给定关系模式下关系代数Q₁与Q₂不等价及Q₂无效的示例咨询

First, let’s recap the context: we have a relation R(A, C) (with A as the primary key), and two queries:

• Q₁ = π_A (σ_{C<10}(R))
• Q₂ = σ_{C<10}(π_A(R))

The key issue here is that Q₂ is a completely invalid expression—it can’t even be executed, let alone produce a result equivalent to Q₁. Let’s use a concrete data example to break this down clearly.

Step 1: Sample Data for Relation R #

Let’s populate R with real values to make the example tangible:

How Q₁ Works (Valid and Executable) #

Q₁ follows the logical order: filter first, then project

1. First, the selection operation σ_{C<10}(R) runs: it keeps only rows where C is less than 10. This gives us:

   A	C
   1	5
   3	8

2. Next, the projection π_A(...) drops the C column, leaving only the A values from the filtered rows:

   A
   1
   3

Why Q₂ Fails (Invalid Expression) #

Q₂ reverses the order fatally: project first, then filter

1. First, π_A(R) runs: this operation keeps only the A column from R, discarding C entirely. The intermediate result looks like this:

   A
   1
   2
   3
   4

2. Now, we try to run σ_{C<10}(...) on this result—but there’s a critical problem: this intermediate table has no C column! The database has no way to check the condition C<10 because the attribute C was already discarded by the earlier projection.

In short, Q₂ tries to filter on an attribute that no longer exists in the data set. Databases will throw an error if you attempt to execute Q₂, since the selection condition references a non-existent attribute. That’s why the two queries aren’t equivalent—one works as intended, the other can’t run at all.

内容的提问来源于stack exchange，提问作者user8779054