Slideshow:
Associative (Relational ALgebra) operators
The following relational algebre operators are associative:
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Comment: they were also commutative !
Union (∪) is associative for sets and bags
Intersection
(∩) is associative for sets and bags
Cartesian product
(×) is associative
Note: sets are equal when attribute values are identified by attribute name
Join
(⋈) is associative
with a caveat
Note: the join attributes must exist across every join operation
Join
(⋈) is associative
with a caveat
Join
(⋈) is associative
with a caveat
S(b,c) ⋈ T(a,d) = empty !!
Join
(⋈) operation with
no common attributes
≡
Cartesian Product !!
- Cartesian product laws:
(R ×Set S) ×Set T = R ×Set (S ×Set T) (R ×Bag S) ×Bag T = R ×Bag (S ×Bag T)
- Union laws:
(R ∪Set S) ∪Set T = R ∪Set (S ∪Set T) (R ∪Bag S) ∪Bag T = R ∪Bag (S ∪Bag T)
- Intersection laws:
(R ∩Set S) ∩Set T = R ∩Set (S ∩Set T) (R ∩Bag S) ∩Bag T = R ∩Bag (S ∩Bag T)
- Join laws:
(R ⋈Set S) ⋈Set T = R ⋈Set (S ⋈Set T) (R ⋈Bag S) ⋈Bag T = R ⋈Bag (S ⋈Bag T) Only if the join attributes exists across the join operation
- Example where
⋈ is
associative:
R(a,b) S(b,c) T(c,d) ( R(a,b) ⋈ S(b,c) ) ⋈ T(c,d) = R(a,b) ⋈ ( S(b,c) ⋈ T(c,d) )
- Counter example:
R(a,b) S(b,c) T(a,d) ( R(a,b) ⋈ S(b,c) ) ⋈ T(a,d) ≠ R(a,b) ⋈ ( S(b,c) ⋈ T(a,d) )
- Comment:
- If we degenerate (= replace) a
join with
no common attributes into
a
cartesian product, then:
( R(a,b) ⋈ S(b,c) ) ⋈ T(a,d) = R(a,b) ⋈ ( S(b,c) × T(a,d) )
Example:
R(a,b) S(b,c) T(a,d) ======================================= (a1,b1) (b1,c3) (a1,d4) (a2,b5) (b5,c7) (a2,d6) ( R(a,b) ⋈ S(b,c) ) ⋈ T(a,d) = {(a1,b1,b1,c3) , (a2,b5,b5,c7)} ⋈ T = {(a1,b1,b1,c3,a1,d4) , (a2,b5,b5,c7,a2,d6)} R(a,b) ⋈ ( S(b,c) × T(a,d) ) = R ⋈ {(b1,c3,a1,d4), (b1,c3,a2,d6), (b5,c7,a2,d6), (b5,c7,a2,d6)} = {(a1,b1,b1,c3,a1,d4), (a2,b5,b5,c7,a2,d6)}
- If we degenerate (= replace) a
join with
no common attributes into
a
cartesian product, then: