Chapter 11: Problem 3

Why can we not guarantee that BCNF relation schemas will be produced by dependency-preserving decompositions of non-BCNF relation schemas? Give a counterexample to illustrate this point.

Short Answer

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BCNF and dependency-preservation can contradict each other in the context of decomposition of a non-BCNF schema. This is due to the strict condition of BCNF where 'every determiner must be a candidate key', which does not always guarantee the maintenance of all functional dependencies post-decomposition. A counterexample is a schema R with attributes {A, B, C} and functional dependencies {A -> B, B -> C}, which upon BCNF decomposition into R1 {A, B} and R2 {B, C} loses the dependency {A -> C} of the original schema.

Step by step solution

Understanding BCNF

BCNF, or Boyce-Codd Normal Form, is a standard for the design of a relational database schema that avoids certain anomalies when inserting, deleting, and updating data. A table is in BCNF if it is in 3NF (Third Normal Form) and for every non-trivial functional dependency \(X \rightarrow Y, X\) is a superkey. Superkey basically means a set of attributes that uniquely identifies tuples in a relation.

Understanding Dependency-Preserving decomposition

Dependency-preserving decomposition ensures minimizing redundancy and chances of anomalies while maintaining all of the functional dependencies after normalization, which is really important in relational databases. This lets one enforce constraints (functional dependencies) of the original schema in the decomposed schema.

Identifying potential conflict

The conflict between BCNF and dependency-preservation comes into play when performing decompositions on a non-BCNF schema. BCNF may not be able to maintain dependency preservation upon decomposing non-BCNF schemas because of its strict condition of 'every determiner being a candidate key'. This is because BCNF does not necessarily take into account the need to maintain all functional dependencies at the time of decomposition.

Providing a counterexample

Let's take a schema R with attributes \( \{A, B, C\} \) and functional dependencies \( \{A \rightarrow B, B \rightarrow C\} \). This schema is not in BCNF since \(B\), which is not a superkey, determines \(C\). However, if we decompose it into BCNF, we might get two schemas: \(R1\{A, B\}\) with \({A \rightarrow B}\) and \(R2\{B, C\}\) with \({B \rightarrow C}\). We can observe that the dependency \({A \rightarrow C}\) from the original schema is lost in the decomposed one. Hence, this is a counterexample where a BCNF decomposition is not dependency-preserving.

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Why can we not guarantee that BCNF relation schemas will be produced by dependency-preserving decompositions of non-BCNF relation schemas? Give a counterexample to illustrate this point.

Short Answer

Step by step solution

Understanding BCNF

Understanding Dependency-Preserving decomposition

Identifying potential conflict

Providing a counterexample

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