## Questions on Minimal Cover or Irreducible Sets or Canonical Cover –

```
Question 1 :Find the minimal cover or irreducible set from the
following functional dependencies :
AB → CD
BC → D
```

Solution : Step 1 : Union Simplification : AB → C ..(1) AB → D ..(2) BC → D ..(3) Step 2 : Removal of Redundant set of FDs - (i) Find the closure of LHS of (1), i.e. Compute (AB)+ from (2) and (3) by hiding (1). If we get RHS of (1) in the closure,i.e. 'C', then (1) is redundant, otherwise non redundant. (AB)+ = {ABD} as (AB)+ will not determine C, So AB → C is non redundant. (ii) Similarly for (2), Compute (AB)+ from (1) and (3) by hiding (2). (AB)+ = {ABCD} as (AB)+ will determine D, therefore, AB → D is redundant. Now, Remaining FDs = {AB→C, BC→D} (iii) For (3), Compute (BC)+ from (1) by hiding (3). (BC)+ = {BC} as (BC)+ will not determine D, so, BC → D is also non redundant. Step 3 : LHS Simplification or Removal of Extraneous Attributes - FDs remained after step 2 AB→C .. (3) BC→D .. (4) AB → C (i) Remove A from AB → C, and find B+ from (4). If the attribute A exists in the closure, then A is said to be redundant. B+ = {B} ⇒ A is Non Redundant (ii) Remove B from AB → C and find A+ from (4). A+ = {A} ⇒ B is Non Redundant. BC → D (i) Remove B from BC → D, and find C+ from (3). C+ = {C} ⇒ B is Non Redundant (ii) Remove C from BC → D and find B+ from (3). B+ = {B} ⇒ C is Non Redundant Apply step 2 again if we find any redundant attribute, otherwise skip So Finally, Irreducible set will be AB→C BC→D

The another Example will give you more clarification about solving minimal cover questions.

```
Question 2 :
Find the minimal cover or irreducible set from the following
functional dependencies :
ABCD → E ..(1)
E → D ..(2)
AC → D ..(3)
A → B ..(4)
```

Solution : step 1 : No need for union simplification as FDs are already simplified. Step 2 : Removal of redundant set of FDs - (i) For (1), Compute (ABCD)+ from (2) (3) and (4) by hiding (1). (ABCD)+ = {ABCD} ⇒ ABCD → E is non redundant. (ii) For (2), Compute (E)+ from (1) (3) and (4) by hiding (2). (E)+ = {E} ⇒ E → D is non redundant. (iii) For (3), Compute (AC)+ from (1) (2) and (4) by hiding (3). (AC)+ = {ACB} ⇒ AC → D is non redundant. (iv) For (4), Compute (A)+ from (1) (2) and (3) by hiding (4). (A)+ = {A} ⇒ A → B is non redundant. Step 3 : LHS Simplification or Removal of Extraneous Attributes - FDs remained after step 2 is same as in step 1. ABCD → E (i) Remove A from ABCD → E, and find BCD+ from (2) (3) and (4). If attribute A exists in the closure, then A is said to be redundant. (BCD)+ = {BCD} ⇒ A is Non Redundant (ii) Remove B from ABCD → E and find ACD+ from (2) (3) and (4). (ACD)+ = {ACDB} ⇒ B is Redundant. ⇒ FD will now reduced to ACD → E (iii) Remove C from ACD → E, and find AD+ from (2) (3) and (4). (AD)+ = {ADB} ⇒ C is Non Redundant. (iv) Remove D from ACD → E, and find AC+ from (2) (3) and (4). (AC)+ = {ACDB} ⇒ D is Redundant. ⇒ FD will now reduced to AC → E AC → D (i) Remove A from AC → D and find C+ from (1) (2) and (4). (C)+ = {C} ⇒ A is non redundant. (ii) Remove C from AC → D and find A+ from (1) (2) and (4). (A)+ = {AB} ⇒ C is non redundant. FDs remained after step 3 are : AC → E ..(5) E → D ..(6) AC → D ..(7) A → B ..(8) Step 4 : Applying step 2 again to find redundant FDs - (i) For (5), Compute (AC)+ from (6) (7) and (8) by hiding (5). (AC)+ = {ACDB} ⇒ AC → E is non redundant. (ii) For (6), Compute (E)+ from (5) (7) and (8) by hiding (6). (E)+ = {E} ⇒ E → D is non redundant. (iii) For (7), Compute (AC)+ from (5) (6) and (8) by hiding (7). (AC)+ = {ACEDB} ⇒ AC → D is redundant. So, excluded. (iv) For (8), Compute (A)+ from (5) and (6) by hiding (8). (A)+ = {AB} ⇒ A → B is non redundant. So Finally, Irreducible set will be AC → E E → D A → B

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