Relational Schema Practice Questions

The fundamental rule for a superkey is that it must contain a candidate key. For the given relation $R(A, B, C, D)$, the candidate keys are $\{A, B\}$ and $\{A, C\}$. Therefore, any superkey of $R$ must contain either $\{A, B\}$ or $\{A, C\}$.

Let's list all possible superkeys based on this rule:

1. Superkeys containing $\{A, B\}$: These are formed by taking $\{A, B\}$ and adding any subset of the remaining attributes ({C, D}$).

   • $\{A, B\}$
   • $\{A, B, C\}$
   • $\{A, B, D\}$
   • $\{A, B, C, D\}$

2. Superkeys containing $\{A, C\}$: These are formed by taking $\{A, C\}$ and adding any subset of the remaining attributes ({B, D}$).

   • $\{A, C\}$
   • $\{A, C, D\}$
   • $\{A, B, C\}$ (already listed above)
   • $\{A, B, C, D\}$ (also already listed above)

Combining the unique superkeys from both lists, we get:
$\{A, B\}$, $\{A, C\}$, $\{A, B, C\}$, $\{A, B, D\}$, $\{A, C, D\}$, $\{A, B, C, D\}$.

Counting these unique sets, the total number of superkeys is 6.

R1(A, B, C, D, E) with FDs: {D → E, EA → B, EB → C}
Candidate keys for R1:
(DA)+ = {DAEBC}. So DA is a candidate key.
R1 is in 1NF (all attributes are atomic).
Check for 2NF: EA → B and EB → C.
Since DA is a candidate key, D and A are prime attributes. E, B, C are non-prime attributes.
EA → B: E is a non-prime attribute. This is a partial dependency because E is part of the candidate key (DA) and determines B. Also, (EA) is not a candidate key. Therefore, R1 is NOT in 2NF. So, R1 is NOT in 3NF either. Hence (C) is TRUE.

R2(A, B, C, D) with FDs: {A → D, A → B, C → A}
Candidate keys for R2:
(C)+ = {CADB}. So C is a candidate key.
Check for 2NF: All attributes are prime as C is the only candidate key. No partial dependencies possible. So R2 is in 2NF.
Check for 3NF: For C → A, C is a candidate key, so it's OK. For A → D and A → B:
A is a prime attribute (part of candidate key C+), and D and B are prime attributes. Since C is the only candidate key, A is not a superkey.
This implies a transitive dependency for C → A → D and C → A → B (if D and B were non-prime). However, A, B, D are also prime attributes.
A → D: Here A is not a candidate key. D is a prime attribute. So this is a partial dependency for any candidate key containing A.
Thus, R2 is NOT in 3NF. Hence (D) is TRUE.

A functional dependency $F: X \rightarrow Y$ is useful if it satisfies three conditions:

1. $X$ is not the empty set.
2. $Y$ is not the empty set.
3. The intersection of $X$ and $Y$ is the empty set ($X \cap Y = \emptyset$).

Given a relation $R$ with 4 attributes, let $A = \{A_1, A_2, A_3, A_4\}$.
The total number of possible subsets for $X$ (left-hand side) and $Y$ (right-hand side) is $2^4 = 16$.
However, $X$ and $Y$ cannot be empty, so $2^4 - 1 = 15$ non-empty subsets.

For each possible non-empty subset $X \subseteq A$:
The non-empty subsets $Y \subseteq A$ such that $X \cap Y = \emptyset$.
The attributes in $Y$ must be from $A \setminus X$.
Let $|X| = k$. Then $|A \setminus X| = 4-k$.
The number of non-empty subsets of $A \setminus X$ is $2^{(4-k)} - 1$.

We sum this over all possible non-empty subsets $X$:

• If $|X|=1$: There are $\binom{4}{1}$ such subsets ($X$). For each, $A \setminus X$ has 3 attributes. Number of $Y$ sets is $2^3 - 1 = 7$.
  So $4 \times 7 = 28$ FDs.
  (e.g., for $X=\{A_1\}$, $Y$ can be any non-empty subset of $\{A_2, A_3, A_4\}$).
• If $|X|=2$: There are $\binom{4}{2}$ such subsets ($X$). For each, $A \setminus X$ has 2 attributes. Number of $Y$ sets is $2^2 - 1 = 3$.
  So $6 \times 3 = 18$ FDs.
  (e.g., for $X=\{A_1, A_2\}$, $Y$ can be any non-empty subset of $\{A_3, A_4\}$).
• If $|X|=3$: There are $\binom{4}{3}$ such subsets ($X$). For each, $A \setminus X$ has 1 attribute. Number of $Y$ sets is $2^1 - 1 = 1$.
  So $4 \times 1 = 4$ FDs.
  (e.g., for $X=\{A_1, A_2, A_3\}$, $Y$ can be $\{A_4\}$).
• If $|X|=4$: There are $\binom{4}{4}$ such subsets ($X$). For each, $A \setminus X$ has 0 attributes. Number of $Y$ sets is $2^0 - 1 = 0$.
  So $1 \times 0 = 0$ FDs.

