Download - Well-designed XML Data
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Well-designed XML Data
Marcelo Arenas and Leonid Libkin
University of Toronto
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Outline
Part 1 - Database Normalization from the 1970s and 1980s
Part 2 - Classical theory revisited: normalizing XML documents
Part 3 - Classical theory re-done: new justifications for normalization
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Part 1: Classical Normalization
Design: decide how to represent the information in a particular data model.
• Even for simple application domains there is a large number of ways of representing the data of interest.
We have to design the schema of the database.
• Set of relations.
• Set of attributes for each relation.
• Set of data dependencies.
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Designing a Database: An Example
Attributes: number, title, section, room.
Data dependency: every course number is associated with only one title.
Relational Schema:
R(number, title, section, room),
number title
GOOD alternative:
S(number, title), number title
T(number, section, room),
BAD alternative:
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Problems with BAD: Update Anomaly
number title section room
CSC258 Computer Organization 1 LP266
CSC258 Computer Organization 2 GB258
CSC258 Computer Organization 3 GB248
CSC434 Database Systems 1 GB248
Title of CSC258 is changed to Computer Organization I.
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Problems with BAD: Update Anomaly
number title section room
CSC258 Computer Organization 1 LP266
CSC258 Computer Organization 2 GB258
CSC258 Computer Organization 3 GB248
CSC434 Database Systems 1 GB248
Title of CSC258 is changed to Computer Organization I.
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Problems with BAD: Update Anomaly
number title section room
CSC258 Computer Organization I
1 LP266
CSC258 Computer Organization I
2 GB258
CSC258 Computer Organization I
3 GB248
CSC434 Database Systems 1 GB248Title of CSC258 is changed to Computer Organization I.The instance stores redundant information.
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Deletion Anomaly
number title section room
CSC258 Computer Organization I
1 LP266
CSC258 Computer Organization I
2 GB258
CSC258 Computer Organization I
3 GB248
CSC434 Database Systems 1 GB248CSC434 is not given in this term.
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Deletion Anomaly
number title section room
CSC258 Computer Organization I
1 LP266
CSC258 Computer Organization I
2 GB258
CSC258 Computer Organization I
3 GB248
CSC434 Database Systems 1 GB248CSC434 is not given in this term.
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Deletion Anomaly
number title section room
CSC258 Computer Organization I
1 LP266
CSC258 Computer Organization I
2 GB258
CSC258 Computer Organization I
3 GB248
CSC434 is not given in this term.
Additional effect: all the information about CSC434 was deleted.
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Insertion Anomaly
number title section room
CSC258 Computer Organization I
1 LP266
CSC258 Computer Organization I
2 GB258
CSC258 Computer Organization I
3 GB248
A new course is created: (CSC336, Numerical Methods)
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Insertion Anomaly
number title section room
CSC258 Computer Organization I
1 LP266
CSC258 Computer Organization I
2 GB258
CSC258 Computer Organization I
3 GB248
A new course is created: (CSC336, Numerical Methods)
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Insertion Anomaly
number title section room
CSC258 Computer Organization I
1 LP266
CSC258 Computer Organization I
2 GB258
CSC258 Computer Organization I
3 GB248
CSC336 Numerical Methods ? ?A new course is created: (CSC336, Numerical Methods)The instance stores attributes that are not directly related.
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Avoiding Update Anomalies
number
title
CSC258
Computer Organization
CSC434
Database Systems
number
section room
CSC258
1 LP266
CSC258
2 GB258
CSC258
3 GB248
CSC434
1 GB248Title of CSC258 is changed to Computer Organization I.
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Avoiding Update Anomalies
number
title
CSC258
Computer Organization
CSC434
Database Systems
number
section room
CSC258
1 LP266
CSC258
2 GB258
CSC258
3 GB248
CSC434
1 GB248Title of CSC258 is changed to Computer Organization I.
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Avoiding Update Anomalies
number
title
CSC258
Computer Organization I
CSC434
Database Systems
number
section room
CSC258
1 LP266
CSC258
2 GB258
CSC258
3 GB248
CSC434
1 GB248Title of CSC258 is changed to Computer Organization I.CSC434 is not given in this term.
The instance does not store redundant information.
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Avoiding Update Anomalies
number
title
CSC258
Computer Organization I
CSC434
Database Systems
number
section room
CSC258
1 LP266
CSC258
2 GB258
CSC258
3 GB248
CSC434
1 GB248CSC434 is not given in this term.
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Avoiding Update Anomalies
number
title
CSC258
Computer Organization I
CSC434
Database Systems
number
section room
CSC258
1 LP266
CSC258
2 GB258
CSC258
3 GB248
CSC434 is not given in this term.
