Slideshow:
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We must now find:
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best ⋈1 = null;
best ⋈2 = null;
best_cost = ∞;
for ( ⋈1 = {one-pass, 2-pass, 3-pass, ...} ) do
{
for ( ⋈2 = {one-pass, 2-pass, 3-pass, ...} ) do
{
if ( memory-requirement(⋈1, ⋈2) ≤ M )
{
if ( cost(⋈1, ⋈2)) < best_cost )
{
best ⋈1 = ⋈1;
best ⋈2 = ⋈2;
}
}
}
}
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We will illustrate the search algorithm with a few examples next....
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B(R) × B(S)
B(R ⋈ S) = ----------------------- = k blocks
max( V(R,x), V(S,x) )
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We will vary the the value k and study the effect.
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