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Solution

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.