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
above query.

Here is my answer:
á T.branch_name (T.assets > S.assets v S.branch_city = “Brooklyn”) (branch)

To write an efficient relational algebra expression equivalent to the given query, we can break it down into smaller parts and then combine them using the appropriate relational algebra operations.

1. Start with the branch table: branch
2. Select rows where the branch_city is "Brooklyn": σ(S.branch_city = "Brooklyn")(branch)
3. Rename the resulting table S: ρ(S)(σ(S.branch_city = "Brooklyn")(branch))
4. Join the branch table with itself: branch ⨝ branch
5. Rename the result table T: ρ(T)(branch ⨝ branch)
6. Apply a selection condition where T.assets > S.assets: σ(T.assets > S.assets)(ρ(T)(branch ⨝ branch))
7. Finally, project only the branch_name column from the result: π(T.branch_name)(σ(T.assets > S.assets)(ρ(T)(branch ⨝ branch)))

Putting it all together, the efficient relational algebra expression for the given query is:

π(T.branch_name)(σ(T.assets > S.assets)(ρ(T)(branch ⨝ branch)))

Note: The notation used here follows the classic relational algebra notation, where σ represents selection, π represents projection, ⨝ represents the join operation, and ρ represents renaming.