Total number of useful functional dependencies = $28 + 18 + 4 + 0 = 50$.
===END_Q56===

Let's analyze each statement using the Armstrong's axioms for functional dependencies (FDs):

1. Reflexivity: If $Y \subseteq X$, then $X \rightarrow Y$.
2. Augmentation: If $X \rightarrow Y$, then $XZ \rightarrow YZ$.
3. Transitivity: If $X \rightarrow Y$ and $Y \rightarrow Z$, then $X \rightarrow Z$.

Given options:
(a) $(X, Y) \rightarrow (Z, W)$ implies $X \rightarrow (Z, W)$: This is FALSE. This is a projection rule, but not a valid inference from Armstrong's axioms without further information. For example, if $X=\{A\}, Y=\{B\}, Z=\{C\}, W=\{D\}$, $AB \rightarrow CD$ does not imply $A \rightarrow CD$. $A$ alone might not determine $C$ or $D$.
(b) $(X, Y) \rightarrow (Z, W)$ implies $(X, Y) \rightarrow Z$: This is TRUE. This is an application of decomposition rule (derived from Armstrong's axioms): If $X \rightarrow YZ$, then $X \rightarrow Y$ and $X \rightarrow Z$. Here, $Y$ in the rule is $Z$ in the question, and $Z$ in the rule is $W$ in the question. So, $(X,Y) \rightarrow (ZW)$ implies $(X,Y) \rightarrow Z$.
(c) $((X, Y) \rightarrow Z \text{ and } W \rightarrow Y)$ implies $(X, W) \rightarrow Z$: This is TRUE.
Given: 1. $XY \rightarrow Z$ and 2. $W \rightarrow Y$.
From (2) by Augmentation: $XW \rightarrow XY$.
From $XW \rightarrow XY$ and $XY \rightarrow Z$ by Transitivity: $XW \rightarrow Z$.
This statement holds.
(d) $(X \rightarrow Y \text{ and } Y \rightarrow Z)$ implies $X \rightarrow Z$: This is TRUE. This is the Transitivity rule directly.

Therefore, statements (b), (c), and (d) are TRUE.

The degree or arity of a relation in relational database systems refers to the number of attributes (columns) in its relation schema.
This is a fundamental definition in relational database theory.
The other options refer to the number of rows (tuples), total entries, or distinct domains, which are different concepts.
Therefore, the number of attributes directly corresponds to the degree or arity.

Given relation $R(A, B, C, D, E)$ with functional dependencies:

1. $AB \rightarrow C$
2. $BC \rightarrow D$
3. $C \rightarrow E$

To find superkeys, first find the candidate keys. A candidate key is a minimal superkey.
Let's check the attributes that do not appear on the right side of any FD. These attributes must be part of any candidate key.
Right side attributes: C, D, E.
Left side attributes: A, B.
Attributes that only appear on the left side or not at all are A, B. So, A and B must be part of any candidate key. Let's find the closure of AB.
$(AB)^+$:
$AB \rightarrow AB$ (reflexivity)
Using $AB \rightarrow C$, we get $ABC$.
Using $BC \rightarrow D$, from $ABC$, we get $ABCD$.
Using $C \rightarrow E$, from $ABCD$, we get $ABCDE$.
So, $(AB)^+ = \{A, B, C, D, E\}$. Since $(AB)^+$ contains all attributes, AB is a candidate key.
Since A and B are the only attributes that do not appear on the RHS of any FD, AB must be a candidate key, and it is the only candidate key because it is minimal.

Now, we need to find the number of superkeys.
A superkey is any set of attributes that contains a candidate key.
The only candidate key is $AB$.
Any superset of $AB$ will be a superkey.
The set of all attributes is $U = \{A, B, C, D, E\}$.
The number of superkeys is the number of subsets of $U$ that contain $\{A, B\}$.
This is equivalent to finding all subsets of the remaining attributes $\{C, D, E\}$ and adding $\{A, B\}$ to each of them.
The set $\{C, D, E\}$ has $2^3 = 8$ subsets:
$\emptyset, \{C\}, \{D\}, \{E\}, \{C, D\}, \{C, E\}, \{D, E\}, \{C, D, E\}$.
Adding $\{A, B\}$ to each of these subsets yields the superkeys:
$\{A, B\}, \{A, B, C\}, \{A, B, D\}, \{A, B, E\}, \{A, B, C, D\}, \{A, B, C, E\}, \{A, B, D, E\}, \{A, B, C, D, E\}$.
There are 8 superkeys.

