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ponents
), the set L of the components of the
warehouse is selected. L constitutes the set of the
components needed in the DW. We suppose that
the set of the components of the warehouse is a
subset of all the components of the various DBs. In
other words we have: L ⊆ ∪
k∈D
(C
k
). Consequently,
L=∪
k∈D
(L
k
) where L
k
=L ∩ (C
k
).
We suppose that
L ≠
∅ and some
L
k
can be
empty with
C
k
∩ C
k'
= ∅ and
L
k
∩ L
k'
= ∅ for any
k≠k'
. Thus, for all
x
∈
L,
a single
k
∈
D
exists
such that
x
∈
L
k
.
Example 2:
Considering example 1, the set L
of the components of the warehouse is
L
=
{x
1
, x
5
,
y
1
, y
2
, y
7
} with L
1
=
{x
1
, x
5
}
and
L
2
=
{y
1
, y
2
, y
7
}
The second step (
The decomposition of the
whole components of the warehouse
) consists
in replacing recursively all the composite ele-
ment by their components. We defined as simple
component, every component c such that
comp
(
c
)
=∅. It thus acts, in this step, to recursively replace
every non simple component c of L (
comp
(
c
) ≠
∅) by all the elements of comp(c), until there is
only simple components.
Let DB = (C, ref, comp) be a DS. Let us de-
note that
C
S
is the set of simple components. The
decomposition of the composed components is
made using the recursive function (Hamdoun,
2006) decomp: C → P(C
S
). It is defined as
follows:if
decomp
(
c
):=
{
c
}
then
comp
(
c
) = ∅,
else
decomp
(
c
):= ∪
c'∈ comp(c)
(decomp (c')
In fact, decomp(c) makes simple the set of
components which took part in the construction
of c.
Example 3:
By considering example 2, the
result of the decomposition is the following set:L'
= {x
2
, x
3
, x
5
, y
1
, y
4
, y
3
, y
8
, y
9
, y
10
}.
The third step consists in filtering the compo-
nents set. This one is refined using the two relations
S and I of the selected environment. Indeed, two
sub-steps of filtering are carried out on the set L
according to these relations. The first filtering op-
eration, using the partial order I, consists in leaving
the components corresponding
to the maximum
elements
with respect to the partial order I. Recall
that: X is a maximum element in a partial order
if it does not exist
y
such that
y > x
.
Example 4:
According to the example 3, L'
becomes L'={x
2
, x
3
, x
5
, y
1
, y
3
, y
8
, y
9
, y
10
}.
In this second filtering operation, for each
equivalence class of the relation S, only one repre-
sentative is left and others are deleted. The choice
of this representative element is arbitrary and does
not intervene in the integration process.
Example 5:
By supplementing example 4,
L' becomes the following set:L'={x
2
, x
3
, x
5
, y
1
,
y
8
, y
9
, y
10
}. The set L', result of the filtering, is
thus formed: L'⊆L.L' can be written as follows:
L'=∪
k∈D
(L'
k
) avec L'
k
=L' ∩ (L
k
).
Note: In the realization of these two filter-
ing operations the order of processing is crucial
(filtering by using I then filtering by using S).
The following example illustrates the two results
obtained if the order is not respected.
Example 6:
Let us suppose that L'={x
2
, y
3
,
y
5
} and assume x
2
S
y
3
and y
5
I
x
2
. The result of
filtering by using I is L'={x
2
, y
3
} then by apply-
ing S we obtain L'={y
3
}. The result of filtering
by using S leads to L'={y
3
, y
5
} ; then by applying
I we obtain L'={y
3
, y
5
}.
The fourth step (
The generation of the total
schema of the DW
) is made up applying the four
following sub-steps:
a)
Construction of the graphs associated with
the sources: In each graph
G
k
corresponding
to a data source
DB
k
, the nodes represent all
the components
c
∈
C
k
where not exists
c'
∈
C
k
with
c
∈
comp
k
(
c'
). The edges rely each
couple of nodes (n(c), n(c')) corresponding
to a couple of components (
c, c'
) such that
∃ X ⊂ comp
k
(c), Y ⊂ comp
k
(c
'
) with X ref
k
Y.
Example 7:
By supplementing example 5 we
obtain the two graphs shown in Figure 3.
b)
Detection of the connected sub-graphs:
a graph is qualified as connected when it
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