Slideshow:
The cartesian product operator
Cartesian product simply concatenate all combinations of tuples...
The one-pass cartesian product
algorithm
Requirement: the entire relation S must be stored in M−1 buffers
The one-pass cartesian product
algorithm
IO cost and buffer requirement for Cartesian Product (×)
- (Cartesian) Product operator:
R × S - Example (Cartesian) Product:
R = {1, 2}; S = {2, 4}; R × S = { (1,2), (1,4), (2,2), (2,4) }
- Phase 1:
use M−1 buffers to
read the
smaller relation in
memory:
- Phase 2: use
1 buffer to
read the
larger relation -
one block at a time -
and output
R × S:
- Cost:
read S and
R once:
- Cost(one-pass ×) = B(R) + B(S) (Clustered relations)
- Constraint:
store S
completely
in memory and
use 1 buffer to
read
R:
- M(one-pass ×) ≥ B(S) + 1 (Clustered relations)