The title of CSC434 is not removed from the instance.
A new course is created: (CSC336, Numerical Methods)
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Avoiding Update Anomalies
number
title
CSC258
Computer Organization I
CSC434
Database Systems
number
section room
CSC258
1 LP266
CSC258
2 GB258
CSC258
3 GB248
A new course is created: (CSC336, Numerical Methods)
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Avoiding Update Anomalies
number
title
CSC258
Computer Organization I
CSC434
Database Systems
CSC336
Numerical Methods
number
section room
CSC258
1 LP266
CSC258
2 GB258
CSC258
3 GB248
A new course is created: (CSC336, Numerical Methods)No information about sections has to be provided.Each relation stores attributes that are directly related.
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Normalization Theory
Main idea: a normal form defines a condition that a well designed database should satisfy.
Normal form: syntactic condition on the database schema.• Defined for a class of data dependencies.
Main problems:
• How to test whether a database schema is in a particular normal form.
• How to transform a database schema into an equivalent one satisfying a particular normal form.
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BCNF: a Normal Form for FDs
Functional dependency (FD) over R(A1, …, An) : X Y , X, Y {A1, …, An}.
X Y : two rows with the same X-values must have the same Y-values.• number title : two rows with the same course
number must have the same title.
Key dependency : X A1 An
• X is a key: two distinct rows must have distinct X-values.
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BCNF: a Normal Form for FDs
is a set of FD over R(A1, …, An).
Relation schema R(A1, …, An), is in BCNF if for every X Y in , X is a key.
A relational schema is in BCNF if every relation schema is in BCNF.
In BCNF: S(number, title), number title
T(number, section, room),
Not in BCNF:
R(number, title, section, room),
number title
In BCNF: S(number, title), number title
T(number, section, room),
In BCNF: S(number, title), number title
T(number, section, room),
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Normalization Theory Today
Normalization theory for relational databases was developed in the 70s and 80s.
Why do we need normalization theory today?• New data models have emerged: XML.
• XML documents can contain redundant information.
Redundant information in XML documents:• Can be discovered if the user provides semantic
information.
• Can be eliminated.
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XML Documents
courses
coursecourse
@cno @cno taken_bytaken_by
student student
@sno@name @grade@grade @name@sno
student. . .
“st1” “st1” “A+”“B+”
“CSC258” “CSC434”
“Fox”“Fox”
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XML Databases
D : : Two students with the same @sno value must have the same name.
courses course*
course @cno
course taken_by
taken_by
student*
student @sno, @name, @grade
student ε
XML Schema: (D, )
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Redundancy in XML
courses
coursecourse info
@cno @cno taken_bytaken_by
student studentstudent
@name@sno
. . .
“CSC258” “CSC434” “st1” “Fox”
@sno@name @grade@grade @name@sno“st1” “st1” “A+”“B+” “Fox”“Fox”
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XML Database Normalization
DTD: Data dependency:
Two students with the same @sno value must have the same name.
courses course*
course @cno
course taken_by
taken_by
student*
student @sno,@name,@grade
student ε
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XML Database Normalization
DTD:
, info* @sno is the identifier of info elements.
courses course*
course @cno
course taken_by
taken_by
student*
student @sno,@grade
student εinfo @sno,
@name
Data dependency:
Two students with the same @sno value must have the same name.
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A “Non-relational” Example
DBLP
conf conf
@title issueissue
article articlearticle
@year@title @title @year
@year
“ICDT”
@year
@year@title“1999”
“1999”
“1999” “2001”
“2001”
“. . .” “. . .” “. . .”
. . .
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XNF: XML Normal Form
Proposed in [AL02].
It eliminates two types of anomalies.
It was defined for XML functional dependencies:
DBLP.conf.@title DBLP.confDBLP.conf.issue
DBLP.conf.issue.article.@year
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Part 3: What was Missing? Justification!
What is a good database design?
• Well-known solutions: BCNF, 4NF, …
But what is it that makes a database design good?
• Elimination of update anomalies.
• Existence of algorithms that produce good designs: lossless decomposition, dependency preservation.
Previous work was specific for the relational model.
• Classical problems have to be revisited in the XML context.
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Justification of Normal Forms
Problematic to evaluate XML normal forms.
• No XML update language has been standardized.
• No XML query language yet has the same “yardstick” status as relational algebra.
• We do not even know if implication of XML FDs is decidable!
We need a different approach.
• It must be based on some intrinsic characteristics of the data.
• It must be applicable to new data models.
• It must be independent of query/update/constraint issues.
