Dr. Alexandra I. Cristea CS 319: Theory of Databases: C7

… previous TC, DC

Equivalence of queries RA TC DC Datalog SQL

Equivalence of TC and RA For every relational algebra (RA) query there is an equivalent tupel calculus (TC) query (and v.v.). Translation RA  TC is fairly straightforward and can be used to obtain TC queries from RA queries. Translation TC  RA is very artificial and does not result in readable RA queries. We will show the RA to TC translation.

Translation of RA to TC Step 1: Reduce the algebra expression E to the basic operators Step 2: If E is a single relation r then E translates to { t | t  r }. If r has attributes B 1,...,B n then it can also be written as {t |  s  r ( t[B 1 ]=s[B 1 ] ...  t[B n ]=s[B n ] ) }.

Translation of RA to TC Step 3: Apply the following substeps recursively (if the algebra expressions are base relations an  may need to be added): –Renaming: let { t | f(t) } be a TC expression equivalent to an RA expression E that uses attributes B 1,...,B n : Let E 1 =  x(A1,...,An) (E), then the translation to TC is { s |  t  r (f(t)  s[A 1 ]=t[B 1 ] ...  s[A n ]=t[B n ]) }

Translation of RA to TC –Cartesian Product: if E 1 (over A 1,...,A n ) translates to { t | f(t) } and E 2 over B 1,...,B m translates to { s | g(s) } then E 1  E 2 translates to { r | f(t)  g(s)  t[A 1 ]=r[A 1 ]  t[A n ]=r[A n ]  s[B 1 ]=r[B 1 ]  s[B m ]=r[B m ] )) } (attribute names must be unique in t)

Translation of RA to TC –Selection: let { t | f(t) } be a TC expression equivalent to E, then E 1 =  A  B (E) or E 1 =  A  c (E) (where  is , , ,...) is translated to { t | f(t)  t[A]  t[B] } and { t | f(t)  t[A]  c } resp. (replace t[.] with basis tuples from quantificators) –Projection: if E translates to { t | f(t) } then  B1,...,Bn (E) translates to { t | f'(t) } where f' is f with references to t[B 1 ]... t[B n ] only (no other attributes of t are used in f).

Translation of RA to TC –Union: let { t | f(t) } be a TC expression equivalent to E 1 and { t | g(t) } be a TC expression equivalent to E 2 then E 1  E 2 is translated to { t | f(t)  g(t) } –Difference: let { t | f(t) } be a TC expression equivalent to E 1 and { t | g(t) } be a TC expression equivalent to E 2 then E 1  E 2 is translated to { t | f(t)  g(t) }

Example List all drinkers who visit a bar that serves a beer they like. RA:  V.d (V  S  L) –Step 1:  V.d (  V.d, V.k, S.b (  V.k=S.k  S.b=L.b  V.d=L.d (V  S  L))) –Step 2: The relations are translated to { v | v  V } { s | s  S } { l | l  L } –Step 3: We first translate the product, then the selection and then the projection.

Example (cont.) –Cartesian product: { t |  v  V (  s  S (  l  L (  t[vd]=v[d]  t[vk]=v[k]  t[sk]=s[k]  t[sb]=s[b]  t[ld]=l[d]  t[lb]=l[b] ))) } (attribute names must be unique) –Selection: { t |  v  V (  s  S (  l  L (  t[vd]=v[d]  t[vk]= v[k]  t[sk]=s[k]  t[sb]=s[b]  t[ld]=l[d]  t[lb]=l[b]  v[k]=s[k]  s[b]=l[b]  v[d]=l[d] ))) }

Example (cont.) –Projection: { t |  v  V (  s  S (  l  L (  t[vd]=v[d]  v[k]=s[k]  s[b]=l[b]  v[d]=l[d] ) ) ) }

Summary We have learned query equivalence

… to follow Exam discussion

Beer question n (tupel calculus) Give all beers that are served in bars where none of the visitors like any other beer (than the ones served in that bar). Is this a correct answer?

A faulty answer Consider Or after a rewrite Suppose there are 2 beers a, b, 1 bar k, 2 drinkers x, y; everyone serves/visits/likes everything Then the correct answer should result in {a, b}. The tupel calculus expression does not contain a, because every visitor of k also likes b.

Beer question n (RA) Give all beers that are served in bars where none of the visitors like any other beer (than the ones served in that bar).  b (S  (  k (S) -  k ( V  L - V  L  S))) This is not so easy to translate to the earlier TC expression

Common tupel calculus mistakes Explain what is wrong in the following queries: Translate L&V quantor as and not as