Consider a database with three relation instances shown below. The primary…

Simple aur proper explanation:

Comparison Calculation

We calculate the comparisons step by step.

1. First inner query: Red cars

SELECT R.did
FROM Cars C, Reserves R
WHERE R.cid = C.cid AND C.color = 'red'

• First, we scan all 4 tuples of the Cars relation to find red cars.
  Comparisons = 4

• There are 2 red cars. For each red car, all 10 tuples of Reserves are checked to match R.cid = C.cid.
  Comparisons = 2 × 10 = 20

So, for the first inner query:

4 + 20 = 24 comparisons

The resulting did values are:

{22, 22, 31, 31, 64}

After removing duplicates:

{22, 31, 64}

2. Second inner query: Green cars

SELECT R.did
FROM Cars C, Reserves R
WHERE R.cid = C.cid AND C.color = 'green'

• Scan all 4 tuples of Cars to find green cars.
  Comparisons = 4

• There is 1 green car, so all 10 tuples of Reserves are checked.
  Comparisons = 1 × 10 = 10

So, for the second inner query:

4 + 10 = 14 comparisons

The resulting did values are:

{22, 31, 74}

3. Intersection of the two sets

First set:

{22, 31, 64}

Second set:

{22, 31, 74}

Common values are:

{22, 31}

Using linear search:

• For 22, it matches on the first comparison → 1 comparison

• For 31, it matches on the second comparison → 2 comparisons

• For 74, all 3 values of the first set are checked → 3 comparisons

So:

1 + 2 + 3 = 6 comparisons

4. Finding these drivers in the Drivers table

Now we need to find did = 22 and did = 31 in the Drivers table using linear search.

• 22 is found on the first try → 1 comparison

• 31 is found on the third try → 3 comparisons

So:

1 + 3 = 4 comparisons

Total Comparisons

First inner query = 24
Second inner query = 14
Intersection = 6
Drivers table search = 4

24+14+6+4=48 {24 + 14 + 6 + 4 = 48}

Therefore, the total number of comparisons is 48, so Option B is the correct answer.