Round Colour Space for Pentachromacy
Circularity of Multidimensional Hue
Alfredo Restrepo Palacios
Laboratorio de Se˜nales, Dept. Ing. El´ectrica y Electr´onica, Universidad de los Andes,
Carrera 1 No. 18A-70; of. ML-427, Bogot´a 111711, Colombia
Keywords:
Pentachromacy, Hypercube, Chromatic Saturation, Luminance, Equatorial Sphere, S
3
, Permutations, S
5
,
Hamiltonian Circuit, Cayley Graph.
Abstract:
We generalize previous results to dimension 5 and further. The geometry of the 5-hypercube [0,1]
5
gives a
model for colour vision in the case of 5 photoreceptor types and a colour space corresponding to the combina-
tion of ve primary lights. In particular, we focus on the (topologically spherical) boundary of the hypercube
and on an equatorial sphere within the boundary, roughly orthogonal to the achromatic segment. In the poly-
topal and double-cone type spaces, we consider a tridimensional hue component; in the round Runge space
we consider a 4-dimensional colourfulness component.
1 INTRODUCTION
The three properties of colour, luminance, hue and
chromatic saturation can be seen as geometric prop-
erties of points in the RGB cube; properties that de-
pend on the position and orientattion of the points in
the cube with respect to both the
black point
and the
achromatic segment
(Restrepo, 2011). This approach
generalises to dimension 4 (Restrepo, 2012a), (Re-
strepo, 2012b), (Restrepo, 2013b), (Restrepo, 2013a)
and to dimensions 5 and further. A model for colour
vision in the case of 5 photoreceptor types, or a colour
space corresponding to the additive combination of
ve primary lights is presented here. The approach
allows to do pentachromatic colour image processing
and the study of pentachromatic metamerism.
Pentachromacy is the case of the visual system
of many animals, e.g. pigeons (David M. Hunt
and Davies, 2009) and dragonflies and flies (Kelber,
2006).
For the visualization of multispectral images with
ve bands, pentachromacyis likewise relevant as pen-
tachromatic colour processing followed by the RGB
visualisation of three out of the ve channels makes
explicit important aspects of the image (Restrepo and
Maldonado, 2015). Likewise, in the screen ilumi-
nation industry, it is also useful to have models for
more than three primary lights (Shmuel Roth, 2010),
(Roger P. A. Delnoij, 2012) in the visible spectrum.
2 GEOMETRY AND COLOUR
We extend the geometric characterisation of trichro-
matic colour to a model for pentachromatic colour
that is based on the geometry of the hypercube [0,1]
5
.
This interpretation provides a basis for the processing
and visualisation of pentachromatic images as well as
a plausible model for the study of the colour vision
systems of pentachromatic animals.
Let the interval [0,1] model the set of possible in-
tensities of each of ve primary lights in an additive
colour combination or, of the possible response levels
of each of five photoreceptors. In this way the
cubic
colour space
[0,1]
5
R
5
models the set of possible
primary combinations or of photoreceptor responses;
thus,
colours
are modelled as points in the hypercube.
The points of R
5
are denoted either as (v,w,x, y, z) or
as p = (p
0
, p
1
, p
2
, p
3
, p
4
). The position of the colour
points, relative to an
equatorial
3-sphere σ in the 4-
spherical boundaryΣ := [0,1]
5
of the 5-cube; σ is the
basis for the definition of tridimensional
pentachro-
matic hue
while Σ is the basis for the definition of
4-dimensional
pentachromatic kolor
.
The boundary Σ of the 5-cube is the set of
the colour points having at least one 0-valued co-
ordinate or one 1-valued coordinate. The points
(p
0
, p
1
, p
2
, p
3
, p
4
) Σ are classified into 5 × 2 = 10
4-cubes depending on which coordinate p
i
is equal to
1 or equal to 0; e.g. {0vwxyz} and {1vwxyz}. As a
matter of fact, Σ is a PL (piecewise linear) 4-sphere,
98
Restrepo Palacios A..
Round Colour Space for Pentachromacy - Circularity of Multidimensional Hue.
DOI: 10.5220/0005301900980105
In Proceedings of the 10th International Conference on Computer Vision Theory and Applications (VISAPP-2015), pages 98-105
ISBN: 978-989-758-089-5
Copyright
c
2015 SCITEPRESS (Science and Technology Publications, Lda.)
a topological S
4
, that in addition to these 10 4-cubes,
can be
celled
into 40 3-cubes, 80 squares, 80 edges
and 32 points; these numbers result as follows: by
fixing 2 of the coordinates p
i
and p
j
of points in the
5-cube with values in {0,1}, you get the 2
2
×
5
2
= 40
3-cubes of Σ; fixing 3 coordinates in {0, 1}, you get
the 2
3
×
5
3
= 80 squares of Σ; fixing 4 coordinates,
2
4
×
5
4
= 80 edges and, fixing all 5 coordinates,
2
5
= 32 points or vertices.
Each of the 32 vertices of Σ, that is points
(p
0
, p
1
, p
2
, p
3
, p
4
) with p
i
{0,1} is a vertex of
5
2
= 10 3-cubes in the cell complex; e.g. (00000)
{00xyz}. Each of the 80 edges is an edge of
4
2
=
6 3-cubes; for example, {v0000} {v00yz}. Each of
the 80 squares is a face of 3 3-cubes; for example,
{vw000} {vwx00},{vw0y0},{vw00yz}.
