15 - Canonical Cover & Extraneous Attributes

🎯 Learning Objectives

• Compute the canonical cover of a set of functional dependencies
• Identify and remove extraneous attributes
• Find candidate keys using attribute closure techniques
• Count superkeys from candidate keys
• Understand why canonical cover matters for checking FDs efficiently

📋 Prerequisites

• 14 - Functional Dependencies — FDs, Armstrong's axioms, attribute closure

📖 Core Content

15.1 Intuition: Minimal FD Set

• No redundant FDs (each FD is necessary)
• No extraneous attributes on either side
• Only single attributes on the RHS

Why This Matters: Canonical cover reduces the number and complexity of FDs the system needs to check. It's also heavily tested in exams (~12% of all questions involve FDs and canonical cover).

15.2 Algorithm to Compute Canonical Cover

sql
canonical_cover(F):
    // Step 1: Decompose RHS to single attributes
    Replace each FD X → {A₁, A₂, ..., Aₙ} with X → A₁, X → A₂, ..., X → Aₙ
    // Step 2: Remove extraneous attributes from LHS
    For each FD X → A:
        For each attribute B in X:
            If (X - {B}) → A can be derived from F, remove B from X
    // Step 3: Remove redundant FDs
    For each FD X → A:
        If F - {X → A} implies X → A, remove X → A
    Return Fc

15.3 Extraneous Attributes

On the LHS (Left-Hand Side)

• If $A$ is extraneous, then $B \rightarrow C$ can be derived from $F$
• Check: Compute $\{B\}^+$ using $F$. If $C \in \{B\}^+$, then $A$ is extraneous

On the RHS (Right-Hand Side)

• If $B$ is extraneous, then $A \rightarrow B$ can be derived from $F - \{A \rightarrow BC\} \cup \{A \rightarrow C\}$
• Check: Does $F$ imply $A \rightarrow B$? Compute $\{A\}^+$ using $F' = F - \{A \rightarrow BC\} \cup \{A \rightarrow C\}$. If $B \in \{A\}^+$, then $B$ is extraneous.

15.4 Worked Examples: Canonical Cover

Example 1: Simple Canonical Cover

• Is $A$ extraneous? Compute $\{B\}^+$ using F.
  • result = {B}
  • $B \rightarrow C$: B ⊆ result → result = {B, C}
  • C ∈ $\{B\}^+$? Yes! So A is extraneous.
  • Remove A: $B \rightarrow C$ remains (already exists in F)
• Is $B$ extraneous? Compute $\{A\}^+$ using F.
  • result = {A}
  • $A \rightarrow B$ → {A, B}
  • $A \rightarrow C$ → {A, B, C}
  • C ∈ $\{A\}^+$? Yes! So B is extraneous.
  • Remove B: $A \rightarrow C$ remains (already exists in F) After checking, $AB \rightarrow C$ is redundant (covered by $A \rightarrow C$ and $B \rightarrow C$). Remove it. $F = \{A \rightarrow B, A \rightarrow C, B \rightarrow C\}$ Step 3: Remove redundant FDs Check $A \rightarrow B$: Can we derive it from the remaining FDs? $F - \{A \rightarrow B\} = \{A \rightarrow C, B \rightarrow C\}$ Compute $\{A\}^+$ using this set: {A} → $A \rightarrow C$ → {A, C}. B ∉ {A, C}. So $A \rightarrow B$ is NOT redundant. Check $A \rightarrow C$: $F - \{A \rightarrow C\} = \{A \rightarrow B, B \rightarrow C\}$ Compute $\{A\}^+$: {A} → $A \rightarrow B$ → {A, B} → $B \rightarrow C$ → {A, B, C}. C ∈ {A, C}! So $A \rightarrow C$ is redundant. Check $B \rightarrow C$: $F - \{B \rightarrow C\} = \{A \rightarrow B, A \rightarrow C\}$ Compute $\{B\}^+$: {B}. C ∉ {B}. So $B \rightarrow C$ is NOT redundant. Canonical cover: $F_c = \{A \rightarrow B, B \rightarrow C\}$

