\(\require{color}\) \(\newcommand{\AntiJoin}{ \ \bar{\Join} \ } \)

Translating SQL to Relational Algebra

query-processing-1st.png

Translating SQL to RA expression is the second step in Query Processing Pipeline

  • Input: Logical Query Plan - expression in Extended Relational Algebra
  • Output: Optimized Logical Query Plan - also in Relational Algebra


Union, Intersection, Difference

Translation is straightforward

(SELECT * FROM R1) INTERSECT (SELECT * FROM R2)

Is $R_1 \cap R_2$


UNION $\to R_1 \cup R_2$

EXCEPT $\to R_1 - R_2$


Select-From-Where No Subqueries

Query

SELECT movieTitle
FROM StarsIn, MovieStarM
WHERE starName = M.name AND M.birthdate = 1960
  • in the from clause we have all relations we need
  • so we make a Cartesian Product for all relations there
  • if there is an alias - we do Renaming
  • then we filter the Cartesian Product
  • then translate the where clause too

So we get:

[math]\pi_\text{movieTitle} \sigma_{\text{starName = M.name } \land \text{M.birthdate = 1960}}(\text{StartsIn} \times \rho_M (\text{MovieStar}))[/math]

(Maybe not the most efficient way, but it will be optimized further)

Normalization Step

Suppose we have subqueries in the "Where" clause

SELECT movieTitle FROM StarsIn
WHERE starName IN (
    SELECT name
    FROM MovieStar
    WHERE birthdate=1960)

Here we may have different constraints:

  • $\text{in}, \leqslant, <, \geqslant, >, =, \neq$, etc
  • whenever we have such constraints, we may replace them with quantifiers $\forall$ and $\exists$
  • or with EXISTS and IN or NOT EXISTS
  • so we first translate a SQL query to the equivalent SQL with EXISTS or NOT EXISTS


Example 1: IN

SELECT movieTitle FROM StarsIn
WHERE starName IN (
    SELECT name
    FROM MovieStar
    WHERE birthdate=1960)

to

SELECT movieTitle FROM StarsIn
WHERE EXISTS (
    SELECT name
    FROM MovieStar
    WHERE birthdate=1960 AND name=starName)


Example 2: $\geqslant$

SELECT name FROM MovieExec
WHERE netWorth >= (
    SELECT E.netWorth
    FROM MovieExec E)

to

SELECT name FROM MovieExec
WHERE NOT EXISTS (
    SELECT E.netWorth
    FROM MovieExec E
    WHERE netWorth < E.netWorth)


Example 3: aggregated attributes

SELECT C FROM S
WHERE C IN (
    SELECT SUM(B) FROM R
    GROUP BY A)

to

SELECT C FROM S
WHERE EXISTS (
    SELECT SUM(B) FROM R
    GROUP BY A
    HAVING SUM(B) = C)

(note that in this case we use "HAVING" and not "WHERE")


So the first step when processing these kinds of queries is normalization step:

  • translate a query into EXISTS/NOT EXISTS form

Hence we can assume that all queries are in this form

  • We then apply the next step: for correlated queries


Correlated Queries

  • A subquery can refer to attributes of relations that are introduces in the outer query
  • def: we call such queries correlated subqueries
  • the outer relation is called the context relation - a correlated subquery uses its attributes
  • a parameter - is a set of attributes of all context relations of a subquery

Example:

correlated-subqueries-1.png

here

  • the subquery refers to S.starName, so it's correlated
  • S is the context relation for the subquery
  • S.starName is a parameter to the correlated subquery

EXISTS in the Where Clause (by example)

SELECT S.movieTitle, M.studioName
FROM StarsIn S, Movie M
WHERE S.movieYear >= 2000
AND S.movieTitle = M.title
AND EXISTS (
    SELECT name
    FROM MovieStar
    WHERE birthdate = 1960 AND name = S.starName)


