Finding Candidate Keys for Relational Schema S

We are given a relational schema $\text{S} = (\text{U, V, W, X, Y, Z})$ and a set of functional dependencies (FDs): $\{\text{U} \rightarrow \text{V, VW} \rightarrow \text{X, Y} \rightarrow \text{W, X} \rightarrow \text{U}\}$. Our goal is to find the candidate keys for this schema based on these FDs.

A candidate key is a minimal superkey. A superkey is a set of attributes that uniquely identifies all other attributes in the relation. In terms of functional dependencies, a set of attributes K is a superkey if the closure of K, denoted as $\text{K}^{+}$, contains all attributes in the schema S. Minimality means that no proper subset of K is also a superkey.

Identifying Attributes Not on the Right-Hand Side

Attributes that do not appear on the right-hand side (RHS) of any functional dependency must be part of every candidate key. Let's list the attributes on the RHS of the given FDs:

• $\text{U} \rightarrow \text{V}$: V is on the RHS.
• $\text{VW} \rightarrow \text{X}$: X is on the RHS.
• $\text{Y} \rightarrow \text{W}$: W is on the RHS.
• $\text{X} \rightarrow \text{U}$: U is on the RHS.

The attributes on the RHS are {U, V, W, X}. The attributes in the schema S are {U, V, W, X, Y, Z}. The attributes not on the RHS are {Y, Z}. Therefore, any candidate key must contain both Y and Z.

Calculating Closures to Find Superkeys

Since YZ must be in every candidate key, we start by considering sets that include YZ and adding other attributes from {U, V, X} (the attributes that can be determined from YZ and other FDs). Let's check sets formed by adding one attribute from {U, V, X} to YZ.

• Closure of UYZ:
  $(\text{UYZ})^{+} = \text{UYZ}$
  Using $\text{U} \rightarrow \text{V}$: $\text{UYZ} \rightarrow \text{UVYZ}$
  Using $\text{Y} \rightarrow \text{W}$: $\text{UVYZ} \rightarrow \text{UVWYZ}$
  Using $\text{VW} \rightarrow \text{X}$: $\text{UVWYZ} \rightarrow \text{UVWXYZ}$
  Using $\text{X} \rightarrow \text{U}$: $\text{UVWXYZ}$ (U is already in the set)
  $(\text{UYZ})^{+} = \{\text{U, V, W, X, Y, Z}\}$. Since the closure contains all attributes in S, UYZ is a superkey.
• Closure of VYZ:
  $(\text{VYZ})^{+} = \text{VYZ}$
  Using $\text{Y} \rightarrow \text{W}$: $\text{VYZ} \rightarrow \text{VWYZ}$
  Using $\text{VW} \rightarrow \text{X}$: $\text{VWYZ} \rightarrow \text{VWXYZ}$
  Using $\text{X} \rightarrow \text{U}$: $\text{VWXYZ} \rightarrow \text{UVWXYZ}$
  Using $\text{U} \rightarrow \text{V}$: $\text{UVWXYZ}$ (V is already in the set)
  $(\text{VYZ})^{+} = \{\text{U, V, W, X, Y, Z}\}$. Since the closure contains all attributes in S, VYZ is a superkey.
• Closure of XYZ:
  $(\text{XYZ})^{+} = \text{XYZ}$
  Using $\text{X} \rightarrow \text{U}$: $\text{XYZ} \rightarrow \text{UXYZ}$
  Using $\text{U} \rightarrow \text{V}$: $\text{UXYZ} \rightarrow \text{UVXYZ}$
  Using $\text{Y} \rightarrow \text{W}$: $\text{UVXYZ} \rightarrow \text{UVWXYZ}$
  Using $\text{VW} \rightarrow \text{X}$: $\text{UVWXYZ}$ (V, W, X are already in the set)
  $(\text{XYZ})^{+} = \{\text{U, V, W, X, Y, Z}\}$. Since the closure contains all attributes in S, XYZ is a superkey.

Thus, UYZ, VYZ, and XYZ are superkeys.

Checking Minimality for Candidate Keys

Now we need to check if these superkeys are minimal. We do this by calculating the closure of their proper subsets. If the closure of any proper subset contains all attributes, the original set is not minimal and thus not a candidate key.

