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Full reducers

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• Full reducer
  • Full reducer:

    Full reducer = a fixed, finite sequence of semi-join operations that will eliminate all dangling tuples from the (participating) relations

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  • Example:

    The tuple (a4,b4,c4) is a dangling tuple: The semi-join R ⋉ S removes all dangling tuples: Therefore: R ⋉ S is a full reducer for R

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  • Similarly:

    The semi-join S ⋉ R is a full-reducer for S

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  • Therefore:

    R ⋉ S S ⋉ R

    is a full reducer since it eliminates all dangling tuples from the (participating) relations

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• Preliminary findings ---- Full reducers for more than 2 relations
  • Consider the following 3 relations:

    R(A,B) S(B,C) T(C,A) ========================================= (1,1) (1,1) (1,2) (2,2) (2,2) (2,3) (3,3) (3,3) (3,4) ... ... ... (n-1,n-1) (n-1,n-1) (n-1,n) (n,n) (n,n)

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  • Observation:

    R(A,B) ⋈ S(B,C) ⋈ T(C,A) = ∅

    Reason:

    R(A,B) ⋈ S(B,C) is the following relation: R(A,B) ⋈ S(B,C): A B C ==================== (1 , 1 , 1) (2 , 2 , 2) (3 , 3 , 3) .... (n , n , n) Notice that: ∀ t ∈ R(A,B) ⋈ S(B,C): t(A) = t(C) So, when we join R(A,B) ⋈ S(B,C) with the relation T(C,A): T(C,A) =========== (1,2) (2,3) (3,4) ... (n-1, n) ∀ t ∈ T(C,A): t(A) ≠ t(C) The result is: the empty set !!!

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  • Conclusion:

    Every tuple in R, S and T is a dangling tuple !!!

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• Full reducer for more than 2 relations
  • Consider the following 3 relations stored on 3 different sites:

    R(A,B) S(B,C) T(C,A) ========================================= (1,1) (1,1) (1,2) (2,2) (2,2) (2,3) (3,3) (3,3) (3,4) ... ... ... (n-1,n-1) (n-1,n-1) (n-1,n) (n,n) (n,n)

    Recall that:

    Since R ⋈ S ⋈ T = ∅: all tuples in R, S and T are dangling in the 3-way join !!!

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  • Suppose we use semi-join (⋉) to eliminate the dangling tuples from the relations:

    R ⋉ S    (= R ⋈ πB(S)) R(A,B) S(B,C) πB(S) R ⋉ S =================================================== (1,1) (1,1) (1) (1,1) (2,2) (2,2) (2) (2,2) (3,3) (3,3) (3) (3,3) ... ... ... ... (n-1,n-1) (n-1,n-1) (n-1) (n-1,n-1) (n,n) (n,n) (n) (n,n) Eliminates a grant total of 0 dangling tuples !!! R ⋉ T    (= R ⋈ πA(T)) R(A,B) T(C,A) πA(T) R ⋉ T =================================================== (1,1) (1,2) (2) (2,2) (2,3) (3) (2,2) (3,3) (3,4) (4) (3,3) ... ... ... ... (n-1,n-1) (n-1,n) (n) (n-1,n-1) (n,n) (n,n) Eliminates a grant total of 1 dangling tuple !!! S ⋉ T    (= S ⋈ πC(T)) S(B,C) T(C,A) πC(T) S ⋉ T =================================================== (1,1) (1,2) (1) (1,1) (2,2) (2,3) (2) (2,2) (3,3) (3,4) (3) (3,3) ... ... ... ... (n-1,n-1) (n-1,n) (n-1) (n-1,n-1) (n,n) Eliminates a grant total of 1 dangling tuple !!!

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  • Conclussions:

    Each semi-join operation removes at most one (1) dangling tuple from a relation We can (and must) repeatedly use more semi-join operations to remove more dangling tuples in the above example

    In fact:

    We will need to use 3×n − 1 semi-join operations to remove all (dangling) tuples from the relations....

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• Moral of this example
  • Fact:

    For some join operations with more than 2 relations: There is no fixed finite sequence of semi-join operations that guarantee to eliminate all dangling tuples from a relation I.e.: There is no full reducer available !!!