For decomposition $D_1$, the relations are $R_1(P,Q,S,T)$, $R_2(P,T,X)$, $R_3(Q,Y)$, and $R_4(Y,Z,W)$.
Joining $R_1$ and $R_2$: common attributes are $\{P,T\}$. Since $(P,T)^+ = \{P,T,X,Y,Z,W\}$, $\{P,T\}$ is a superkey for $R_2(P,T,X)$, making the join lossless.
The subsequent joins with $R_3(Q,Y)$ (common attribute $Q$, $Q \rightarrow Y$) and $R_4(Y,Z,W)$ (common attribute $Y$, $Y \rightarrow ZW$) are also lossless as the common attributes form a superkey for one of the relations. Thus, $D_1$ is a lossless decomposition.

For decomposition $D_2$, the relations are $R_1(P,Q,S)$, $R_2(T,X)$, $R_3(Q,Y)$, and $R_4(Y,Z,W)$.
Joining $R_1(P,Q,S)$ and $R_2(T,X)$, there are no common attributes.
A decomposition is lossless if and only if for every pair of relations $R_i, R_j$ in the decomposition, the intersection of their schemas is a superkey for at least one of them. Since the intersection of $R_1$ and $R_2$ is empty, it cannot be a superkey, making the join lossy. Thus, $D_2$ is a lossy decomposition.

Statement S1: A relation scheme can have multiple foreign keys. For example, a relation Enrollment could have foreign keys referencing Student and Course relations. Thus, S1 is false.
Statement S2: A foreign key in a relation scheme R can refer to tuples within the same relation R. This is known as a self-referencing foreign key. For example, an Employee relation might have a ManagerID foreign key that references EmployeeID within the same Employee relation. Thus, S2 is false.
Since both statements S1 and S2 are false, option (d) is correct.

The database three-level architecture consists of External, Conceptual, and Internal levels. Logical Data Independence defines the ability to modify the conceptual schema without necessitating changes to external schemas or application programs. Conversely, Physical Data Independence relates to modifying physical storage without impacting the conceptual level. As the question specifies immunity of the external schemas to changes in the conceptual schema, it corresponds to Logical Data Independence.

A candidate key must uniquely identify each tuple in a relation.

• $\text{{Last Name}}$ is invalid because "Smith" appears twice.
• $\text{{Room}}$ is invalid because "33" occurs for two different entries.
• $\text{{Shift}}$ is invalid because "Morning" is associated with multiple rows.
• The composite attribute $\text{{Room, Shift}}$ produces unique combinations: $\{(234, \text{Morning}), (33, \text{Afternoon}), (33, \text{Evening}), (222, \text{Morning})\}$. Since these pairs are distinct for every row, $\text{{Room, Shift}}$ is a candidate key.

A data dictionary acts as a centralized repository of metadata, serving as the definitive catalog for database structures. It maps every $Field\_Name$ to its specific characteristics to ensure system data integrity. Specifically, it defines the $Field\_Format$ (the structural layout, such as pattern constraints) and the $Field\_Type$ (the data category, such as $INTEGER$ or $VARCHAR$) for every record. Since a comprehensive schema description requires defining the name, format, and type for every database attribute, these components collectively define its purpose. Therefore, the data dictionary identifies all these listed parameters.

A superkey is a set of attributes that contains a candidate key. In this relational schema, the primary key is given as VY. Since a primary key is always a candidate key, any attribute set that is a superset of {V, Y} will be a superkey.

We examine the options:

• VXYZ contains VY.
• VWXY contains VY.
• VWXYZ contains VY.

The attribute set VWXZ does not contain the attribute Y, and therefore it is not a superset of the primary key VY. Thus, VWXZ is not a superkey.

The design of a logical schema in database management systems aims to optimize both data integrity and usability. First, avoiding data inconsistency (A) is achieved through normalization, which minimizes data redundancy and eliminates update anomalies. Second, being able to construct queries easily (B) is ensured by organizing data into clear, intuitive entity relationships that simplify relational operations. Finally, being able to access data efficiently (C) relies on a structured logical design that allows query optimizers to traverse relationships and perform joins effectively. Since all these objectives are fundamental to effective database modeling, they represent the complete set of goals.

In database management systems, the TEXT data type is designed to store large volumes of character data that often exceed the limits of standard VARCHAR(n) types. Unlike fixed-length or defined-length types where the maximum size is explicitly set at the schema level, the storage capacity of a TEXT field is implementation-specific. Limits are dictated by the underlying database engine, page size, or memory constraints rather than a universal constant like $127$ or $255$. Since the capacity is defined by the system architecture rather than a static constraint, the maximum length is inherently variable.

