Question: We can store the entire smaller relation in memory read
The indexed nested-loop join algorithm described in Section 13.5.3 can be inefficient if the index is a secondary index, and there are multiple tuples with the same value for the join attributes. Why is it inefficient? Describe a way, using sorting, to reduce the cost of retrieving tuples of the inner relation. Underwhat conditions would this algorithm be more efficient than hybrid merge–join?
Answer to relevant QuestionsSuppose that a B + - tree index on branch-city is available on relation branch, and that no other index is available. List different ways to handle the following selections that involve negation? a. σ ¬ ((branch-city Consider the relations r1 (A, B, C), r2 (C, D, E), and r3 (E, F), with primary keys A, C, and E, respectively. Assume that r1 has 1000 tuples, r2 has 1500 tuples, and r3 has 750 tuples. Estimate the size of r1 Θ r2 ...SQL allows relations with duplicates. a. Define versions of the basic relational-algebra operations σ, Π, ×, Π, −, ∪, and ∩ that work on relationswith duplicates, in a way consistent with ...List the ACID properties. Explain the usefulness of each. Consider the precedence graph of Figure is the corresponding schedule conflict serializable? Explain your answer.
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