Solution

The correct answer is option 1) π_eid( π_eid,cid(Own) ÷ π_cid(Comp) )

Key Points

• Relational Division: ✅ The expression π_eid( π_eid,cid(Own) ÷ π_cid(Comp) ) returns exactly those eids for which every brand cid in Comp appears with that employee in Own. This is the canonical way to express “owns all brands”.
• Why the other options are wrong:
  • Option 2: ❌ πeid( πeid(Own) × πcid(Comp) ) is a Cartesian product of all eids with all cids; projecting eid just gives all employees that appear in Own, not only those who own every brand.
  • Option 3: ❌ πeid( πeid,cid(Own) × πcid(Comp) ) is also a Cartesian product; after projection it again returns all eids in Own, ignoring the “for all brands” requirement.
  • Option 4: ❌ πeid( πeid(Own) × ( πcid,cName(Own) ÷ πcid(Comp) ) ) is ill-typed (the relation Own doesn’t have cName), and even if fixed, the product with πeid(Own) does not enforce the universal condition per employee.

Additional Information

• Division-free equivalent (anti-join idea): Let A = πcid(Comp) and E = πeid(Own). Compute the “missing pairs” M(eid,cid) = (E × A) − πeid,cid(Own). Then result is E − πeid(M) (employees for whom no brand is missing).
• Group-by intuition: Count distinct cid owned by each eid and keep those whose count equals |πcid(Comp)| (this is how SQL mirrors division).

Hence, the correct answer is: option 1) π_eid( π_eid,cid(Own) ÷ π_cid(Comp) )