To determine if a set of attributes is a key of $R = \{A, B, C, D, E, F\}$, we must calculate its closure to see if it determines all attributes in the relation. The given dependencies are $C \rightarrow F$, $E \rightarrow A$, $EC \rightarrow D$, and $A \rightarrow B$.

1. For option B, calculate the closure $\{EC\}^+$:
2. $\{EC\}^+ = \{E, C\}$
3. $\{E, C\} \xrightarrow{C \rightarrow F} \{E, C, F\}$
4. $\{E, C, F\} \xrightarrow{E \rightarrow A} \{E, C, F, A\}$
5. $\{E, C, F, A\} \xrightarrow{A \rightarrow B} \{E, C, F, A, B\}$
6. $\{E, C, F, A, B\} \xrightarrow{EC \rightarrow D} \{E, C, F, A, B, D\} = \{A, B, C, D, E, F\}$
7. Since $\{EC\}^+$ contains all attributes of $R$, $EC$ is a superkey (and in this case, a key). Other options do not produce the full attribute set.

In the relational model, a domain $D_i$ is defined as a set of atomic values.
A relation $R$ is a set of ordered $n$-tuples $(d_1, d_2, \dots, d_n)$, where each element $d_i$ belongs to the domain $D_i$.
The Cartesian product of $n$ domains, denoted as $D_1 \times D_2 \times \dots \times D_n$, represents the set of all possible $n$-tuples that can be formed from these domains.
Since a relation $R$ consists of a specific selection of these possible tuples, it is mathematically defined as a subset of the Cartesian product.
Thus, $R \subseteq D_1 \times D_2 \times \dots \times D_n$.

To find the key for the relation R, we need to determine a minimal set of attributes that can functionally determine all other attributes in R. We calculate the closure of candidate sets:

1. Start with {E, F}:
   ${EF}^+ = {EFG}$ (from $EF \rightarrow G$)
   ${EF}^+ = {EFGIJ}$ (from $F \rightarrow IJ$)
   Since H is not derived, {E,F} is not a key.

2. Consider adding H to {E,F}, forming {E,F,H}:
   ${EFH}^+ = {EFHGIJ}$ (from $EF \rightarrow G$ and $F \rightarrow IJ$)
   ${EFH}^+ = {EFHGIJKL}$ (from $EH \rightarrow KL$)
   ${EFH}^+ = {EFHGIJKLM}$ (from $K \rightarrow M$)
   ${EFH}^+ = {EFHGIJKLM N}$ (from $L \rightarrow N$)
   Since ${EFH}^+$ contains all attributes of R, {E,F,H} is a superkey. Removing any attribute from {E,F,H} (e.g., H from ${EH}^+$) would not allow it to determine all attributes, making it a minimal superkey and thus a candidate key.

The problem states that E is the key for the relation schema R(E, F, G, H).
A super key is any set of attributes that uniquely identifies tuples in a relation. It must contain at least one candidate key. Since E is the key, it is also a candidate key.

Any set of attributes that includes E will be a super key. The remaining attributes are F, G, and H.
We need to find all possible subsets of these remaining attributes and combine them with E to form super keys. The number of subsets for a set with $n$ elements is $2^n$.
Here, the set of remaining attributes is {F, G, H}, which has 3 elements. So, there are $2^3 = 8$ possible subsets:
{}, {F}, {G}, {H}, {F, G}, {F, H}, {G, H}, {F, G, H}.
Combining E with each of these subsets gives the super keys:
{E}, {E, F}, {E, G}, {E, H}, {E, F, G}, {E, F, H}, {E, G, H}, {E, F, G, H}.
Therefore, the maximum number of super keys is 8.

[CORRECT_OPTION: N/A]

A candidate key is a minimal superkey, meaning no proper subset of the key is also a superkey. Given the set of superkeys $\{ \{x\}, \{x, y\}, \{a, b\}, \{a, b, c\}, \{x, y, z\} \}$, we evaluate each:

1. $\{x\}$ is minimal as it has no proper subsets.
2. $\{x, y\}$ is not minimal because $\{x\} \subset \{x, y\}$ is a superkey.
3. $\{a, b\}$ is minimal because its proper subsets $\{a\}$ and $\{b\}$ are not superkeys.
4. $\{a, b, c\}$ is not minimal because $\{a, b\} \subset \{a, b, c\}$ is a superkey.
5. $\{x, y, z\}$ is not minimal because $\{x\} \subset \{x, y, z\}$ is a superkey.

Thus, the candidate keys are $\{x\}$ and $\{a, b\}$.