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 sailors who have reserved a red and a green boat?

We need the names of sailors who have reserved both types of boats:

Red boat

Green boat

First, find sailors who reserved a red boat.

Red boats are selected by:

$\sigma_{color='red'}(Boats)$

Now join this with Reserves table, because Reserves table contains the information about which sailor reserved which boat.

$\pi_{sid}((\sigma_{color='red'}(Boats))\bowtie Reserves)$

This gives the $sid$ of sailors who reserved a red boat.

Rename this result as:

$Tempred$

So,

$Tempred=\pi_{sid}((\sigma_{color='red'}(Boats))\bowtie Reserves)$

Now, find sailors who reserved a green boat.

Green boats are selected by:

$\sigma_{color='green'}(Boats)$

Join with Reserves table and project $sid$:

$\pi_{sid}((\sigma_{color='green'}(Boats))\bowtie Reserves)$

Rename this result as:

$Tempgreen$

So,

$Tempgreen=\pi_{sid}((\sigma_{color='green'}(Boats))\bowtie Reserves)$

Now, sailors who reserved both red and green boats will be obtained by intersection:

$Tempred\cap Tempgreen$

This gives common $sid$ values.

Finally, join this result with Sailors table to get sailor names:

$\pi_{sname}((Tempred\cap Tempgreen)\bowtie Sailors)$

Therefore, option (a) correctly finds sailors who reserved both a red and a green boat.