Our approach is based on information theory.
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Information Theory
Entropy measures the amount of information provided by a certain event.
Assume that an event can have n different outcomes with probabilities p1, …, pn.
Amount of information gained by knowing that event i occurred :Average amount of information gained (entropy) :
Entropy is maximal if each pi = 1/n :
ip
1log
n
i ii p
p1
1log
nlog
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Entropy and Redundancies
Database schema: R(A,B,C), A B
Instance I:
Pick a domain properly containing adom(I) :
• Probability distribution: P(4) = 0 and P(a) = 1/5, a ≠ 4
• Entropy: log 5 ≈ 2.322
A B C
1 2 3
1 2 4
A B C
1 2 3
1 2 4
A B C
1 2
1 2 4
A B C
1 2 3
1 2 4
A B C
1 3
1 2 4
Pick a domain properly containing adom(I) : {1, …, 6}
• Probability distribution: P(2) = 1 and P(a) = 0, a ≠ 2
• Entropy: log 1 = 0
{1, …, 6}
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Entropy and Normal Forms
Let be a set of FDs over a schema S.
Theorem (S,) is in BCNF if and only if for every instance of (S,) and for every domain properly containing adom(I), each position carries non-zero amount of information (entropy > 0).
This is a clean characterization of BCNF , but the measure is not accurate enough ...
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Problems with the Measure
The measure cannot distinguish between different types of data dependencies.
It cannot distinguish between different instances of the same schema:
A B C
1 2 3
1 2 4
1 5
A B C
1 2 3
1 4
entropy = 0
R(A,B,C), A B
entropy = 0
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A General Measure
Instance I of schema R(A,B,C), A B :
A B C
1 2 3
1 2 4
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A General Measure
Instance I of schema R(A,B,C), A B :
Initial setting: pick a position p Pos(I) and pick k such that adom(I) {1, …, k}. For example, k = 7.
A B C
1 2 3
1 2 4
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A General Measure
Instance I of schema R(A,B,C), A B :
Initial setting: pick a position p Pos(I) and pick k such that adom(I) {1, …, k}. For example, k = 7.
A B C
1 2 3
1 2 4
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A General Measure
Instance I of schema R(A,B,C), A B :
Initial setting: pick a position p Pos(I) and pick k such that adom(I) {1, …, k}. For example, k = 7.
A B C
1 3
1 2 4
Computation: for every X Pos(I) – {p}, compute probability distribution P(a | X), a {1, …, k}.
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A General Measure
Instance I of schema R(A,B,C), A B :
A B C
1 3
1 2 4
Computation: for every X Pos(I) – {p}, compute probability distribution P(a | X), a {1, …, k}.
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A General Measure
Instance I of schema R(A,B,C), A B :
A B C
3
1 2
Computation: for every X Pos(I) – {p}, compute probability distribution P(a | X), a {1, …, k}.
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A General Measure
Instance I of schema R(A,B,C), A B :
A B C
3
1 2
Computation: for every X Pos(I) – {p}, compute probability distribution P(a | X), a {1, …, k}.
P(2 | X) =
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A General Measure
Instance I of schema R(A,B,C), A B :
A B C
2 3
1 2
Computation: for every X Pos(I) – {p}, compute probability distribution P(a | X), a {1, …, k}.
P(2 | X) =
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A General Measure
Instance I of schema R(A,B,C), A B :
A B C
1 2 3
1 2 1
Computation: for every X Pos(I) – {p}, compute probability distribution P(a | X), a {1, …, k}.
P(2 | X) =
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A General Measure
Instance I of schema R(A,B,C), A B :
A B C
4 2 3
1 2 7
Computation: for every X Pos(I) – {p}, compute probability distribution P(a | X), a {1, …, k}.
P(2 | X) =
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A General Measure
Instance I of schema R(A,B,C), A B :
A B C
1 2 3
1 2 3
Computation: for every X Pos(I) – {p}, compute probability distribution P(a | X), a {1, …, k}.
P(2 | X) = 48/
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A General Measure
Instance I of schema R(A,B,C), A B :
A B C
3
1 2
Computation: for every X Pos(I) – {p}, compute probability distribution P(a | X), a {1, …, k}.
P(2 | X) = 48/
For a ≠ 2, P(a | X) =
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A General Measure
Instance I of schema R(A,B,C), A B :
A B C
a 3
1 2
Computation: for every X Pos(I) – {p}, compute probability distribution P(a | X), a {1, …, k}.
