به انگلیسی

Asking questions with sets of rows

Computing · Data & Databases · ages 18-19

Name ______________________ Date ____________

Students has name and age attributes. In π_name(σ_age>10(Students)), which operation acts first?
- Project names first, then test the removed age attribute
- Run both operations together without an intermediate relation
- Select by age, then project names
R has attributes id, name and course. What is the result schema of π_id,course(R)?
- id, name and course
- id and course
- name and course only
- id and name only
R(a,b) = {(1,4),(2,5)}. Which display contains exactly the same relation?
- S(a,b) = {(4,1),(5,2)}
- S(a,b) = {(1,4),(2,6)}
- S(a,b) = {(2,5),(1,4)}
- S(a,b) = {(1,4),(2,5),(3,6)}
Which sentence describes a relation in the mathematical relational model?
- A set of tuples over named attributes
- A list of values whose row positions define identity
- A collection of rows with no fixed set of attributes
Compatible relations R and S share 2 of R's 4 tuples. How many tuples are in R - S?
Answer: ______________
Students(id,name,age) and Enrolled(id,course) share only id. Which expression returns name/course pairs for enrolled students over 21? Here ⋈ is natural join.
- π_name,course(σ_age>21(Students) ⋈ Enrolled)
- π_name,course(Students ⋈ Enrolled), retaining students of every age
- π_name,course(π_name(Students) ⋈ Enrolled), removing the matching id
- π_name,course(Students ∪ Enrolled), pooling the different schemas
Compatible relations R and S contain 6 and 5 tuples. Exactly 3 tuples occur in both. How many tuples are in R ∪ S?
Answer: ______________
R(id,name) = {(10,Uma),(11,Uma),(12,Sol)}. How many tuples are in π_name(R) under pure relational algebra?
- 2 tuples
- 3 tuples
- 1 tuple
- 0 tuples
Students(id,age) and Enrolled(id,course) share only id. Which expression always equals σ_age>21(Students ⋈ Enrolled)?
- Students ⋈ σ_age>21(Enrolled), testing age in the enrollment input
- π_id(σ_age>21(Students)) ⋈ Enrolled, removing age from the result
- σ_age>21(Students) ⋈ Enrolled
- σ_age>21(Students) ∪ Enrolled, replacing matching pairs with union
R(a,b) = {(2,7)} and S(a,b) = {(2,9)}. Kim claims π_a(R ∩ S) = π_a(R) ∩ π_a(S). Why does this example disprove the claim?
- Intersection cannot compare tuples containing more than one attribute
- Projection hides the differing b values and creates a shared a value
- Projection changes the original a values before comparing their membership

Answer key

For grown-ups. Fold this page away before handing over the rest.

Asking questions with sets of rows W1-mt_unDBd_Ug8T-s1

Select by age, then project names · The inner selection receives Students first. Projection then retains the names from the surviving rows.
id and course · Projection retains exactly its named attributes: id and course. It removes name.
S(a,b) = {(2,5),(1,4)} · The matching display contains the same complete tuples under the same attributes. Only their written order changes.
A set of tuples over named attributes · A relation has named attributes and a set of tuples. Each distinct tuple appears once, without a meaningful row order.
2 · Remove the 2 shared tuples from R's 4 tuples: 4 - 2 = 2.
π_name,course(σ_age>21(Students) ⋈ Enrolled) · Filter while age is available, join enrollment using id, then project the required name and course attributes.
8 · Adding 6 + 5 counts the shared tuples twice. Subtract their 3 extra counts: 11 - 3 = 8.
2 tuples · Dropping id makes the two Uma tuples identical. The projected relation contains (Uma) and (Sol).
σ_age>21(Students) ⋈ Enrolled · The age test uses only Students attributes. Filtering that input first retains exactly the same matching pairs and output attributes.
Projection hides the differing b values and creates a shared a value · The full tuples differ, so the left result is empty. Both projections contain 2, so the right result contains 2.