Algorithm

  • it's recursive: translate the subqueries first
    • [math]\pi_\text{name} \sigma_{ \begin{subarray}{l} \text{birthDate = 1960 } \land \\ \text{name = S.starName} \\ \end{subarray} } (\text{MovieStar})[/math]
    • problem: cannot find S.starName in the input relation
    • so it must be a correlated query
    • we therefore need to recognize that this is a context relation's parameter
  • so we need to add the context relations and parameters
    • [math]\pi_{ \begin{subarray}{l} \color{blue}{\text{S.movieTitle}}, \\ \color{blue}{\text{S.movieYear}}, \\ \color{blue}{\text{S.starName}}, \\ \text{name} \\ \end{subarray} } \sigma_{ \begin{subarray}{l} \text{birthDate = 1960 } \land \\ \text{name = S.starName} \\ \end{subarray} } (\text{MovieStar} {\color{red}{\times \rho_S(\text{StarsIn}) }})[/math]
  • next, we translate the "from" clause
    • $\rho_S(\text{StarsIn}) \times \rho_M(\text{Movie})$
  • now we need to synchronize the subresult by join
    • from the subquery we need to keep only the parameter attributes (the blue ones) - can remove $\text{name}$
    • join: if something exists, we will join on it
    • [math]\big[ \rho_S(\text{StarsIn}) \times \rho_M(\text{Movie}) \big] \Join \big[ \pi_{ \begin{subarray}{l} \color{blue}{\text{S.movieTitle}}, \\ \color{blue}{\text{S.movieYear}}, \\ \color{blue}{\text{S.starName}} \end{subarray} } \sigma_{ \begin{subarray}{l} \text{birthDate = 1960 } \land \\ \text{name = S.starName} \\ \end{subarray} } (\text{MovieStar} {\color{red}{\times \rho_S(\text{StarsIn}) }}) \big][/math]
  • note that we have $\rho_S(\text{StarsIn})$ on the both sides of the join
    • can just drop it (it won't affect the join)
    • [math]\big[ \rho_M(\text{Movie}) \big] \Join \big[ \pi_{ \begin{subarray}{l} \text{S.movieTitle}, \\ \text{S.movieYear}, \\ \text{S.starName} \end{subarray} } \sigma_{ \begin{subarray}{l} \text{birthDate = 1960 } \land \\ \text{name = S.starName} \\ \end{subarray} } (\text{MovieStar}) \big][/math]
  • finally we translate "WHERE" and "SELECT"
    • [math]\pi_{ \begin{subarray}{l} \text{S.movieTitle}, \\ \text{M.studioName} \end{subarray} } \sigma_{ \begin{subarray}{l} \text{S.movieYear $\geqslant$ 2000 } \land \\ \text{S.movieTitle = M.title} \\ \end{subarray} } \big[ \rho_M(\text{Movie}) \Join \pi_{ \begin{subarray}{l} \text{S.movieTitle}, \\ \text{S.movieYear}, \\ \text{S.starName} \end{subarray} } \sigma_{ \begin{subarray}{l} \text{birthDate = 1960 } \land \\ \text{name = S.starName} \\ \end{subarray} } (\text{MovieStar}) \big][/math]


NOT EXISTS in the Where Clause (by example)

SELECTS.movieTitle, M.studioName
FROM StarsIn S, Movie M
WHERE S.movieYear >= 2000
AND S.movieTitle = M.title
AND NOT EXISTS (
    SELECT name
    FROM MovieStar
    WHERE birthdate = 1960 AND name = S.starName)


