Information Technology Reference
In-Depth Information
Notice that:
•
μ
P
(
p
N
)=
ϕ
(
p
N
)
p
N
)
S
(
p
)) =
λ
(
h
)
ϕ
(
,
μ
F
(
S
(
p
)) =
εϕ
(
, and as it is sup-
posed that it is 0
<
ε
<
1, it follows
μ
F
(
S
(
p
))
≤
μ
P
(
S
(
p
))
,forall
p
≤
N
.
•
If
p
is fixed, and
λ
(
h
1
)
≤
λ
(
h
2
)
, it follows
μ
h
1
(
S
(
p
))
≤
μ
h
2
(
S
(
p
))
.
0
N
•
μ
h
(
S
(
0
)) =
λ
(
h
)
ϕ
(
N
)=
0,
μ
h
(
S
(
N
)) =
λ
(
h
)
ϕ
(
N
)=
λ
(
h
)
,and
μ
P
(
S
(
N
))
=
1,
μ
F
(
S
(
N
)) =
ε
.
P
N
)
≤
ϕ
(
P
N
)
≤
•
From
λ
(
h
)
≤
1 it follows
λ
(
h
)
ϕ
(
1, by
p
≤
N
.
Hence,
μ
h
(
S
(
p
))
≤
1.
•
Obviously 0
≤
μ
h
(
S
(
p
))
and
μ
h
(
S
(
p
))
∈
[
0
,
1
]
. Exactly, from
ε
≤
λ
(
h
)
p
N
)
≤
μ
h
(
p
N
)
≤
follows
εϕ
(
S
(
p
))
,andas
ε
is small and
ϕ
(
1, this lower
bound is actually a small number.
c. By choosing
ϕ
, different models for the degree up to which
S
(
p
)
can be
considered to be a heap are obtained. For instance,
•
Linear model
(
ϕ
=
id
)
,
μ
h
(
S
(
p
)) =
λ
(
h
)
p
/
N
p
N
)
x
2
2
,
•
Quadratic model
(
ϕ
(
x
)=
)
,
μ
h
(
S
(
p
)) =
λ
(
h
)(
•
etc.
Finally, provided '
S
(
p
)
is not a heap' can be represented by a strong-negation
N
ϕ
1
,
μ
h
=
N
ϕ
1
◦
μ
h
,is
p
N
)))
≥
λ
(
p
N
))
⇔
))
⇔
ϕ
−
1
1
μ
h
(
S
(
p
))
≥
μ
h
(
S
(
p
(
1
−
ϕ
1
(
λ
(
h
)
ϕ
(
h
)
ϕ
(
⇔
ϕ
−
1
1
2
(
ϕ
−
1
(
)
P
N
)
⇔
ϕ
−
1
(
1
/
2
)
1
1
≥
ϕ
(
)
·
N
≥
p
λ
(
h
)
λ
(
h
)
(
ϕ
−
1
(
1
/
2
)
,
ϕ
−
1
with which the kernel is
[
0
)
·
N
]
. For instance,
1
λ
(
h
)
x
2
, the threshold between heap and not heap, is
•
If,
ϕ
1
=
id
,
ϕ
(
x
)=
N
0
.
5
λ
(
N
71
N
1
1
λ
(
)
=
)
=
0
.
that, for
h
=
P
is 0
.
71
N
.
h
2
λ
(
h
h
)
.
0
5
N
N
2
.
2. Let us consider the case in which the prototype is a circular cone, C, of radius
r and height h, whose volume is given by V
•
If
ϕ
1
=
ϕ
=
id
,is
N
)
=
that, for
h
=
P
is
λ
(
h
2
λ
(
h
)
1
3
r
2
=
(
π
×
×
)
H
, and suppose the
3
.
heaps are in the unit cube of
R
3
2
,radius
r
If the cone
C
is in
[
0
,
1
]
with its base in
[
0
,
1
]
=
0
.
5 and height
H
=
1,
≤
σ
(
)
≤
let us take the same
F
, as well as a perceptive similarity index 0
h
1. For
1
4
, that represents that
the volume of
C
is around four times de volume of figure (c). Since
Vol
σ
(
)=
instance, the heap in figure (c) of 15.4.2 could have
h
(
)=
π
/
C
2,
σ
(
h
)=
1
/
4 means an evaluation of the volume of
h
around
π
/
48 cubic units.
Thus, we can take
p
N
)
,
with which all has been said in the former paragraph can be repeated exactly by just
changing
μ
h
(
S
(
p
)) =
σ
(
h
)
ϕ
(
.
Hence, no actual difference with the case of the pyramid appears.
λ
(
h
)
by
σ
(
h
)