Example 2: More Complex Canonical Cover

• Is C extraneous? Compute $\{D\}^+$ using F.
  • result = {D}
  • $B \rightarrow D$: B not in result. $E \rightarrow A$: E not in result.
  • Actually, $\{D\}^+ = \{D\}$. E ∉ $\{D\}^+$, so C is NOT extraneous.
• Is D extraneous? Compute $\{C\}^+$ using F.
  • result = {C}. C doesn't appear on any LHS except in CD.
  • $\{C\}^+ = \{C\}$. E ∉ $\{C\}^+$, so D is NOT extraneous. Step 3: Remove redundant FDs Check $A \rightarrow B$: Can $B$ be derived from others? $F - \{A \rightarrow B\} = ...$ Need to check. Honestly computing by hand is complex. The key insight: since $E \rightarrow A$ and all A→... FDs, much is interconnected. A simpler approach might be needed. Let me try a different method. Given $F = \{A \rightarrow BC, CD \rightarrow E, B \rightarrow D, E \rightarrow A\}$ Step 1: Decompose RHS $F = \{A \rightarrow B, A \rightarrow C, CD \rightarrow E, B \rightarrow D, E \rightarrow A\}$ Step 2: Check extraneous LHS. Check $CD \rightarrow E$:
• Is C extraneous? Can we derive $D \rightarrow E$?
  • $\{D\}^+$ with F = {D}. $B \rightarrow D$ doesn't help (B not in result). CD → E needs C. So $\{D\}^+ = \{D\}$. E ∉ $\{D\}^+$. C is NOT extraneous.
• Is D extraneous? Can we derive $C \rightarrow E$?
  • $\{C\}^+$ = {C}. E ∉ $\{C\}^+$. D is NOT extraneous. Step 3: Remove redundant FDs Check $A \rightarrow B$ in $F' = F - \{A \rightarrow B\} = \{A \rightarrow C, CD \rightarrow E, B \rightarrow D, E \rightarrow A\}$:
• $\{A\}^+$ using F':
  • result = {A}
  • $A \rightarrow C$ → {A, C}
  • $E \rightarrow A$: E not in result
  • $CD \rightarrow E$: C and D? D not in result
  • $B \rightarrow D$: B not in result
  • $\{A\}^+ = \{A, C\}$. B ∉ $\{A, C\}$. So $A \rightarrow B$ is NOT redundant. Check $A \rightarrow C$ in $F' = \{A \rightarrow B, CD \rightarrow E, B \rightarrow D, E \rightarrow A\}$:
• $\{A\}^+$: {A} → $A \rightarrow B$ → {A, B} → $B \rightarrow D$ → {A, B, D}. C not in result.
• So $A \rightarrow C$ is NOT redundant. Check $CD \rightarrow E$ in $F' = \{A \rightarrow B, A \rightarrow C, B \rightarrow D, E \rightarrow A\}$:
• $\{C, D\}^+$: {C, D}. No FD has just {C, D} on LHS except CD→E which we removed.
• So $CD \rightarrow E$ is NOT redundant. Check $B \rightarrow D$ in $F' = \{A \rightarrow B, A \rightarrow C, CD \rightarrow E, E \rightarrow A\}$:
• $\{B\}^+$: {B}. No FD applies. D not in result.
• So $B \rightarrow D$ is NOT redundant. Check $E \rightarrow A$ in $F' = \{A \rightarrow B, A \rightarrow C, CD \rightarrow E, B \rightarrow D\}$:
• $\{E\}^+$: {E}. A not in result.
• So $E \rightarrow A$ is NOT redundant. Canonical cover: $F_c = \{A \rightarrow B, A \rightarrow C, CD \rightarrow E, B \rightarrow D, E \rightarrow A\}$ Hmm, same as the original (after decomposition). No redundancy.

