Candidate key finding using closure in DBMS Theory - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to compute the closure of attribute set A under functional dependencies F.

DBMS Theory

closure = set(A)
changed = True
while changed:
    changed = False
    for (X, Y) in F:
        if set(X).issubset(closure) and not set(Y).issubset(closure):
            closure.update([1])
            changed = True

Drag options to blanks, or click blank then click option'

AF

BX

Cclosure

DY

Attempts:

3 left

💡 Hint

Common Mistakes

Adding X instead of Y to the closure.

Updating closure with the whole FD instead of Y.

The closure of A is expanded by adding attributes Y from functional dependencies X -> Y when X is a subset of the closure.

2fill in blank

medium

Complete the code to check if attribute set A is a candidate key by comparing its closure with all attributes R.

DBMS Theory

if closure == [1]:
    print("A is a candidate key")
else:
    print("A is not a candidate key")

Drag options to blanks, or click blank then click option'

AR

BA

CF

Dclosure

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing closure with A or F instead of R.

Using assignment instead of equality check.

A candidate key's closure must include all attributes in the relation R.

3fill in blank

hard

Fix the error in the code that computes closure by replacing the incorrect operator.

DBMS Theory

closure = set(A)
for (X, Y) in F:
    if set(X) [1] closure:
        closure.update(Y)

Drag options to blanks, or click blank then click option'

A==

B<=

C>

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using equality or inequality instead of subset check.

Using greater than operator which is incorrect.

To check if X is a subset of closure, use the subset operator <=.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps attribute sets to their closures.

DBMS Theory

closures = { [1]: compute_closure([2], F) for [1] in subsets(R)}

Drag options to blanks, or click blank then click option'

AX

BF

CR

Dclosure

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables for key and argument.

Using F or R incorrectly in place of the variable.

The comprehension iterates over subsets of R using variable X and computes closure for each X.

5fill in blank

hard

Fill all three blanks to filter candidate keys from all subsets by checking closure equality.

DBMS Theory

candidate_keys = [[1] for [2] in subsets(R) if compute_closure([2], F) == [3]]

Drag options to blanks, or click blank then click option'

AX

Bsubset

CR

DF

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inconsistently.

Comparing closure with F instead of R.

We iterate over subsets named subset, check if their closure equals R, and collect those subsets as candidate keys.

Practice

1. What does the closure of a set of attributes represent in candidate key finding?

easy

A. All attributes functionally determined by the set

B. The number of attributes in the set

C. The total number of functional dependencies

D. The primary key of the table

Solution

1. Step 1: Understand closure definition

   The closure of a set of attributes includes all attributes that can be functionally determined from that set using the given functional dependencies.

2. Step 2: Relate closure to candidate keys

   If the closure covers all attributes in the relation, the set is a superkey; minimal such sets are candidate keys.

3. Final Answer:

   All attributes functionally determined by the set -> Option A

4. Quick Check:

   Closure = attributes reachable from set [OK]

Hint: Closure shows all attributes a set can determine [OK]

Common Mistakes:

• Confusing closure with number of attributes
• Thinking closure is the primary key itself
• Ignoring functional dependencies in closure

2. Which of the following is the correct notation to represent the closure of attribute set {A, B}?

easy

A. {A, B}+

B. {A, B}*

C. closure(A, B)

D. closure{A, B}

Solution

1. Step 1: Recall standard closure notation

   The closure of a set of attributes X is denoted as X⁺, meaning all attributes functionally determined by X.

2. Step 2: Match notation to options

   {A, B}+ uses the correct notation {A, B}+. Other options are not standard or incorrect.

3. Final Answer:

   {A, B}+ -> Option A

4. Quick Check:

   Closure notation uses plus sign [OK]

Hint: Closure notation always uses plus sign after attribute set [OK]

Common Mistakes:

• Using star (*) instead of plus (+)
• Writing closure as a function call
• Misplacing braces or parentheses

3. Given relation R(A, B, C, D) and functional dependencies:
A -> B, B -> C, C -> D. What is the closure of {A}?

medium

A. {A, B}

B. {A, B, C}

C. {A, B, C, D}

D. {A}

Solution

1. Step 1: Start with attribute set {A}

   Initially, closure({A}) = {A}.

2. Step 2: Apply functional dependencies stepwise

   From A -> B, add B: closure = {A, B}. Then B -> C adds C: closure = {A, B, C}. Finally, C -> D adds D: closure = {A, B, C, D}.

3. Final Answer:

   {A, B, C, D} -> Option C

4. Quick Check:

   Closure includes all reachable attributes [OK]

Hint: Apply dependencies repeatedly until no new attributes added [OK]

Common Mistakes:

• Stopping after first dependency
• Ignoring transitive dependencies
• Confusing closure with original set

4. Identify the error in this closure calculation for relation R(A, B, C) with FDs:
A -> B, B -> C. Closure of {B} is given as {B, A}. What is wrong?

medium

A. Closure should include all attributes

B. Closure cannot include B itself

C. Closure should be empty

D. Closure should include C, not A

Solution

1. Step 1: Analyze given closure {B, A}

   Starting with {B}, applying B -> C adds C, so closure should be {B, C}.

2. Step 2: Check if A can be added

   There is no FD that allows B to determine A, so A should not be in closure.

3. Final Answer:

   Closure should include C, not A -> Option D

4. Quick Check:

   Closure only adds attributes functionally determined [OK]

Hint: Only add attributes on right side of FDs from closure set [OK]

Common Mistakes:

• Adding attributes not functionally dependent
• Confusing direction of dependencies
• Assuming closure includes all attributes

5. For relation R(A, B, C, D, E) with FDs:
A -> BC, CD -> E, B -> D. Which of the following is a candidate key?

hard

A. {A, D}

B. {A}

C. {A, C}

D. {B, C}

Solution

1. Step 1: Compute closure of {A}

   Start with {A}. From A -> BC, add B and C: closure = {A, B, C}. From B -> D, add D: closure = {A, B, C, D}. From CD -> E, since C and D are in closure, add E: closure = {A, B, C, D, E}.

2. Step 2: Check minimality of {A}

   Since closure({A}) includes all attributes, {A} is a superkey. It is minimal because it has only one attribute.

3. Final Answer:

   {A} -> Option B

4. Quick Check:

   Minimal set with full closure is candidate key [OK]

Hint: Check closure of smallest sets covering all attributes [OK]

Common Mistakes:

• Not checking transitive dependencies
• Assuming larger sets are always candidate keys
• Ignoring minimality condition