Candidate key finding using closure in DBMS Theory - Time & Space Complexity

Time Complexity: Candidate key finding using closure

O(2^n * m)

Understanding Time Complexity

Finding candidate keys using closure involves checking attribute combinations to see if they determine all other attributes.

We want to understand how the time needed grows as the number of attributes increases.

Scenario Under Consideration

Analyze the time complexity of the following pseudocode for candidate key finding using closure.

FOR each subset S of attributes in relation R
  Compute closure of S
  IF closure of S includes all attributes in R
    Mark S as a candidate key
  END IF
END FOR
    

This code tries every possible attribute subset, computes its closure, and checks if it covers all attributes.

Identify Repeating Operations

Look for repeated steps that take most time.

• Primary operation: Computing closure for each subset of attributes.
• How many times: Once for every subset, which grows as the number of attributes increases.

How Execution Grows With Input

As the number of attributes (n) grows, the number of subsets grows very fast.

Input Size (n)	Approx. Operations
3	8 subsets to check
5	32 subsets to check
10	1024 subsets to check

Pattern observation: The number of subsets doubles with each added attribute, causing exponential growth.

Final Time Complexity

Time Complexity: O(2^n * m)

This means the time grows exponentially with the number of attributes, where n is attributes count and m is number of functional dependencies.

Common Mistake

[X] Wrong: "Checking only a few subsets is enough to find all candidate keys quickly."

[OK] Correct: Candidate keys can be any subset, so skipping subsets risks missing keys and leads to incorrect results.

Interview Connect

Understanding this complexity helps you explain why candidate key discovery can be slow and shows your grasp of database fundamentals.

Self-Check

What if we limited the subsets to only those with a fixed size? How would the time complexity change?

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