of Decomposition relational schemes
Desirable properties of decompositions Dependency preserving decompositions Lossless join decompositions
05/15/09
1
CS319 Theory of Databases
Desirable properties of decompositions 1
Lossless decompositions A decomposition of the relation scheme R into subschemes R1, R2, ..., Rn is lossless if, given tuples r1, r2, ..., rn in R1, R2, ..., Rn respectively, such that ri and rj agree on all common attributes for all pairs of indices (i,j), the - uniquely defined - tuple derived by joining r1, r2, ..., rn is in R. Terminology: "lossless join" decomposition
05/15/09 2 CS319 Theory of Databases
Desirable properties of decompositions 3
Dependency preserving decompositions A decomposition of the relation scheme R into subschemes R1, R2, ..., Rn is dependency preserving if all the FDs within R can be derived from those within the relations R1, R2, ..., Rn. If F is the set of dependencies defined on R, then the requirement is that the set G of dependencies that can be obtained as projections of dependencies in F+ onto R1, R2, ..., Rn together generate F+. Note carefully that it is not enough to check whether projections of dependencies in F onto R1, R2, ..., Rn together generate F+. 3 05/15/09 CS319 Theory of Databases
Desirable properties of decompositions 4
An illustrative example
SCAIP1
Replace SADDRESS by CITY and AGENT fields in SUPPLIERS(SNAME, SADDRESS, ITEM, PRICE) Semantics: Each supplier is based in a city, and the enterprise responsible for setting up the database has an agent for each city. Derive in this way a new relation SCAIP(S, C, A, I, P) where S is SNAME, C is CITY, A is AGENT etc.
05/15/09 4 CS319 Theory of Databases
Desirable properties of decompositions 5
An illustrative example
SCAIP2
Derive in this way a relation SCAIP(S, C, A, I, P) where S is SNAME, C is CITY, A is AGENT etc.
The set F of functional dependencies is generated by: S C, C A, S I P ... each supplier sited in one city ... each city has one agent serving it ... each supplier sells each given item at fixed price
05/15/09 5 CS319 Theory of Databases
Desirable properties of decompositions 6
An illustrative example
SCAIP3 F { S C, C A, S I P }
Consider decomposition { SCA, SIP }: This is lossless: Suppose that the tuples tSCA and tSIP are in the relations SCA and SIP respectively. If tSCA and tSIP agree on S, then their join is a tuple tSCAIP (s,c,a,i,p), where c and a are determined by the attribute s and the i and p attributes are such that p is determined by s and i. Any tuple that satisfies these two FDs is in the relation SCAIP.
05/15/09 6 CS319 Theory of Databases
Desirable properties of decompositions 7
An illustrative example Consider decomposition { SCA, SIP }: Also dependency preserving: the sets of dependencies { S C, C A } and { S I P } are included in the projections of F+ onto SCA and SIP. This means that the FDs in F, from which all dependencies are generated, are explicit in the sub-schemes SCA and SIP in this case.
05/15/09 7 CS319 Theory of Databases
SCAIP4 F { S C, C A, S I P }
Desirable properties of decompositions 8
An illustrative example
SCAIP5 F { S C, C A, S I P }
In decomposition {SIP, SCA}, have problems with SCA. E.g. update anomaly if want to store an agent for a city in which no supplier is currently located Get around this by decomposing SCA further: decompose as {SC, CA} decompose as {SC, SA} decompose as {CA, SA}
05/15/09 8 CS319 Theory of Databases
Desirable properties of decompositions 9
An illustrative example decompose as { SC, CA } this is both lossless join and dependency preserving decompose as { SC, SA } In this case the images of the FDs in F+ on SC and SA are {S C} and {S A} respectively, but the dependency C A can't be inferred. So this decomposition is not dependency preserving.
05/15/09 9 CS319 Theory of Databases
SCAIP6 F { S C, C A, S I P }
Desirable properties of decompositions 10
An illustrative example decompose as {CA, SA} In this case, have possibility that Fred is agent for Hull and York, and PVC based in Hull. Then: (Hull, Fred) * (PVC, Fred) = (PVC, Hull, Fred) (York, Fred) * (PVC, Fred) = (PVC, York, Fred) The second join is not in the relation SCA. So this decomposition is not lossless join.
