Question

Suppose a relation R(A,B) has 5 tuples (1,2), (1,3), (3,2), (3,4), (4,2). A result of γ A , AVG(B) -> AVG_B(R) is __________

Select one:
a. A relation has 3 tuples (1,2), (3,2), (4,1)
b. A relation has 3 tuples (1,2), (3,2), (4,2)
c. A relation has 3 tuples (1,2.5), (3,3), (4,2)
d. A relation has 3 tuples (1,5), (3,6), (4,2)

Answer 1

The result of γ A , AVG(B) -> AVG_B(R) for the given tuples is a relation with 3 tuples: (1,2.5), (3,3), (4,2), which calculates the average of column B for each unique value in column A.

Step-by-step explanation:

The question is about the aggregation operation in a database, specifically using the group by operator (denoted by γ) and the average function (denoted by AVG). Given the relation R(A,B) with tuples (1,2), (1,3), (3,2), (3,4), (4,2), we are asked to group by column A and calculate the average of column B for each group. The group by operator creates groups of rows in a table that have the same values in specified columns and allows aggregate functions like AVG to be applied to each group.

To find γ A , AVG(B) -> AVG_B(R), we compute the average of B for each unique A value. For A=1, the average of B is (2+3)/2 = 2.5. For A=3, the average of B is (2+4)/2 = 3. For A=4, B is just 2 since there is only one tuple. Therefore, the correct answer is a relation with 3 tuples: (1,2.5), (3,3), (4,2).