select T.branch_name
from branch T, branch S
where T.assets > S.assets and S.branch_city = "Brooklyn"
Write an efficient relational algebra
expression that is equivalent to this query.
Here is my answer:
á T.branch_name (T.assets > S.assets v S.branch_city = “Brooklyn”) (branch)
To write an equivalent relational algebra expression for the given SQL query, we can break it down step by step:
1. First, we need to perform a Cartesian product (×) between the Branch table (T) and itself, giving us a temporary relation X. This will create all possible combinations of branch records.
X ← T × T
2. Next, we need to apply a selection (σ) operation on the temporary relation X, with two conditions: T.assets > S.assets and S.branch_city = "Brooklyn".
Y ← σ(T.assets > S.assets ∧ S.branch_city = "Brooklyn")(X)
3. Finally, we project (π) the branch_name attribute from the resulting relation Y to get the desired output.
Z ← π(T.branch_name)(Y)
Therefore, the resulting efficient relational algebra expression that is equivalent to the given SQL query is:
Z ← π(T.branch_name)(σ(T.assets > S.assets ∧ S.branch_city = "Brooklyn")(T × T))
This expression represents the same set of operations as the original SQL query and will yield the expected result.