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Attribute Closure Algorithm and its Utilization
  • Difficulty Level : Easy
  • Last Updated : 18 Sep, 2020

Closure of a set F of FDs is the set F+ of all FDs that can be inferred from F. It is also known as complete set of Functional Dependency. It is denoted by F+.

Algorithm : Attribute Closure set

Algorithm to compute a+, the closure of a under F
Result:= a;
while (changes to Result) do  
    for each B → Y in F do
        Begin
            if B ⊆  Result  then  Result := Result ∪  Y  
        End

Utilization of Attribute Closure –

  1. Test whether attribute is superkey or not.

    Example :
    Let R = (A, B, C, G, H, I) and set of FD are F = { A → B, A → C, CG → H, CG → I, B → H}

    Find out (AG)+.



    Result = {A, G}

    First loop :

    A → B includes B in the Result as A⊆ Result (which is AG), so Result := Result ∪ B. 
    Hence Result = {A, G, B}
    A → C causes Result to become ABCG.
    CG → H causes the Result to become ABCGH.
    CG → I causes the Result to become ABCGHI.
    B → H causes Result to become ABCGHI.

    Second loop :

    A → B causes the Result to be unchanged i.e. ABCGHI (B is already part of the Result).
    A → C causes Result to be unchanged.
    CG → H causes the Result to be unchanged.
    CG → I causes the Result to be unchanged.
    B → H causes Result to be unchanged.

    At the end of the second loop, the Result does not change so exit from the loop.

    (AG)+ = {A, B, C, G, H, I}

    Conclusion: AG is a super key as all other attributes can be determined by it.

  2. Check whether Fd exists :

    Example :
    Let R = (A, B, C, G, H, I) and set of FD are F = { A →B, A → C, CG → H, CG → I, B → H}

    Check HB → I holds or not

    Result = {H, B}



    First loop :

    In A → B as A ⊆ Result (which is HB) so nothing will be added.
    In A → C nothing added.
    CG → H (CG ⊆ Result (which is HB) so nothing added)
    CG → I nothing added.
    B → H nothing added.

    At the end of the first loop, the Result does not change so exit from the loop.

    Conclusion: HB → I does not hold.

  3. An alternate way to find Closure of FDs (F+).

    Example :
    Given a relation R (A, B, C, D, E, F) and set of FDs F: {A → BC, E → CF, B → E, CD → EF}

    Find out closure of {A, B}+

    Step-1: Result = AB

    Step-2: First loop

    Result = ABC for A → BC, A ⊆ Result so Result = Result ∪ BC.
    Result = ABC for E→ CF, E ∉ Result so Result = Result.
    Result = ABCE for B → E, B ⊆Result so Result = Result ∪ E.
    Result = ABCE for CD → EF, CD ∉ Result so Result = Result.

    The Result before step2 is AB and after step 2 is ABCE which is different so repeat the same as step 2.

    Step-3: Second loop

    Result = ABCE for A → BC, A ⊆ Result so Result = Result ∪ BC.
    Result = ABCEF for E → CF, E ⊆ Result so Result = Result ∪ CF.
    Result = ABCEF for B → E, B ⊆ Result so Result = Result ∪ E.
    Result = ABCEF for CD → EF, CD ∉ Result so Result = Result.

    The Result before step 3 is ABCE and that after step 3 is ABCEF which is different, so repeat the same as step 3.

    Step-4: Third loop

    Result = ABCEF for A → BC, A ⊆ Result so Result = Result ∪ BC.
    Result = ABCEF for E → CF, E ⊆ Result so Result = Result ∪ CF.
    Result = ABCEF for B → E, B ⊆ Result so Result = Result ∪ E.
    Result = ABCEF for CD → EF, CD ∉ Result so Result = Result.

    The Result before step4 is ABCEF and after step 3 is ABCEF which is the same so stop.

    So Closure of {A, B}+ is {A, B, C, E, F}.

    Conclusion: For every attribute/set of attributes we can find closure. This Results in F+.

    My Personal Notes arrow_drop_up
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