P(2 | X) = 48/
For a ≠ 2, P(a | X) =
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A General Measure
Instance I of schema R(A,B,C), A B :
A B C
2 a 3
1 2 7
Computation: for every X Pos(I) – {p}, compute probability distribution P(a | X), a {1, …, k}.
P(2 | X) = 48/
For a ≠ 2, P(a | X) =
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A General Measure
Instance I of schema R(A,B,C), A B :
A B C
1 a 3
1 2 6
Computation: for every X Pos(I) – {p}, compute probability distribution P(a | X), a {1, …, k}.
P(2 | X) = 48/
For a ≠ 2, P(a | X) = 42/
(48 + 6 42) = 0.16
(48 + 6 42) = 0.14
Entropy ≈ 2.8057 (log 7 ≈ 2.8073)
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A General Measure
Instance I of schema R(A,B,C), A B :
A B C
1 3
1 2 4
Value : we consider the average over all sets X Pos(I) – {p}.
•Average: 2.4558 < log 7 (maximal entropy)
•It corresponds to conditional entropy.
•It depends on the value of k ...
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A General Measure
Previous value:
For each k, we consider the ratio:
• How close the given position p is to having the maximum possible information content.
General measure:
)|( pInf kI
k
pInf kI
log
)|(
k
pInfpInf
kI
kI log
)|(lim)|(
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Basic Properties
The measure is well defined:
For every set of first order constraints defined over a schema S, every I inst(S,), and every p Pos(I): exists.
Bounds:
)|( pInf I
1)|(0 pInf I
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Basic Properties
The measure does not depend on a particular representation of constraints. If 1 and 2 are equivalent:
It overcomes the limitations of the simple measure: R(A,B,C), A B
)|()|( 21 pInfpInf II
A B C
1 2 3
1 2 4
1 5
A B C
1 2 3
1 4
0.875 0.781
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Well-Designed Databases
Definition A database specification (S,) is well-designed if for every I inst(S,) and every p Pos(I), = 1.
In other words, every position in every instance carries the maximum possible amount of information.
We would like to test this definition in the relational world ...
)|( pInf I
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Relational Databases
is a set of data dependencies over a schema S:
= : (S,) is well-designed.
is a set of FDs: (S,) is well-designed if and only if (S,) is in BCNF.
is a set of FDs and MVDs: (S,) is well-designed if and only if (S,) is in 4NF.
is a set of FDs and JDs:
• If (S,) is in PJ/NF or in 5NFR, then (S,) is well-designed. The converse is not true.
• A syntactic characterization of being well-designed is given in [AL03].
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Relational Databases
If (S,) is in DK/NF, then (S,) is well-designed. The converse is not true.
The problem of verifying whether a relational schema is well-designed is undecidable.
If the schema contains only universal constraints (FDs, MVDs, JDs, …), then the problem is co-NEXPTIME-complete.
• If each relation in S has at most m attributes, then the problem is -complete.
Now we would like to apply our definition in the XML world ...
P2
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XML Databases
XML schema: (D,).
• D is a DTD.
• is a set of data dependencies over D.
We would like to evaluate XML normal forms.
The notion of being well-designed extends from relations to XML.
• The measure is robust; we just need to define the set of positions in an XML tree T: Pos(T).
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Positions in an XML Tree
DBLP
conf conf
@title issueissue
article articlearticle
@year@title @title @year
“ICDT”
@year@title“1999” “1999” “2001”“. . .” “. . .” “. . .”
“ICDT”
“1999” “1999” “2001”“. . .” “. . .” “. . .”
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Well-Designed XML Data
We consider k such that adom(T) {1, …,k}.
For each k :
We consider the ratio:
General measure:
)|( pInf kT
k
pInfpInf
kT
kT log
)|(lim)|(
kpInf kT log/)|(
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XNF: XML Normal Form
For arbitrary XML data dependencies:
Definition An XML specification (D,) is well-designed if for every T inst(D,) and every p Pos(T), = 1.
For functional dependencies:
Theorem An XML specification (D,) is in XNF if and only if (D,) is well-designed.
)|( pInfT
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Normalization Algorithms: BCNF
Relation schema: R(X,Y,Z), • Not in BCNF: X Y and X A, for every A
Z.
Basic decomposition: replace R(X,Y,Z) by S(X,Y) and T(X,Z).
