Exercise #57

This exercise is somewhat similar to the exercise 56, i.e. here it is possible to suppose the same mistakes in calculation of the number of sunken ships. However the situation is also aggravated with definition of the total number of the ships in a class. Let’s consider the solution that the check system doesn’t accept. 

Exercise 3.13.1

SELECT c.class, SUM(outc)
FROM Classes c LEFT JOIN
Ships s ON c.class = s.class LEFT JOIN
(SELECT ship, 1 outc
FROM Outcomes
WHERE result = 'sunk') o ON s.name = o.ship OR
c.class = o.ship
GROUP BY c.class
HAVING COUNT(*) > 2 AND
SUM(outc) IS NOT NULL;

The first left join gives all classes repeating so many times as the number of ships available in the Ships table. If any class doesn’t have ships in this table, it will be noted one time, and it gives us an opportunity to consider the leading ships of the class in the Outcomes table, if any.

Next, one left join is being worked out with the set of sunken ships on the predicate

ON s.name = o.ship OR c.class = o.ship

In the calculating column 1 is being inserted, if the name of the sunken ship coincides either with the name of the ship, or with the name of the class from the set had been got earlier. So, here we do try to consider the leading (head) ships.

Finally, the grouping by the classes with selection by the number of ships (rows) in the class is being worked out, and the sum of the sunken ships (units in the column “outs”) is being calculated. Author of this solution offers the rational way to calculate in one grouping both the total number of ships, and the quantity of the sunken ships in the class. The predicate,

SUM(outc) IS NOT NULL

in accordance with the terms of the task, removes from the result such classes that don’t have any sunken ships.

Those who read the analysis of the previous tasks, have already guessed, what the problem is. That’s right, the problem is in the predicate of the second join. But not only in this.
в этом.

Let’s consider the next variant of data. Let for some class class_N in the Ships table we have two ships: ship_1 and ship_2. Besides, in the Outcomes table there is the sunken ship ship_1 and survived the leading ship – class_N.

The first join gives:

We work out the second join:

In the result this class will not get into the resulting set at all, because the condition COUNT(*) > 2 won’t be held, but actually there are three ships. The reason of the mistake lies in the fact that we perform the join only on the sunken ships, simultaneously counting the total number of ships.

Now let’s change a little data in the example. And let the leading ship class_N to be also sunk. Then the result of the join is:

The last two rows will be got in the result of joining the row of the sunken leading ship, as the predicate c.class=o.ship gives “true”. So, instead of one row for the leading ship we get a row for every ship of the class from the Ships table. Totally, instead of

You may try to correct this solution or to use another way on the basis of the inner join and union.

As it will seem surprising, but three absolutely different solutions presented below contain the same mistake, at least, they return the same result on the checking database of a site.

Exercise 3.13.2

SELECT class, SUM(sunk)
FROM (SELECT class, COUNT(*) AS sunk
FROM Ships a JOIN
Outcomes b ON a.name = b.ship AND
a.class <> b.ship
WHERE result = 'sunk'
GROUP BY class
UNION ALL
SELECT class, '1'
FROM Classes a JOIN
Outcomes b ON a.class = b.ship
WHERE result = 'sunk'
UNION ALL
SELECT class, '0'
FROM classes
) t
-- classes the number of ships in which great than 2:
WHERE class IN (SELECT t1.class
FROM (SELECT a.class
FROM Classes a LEFT JOIN
Ships b ON a.class = b.class
) t1 LEFT JOIN (SELECT DISTINCT ship
FROM Outcomes
WHERE ship NOT IN (SELECT name
FROM Ships
)
) t2 ON t1.class = t2.ship
GROUP BY t1.class
HAVING COUNT(*) > 2
)
GROUP BY class
HAVING SUM(sunk) > 0

Exercise 3.13.3

SELECT a.class AS cls, a.num_sunks AS sunk
FROM (SELECT c.class, COUNT (o.ship) AS num_sunks
FROM Outcomes o LEFT JOIN
Ships s ON o.ship = s.name LEFT JOIN
Classes c ON s.class = c.class
WHERE o.result = 'sunk'
GROUP BY c.class) a,
(SELECT c1.class
FROM Ships s1, Classes c1
WHERE s1.class = c1.class
GROUP BY c1.class
HAVING COUNT(name) >= 3
) B
WHERE a.class = b.class

Exercise 3.13.4

SELECT class, COUNT(result) AS sunk
FROM (SELECT class, result, name
FROM Ships LEFT JOIN
Outcomes ON ship=name AND
class IS NOT NULL AND
result = 'sunk'
) T
GROUP BY class
HAVING COUNT(class) > 2 AND
COUNT(result) > 0

Analyse subtleties of the above-mentioned solutions, the most beautiful of which, certainly, is 3.13.4. Only one join for which at once it is counted up both quantity of sunken ships, and the total number of the ships in a class. These solutions have a common mistake we have talked above about: the head ships which are present in Outcomes table and are absent in Ships table have been not taken into account.

T&S

To solve the problem on SQL-EX.RU