Algorithm

  • Same as before: we translate the subquery
    • [math]\pi_\text{name} \sigma_{ \begin{subarray}{l} \text{birthDate = 1960 } \land \\ \text{name = S.starName} \\ \end{subarray} } (\text{MovieStar})[/math]
  • Then we add context relations and context parameters
    • [math]\pi_{ \begin{subarray}{l} \color{blue}{\text{S.movieTitle}}, \\ \color{blue}{\text{S.movieYear}}, \\ \color{blue}{\text{S.starName}}, \\ \text{name} \\ \end{subarray} } \sigma_{ \begin{subarray}{l} \text{birthDate = 1960 } \land \\ \text{name = S.starName} \\ \end{subarray} } (\text{MovieStar} {\color{red}{\times \rho_S(\text{StarsIn}) }})[/math]
  • And same for the FROM clause
    • $\rho_S(\text{StarsIn}) \times \rho_M(\text{Movie})$
  • Then we need to synchronize the results, but this time with Anti-Join ($\AntiJoin$)
    • [math]\big[ \rho_S(\text{StarsIn}) \times \rho_M(\text{Movie}) \big] \AntiJoin \big[ \pi_{ \begin{subarray}{l} \text{S.movieTitle}, \\ \text{S.movieYear}, \\ \text{S.starName} \end{subarray} } \sigma_{ \begin{subarray}{l} \text{birthDate = 1960 } \land \\ \text{name = S.starName} \\ \end{subarray} } (\text{MovieStar} \times \rho_S(\text{StarsIn}) ) \big][/math]
    • note that here the simplification is not possible: the semantics of Anti-Join is different from Join
    • so we cannot remove $\rho_S(\text{StarsIn})$ from both parts
  • the last step is the same: we translate "WHERE" and "SELECT"
    • [math]\pi_{ \begin{subarray}{l} \text{S.movieTitle}, \\ \text{M.studioName} \end{subarray} } \sigma_{ \begin{subarray}{l} \text{S.movieYear $\geqslant$ 2000 } \land \\ \text{S.movieTitle = M.title} \\ \end{subarray} } \bigg[ \big[ \rho_S(\text{StarsIn}) \times \rho_M(\text{Movie}) \big] \AntiJoin \big[ \pi_{ \begin{subarray}{l} \text{S.movieTitle}, \\ \text{S.movieYear}, \\ \text{S.starName} \end{subarray} } \sigma_{ \begin{subarray}{l} \text{birthDate = 1960 } \land \\ \text{name = S.starName} \\ \end{subarray} } (\text{MovieStar} \times \rho_S(\text{StarsIn}) ) \big] \bigg][/math]


EXISTS Subqueries in WHERE Combined with Other

So far we've considered only queries of the following form:

SELECT ... FROM ...
WHERE ... AND
      EXISTS (...) AND
      ... AND
      NOT EXISTS (...)

I.e. EXISTS and NOT EXISTS are in the "WHERE" clause joined by "AND"


What about the following query?

SELECT ... FROM ...
WHERE
    A = B AND NOT (EXISTS (...) AND C < 6)


SELECT ... FROM ...
WHERE
    (A = B AND NOT (EXISTS (...))) OR
    (A = B AND C >= 6)
  • Then we distribute OR (to UNION)
(SELECT ... FROM ...
  WHERE
    A = B AND NOT EXISTS (...))
UNION
(SELECT ... FROM ...
  WHERE
    A = B AND C >= 6)

As we've seen, UNION is translated as $\cup$


Union In Subqueries

We may have UNOIN in subqueries

SELECT S1.C, S2.C
FROM S S1, S S2
WHERE EXISTS (
  (SELECT R1.A, R1.B FROMR R1
   WHERE A = S1.C AND B = S2.C) -- (1)
  UNION
  (SELECT R2.A, R2.B FROMR R2
   WHERE B = S1.C) -- (2)
)
  • Recall that to be able to UNION two relations, they must have the same schema
  • But in this case:
    • (1) has 2 context relations $S_1$ and $S_2$
    • (2) has only 1 context relation $S_1$
  • $\Rightarrow$ When translating, need to add $S_2$ to (2) as well
  • and make sure that they have the same name

[math]\bigg( \underbrace{ \pi_{ \begin{subarray}{l} S_1.C, \ S_2.C, \\ R_1.A \ {\color{blue} \to \ A}, \\ R_1.B \ {\color{blue} \to \ B} \end{subarray} } \sigma_{ \begin{subarray}{l} A = S_1.C \ \land \\ B = S_2.C \\ \end{subarray} } \big[\rho_{R_1}(R) \times \rho_{S_1}(S) \times \rho_{S_2}(S) \big] }_{(1)} \bigg) \ {\color{blue} \cup } \ \bigg( \underbrace{ \pi_{ \begin{subarray}{l} S_1.C, \ S_2.C, \\ R_1.A \ {\color{blue} \to \ A}, \\ R_1.B \ {\color{blue} \to \ B} \end{subarray} } \sigma_{B = S_1.C} \big[\rho_{R_1}(R) \times \rho_{S_1}(S) {\color{blue} \times \rho_{S_2}(S) } \big] }_{(2)} \bigg)[/math]


Translating Joins

Joins

(SELECT * FROM R R1) JOIN (SELECT * FROM R R1) ON R1.A = R2.B

We translate as follows:

  • $\rho_{R_1}(R) \Join_{R_1.A = R_2.B} \rho_{R_2}(R)$


Group and Having

Suppose we have the following query:

SELECT name, SUM(length)
FROM MovieExec, Movie
WHERE cert = producer
GROUP BY name
HAVING MIN(year) < 1930

We translate it as

  • [math]\pi_{ \begin{subarray}{l} \text{name}, \\ \text{SUM(length)} \end{subarray} } {\color{blue} \sigma_{\text{MIN(year)} \lt 1930} \gamma_{ \begin{subarray}{l} \text{name}, \\ \text{MIN(year)}, \\ \text{SUM(length)} \end{subarray} } } \sigma_{\text{cert = producer}} (\text{MovieExec} \times \text{Movie})[/math]
  • here the translate the HAVING clause as $\sigma$ before the $\gamma$
  • also note that SUM(length) goes to $\gamma$


Exercises

Exercises from Database Systems Architecture (ULB)

  • the exercises: [1]
  • the proposed solutions [2]


Exercise 1

The given relations:

  • Student(snum, sname, major, level, age)
  • Class(name, meets_at, room, fid)
  • Enrolled(snum, cname)
  • Faculty(fid, fname, deptid)


SELECT C.name
FROM Class C
WHERE C.room = 'R128' OR 
      C.name IN (
          SELECT E.cname
          FROM Enrolled E
          GROUP BY E.cname
          HAVING COUNT(*) >= 5)

First we distribute OR

SELECT C.name
FROM Class C
WHERE C.room = 'R128'

UNION 

SELECT C.name
FROM Class C
WHERE C.name IN (
    SELECT E.cname
    FROM Enrolled E
    GROUP BY E.cname
    HAVING COUNT(*) >= 5)

for the subquery we replace IN to EXISTS

SELECT C.name
FROM Class C
WHERE EXISTS (
    SELECT E.cname
    FROM Enrolled E
    WHERE E.cname = C.name
    GROUP BY E.cname
    HAVING COUNT(*) >= 5)


Now we translate the subquery

  • [math]q_1 = \pi_{\text{E.name, C.*}} \sigma_{\text{cat} \geqslant 5} \gamma_{ \begin{subarray}{l} \text{E.cname}, \\ \text{count(*) $\to$ cnt}, \\ \text{C.*} \end{subarray} } \sigma_{\text{E.cname = C.name}} \big( \rho_E(\text{Enrolled}) \times \rho_C(\text{Class}) \big)[/math]
  • note that we use [math]\gamma_{ \begin{subarray}{l} \text{E.cname}, \\ \text{count(*) $\to$ cnt}, \\ \text{C.*} \end{subarray} }[/math] and not [math]\gamma_{ \begin{subarray}{l} \text{E.cname}, \\ \text{count(*) $\to$ cnt} \end{subarray} }[/math], because in the second case it will return only the two specified columns

Next, we need to synchronize (or "decorrelate") the subquery $q_1$ and the outer query

  • [math] \pi_{\text{C.name}} \Big[ \rho_C(\text{Class}) \Join \pi_{\text{C.*}} \pi_{\text{E.name, C.*}} \sigma_{\text{cat} \geqslant 5} \gamma_{ \begin{subarray}{l} \text{E.cname}, \\ \text{count(*) $\to$ cnt}, \\ \text{C.*} \end{subarray} } \sigma_{\text{E.cname = C.name}} \big( \rho_E(\text{Enrolled}) \times \rho_C(\text{Class}) \big) \Big][/math]
  • add $\pi_{\text{C.*}}$ because we need only these values - E.name was used for EXISTS part only
  • since we have $\rho_C(\text{Class})$ on both sides of the Join - we can drop the first one (as well as the Join)
  • and we also can merge successive projections
  • so we get:
  • [math]\pi_{\text{C.name}} \sigma_{\text{cat} \geqslant 5} \gamma_{ \begin{subarray}{l} \text{E.cname}, \\ \text{count(*) $\to$ cnt}, \\ \text{C.*} \end{subarray} } \sigma_{\text{E.cname = C.name}} \big( \rho_E(\text{Enrolled}) \times \rho_C(\text{Class}) \big)[/math]


Now we do the union (easy)

