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x n
x d = 0 Pr ( G t , x d ) · g t 1 ( v i x d )
g t (
v i )=
Therefore,
x d = 0 Pr ( G t , x d ) · g t 1 ( v i x d ) g t 1 ( v i x d ) |
x n
g t (
|
g t (
v i )
v i ) | = |
Let x c (1
. Suppose x c 1
x c
c
ρ
) be the
ρ
-quantiles of f t 1 (
x
)
v i
x d
(1
), there are two cases:
First, if v i
c
ρ
x c 1 +
x c
c
1
c
1
ρ
c
ρ
, then g t 1 (
x d
v i
x d )=
and
g t 1 (
v i
x d )
.
2
ρ
1
ρ
g t 1 (
Thus, 0
g t 1 (
v i
x d )
v i
x d )
.
x c 1 +
x c
, then g t 1 (
c
ρ
c 1
c
ρ
Second, if v i
x d >
v i
x d )=
and
ρ
g t 1 (
v i
x d )
.
2
1
g t 1 (
Thus,
ρ
g t 1 (
v i
x d )
v i
x d )
0.
g t 1 (
1
ρ
In both cases,
|
g t 1 (
v i
x d )
v i
x d ) |≤
. Therefore,
x n
x d = 0 Pr ( G t , x d ) ·
1
ρ =
1
ρ .
g t (
|
g t (
v i )
v i ) |≤
Using the above approximation method, the overall complexity of computing the
approximate k equi-depth histogram answer is O
2
(
m
ρ
)
, where m is the number of
connected components.
7.6.2.3 Sum/Average Queries
The query answering algorithm for count queries can be extended to evaluate sum
and average queries with minor changes.
To answer a sum query Q sum , we apply the same preprocessing techniques dis-
cussed in Sections 7.3 and 7.4. The only difference is that the private vertices in
the same clique with the same value in attribute
are compressed as one vertex.
When scanning each clique, the sum distribution of its subtree are computed and
passed to its parent clique. The overall complexity is O
A
n 2
, where n is the number
of values in v min , the smallest value of any linkage in attribute
(
)
A
, and v max , the sum
of the values of all linkages in attribute
A
. The average query can be evaluated
similarly.
 
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