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(
)
(
)
d
M
c
,
y
=
c
−
c
−
1
−
λ
L
+
c
−
c
+
λ
L
1
1
2
2
(4)
(
)
=
1
−
λ
L
+
λ
L
(
)
=
1
−
λ
L
+
λ
L
=
L
Thus every point on the side has the same distance to the center and this holds for
all sides. As a consequence, the length of the side and the length of the diagonal are
equal.
i
q
c
i
c
p
Fig. 1.
Complementary element in a Manhattan shape-space
One way to achieve a unique complement is to require that the three points
i
,
c
and
i
c
lie on a straight line. This results in an additional condition:
(v)
for all
j
= 1, …,
n
c
=
(
i
+
i
)
2
j
j
c
j
n
Notice that this holds only for shape-spaces based on infinite sets, in particular
.
Things are different and simpler for finite shape-spaces, in particular Hamming
spaces, because here the complement is determined uniquely in a natural way as will
be shown in the next section.
ℜ
3 Finite Shape-Spaces
If
V
is a finite set of
n
elements, the power set of
V
, 2
V
, forms a complete lattice with
respect to set inclusion. This lattice is isomorphic to the lattice formed by the set
H
of
binary strings of length
n
with
1
as top and
0
as bottom element and an appropriate
partial order on the set. To define such an order we need a function that counts the
number of 1's in a string. This is equivalent to computing the sum of the digits of the
number; the desired function will be called
ones
, i.e.
n
()
∑
=
(5)
ones
x
=
x
i
i
1
By means of
ones
the partial order
p
on
H
is defined as follows:
