob. 35ECh. 7 - Prob. 36ECh. 7 - Prob. 37ECh. 7 - Prob. 38ECh. 7 - Prob. 39ECh. 7 - Prob. 40ECh. 7 - Prob. 41ECh. 7 - Prob. 42ECh. 7 - Prob. 43E
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- Consider a database containing two relations: R(A,B,C) and S(D,E,F). For each of three (your choice) of the 12 rewriting rules for relational algebra expression, give an example pair of relational algebra expressions involving R and S such that the rule rewrites one expression to the other expression, and give another example pair to show the rewriting rule cannot be used since some of the conditions for the rule arenot satisfiedarrow_forwardFor relation (R), its attributes and its functional dependencies (F): R(A, B, C, D, E) C D → B A → D E → C 1. Decompose the above relation using the Boyce-Codd Normal Form (BCNF) decomposition. 2. What is the key for the relation R?arrow_forwardLet R be a relational schema with attributes {M, N, O, P, Q, R} and let F be the set of functional dependencies F = {MN→OP, O→Q, QR→N, R→P}. Prove whether R is in 3NF.arrow_forward
- Consider a many-to-one relationship R between entity sets A and B. Supposethe relation created from R is combined with the relation created from A. InSQL, attributes participating in a foreign key constraint can be null. Explainhow a constraint on total participation of A in R can be enforced using not nullconstraints in SQL.arrow_forwardGive an SQL schema definition for the employee database of Figure 3.19.Choose an appropriate domain for each attribute and an appropriate primarykey for each relation schema. Include any foreign-key constraints that might be appropriate.arrow_forwardGiven relations P and Q, both over attributes A and B, write a query in relational algebra under bag semantics that returns P if Q is empty and returns Q if Q is not empty.arrow_forward
- Discuss the alternative ways that normalization can be used to support database design. How does normalization eradicate update anomalies from a relation?arrow_forwardDatabase Systems For each part of this problem you will need to construct a single SQL query which will check whether a certain condition holds on a specific instance of a relation, in the following way: your query should return an empty result if and only if the condition holds on the instance. (If the condition doesn’t hold, your query should return something non-empty). Note our language here: the conditions that we specify cannot be proved to hold in general without knowing the externally-defined functional dependencies; so what we mean is, check whether they could hold in general for the relation, given a specific set of tuples. You may assume that there will be no NULL values in the tables, and you may assume that the relations are sets rather than multisets, but otherwise your query should work for general in- stances. A is a superkey for a relation T (A, B, C, D). The combinations of two attributes in the relation T (A, B, C, D) are each keys. A tuple-generating…arrow_forwardConsider a relation with schema R(A,B,C,D,E,F,G). Choose the key for the relation for the following sets of functional dependencies. BCD→A BC→E A→F F→ G C→ D A→ G ? A) B B) C C) BC D) All combinations of attributes are keyarrow_forward
- Explain the concept of functional dependency in the context of database normalization and how it relates to the Second Normal Form (2NF).arrow_forwardConsider the relation r(A, B, C) with the attribute A as the index. Give an example of a query that can be answered only by looking at the index, rather than the tuples in the relationship. (Index-only plans are query plans that just utilise the index and do not access the actual relation.)arrow_forwardConsider a relation r(A, B, C), with an index on attribute A. Give an example of a query that can be answered by using the index only, without looking at the tuples in the relation. (Query plans that use only the index, without accessing the actual relation, are called index-only plans.)arrow_forward
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