• For UYZ:
  • $(\text{UY})^{+} = \text{UY} \xrightarrow{\text{U}\rightarrow\text{V}} \text{UVY} \xrightarrow{\text{Y}\rightarrow\text{W}} \text{UVWY} \xrightarrow{\text{VW}\rightarrow\text{X}} \text{UVWXY} \xrightarrow{\text{X}\rightarrow\text{U}} \text{UVWXY}$. Misses Z. Not a superkey.
  • $(\text{UZ})^{+} = \text{UZ} \xrightarrow{\text{U}\rightarrow\text{V}} \text{UVZ}$. Misses W, X, Y. Not a superkey.
  • $(\text{YZ})^{+} = \text{YZ} \xrightarrow{\text{Y}\rightarrow\text{W}} \text{YZW}$. Misses U, V, X. Not a superkey.
  Since no proper subset of UYZ is a superkey, UYZ is a candidate key.
• For VYZ:
  • $(\text{VY})^{+} = \text{VY} \xrightarrow{\text{Y}\rightarrow\text{W}} \text{VWY} \xrightarrow{\text{VW}\rightarrow\text{X}} \text{VWXY} \xrightarrow{\text{X}\rightarrow\text{U}} \text{UVWXY}$. Misses Z. Not a superkey.
  • $(\text{VZ})^{+} = \text{VZ}$. Misses U, W, X, Y. Not a superkey.
  • $(\text{YZ})^{+} = \text{YZ} \xrightarrow{\text{Y}\rightarrow\text{W}} \text{YZW}$. Misses U, V, X. Not a superkey.
  Since no proper subset of VYZ is a superkey, VYZ is a candidate key.
• For XYZ:
  • $(\text{XY})^{+} = \text{XY} \xrightarrow{\text{X}\rightarrow\text{U}} \text{UXY} \xrightarrow{\text{U}\rightarrow\text{V}} \text{UVXY} \xrightarrow{\text{Y}\rightarrow\text{W}} \text{UVWXY}$. Misses Z. Not a superkey.
  • $(\text{XZ})^{+} = \text{XZ} \xrightarrow{\text{X}\rightarrow\text{U}} \text{UXZ} \xrightarrow{\text{U}\rightarrow\text{V}} \text{UVXZ}$. Misses W, Y. Not a superkey.
  • $(\text{YZ})^{+} = \text{YZ} \xrightarrow{\text{Y}\rightarrow\text{W}} \text{YZW}$. Misses U, V, X. Not a superkey.
  Since no proper subset of XYZ is a superkey, XYZ is a candidate key.

Analyzing Other Sets from Options

Let's check the closure of other sets mentioned in the options to confirm they are not candidate keys (or even superkeys):

• Closure of UY: $(\text{UY})^{+} = \{\text{U, V, W, X, Y}\}$. Misses Z. Not a superkey.
• Closure of VY: $(\text{VY})^{+} = \{\text{U, V, W, X, Y}\}$. Misses Z. Not a superkey.
• Closure of XY: $(\text{XY})^{+} = \{\text{U, V, W, X, Y}\}$. Misses Z. Not a superkey.
• Closure of VWZ:
  $(\text{VWZ})^{+} = \text{VWZ} \xrightarrow{\text{VW}\rightarrow\text{X}} \text{VWXZ} \xrightarrow{\text{X}\rightarrow\text{U}} \text{UVWXZ} \xrightarrow{\text{U}\rightarrow\text{V}} \text{UVWXZ}$. Misses Y. Not a superkey.

Summary of Candidate Keys

Based on our analysis, the candidate keys for the given relational schema and functional dependencies are {UYZ, VYZ, XYZ}.

Comparing with Given Options

Let's compare our findings with the provided options:

• Option 1: UY, VY - Neither UY nor VY are superkeys. Incorrect.
• Option 2: UY, VY, XY - UY, VY, and XY are not superkeys. Incorrect.
• Option 3: UYZ, VYZ, VWZ - UYZ and VYZ are candidate keys, but VWZ is not a superkey. Incorrect.
• Option 4: UYZ, VYZ, XYZ - UYZ, VYZ, and XYZ are all candidate keys we identified. Correct.

The candidate keys among the given options are UYZ, VYZ, and XYZ.

Revision Table: Relational Database Key Concepts

Additional Information: Database Keys

In database design, various types of keys play crucial roles in maintaining data integrity and establishing relationships between tables.

• Primary Key: This is one of the candidate keys chosen by the database designer to uniquely identify each tuple (row) in a relation. It must be unique and non-null. There can be only one primary key per table.
• Alternate Keys: These are the candidate keys that are not chosen as the primary key. A relation can have multiple alternate keys, or none if there's only one candidate key that becomes the primary key.
• Foreign Key: This is a set of attributes in one relation (the referencing relation) that refers to the primary key or a candidate key of another relation (the referenced relation). Foreign keys are used to link tables and maintain referential integrity. They establish relationships between different tables in a relational database.

Understanding these different types of keys and how to find candidate keys using functional dependencies is fundamental for database design and normalization.