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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
)
 
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