Decomposition
Liangsheng Deng
Section 2
11/15/2005
Recall Functional Dependencies
 A->B holds iff for any pairs of tuples t1 and t2
such that if t1[A] = t2[A], then t1[B] = t2[B].
 Armstong’s axioms:



Reflexivity rule. If a is a set of attributes and b is a
subset of a, then a->b holds.
Augmentation rule. If a->b holds and c is a set of
attributes, then ac->bc holds.
Transitivity rule. If a->b holds and b->c holds, then
a->c holds.
Closure of a Set of FD
 Given f, a set of FD of a relational
schema, we can use Armstrong axioms to
derive other possible FD implied by f.
The closure of f is the set of all FD implied
by f.
 Determine whether S, a set of attributes,
is a superkey by computing the closure of
S.
Closure Computing Example
 Given R(A, B, C, D, E, H) with
FD1. BD->A, FD2. AC->H, FD3. D->C,
FD4. E->D.
Is BE a superkey of R?
{BE}+ = {BDE} because E->D
{BE}+ = {BCDE} because D->C
{BE}+ = {ABCDE} because BD->A
{BE}+ = {ABCDEH} because AC->H
BE is a superkey of R.
Lossless Decomposition
The purpose of using lossless decomposition:
 Reduce unnecessary redundancy.
 Retrieve information efficiently
How do we decide a decomposition is a
lossless decomposition?
Let R be a relation schema.
Let F be a set of functional dependencies on R.
Let R1 and R2 form a decomposition of R.
The decomposition is a lossless-join decomposition of R if at
least one of the following functional dependencies holds.
R1 ∩ R2 -> R1 or
R2 ∩ R1 -> R2
Lossless Decomposition Testing (1)
 Given R1(A, B), R2(A, C, D) and
F = {A->C; D->B; AC->B}.
Is the join of R1 and R2 lossless?
 R1 ∩ R2 = {A}
If we can prove A->B, then the answer is yes.
Proof:
A->AC because A->C
A->B because A->AC and AC->B
So, it is a lossless cecomposition.
Lossless Decomposition Testing (2)
 To test whether R1 ∩ R2 -> R1 or R2, we can
test whether R1 ∩ R2 is a superkey of
R1 or R2 by computing the closure of R1 ∩ R2.
 Same question in the previous slide.
 {R1 ∩ R2}+ = {A}+,
{A}+ = {AC} because A->C
{A}+ = {ABC} because AC->B
Conclusion: A is a superkey of R1.
Therefore, it is a lossless decomposition.
Lossless Decomposition Testing (3)
 For the case of decomposition of a schema into
more than 2 schemas, we can use the chase
matrix algorithm.
 For example, F = {A->C, B->C, C->D,
DE->C, CE->A}, R1(AD), R2(AB), R3(BE),
R4(CDE) and R5(AE).
R1(AD)
R2(AB)
R3(BE)
R4(CDE)
R5(AE)
A
a1
a1
b31
b41
a1
B
b12
a2
a2
b42
b52
C
b13
b23
b33
a3
b53
D
a4
b24
b34
a4
b54
E
b15
b25
a5
a5
a5
Lossless Decomposition Testing (3)
Cont.
A->C, we get
R1(AD)
R2(AB)
R3(BE)
R4(CDE)
R5(AE)
A
a1
a1
b31
b41
a1
B
b12
a2
a2
b42
b52
C
b13
b13
b33
a3
b13
D
a4
b24
b34
a4
b54
E
b15
b25
a5
a5
a5
A
B
C
D
E
a1
a1
b31
b41
a1
b12
a2
a2
b42
b52
b13
b13
b13
a3
b13
a4
b24
b34
a4
b54
b15
b25
a5
a5
a5
B->C, we get
R1(AD)
R2(AB)
R3(BE)
R4(CDE)
R5(AE)
Lossless Decomposition Testing (3)
Cont.
C->D, we get
R1(AD)
R2(AB)
R3(BE)
R4(CDE)
R5(AE)
A
B
C
D
E
a1
a1
b31
b41
a1
b12
a2
a2
b42
b52
b13
b13
b13
a3
b13
a4
a4
a4
a4
a4
b15
b25
a5
a5
a5
A
a1
a1
b31
b41
a1
B
b12
a2
a2
b42
b52
C
b13
b13
a3
a3
a3
D
a4
a4
a4
a4
a4
E
b15
b25
a5
a5
a5
DE->C
R1(AD)
R2(AB)
R3(BE)
R4(CDE)
R5(AE)
Lossless Decomposition Testing (3)
Cont.
CE->A
R1(AD)
R2(AB)
R3(BE)
R4(CDE)
R5(AE)
A
a1
a1
a1
a1
a1
B
b12
a2
a2
b42
b52
C
b13
b23
a3
a3
a3
D
a4
a4
a4
a4
a4
E
b15
b25
a5
a5
a5
The third tuple has a1, a2, a3 ,a4 and a5, so
it is a lossless decomposition.
Dependency Preservation




Let F be a set of functional dependencies on
schema R.
Let {R1,R2,…Rn} be a decomposition of R.
The restriction of F to Ri is the set of all
functional dependencies in F+ that include
only attributes of Ri.
Functional dependencies in a restriction can
be tested in one relation, as they involve
attributes in one relation schema.
Dependency Preservation Cont.
 The set of restrictions F1,F2,…Fn is the
set of dependencies that can be checked
efficiently.
 We need to know whether testing only the
restrictions is sufficient.
 Let F’ = F1 U F2 U….Fn.
 F’ is a set of functional dependencies on
schema R, but in general, F’ != F.
Dependency Preservation Cont.
 A decomposition having the property that
is a dependency-preserving decomposition.
Dependency Preservation Cont.
An Easier Way To Test For Dependency
Preservation.
Rather than compute F+ and F’+ , and see whether they
are equal, we can do this:
Find F – F’, the functional dependencies not checkable
in one relation.
See whether this set is obtainable from F' by using
Armstrong's Axioms.
This should take a great deal less work, as we have
(usually) just a few functional dependencies to work on.
Dependency Preservation Cont.
 For example, given F={AB->C, C->A},
R1(AC) and R2(BC).
 F’ = {C->A}.
F – F’ = {AB->C}.
AB->C can not be derived from F’.
So, it is not a dependency-preserving
decomposition.
Reference
http://www.cs.sfu.ca/CC/354/zaiane/materi
al/notes/Chapter7/node1.html