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the presence of records, and in the presence of records and lists.

Let

T

be any non-empty subset of {lists, sets, multisets} apart from {lists} . The gen­

eralised A rmstrong axioms, i. e.,

X -+ Y y � X' -+

{Y} y � X,

{X, Y}

-+ UN

X, Y reconcilable,

X -+ Y, Y -+ Z X -+ Z

form a minimal, sound and complete set of inference rules for the implication of FDs in

the presence of records and

T. 0

In terms of Figure 1 . 1 , Theorem 5.23 extends the knowledge on the class of FDs and the problem of axiomatisability along the data type dimension covering all combinations of record-, list-, set- and multiset-valued attributes. We would like to do the same extension for the implication problem of FDs.

5 .4 Implication Problem

In view of Theorem 5. 15,

E

f= X -+ Y holds if and only if

E

f-9\ X

-+

Y holds where

9'\ are the generalised Armstrong axioms from Definition 5.4. Given some set

E

one can enumerate all FDs derivable from it. However, the enumeration algorithm is time consuming and therefore impractical. We will now present a provably-correct membership algorithm and prove that it works efficiently, i.e., in polynomial time in the number of subattributes of the underlying nested attribute and the number of FDs given.

5.4.1 The Closure

Similar to the RDM [29] and similar to the case of list-valued attributes in Chapter 3 we introduce the notion of a closure for a set of nested attributes with respect to a given set of FDs. Please note that this notion already played an important role in proving Theorem 5. 15. The closure is defined with respect to the set 9'\ of the generalised Armstrong axioms from Definition 5.4.

5.4. IMPLICATION PROBLEM Sebastian Link Definition 5.24. Let

N

E N A, X

� Sub(N)

a set of subattributes of

N,

and

E

a set of FDs on

N.

The

closure x+ � Sub(N)

of X with respect to

E

is

x+ =

{ Z : X -t { Z} E

E+

} . o

According to Theorem 5.15 the closure

x+

of X is therefore the set of all nested attributes which are functionally determined by X with respect to a given set

E

of FDs. The computation of

x+

is sufficient for deciding whether

E f=

X -t

Y

holds.

Lemma 5.25.

Let N E

N A,

and E a set of FDs on N. Then

X -t

Y

E

E+

if and only if Y

�

x+

Proof.

If X ---+

Y

E

E+,

then X -t { Y } E

E+

for all Y E

Y

by the subset rule. This means

all Y E

Y

are elements of

x+,

i.e. ,

Y � x+.

Assuming that every Y E

_Y

also satisfies Y E

x+

implies that X -t {Y} E

J.;+

for all

Y E

Y.

We infer that X -t

Y

E

E+

by the derivability of the union rule. 0

Let X, Y �

Sub(N) .

We call X a

generalised subset

of

Y,

denoted by X

�gen Y,

if and only if for every X E X there is some Y E

Y

with X ::; Y (Hoare-ordering) . Note that

� gen

is a pre-order (reflexive, transitive) on the powerset

P(Sub(N))

of

Sub(N) .

The distinct sets X = {L[A] , L [-\] } and

Y =

{L[A] } are generalised subsets of one another, i.e.,

� gen

is not symmetric.

The projection of any tuple on a superattribute always determines the projection of this tuple on each of its subattributes. It is therefore sufficient to consider only maximal subattributes with respect to ::; . The set of all maximal elements of some :S-ideal X

�

Sub(N)

is formally defined as Xmax = { X E X : VZ E X.X ::; Z implies X = Z} . It is an immediate consequence of Lemma 5 .26 that

Y � x+

if and only if Y �gen X�ax·

Lemma 5.26.

Let N

E N A,

and

X �

Sub(N) an ideal with respect to ::; . For all Y �

Sub(N) we have

Y

� X

if and only if Y �gen

Xmax

Proof.

We show the

only if

part first. Let Y E

Y

be arbitrary. From

Y �

X follows Y E X.

Consequently there is some Z E Xmax with Y ::; Z. This shows that Y

�gen

Xmax·

It remains to consider the

if

part. Let Y E

_Y

be arbitrary again. Since

Y �gen

Xmax, there is some Z E Xmax with Y ::; Z. Since Xmax

�

X we have

Z

E X. However, X is an ideal with respect to ::; , i.e., Y E X holds as well. This shows

Y

� X. 0

5.4.2 Units of Nested Attributes

In order to solve the implication problem for FDs on some nested attribute

N

we will split

N

into mutually reconcilable subattributes

Ni

and solve the projected implication problems on the

Ni

simultaneously. The idea is to choose the units

Ni

of

N

such that for all subattributes

U, V

E

_Sub(N)

we have that

U

and

V

are reconcilable if and only if for all units

Ni

of

N

the subattributes

U n Ni

and

V n Ni

are comparable with respect to ::; . This motivates the following definition.

Definition 5.27. Let

N

E N A. A nested attribute

Ni

E N A is a

unit of N

if and only if

1 .

Ni

:S

N,

2.

VX, Y ::; Ni,

if

X

and

Y

are reconcilable, then

X ::; Y

or

Y ::; X,

3.

Ni is ::;-maximal with properties 1 . and 2.

The set of all units of

N

is denoted by

U(N) .

D

The property that two subattributes

U, V

E

Sub(N)

are reconcilable is not tran­ sitive: if

N = L(K { M(A, B)}, C)

and

U = L(K { M (A, .\) } , A) ,

W

= L(A, C)

and

V = L(K{M(A, B) } , A),

then

U

and W are reconcilable, W and

V

are reconcilable, but

U

and

V

are not reconcilable. In fact,

U, V

E

Sub(L(K{M(A, B) } , A)) ,

but they are incom­ parable with respect to ::; .

E xAMPLE 5 . 5 . Let

N = L1 (L2(L3 (A, B)) , L4 [L5(C, L6 (D) )], L7(E, L8{ L9 (F, G, H)})).

The units of

N

are

- Ll (L2 (L3(A, B)) , A, L7(.\, .\)) ,

- L1 (.\, £4[£5( C, .\)] , £7(.\, .\)),

- L1 (.\, L4[L5(A, L6(D) )] , L7 (A, .\)),

- L1 (.\, A, L7(E, A))

and

- L1 (.\, A, L1(.\, L8 {L9(F, G, H) })).

Clearly

L1 (A, A, L7(A, L8 {L9 (.\, G, H)}))

also has properties 1 . and 2. of Definition 5.27,

but is not maximal with respect

::;

. D

Next we give an inductive characterisation of units.

k

Lemma 5 .28.

Let N

E N

A. Then U(N) =

U

{L(AN! l · .. , M, . . . , ANk ) : M

E

i=l

U (Ni) and Ni =/= ANJ, if N = L(N1 , . . . , Nk) and N =/= AN, U(N) = {L[M'] : M'

E

U(M) },

In document Evaluación de la estrategia de reducción de la pobreza de Honduras 2000 2015 (página 58-63)