Consider the relational schema of sailors $S$, Reserves $R$ and Boats $B$.Table 1: Sailors $S$

Sid	Sname	Rating	Age
22	Dustin	7	45.0
29	Brutus	1	33.0
31	Lubber	8	55.5
32	Andy	8	25.5
58	Rusty	10	35.5
64	Horatio	7	35.5
71	Zorba	10	16.0
74	Horatio	9	35.0
85	Art	3	25.5
95	Bob	3	63.5

Table 2: Reserves $R$

Sid	Bid	day
22	101	10/10/98
22	102	10/10/98
22	103	10/8/98
22	104	10/7/98
31	102	11/10/98
31	103	11/6/98
31	104	11/12/98
64	101	9/5/98
64	102	9/8/98
74	103	9/8/98

Table 3: Boats $B$

Bid	Bname	Color
101	Interlake	blue
102	Interlake	red
103	Clipper	green
104	Marine	red

Which of the following relational algebra query computes the names of sailor who have reserved all boats?

We need the names of sailors who have reserved all boats.

In relational algebra, when we need to find entities related to all values of another relation, we use division operation.

Here,

$Reserves(Sid,Bid,day)$ tells which sailor reserved which boat.

$Boats(Bid,Bname,Color)$ tells all available boats.

First, take only required columns from Reserves:

$\pi_{sid,bid}(Reserves)$

This gives pairs of sailors and boats reserved by them.

Now take all boat IDs from Boats:

$\pi_{bid}(Boats)$

To find sailors who reserved every boat, apply division:

$(\pi_{sid,bid}(Reserves))/(\pi_{bid}(Boats))$

This gives only those $sid$ values which are associated with all $bid$ values.

Now rename this result as $Tempsids$:

$\rho(Tempsids,((\pi_{sid,bid}Reserves)/(\pi_{bid}Boats)))$

Finally, join $Tempsids$ with Sailors table to get sailor names:

$\pi_{sname}((Tempsids)\bowtie Sailors)$

So, the correct query must use:

$\pi_{sid,bid}Reserves$

and

$\pi_{bid}Boats$

with division operation.

Option (b) correctly represents this logic.