  • Since both parts have the same schema, union is possible
  • The total results is:
  • [math] \pi_\text{C.name} \sigma_\text{C.room = 'R128'} \rho_C(\text{Class}) \cup \pi_{\text{C.name}} \sigma_{\text{cat} \geqslant 5} \gamma_{ \begin{subarray}{l} \text{E.cname}, \\ \text{count(*) $\to$ cnt}, \\ \text{C.*} \end{subarray} } \sigma_{\text{E.cname = C.name}} \big( \rho_E(\text{Enrolled}) \times \rho_C(\text{Class}) \big)[/math]


Exercise with the Count Bug

SELECT F.fname
FROM Faculty F
WHERE 5 > (
    SELECT COUNT(E.snum)
    FROM Class C, Enrolled E
    WHERE C.name = E.cname AND
    C.fid = F.fid)

First translate to an equivalent EXISTS query

SELECT F.fname
FROM Faculty F
WHERE EXISTS (
    SELECT COUNT(E.snum) as CNT
    FROM Class C, Enrolled E
    WHERE C.name = E.cname AND
    C.fid = F.fid
    HAVING CNT < 5)

Remarks

  • note the change in the sign from > to <
  • also we use HAVING instead of WHERE - because GROUP is assumed
  • not all databases will take this kind of query.
    • For instance, MySQL will not (however it's not fully SQL compliant)

Using the rules, we try to translate the query this way:

  • first we translate the subquery
    • [math] \pi_{ \begin{subarray}{l} \text{cnt)}, \\ \text{F.fid}, \\ \text{F.fname}, \\ \text{F.deptid} \end{subarray} } \sigma_{\text{cnt} \lt 5} \gamma_{ \begin{subarray}{l} \text{count(E.snum) $\to$ cnt}, \\ \text{F.*} \end{subarray} } \sigma_{ \begin{subarray}{l} \text{C.name = E.cname } \land \\ \text{C.fid = F.fid} \end{subarray} } \Big[ \rho_C(\text{Class}) \times \rho_E(\text{Enrolled}) \times \rho_F(\text{Faculty}) \Big][/math]
  • then decorrelate it:
    • [math] \rho_F(\text{Faculty}) \Join \bigg( \pi_{\text{F.*}} \pi_{ \begin{subarray}{l} \text{cnt}, \\ \text{F.fid}, \\ \text{F.fname}, \\ \text{F.deptid} \end{subarray} } \sigma_{\text{cnt} \lt 5} \gamma_{ \begin{subarray}{l} \text{count(E.snum) $\to$ cnt}, \\ \text{F.*} \end{subarray} } \sigma_{ \begin{subarray}{l} \text{C.name = E.cname } \land \\ \text{C.fid = F.fid} \end{subarray} } \Big[ \rho_C(\text{Class}) \times \rho_E(\text{Enrolled}) \times \rho_F(\text{Faculty}) \Big] \bigg)[/math]
  • can remove $\rho_F(\text{Faculty})$ and keep only needed projection attributes
    • [math]\pi_{\text{F.name}} \sigma_{\text{cnt} \lt 5} \gamma_{ \begin{subarray}{l} \text{count(E.snum) $\to$ cnt}, \\ \text{F.*} \end{subarray} } \sigma_{ \begin{subarray}{l} \text{C.name = E.cname } \land \\ \text{C.fid = F.fid} \end{subarray} } \Big[ \rho_C(\text{Class}) \times \rho_E(\text{Enrolled}) \times \rho_F(\text{Faculty}) \Big][/math]

Note that this is not the query we want!!!

  • Faculty members who don't teach any class are not output by the expression, but they are output by the original SQL query


Count bug

  • this issue is known as the count bug
  • it occurs when we have subqueries use COUNT without GROUP BY
  • to solve it we need to use right outer join instead of $\times$

[math]\pi_{\text{F.name}} \sigma_{\text{cnt} \lt 5} \gamma_{ \begin{subarray}{l} \text{count(E.snum) $\to$ cnt}, \\ \text{F.*} \end{subarray} } \sigma_{ \begin{subarray}{l} \text{C.name = E.cname } \land \\ \text{C.fid = F.fid} \end{subarray} } \Big[ \rho_C(\text{Class}) \times \rho_E(\text{Enrolled}) \Join^{R}_\text{C.fid = F.fid} \rho_F(\text{Faculty}) \Big][/math]


See also

Sources

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