Dependency preserving

Dependency Preserving
Decomposition

DAVID DENG

CS157B
MARCH 23, 2010

Intro

Decomposition help us eliminate redundancy, root
of data anomalies.

Decomposition should be:
 1. Lossless
 2. Dependency Preserving

What’s Dependency Preservation?

When the decomposition of a relational scheme
preserved the associated set of functional
dependencies.

Formal Definition:
If R is decomposed into R1, R2,…, Rn, then
{F1∪F2∪…∪Fn}+ = F+

Algorithm to check for Dependency Preservation

begin;
for each X  Y in F and with R (R1, R2, …, Rn)
{
let Z = X;
while there are changes in Z
{
from i=1 to n
Z = Z ∪ ((Z ∩ Ri)+ ∩ Ri) w.r.t to F;
}
if Y is a proper subset of Z, current fd is preserved
else decomposition is not dependency preserving;
}
this is a dependency preserving decomposition;
end;

Explain Algorithm Part 1

1. Choose a functional dependency in set F, say you choose
X  Y.
2. Let set Z to the “left hand side” of the functional
dependency, X such Z = X

Starting with R1 in the decomposed set {R1, R2,…Rn)
3. Intersect Z with R1, Z ∩ R1
4. Find the closure of the result from step 3 (Z ∩ R1) using
original set F
5. Intersect the result from step 4 ((Z ∩ R1)+) with R1 again.

Explain Algorithm Part 2

6. Updated Z with new attribute in the result from step 5.
7. Repeat step 3-6 from R2, R3, …, Rn.
8. If there’s any changes between original Z before step 3
and after step 7, repeat step 3-7.

9. Check whether Y is a proper subset of current Z. If it is
not, this decomposition is a violation of dependency
preservation. You can stop now.
10. If Y is a proper subset of current Z, repeat 1-9 until you
check ALL functional dependencies in set F.

Another look at Algorithm

Test each X  Y in F for dependency preservation

result = X
while (changes to result) do
for each Ri in decomposition
t = (result ∩ Ri)+ ∩ Ri
result = result ∪ t
if Y ⊆ result, return true;
else, return false;

[Note: If any false is returned for algorithm, whole
decomposition is not dependency preserving.]

Let’s walk through an example
of using this algorithm.

Example using Algorithm

Given the following:
R(A,B,C,D,E)
F = {ABC, CE, BD, EA}
R1(B,C,D) R2(A,C,E)

Is this decomposition dependency preserving?

Example

R(A,B,C,D,E) F = {ABC, CE, BD, EA}
Decomposition: R1(B,C,D) R2(A,C,E)

Z=AB
For Z ∩ R1 = AB ∩ BCD = B
{B}+ = BD
{B}+ ∩ R1 = BD ∩ BCD = BD
Update Z = AB ∪ BD = ABD, continue

Example


Z=ABD
For Z ∩ R2 = ABD ∩ ACE = A
{A}+ = A
{A}+ ∩ R2 = A ∩ ACE = A
Update Z, Z is still ABD
Since Z changed, repeat checking R1 to R2.

Example


Z=ABD
For Z ∩ R1 = ABD ∩ BCD = BD
{BD}+ = BD
{BD}+ ∩ R1 = BD ∩ BCD = BD
Update Z = ABD ∪ BD = ABD, so Z hasn’t
changed but you still have to continue.

Example


Z=ABD and checking R2 was done 2 slides ago
Z will still be ABD.

Since Z hasn’t change, you can conclude ABC
is not preserved.

Let’s practice with other functional
dependencies.

Example


Z=X=B
For Z ∩ R1 = B ∩ BCD = B
{B}+ = BD
{B}+ ∩ R1 = BD ∩ BCD = BD
Update Z = B ∪ BD = BD
Since Y=D is proper subset of BD, BD is
preserved.

Example


Z=X=C
For Z ∩ R2 = C ∩ ACE = C
{C}+ = CEA
{C}+ ∩ R1 = CEA ∩ ACE = ACE
Update Z = C ∪ACE= ACE
Since Y=E is proper subset of ACE, CE is
preserved.

