Attribute Closure, Super Key & Candidate Key Questions with Solutions (GATE DBMS)

Solution

Step 1: Start with EFH

• EF → G
• F → I, J
• EH → K, L
• K → M
• L → N

👉 EFH⁺ = {E, F, G, H, I, J, K, L, M, N}

👉 Covers all attributes → Super Key

Check minimality:

• Remove E ❌ (cannot get K, L)
• Remove F ❌ (cannot get I, J)
• Remove H ❌ (cannot get K, L)

👉 Minimal → Candidate Key

Solution (Step-by-Step)

We check each option using closure

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Option A: CD → AC

• CD → E
• E → A
• A → C

👉 CD⁺ = {C, D, E, A}

✔ A and C present → IMPLIED

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Option B: BD → CD

• B → D (already have D)

👉 BD⁺ = {B, D}

❌ Cannot get C → NOT IMPLIED

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Option C: BC → CD

• B → D

👉 BC⁺ = {B, C, D}

✔ Contains C and D → IMPLIED

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Option D: AC → BC

• A → B

👉 AC⁺ = {A, B, C}

✔ Contains B and C → IMPLIED

Solution

Step 1: Check AB

• AB → C
• C → D
• D → B

👉 AB⁺ = {A, B, C, D}

👉 Covers all attributes → Super Key

👉 Minimal → Candidate Key

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Step 2: Check AC

• C → D
• D → B

👉 AC⁺ = {A, C, D, B}

👉 Covers all → Candidate Key

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Step 3: Check AD

• D → B
• AB → C

👉 AD⁺ = {A, D, B, C}

👉 Covers all → Candidate Key

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👉 Final Candidate Keys = {AB, AC, AD}

Solution

Step 1: Identify attributes not on RHS

• FDs: AB → C, BC → D
• RHS = {C, D}
• Attributes NOT in RHS = {A, B, E}

👉 These must be part of every candidate key

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Step 2: Check closure of ABE

• AB → C
• Now we have B, C → BC → D

👉 ABE⁺ = {A, B, E, C, D}

👉 Covers all attributes → Super Key

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Step 3: Check minimality

• Remove A ❌ (cannot get C)
• Remove B ❌ (cannot apply AB → C)
• Remove E ❌ (E cannot be derived from anywhere)

👉 Minimal → Candidate Key

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Solution

Step 1: Check AB

• AB → C
• C → D
• CD → BE
• DE → F
• EF → A

👉 AB⁺ = {A, B, C, D, E, F}

👉 Covers all → Candidate Key

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Step 2: Check BC

• C → D
• CD → BE
• DE → F
• EF → A

👉 BC⁺ = {B, C, D, E, F, A}

👉 Covers all → Candidate Key

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Step 3: Check CE

• C → D
• CD → BE
• DE → F
• EF → A

👉 CE⁺ = {C, E, D, B, F, A}

👉 Covers all → Candidate Key

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Step 4: Check EF

• EF → A
• AB → C
• C → D
• CD → BE

👉 EF⁺ = {E, F, A, B, C, D}

👉 Covers all → Candidate Key

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👉 Final Candidate Keys = {AB, BC, CE, EF}

Solution

Step 1: Find attributes not on RHS

• RHS = {C, D, E, F, B}
• Attributes NOT in RHS = {A}

👉 A must be present in every candidate key

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Step 2: Try with A + B

AB → C → DE → F → B

👉 AB⁺ = {A, B, C, D, E, F}

👉 Candidate Key [Learn more about Candidate Key]

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Step 3: Try with A + F

F → B → AB → C → DE

👉 AF⁺ = {A, F, B, C, D, E}

👉 Candidate Key

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Step 4: Try with A + E

E → F → B → AB → C → DE

👉 AE⁺ = {A, E, F, B, C, D}

👉 Candidate Key

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Step 5: Check A + C

C → DE → F → B

👉 AC⁺ = {A, C, D, E, F, B}

👉 Candidate Key

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Solution

Step 1: Identify structure

A → B → C → D → A (cycle)

👉 A, B, C, D are all equivalent (each determines others)

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Step 2: Check attribute E

👉 E is not present in any FD 👉 So E cannot be derived 👉 E must be in every candidate key

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Step 3: Form candidate keys

• AE⁺ = {A, B, C, D, E}
• BE⁺ = {B, C, D, A, E}
• CE⁺ = {C, D, A, B, E}
• DE⁺ = {D, A, B, C, E}

👉 All cover full relation → Candidate Keys

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Solution

Step 1: Start with XY(Find closure of XY if it contains p then Yes otherwise NO)

• XY → W ⇒ add W

👉 Now we have: {X, Y, W}

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Step 2: Use Y → Z

• Y → Z ⇒ add Z

👉 Now we have: {X, Y, W, Z}

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Step 3: Use WZ → P

• WZ → P ⇒ add P

👉 Now we have: {X, Y, W, Z, P}

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👉 P is derived from XY