Example: R(number, title, section, room),
number title
S(number, title), number title
T(number, section, room),
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Normalization Algorithms: BCNF
number title section room
CSC258 Computer Organization 1 LP266
CSC258 Computer Organization 2 GB258
CSC434 Database Systems 1 GB248
number
title
CSC258
Computer Organization
CSC434
Database Systems
number
section room
CSC258
1 LP266
CSC258
2 GB258
CSC434
1 GB248
number, title (R) number, section, room (R)
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Normalization Algorithms: BCNF
number title section room
CSC258 Computer Organization 1 LP266
CSC258 Computer Organization 2 GB258
CSC434 Database Systems 1 GB248
number
title
CSC258
Computer Organization
CSC434
Database Systems
number
section room
CSC258
1 LP266
CSC258
2 GB258
CSC434
1 GB248
S T
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Normalization Algorithms: XNF
The algorithm applies two transformations until theschema is in XNF.
If there is an anomalous FD of the form:
DBLP.conf.issue DBLP.conf.issue.article.@year
then apply the “DBLP example rule”.
Otherwise: choose a minimal anomalous FD and apply the “University example rule”.
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Normalization Algorithms
The information-theoretic measure can also be used for reasoning about normalization algorithms.
For BCNF and XNF decomposition algorithms:
Theorem After each step of these decomposition algorithms, the amount of information in each position does not decrease.
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Future Work
We would like to consider more complex XML constraints and characterize good designs they give rise to.
We would like to characterize 3NF by using the measure developed in this paper.
• In general, we would like to characterize “non-perfect” normal forms.
We would like to develop better characterizations of normalization algorithms using our measure.
• Why is the “usual” BCNF decomposition algorithm good? Why does it always stop?
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Backup Slides
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XNF: XML Normal Form
Given a DTD D and a set of functional dependencies {}:
(D, ) if for any XML tree T conforming to D and satisfying , it is the case that T
(D, )+ = { | (D, ) }
Functional dependency is trivial if it is implied by the DTD alone: (D, )
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XNF: XML Normal Form
XML specification: a DTD D and a set of functional dependencies .
A Relational DB is in BCNF if for every non-trivial functional dependency X Y in the specification, X is a key.
(D, ) is in XNF if:
For each non-trivial FD X p.@l in (D, )+, X p is in (D, )+.
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A Normal Form for FDs and JDs
))()()()(( 21 xRxRxRxR m
iMi
xx
))()()(( 21 jim xxxRxRxR
Let be a set of FDs and JDs over a schema S:
Theorem (S,) is well-designed if and only if for every
R S and every nontrivial JD:
implied by , there exists M {1, ..., m} such that:
1.
2. For every i,j M, implies
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A Normal Form for FDs and JDs (cont’d)
Schema: S = { R(A,B,C) } and = { [AB, AC, BC],
AB C, AC B }.
(S, ) is not in PJ/NF: {AB ABC, AC ABC} does not imply [AB, AC, BC].
(S, ) is not in 5NFR: [AB, AC, BC] is strong-reduced and BC is not a superkey.
(S,) is well-designed.
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Tree Tuples
Paths(D): all paths in a DTD Dcourses.course [email protected].@name
We distinguish two kinds of elements: attributes (@) and element types.
FDs are defined by means of a relational representation of XML documents.
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XML Trees
v1
v2 v3
v0
. . .
courses
coursecourse
@cno
“cs100”
@sno @name@grade @sno @name
@grade
student student
“123” “Fox” “B+” “Smith” “A-”“456”
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Tree Tuples
v1
v2
v0
courses
course
@cno student
“cs100”
t(courses) = v0
t(courses.course) = v1
t(courses.course.@cno) = “cs100”t(courses.course.student) = v2
t(p) = , for the remaining paths
Relational representation: tree tuples - mappings
t : Paths(D) Vertices Strings {}
A tree tuple represents an XML tree:
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XML Tree: set of Tree Tuples
v1
v2 v3
v0
. . .
courses
coursecourse
@cno
“cs100”
@sno@name
@grade @sno@name
@grade
student student
“123” “456”“Fox” “B+” “Smith” “A-”
v1
v2
courses
course
@cno
“cs100”
student
v0
@sno@name
@grade
“123” “Fox” “B+”
v3
@sno@name
@grade
student
“456” “Smith” “A-”
. . .
course
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Functional Dependencies for XML
Expressions of the form: X Y
defined over a DTD D, where X, Y are finitenon-empty subsets of Paths(D).
XML tree T can be tested for satisfaction of X Y
if:
X Y Paths(T) Paths(D)
T X Y if for every pair u, v of tree tuples in T:
u.X = v.X and u.X ≠ implies u.Y = v.Y
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FD: Examples
University DTD: courses course*course @cno, student*student @sno, name, grade
Two students with the same @sno value must have the same name:
courses.course.student.@sno courses.course.student.@name
Every student can have at most one grade in every course:
{ courses.course, courses.course.student.@sno }
courses.course.student.@grade