Example


Z=X=E
For Z ∩ R1 = E ∩ ACE = E
{E}+ = EA
{E}+ ∩ R1 = EA ∩ ACE = EA
Update Z = E ∪ EA= EA
Since Y=A is proper subset of EA, EA is
preserved.

Example


Shortcut:
For any functional dependency, if both LHS
and RHS collectively are within any of the
sub scheme Ri. Then this functional
dependency is preserved.

Exercise #1

Let R{A,B,C,D}
and
F={AB, BC, CD, DA}

Let’s decomposed R into
R1 = AB, R2 = BC, and R3 = CD

Is this a dependency preserving decomposition?

Answer to Exercise #1

R{A,B,C,D} F={AB, BC, CD, DA}
Decomposition: R1 = AB, R2 = BC, and R3 = CD

Yes it is.
You can immediately see that AB, BC, CD are
preserved for R1, R2, R3
The key is to check whether DA is preserved.
Let’s walk through the algorithm.



Z=X=D
For Z ∩ R1 = D ∩ AB = empty set
For Z ∩ R2 = D ∩ BC = empty set
For Z ∩ R3 = D ∩ CD = D
Find {D}+ = DABC
Find {D}+ ∩ R3 = DABC ∩ CD = CD
Update Z to CD. Since Z changed, repeat.



Z = CD
For Z ∩ R1 = CD ∩ AB = empty set
For Z ∩ R2 = CD ∩ BC = C
Find {C}+ = CDAB
Find {C}+ ∩ R2 = CDAB ∩ BC = BC
Update Z = CD ∪ BC = BCD



Z = BCD
For Z ∩ R3 = BCD ∩ CD = CD
Find {CD}+ = CDAB
Find {CD}+ ∩ R3 = CDAB ∩ CD = CD
Update Z is still BCD. Since Z changed, repeat going
trough R1 to R3.



Z = BCD
For Z ∩ R1 = BCD ∩ AB = B
Find {B}+ = BCDA
Find {B}+ ∩ R1 = BCDA ∩ AB = AB
Update Z = BCD ∪ AB = ABCD.
Since Y = A is a subset of ABCD, function DA is
preserved.

Exercise #2

R{A,B,C,D,E)
F={ABD, BE}

Decomposition:
R1{A,B,C} R2{A,D} R3{B,D,E}

Is this a dependency preserving decomposition?


R{A,B,C,D,E) F={ABD, BE}
Decomposition: R1{A,B,C} R2{A,D} R3{B,D,E}

Let’s start with ABD:
Z=A
Z ∩ R1 = A ∩ ABC = A
{A}+ = ABDE
{A}+ ∩ R1 = ABDE ∩ ABC = AB
Update Z = A ∪ AB = AB



Z = AB
Z ∩ R2 = A ∩ AD = A
{A}+ = ABDE
{A}+ ∩ R1 = ABDE ∩ AD = AD
Update Z = AB ∪ AD = ABD
Thus A BD preserved



Based on R3, BE is preserved.
Check B  E:
Z=B
Z ∩ R1 = B ∩ ABC = B
{B}+ = BE
{B}+ ∩ R1 = BE ∩ ABC = B
Update Z = B still the same



Z=B
Z ∩ R2 = B ∩ AD = empty set
Z ∩ R3 = B ∩ BDE = B
{B}+ = BE
{B}+ ∩ R3 = BE ∩ BDE = BE
Update Z = B ∪ BE = BE
Thus BE preserved

End

Reference:
 Yu Hung Chen, “Decomposition”,
http://www.cs.sjsu.edu/faculty/lee/cs157/Decomposition_Yu
HungChen.ppt, SJSU (lol), 2005
 Gary D. Boetticher, “Preserving Dependencies”,
http://www.youtube.com/watch?v=olgwucI3thI, University
of Houston Clear Lake, 2009
 Dr. C. C. Chan, “Example of Dependency Preserving
Decomposition”,
http://www.cs.uakron.edu/~chan/cs475/Fall2000/Example
Dp%20decomposition.htm, University of Akron, Fall 2000
 Tang Nan, “Functional Dependency”,
http://www.se.cuhk.edu.hk/~seg3550/2006/tutorial/t9.ppt

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