05/15/09 10 CS319 Theory of Databases
SCAIP7 F { S C, C A, S I P }
Dependency Preserving Decompositions 1
Let R be a relation scheme, a decomposition of R and F a set of functional dependencies of R. If Z is a set of attributes in R, then Z(F) = { X Y F | XY Z } The decomposition is dependency preserving if F is logically implied by the union of the sets of functional dependencies T(F+), where T ranges over all sub-schemes of .
05/15/09 11 CS319 Theory of Databases
Dependency Preserving Decompositions 2
Illustrative Example R = ABCD and = {AB, BC, CD} F = { A B, B C, C D, D A } Question: is dependency preserving? Certainly { A B, B C, C D } are captured. How about D A? Is also, because F+ { B A, C B, D C } and these FDs are recorded in the sub-schemes AB, BC, CD. Hence the dependency D A is also captured.
05/15/09 12 CS319 Theory of Databases
Dependency Preserving Decompositions 3
Algorithm to check dependency preserving OK := true for each dependency X Y in F do begin Z := X while changes occur in Z do for each sub-scheme T of do Z := Z { A | Z T A is in T(F+) } if not Z Y then OK := false end
05/15/09 13 CS319 Theory of Databases
Dependency Preserving Decompositions 4
Algorithm to check dependency preserving
... while changes occur in Z do for each sub-scheme T of do Z := Z { A | Z T A is in T(F+) } ...
To compute { A | Z T A is in T(F+) } calculate ((Z T)+ T) where the closure (Z T)+ is computed with respect to F over the entire relation scheme R. This avoids need to compute F+.
05/15/09 14 CS319 Theory of Databases
Dependency Preserving Decompositions 5
Illustrating the algorithm in action Consider the relation scheme R = ABCD, the dependencies F = { A B, B C, C D, D A }, and the decomposition = { AB, BC, CD } Clear that A B, B C and C D are preserved ... can prove that the dependency D A is preserved by applying the algorithm Computation of {D}+ over R using F yields {A,B,C,D}
05/15/09 15 CS319 Theory of Databases
Dependency Preserving Decompositions 6
Illustrating the algorithm in action (cont.)
Computation of {D}+ over R using F yields {A,B,C,D}
Z={D} initially. At each iteration of the while-loop, the algorithm introduces a new attribute into Z. For instance, on the first pass, introduce C when T = CD, on second pass, then introduce B when T = BC etc. Hence: Z0 = {D}, Z1 = {C,D}, Z2 = {B,C,D}, Z3 = {A,B,C,D} where Zi is the value of Z after the ith iteration. This proves that dependency D A is preserved.
05/15/09 16 CS319 Theory of Databases
Lossless Join Decompositions 1
Lossless join decomposition Let R be a relation scheme, a decomposition of R and F a set of functional dependencies of R. Suppose that the sub-schemes in are {R1, R2, ... , Rk}. has lossless join if every extensional part r for R that satisfies F is such that r = 1(r) 2(r) ... k(r), where i(r) denotes the projection of r onto Ri. Informally: r is the natural join of its projections onto the sub-schemes R1, R2, ... , Rk.
05/15/09 17 CS319 Theory of Databases
Lossless Join Decompositions 2
Examples (revisited as a reminder) SCAIP = SIP SCA = SIP SC CA SCA SA CA and SCA SA CA lossless lossy
... have possibility that Fred is agent for Hull and York, and that PVC is a supplier based in Hull. Then: (Hull, Fred) * (PVC, Fred) = (PVC, Hull, Fred) (York, Fred) * (PVC, Fred) = (PVC, York, Fred) The second join is not in the relation SCA. So this decomposition is not lossless join.
05/15/09 18 CS319 Theory of Databases
Lossless Join Decompositions 3
Principles of lossless join decomposition Let = { R1, R2, ... , Rk } be a decomposition of R. Define the mapping m ( ) on possible extensions for the relation scheme R [whether or not they satisfy the functional dependencies in R, if there are any], via: m (r) = 1(r) 2(r) ... k(r), where i(r) denotes the projection of r on sub-scheme Ri. Notation: use to ri denote i(r), for 1 i k.
