Conjunctive Query Translation

This describes translation algorithms from Relational Algebra to Conjunctive Queries and vise-versa


Translation to Conjunctive Queries

We can translate a Relational Algebra expression that is in Select-Project-Join form into CQ. Note that it is not possible to translate any other form to it


Translation by Example

Suppose we have the following logical query plan:

$ \pi_{

 \begin{subarray}{l}
   R_1.A, \\
   S_1.B \\
 \end{subarray}

} \sigma_{

 \begin{subarray}{l}
   {\color{grey}{(1)}} \ R_1.A = R_2.A \ \land \\
   {\color{grey}{(2)}} \ R_2.B = 4 \ \land \\
   {\color{grey}{(3)}} \ R_2.A = R_3.A \ \land \\
   {\color{grey}{(4)}} \ R_3.B = S_1.B \ \land \\
   {\color{grey}{(5)}} \ S_1.C = S_2.C \ \land \\
   {\color{grey}{(6)}} \ S_2.B = 4
 \end{subarray}

} \big( \rho_{R_1}(R) \times \rho_{R_2}(R) \times \rho_{R_3}(R) \times \rho_{S_1}(S) \times \rho_{S_2}(S) \big)$

We proceed as follows

  • step 1:
for each relation we create an atom in body with distinct variables
  • step 2:
for all the conditions of the selection part we replace the variables that participate in equations by the same symbol

in this case

  • step 1:
    • $

\begin{array}{l l} Q(x_{R_1.A}, y_{S_1.B}) \leftarrow &

 R(x_{R_1.A}, y_{R_1.B}), \\

& R(x_{R_2.A}, y_{R_1.B}), \\ & R(x_{R_3.A}, y_{R_3.B}), \\ & R(y_{S_1.B}, z_{S_1.C}), \\ & R(y_{S_2.B}, z_{S_2.C}) \\ \end{array} $

    • in the head we put what we want to see in the output, i.e. the variables specified in the projection
    • variable names like $x_{R_1.A}$ are helpful to keep track of the original names, however it could be anything
    • this query now computes $\rho_{R_1}(R) \times \rho_{R_2}(R) \times \rho_{R_3}(R) \times \rho_{S_1}(S) \times \rho_{S_2}(S)$
  • step 2
    we restrict the query so it outputs only the tuples that match the selection condition
    for that for every equality we replace all the occurrence of variables in that inequality to the same variable
    • the name does not matter
    • for constants we put the value of a constant
    • $

\begin{array}{l l} Q(x_{R_1.A}, y_{S_1.B}) \leftarrow &

 R(x_{R_1.A} {\color{grey}{|^{(1)}}}, y_{R_1.B}), \\

& R(x_{R_1.A} {\color{grey}{|^{(1,3)}}}, 4 {\color{grey}{|^{(2)}}}), \\ & R(x_{R_1.A} {\color{grey}{|^{(3)}}}, y_{R_3.B} {\color{grey}{|^{(4)}}}), \\ & R(y_{R_3.B} {\color{grey}{|^{(4)}}}, z_{S_1.C} {\color{grey}{|^{(5)}}}), \\ & R(4 {\color{grey}{|^{(6)}}}, z_{S_1.C} {\color{grey}{|^{(5)}}}) \\ \end{array} $

      • here ${\color{grey}{|^{(i)}}}$ shows what part of the condition was used to do the replacement
  • now this is equivalent to the RA expression


Translation from Conjunctive Queries

Translation from CQ back to Relational Algebra is straightforward

  • the number of elements being joined is equal to the number of atoms
    • i.e. the number of joins is (# of atoms) - 1
  • we then add the selection
    • with an equality condition for each variable name (if repeated 2 or more times)
    • and for each constant
  • finally, only the variables in the head get projected
  • of course, we need to know the relational schema to be able to do that
    • note that in CQ we use positions to denote attributes

Example

Given: $Q(t) \leftarrow \text{MovieStar}({\color{red}{n}}, a, g, {\color{blue}{1940}}), \text{StarsIn}(t, y, {\color{red}{n}})$

We translate it as

  • $\pi_\text{S.movieTitle}

\sigma_{

 \begin{subarray}{l}
   \text{M.name = S.starName }  \land \\
   \text{M.birthDate = 1940} \\
 \end{subarray}

} \big( \rho_M(\text{MovieStar}) \times \rho_S(\text{StarsIn}) \big)$


See Also

Sources

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