Union Operator: Problem with × (cross product) and ⋈ (natural joins) is that they try to match tuples (rows) form the relations (tables). Here, we combine the information vertically.

π_cName College ∪ π_sName Student gives:

Stanford
Susan
Cornell
Mary
John
…

Difference operator

IDs of students who didn't apply anywhere

π_sID Student - π_sID Apply

IDs and names of those students

π_sName (( π_sID Student - π_sID Apply ) ⋈ Student )

Intersection operator

Names that are both a college name and a student name:

π_cNameCollege ∩ π_sName Student

Intersection doesn't add expressive power.

E₁ ∩ E₂ ≣ E₁ - ( E₁ - E₂ )

And if the schema is the same between E₁ and E₂, then

E₁ ∩ E₂ ≣ E₁ ⋈ E₂

Rename Operator (“ρ” rho)

1. ρ_R(A1,…,An)(E) ← General Form
2. ρ_R(E)
3. ρ_A1,…,An(E)

ρ_c(name)(π_cNameCollege) ∪ ρ_c(name)(π_sName Student)

Rename is used for disambiguation in “self-joins”

Pairs of colleges in same state: Stanford, Berkeley; Berlekey, UCLA; …

σ_{s1 = s2} (ρ_c1(n1,s1,e1)( College ) × ρ_c2(n2,s2,e2)( College ) )

or

ρ_c1(n1,s,e1)( College ) ⋈ ρ_c2(n2,s,e2)( College )

This would give us Stanford, Stanford and Berkeley, Berkeley. So add…

σ_{s1 != s2}( ρ_c1(n1,s,e1)( College ) ⋈ ρ_c2(n2,s,e2)( College ) )

We'll still get Stanford, Berkeley and Berkeley, Stanford. So change…

σ_{s1 < s2}( ρ_c1(n1,s,e1)( College ) ⋈ ρ_c2(n2,s,e2)( College ) )

Clever, eh?