05/15/09 19 CS319 Theory of Databases
Lossless Join Decompositions 4
Principles of lossless join decomposition (cont.) Lemma: With R, and ri as above a) r m (r) b) if s = m (r), then i(s) = ri c) m (m (r)) = m (r) The condition on m () specified in part c) identifies it as a closure operation. Cf. closure of an interval of real numbers e.g. 1 < 05/15/09 2 20 CS319 Theory of Databases
Lossless Join Decompositions 5
Proof of lemma a) let t r. Then i(t) ri showing that t 1(r) 2(r) ... k(r) = m (r) b) by part a) r m (r)=s, so that i(s) ri. But if t s, then projection of t onto sub-scheme Ri is in ri by definition of natural join, so that i(s) ri also. c) m (m (r)) = m (s) by definition of s = 1(s) 2(s) ... k(s) = 1(r) 2(r) ... k(r) = m (r) using definition of m and part b).
05/15/09 21 CS319 Theory of Databases
Lossless Join Decompositions 6
Testing for lossless join decomposition assuming all data dependencies in R to be functional Input: A relation scheme R=A1A2 ... An, a set of functional dependencies F, and a decomposition = { R1, R2, ... , Rk } Output: is or is not a lossless join decomposition Construct table of 's and 's, and repeatedly transform the rows by taking account of the FDs until either one row is all 's or no further transformation is possible ...
05/15/09 22 CS319 Theory of Databases
Lossless Join Decompositions 7
Testing for lossless join decomposition (cont.) Construct table of 's and 's, and repeatedly transform the rows by taking account of the FDs until either one row is all 's or no further transformation is possible ... Principle of algorithm: devise a symbolic representation for tuples s1, s2, ... , sk from R1, R2, ... , Rk respectively that are joinable, and for tuples t1, t2, ... , tk in R so that si is projection of ti onto Ri for each i. Impose all those
05/15/09 23 CS319 Theory of conditions on t1, t2, ... , tk that follow from the FDsDatabases in F.
Lossless Join Decompositions 8
Testing for lossless join decomposition (cont.) Construct table of 's and 's, and repeatedly transform the rows by taking account of the FDs until either one row is all 's or no further transformation is possible ... Principle of algorithm: devise a symbolic representation for tuples s1, s2, ... , sk from R1, R2, ... , Rk respectively that are joinable, and for tuples t1, t2, ... , tk in R so that si is projection of ti onto Ri for each i. Impose all those
05/15/09 24 CS319 Theory of Databases
Lossless Join Decompositions 9
Method of testing for lossless join decomposition 1. Construct a table with n columns (corresponding to attributes) with k rows (corresponding to sub-schemes) Initialize the table at row i column j by entering j if attribute Aj appears in Ri and by entering ij otherwise NB 's represent joinable tuples, padded out to R by 's
05/15/09 25 CS319 Theory of Databases
Lossless Join Decompositions 10
Method of testing for lossless join decomposition (cont.) 2. Repeatedly modify the table to take account of all dependencies until no further updates occur i.e. if X Y and two rows agree on all the attributes in X then modify them so that they also agree on all attributes in Y. Explicitly, change attributes in Y thus: if one symbol is an i make the other an i if both symbols are of form *j make both ij or i'j arbitrarily. On termination declare lossless join if and only if one of the rows is 1 2 ... n.
05/15/09 26 CS319 Theory of Databases
Lossless Join Decompositions 11
Illustrative example Verify the decomposition SCAIP = SIP SC CA is a lossless join .... Initial table
S C A I P
SIP 1 12 13 4 5 SC 1 2 23 24 25 CA 31 2 3 34 35
Functional dependencies are S C, C A, S I P
05/15/09 27 CS319 Theory of Databases
Lossless Join Decompositions 12
Illustrative example Functional dependencies are S C, C A, S I P and from these arrive via stage 2 of algorithm at table:
S C A I P SIP 1 2 3 4 5 SC 1 2 3 24 25 CA 31 2 3 34 35
at which point no further dependencies apply. Row 1 shows that the result is lossless
05/15/09 28 CS319 Theory of Databases
Lossless Join Decompositions 13
Principle of the lossless join algorithm illustrated ... Consider the example: in the initial table
S C A I P SIP 1 12 13 4 5 SC 1 2 23 24 25 CA 31 2 3 34 35
... the rows can be seen as representing generic tuples from SIP, SC and CA that are joinable (i.e. agree on all common attributes). The join of these three tuples will necessarily be 1 23 45.
05/15/09 29 CS319 Theory of Databases
Lossless Join Decompositions 14
Principle of the lossless join algorithm illustrated ... Key question: are the functional dependencies enough to ensure that 1 23 4 5 is itself a tuple in the relation SCAIP? After modification to take account of all FDs, suitable tuples matching the template for equality of values in the 3 rows in the table define a valid extensional part for SCAIP: c...