15.5 Superkey Counting

• Number of superkeys = $2^{(N-n)}$ With two candidate keys CK₁ (size n₁) and CK₂ (size n₂):
• Number of superkeys = $2^{(N-n₁)} + 2^{(N-n₂)} - 2^{(N-(n₁+n₂))}$
• (Subtract the overlap counted twice) Example: $R(J, K, L, M, N, O)$ with $F = \{K \rightarrow JL, L \rightarrow K, J \rightarrow O, M \rightarrow N\}$ Find candidate keys:
• Attributes not on RHS: M (not on any RHS). So M is mandatory.
• Compute {M}⁺ = {M, N} (using M→N). Not all attributes.
• Try {M, K}⁺:
  • result = {M, K}
  • K→JL → {J, K, L, M}
  • J→O → {J, K, L, M, O}
  • M→N → {J, K, L, M, N, O} = R
• {M, K} is a superkey. Is K alone with M minimal? {M, K} - M = {K}: K⁺ = {J, K, L} ≠ R. {M, K} - K = {M}: M⁺ = {M, N} ≠ R. So {M, K} is a candidate key. Similarly, {M, L}⁺:
• result = {M, L}
• L→K → {K, L, M}
• K→JL → {J, K, L, M}
• J→O → {J, K, L, M, O}
• M→N → {J, K, L, M, N, O} = R
• {M, L} is another candidate key. Two candidate keys: {M, K} (n₁=2) and {M, L} (n₂=2). They share attribute M. Using formula: $2^{6-2} + 2^{6-2} - 2^{6-(2+2-1)} = 2^4 + 2^4 - 2^{3} = 16 + 16 - 8 = 24$ superkeys.

📐 Key Formulas / Concepts

⚠️ Common Pitfalls

Pitfall 1: Forgetting Step 1 — Decompose RHS First

Pitfall 2: Confusing Extraneous Attribute Check for LHS vs RHS

• LHS extraneous: Remove the candidate attribute from LHS, compute closure. If RHS is in closure → extraneous.
• RHS extraneous: Temporarily remove the attribute from F (not just one FD), compute closure of LHS. If the attribute is in closure → extraneous.

Pitfall 3: Double-Counting Overlapping Superkeys

📝 Practice Questions

Q1. Find the canonical cover of $F = \{A \rightarrow B, B \rightarrow C, A \rightarrow C\}$.

Answer

Step 1: RHS already single attributes. ✓

Step 2: No compound LHS to check. ✓

Step 3: Remove redundant FDs.

• Check $A \rightarrow C$: In $F' = \{A \rightarrow B, B \rightarrow C\}$, compute $\{A\}^+$ = {A, B, C}. C is in closure → redundant. Remove.

Canonical cover: $F_c = \{A \rightarrow B, B \rightarrow C\}$

(We can also keep $A \rightarrow C$ and remove one of the others, but this is the standard minimal form.)

Q2. Find the canonical cover of $F = \{AB \rightarrow C, C \rightarrow A, BC \rightarrow D\}$.

Answer

Step 1: RHS already single attributes. ✓

Step 2: Check extraneous LHS attributes.

Check $AB \rightarrow C$:

• Is A extraneous? Compute $\{B\}^+$: {B}. C ∉ {B}. Not extraneous.
• Is B extraneous? Compute $\{A\}^+$: {A}. C ∉ {A}. Not extraneous.

Check $BC \rightarrow D$:

• Is B extraneous? Compute $\{C\}^+$: {C}. $C \rightarrow A$ → {A, C}. D ∉ {A, C}. Not extraneous.
• Is C extraneous? Compute $\{B\}^+$: {B}. D ∉ {B}. Not extraneous.

Step 3: Check redundant FDs.

• $AB \rightarrow C$ in $\{C \rightarrow A, BC \rightarrow D\}$: $\{AB\}^+$ = {A, B}. C ∉ {A, B}. Not redundant.
• $C \rightarrow A$ in $\{AB \rightarrow C, BC \rightarrow D\}$: $\{C\}^+$ = {C}. A ∉ {C}. Not redundant.
• $BC \rightarrow D$ in $\{AB \rightarrow C, C \rightarrow A\}$: $\{BC\}^+$ = {B, C}. $C \rightarrow A$ → {A, B, C}. D ∉ {A, B, C}. Not redundant.

Canonical cover: $F_c = \{AB \rightarrow C, C \rightarrow A, BC \rightarrow D\}$ (unchanged)

Q3. Given $R(A, B, C, D, E)$ with $F = \{A \rightarrow B, C \rightarrow D, B \rightarrow E\}$, find all candidate keys.

