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Exercise #23 page 3

To not be proofless, I'll show the results of original query (solution 1.15.2) with an extension of output columns set:

Console
Execute
  1. SELECT maker, a.model a_m, b.model b_m, c.model c_m
  2. FROM Product a, PC b, Laptop c 
  3. WHERE ((b.speed >= 750 AND 
  4.         c.speed >= 750
  5.         ) AND 
  6.         a.model = b.model
  7.        ) OR 
  8.        ((b.speed >= 750 AND 
  9.          c.speed >= 750
  10.         ) AND 
  11.          a.model = c.model
  12.        );

Let's consider a couple of rows from the result set: 

maker a_m b_m c_m
B 1121 1121 1752
A 1752 1121 1752

As you can see, model 1121 (PC) belongs to the maker B, but model 1752 (laptop) – to the maker A. So, we don't have a reason to consider that both these makers satisfy the task conditions.

Uniting of demanded models of the PCs and laptops in one set gives only illusion that we receive both types:

Solution 1.15.3

Console
Execute
  1. SELECT maker
  2. FROM (SELECT maker 
  3.       FROM Product INNER JOIN 
  4.            PC ON Product.model = PC.model 
  5.       WHERE type='PC' AND 
  6.             speed >= 750
  7.       UNION ALL
  8.       SELECT maker 
  9.       FROM Product INNER JOIN 
  10.             Laptop ON Product.model = Laptop.model
  11.       WHERE type='laptop' AND 
  12.             speed >= 750
  13.       ) S
  14. GROUP BY maker;

As a result, the list of manufacturers for which there is at least one row in a set from FROM clause will be received. Below shorter variant of the same mistake.

Solution 1.15.4

Console
Execute
  1. SELECT maker
  2. FROM Product
  3. WHERE model IN (SELECT model 
  4.                 FROM PC 
  5.                 WHERE speed >= 750 
  6.                 UNION ALL
  7.                 SELECT model 
  8.                 FROM Laptop 
  9.                 WHERE speed >= 750
  10.                 )
  11. GROUP BY maker;

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