Answer

Step 1: Attributes not on RHS: A, C (both appear only on LHS).

Step 2: Compute $\{A, C\}^+$:

• result = {A, C}
• $A \rightarrow B$ → {A, B, C}
• $B \rightarrow E$ → {A, B, C, E}
• $C \rightarrow D$ → {A, B, C, D, E} = R

Step 3: $\{A, C\}^+$ covers all attributes. Is it minimal?

• Remove A: $\{C\}^+ = \{C, D\} \neq R$
• Remove C: $\{A\}^+ = \{A, B, E\} \neq R$

Candidate key: {A, C}

Q4. How many superkeys does a relation with 7 attributes and one candidate key of size 2 have?

Answer

Using formula: $2^{(N-n)} = 2^{7-2} = 2^{5} = 32$ superkeys.

Any superset of the candidate key is a superkey. With 5 remaining attributes, any subset of them (including empty) can be added to the candidate key, giving $2^5$ possibilities.

Q5. What is the purpose of computing a canonical cover?

Answer

A canonical cover is a minimal set of FDs equivalent to the original set. Its purposes:

1. Efficiency: Fewer FDs to check during updates (reduces overhead)
2. Clarity: Removes redundancy — each FD is necessary
3. Normalization: Used in algorithms to decompose relations into 3NF
4. Equivalence checking: Comparing canonical covers is easier than comparing full FD sets

Q6. Find the canonical cover of $F = \{A \rightarrow BC, B \rightarrow C, AB \rightarrow C\}$.

Answer

Step 1: Decompose: $F = \{A \rightarrow B, A \rightarrow C, B \rightarrow C, AB \rightarrow C\}$

Step 2: Check $AB \rightarrow C$:

• Is A extraneous? $\{B\}^+$ = {B}. $B \rightarrow C$ → {B, C}. C ∈ {B, C}? Yes! A is extraneous.
• Remove A: $B \rightarrow C$ (already exists)
• Is B extraneous? $\{A\}^+$ = {A}. $A \rightarrow B$ → {A, B}. $A \rightarrow C$ → {A, B, C}. C ∈ {A, B, C}? Yes! B is extraneous.
• Both A and B are extraneous → AB→C is fully redundant. Remove.

Step 3: $F = \{A \rightarrow B, A \rightarrow C, B \rightarrow C\}$

• Check $A \rightarrow C$: In $\{A \rightarrow B, B \rightarrow C\}$, $\{A\}^+$ = {A, B, C}. C is in closure → redundant. Remove.

Canonical cover: $F_c = \{A \rightarrow B, B \rightarrow C\}$

Q7. Given $R(A, B, C, D)$ with candidate key $\{A, B\}$, how many superkeys exist?

Answer

$N = 4$ (total attributes), $n = 2$ (candidate key size).

Number of superkeys = $2^{4-2} = 2^2 = 4$.

The superkeys are: {A,B}, {A,B,C}, {A,B,D}, {A,B,C,D}.

Q8. Check if $F_1 = \{A \rightarrow B, B \rightarrow C\}$ and $F_2 = \{A \rightarrow BC\}$ are equivalent.

Answer

Check $F_1$ → $F_2$: Can we derive $A \rightarrow BC$ from $F_1$?

• $A^+$ using $F_1$ = {A, B, C} (via A→B→C). BC ⊆ $A^+$. ✓

Check $F_2$ → $F_1$: Can we derive $A \rightarrow B$ and $B \rightarrow C$ from $F_2$?

• $A \rightarrow B$: $A^+$ = {A, B, C}. B ⊆ $A^+$. ✓
• $B \rightarrow C$: $B^+$ = {B}. C ∉ {B}. ✗

$F_1$ and $F_2$ are NOT equivalent because $B \rightarrow C$ cannot be derived from $F_2$.

🔗 Cross-References

• Next Topic: 16 - Normalization (1NF, 2NF, 3NF)
• Previous Topic: 14 - Functional Dependencies
• Related: 18 - Decomposition (using canonical cover for 3NF decomposition)
• Textbook: Silberschatz, Korth, Sudarshan — Chapter 8 (Relational Database Design) Join Discord Previous14 - Functional DependenciesNext16 - Normalization (1NF-3NF)