Out of the 40 3-cubes of Σ, 20 do not have neither
of the points s := (0 0 0 0 0) or w := (1 1 1 1 1) as
vertices (w and s stand for white and black (schwartz),
respectively.) The union of these 20 cubes is an
equa-
torial
S
3
for Σ, called the
hue sphere
, and denoted as
σ. The points of σ are precisely those points of Σ
having at least one coordinate at value 0 and at least
one coordinate at value 1. The equatorial σ is linked
in S
5
= R
5
{} with the line (circle through )
through the points s and w. The
achromatic segment
is the line segment that joins s and w and is given
by φ := {λs+ (1 λ)w : λ [0, 1]}; it consists of the
colour points having equal coordinates. In a sense that
is made precise below, loosely speaking we say that σ
and φ are orthogonal.
The equatorial σ, of codimension 2 in R
5
, is used
to define pentachromatic hue by giving coordinates to
the points of [0, 1]
5
on the basis of σ, i.e. by locating
the colour point with respect to σ (and φ). There are
other ways to define a pentachromatic hue as there
are other closed 3-manifolds in the cell complex Σ,
that are unions of 3-cubes of Σ but, in a sense, σ is the
canonical choice.
In fact, since each of the 20 3-cubes of σ can be
triangulated into 3! = 6 tetrahedra, according to the
6 possible orderings of the
free
3 coordinates (there
are two
fixed
coordinates p
i
, p
j
, with p
i
, p
j
{0,1},
in each such cube), each of the resulting 120 tetra-
hedra contains the points of σ having one of the 120
possible orderings of their (ve) coordinates, where
the minimal coordinate has value 0 and the maximal
one has value 1. To each point in one such tetra-
hedron of σ, there corresponds a triangle of points
in the hypercube having such an ordering of their
coordinates, this time without the restriction of the
minimal coordinate being 0 and the maximal being
1, and the union of such triangles is the 5-simplex
of the points in the hypercube sharing such order-
ing of their coordinates. For example, the ordering
p
0
p
1
p
2
p
3
p
4
corresponds to the tetrahe-
dron {p
0
= 0 p
1
p
2
p
3
p
4
= 1} in the cube
{p
0
= 0, p
4
= 1} σ. Thus, in this triangulation of σ
into 20×6 = 120 tetrahedra, the interior of each tetra-
hedron precisely corresponds to each of the elements
of the symmetric group S
5
, which in turn corresponds
as well to the interior of the 5-simplex of points in the
5-hypercube having that ordering of their coordinate
values.
We say that a colour point h = [h
0
,h
1
,h
2
,h
3
,h
4
]
is a
hue
if at least one of its coordinates h
i
is 0-valued
and at least one of its coordinates h
j
is 1-valued,
that is, the points of σ are the possible
hues
of the
colour points not on the achromatic segment φ, called
chromatic colour points
. The set of colour points
having a given hue h is the
hue triangle
that is the
cone of the achromatic segment φ and the hue point h
given by {[γ + βh
0
,γ + βh
1
,γ + βh
2
,γ + βh
3
,γ + βh
4
]
: β+ γ 1,0 β,γ 1}. There are 120
hue families
into which the hue points h σ can be classified.
All colours of a given hue family have the same
ordering regarding the relative contributions from
each of the photoreceptors, or primaries. An analogy
with the trichromatic RGB case is for example the
family of the oranges
, where R G B. In the
trichromatic case, a hue is either unary or binary, in
the pentachromatic case, a hue may be unary, binary,
trinary or quaternary, correspondingly depending
on whether it is a vertex, or is in the interior of
a segment, a triangle or a tetrahedron of the PL,
equatorial, hue sphere σ. Hues at the boundary of a
3-cube of σ belong to exactly two families.
The hypercube [0, 1]
5
is triangulated into 120 5-
simplices by taking the topological join of the achro-
matic segment φ and each of the 120 tetrahedra in the
triangulation of σ. (For our purposes, the join of two
subsets A and B of R
N
is the union of the line seg-
ments connecting points of A with points of B. ) Thus,
in each of these 5-simplices, the hues of the colour
points belong to the same hue family.
To each chromatic colour point p =
(p
0
, p
1
, p
2
, p
3
, p
4
) there corresponds a unique
hue
h σ, in fact, p belongs to the unique
hue
triangle
having as one side φ and as opposing vertex
the point h = (h
0
,h
1
,h
2
,h
3
,h
4
) σ with coordinates
h
i
=
p
i
m
ρ
(1)
where m := min{p
i
: i 5} and ρ := max{p
i
: i
5}−min{p
i
: i 5}; for example, the hue of p = [0.1,
0.3, 0.5, 0.7, 0.9] is h = [0, 0.25, 0.5, 0.75, 1].
RoundColourSpaceforPentachromacy-CircularityofMultidimensionalHue
99
The connectivity of the cubes and tetrahedra of
σ is topologically captured by a connectivity graph.
Consider the graph having as nodes the tetrahedra
(equivalently the elements of S
5
) of the chromatic
sphere σ, where two nodes are joined by a branch
precisely when the two corresponding permutations
differ by a transposition of two consecutive elements.
This corresponds to two tetrahedra being connected
by a triangular face. When a colour changes from a
hue family to another, the two corresponding order-
ings differing by a transposition of two consecutive
elements, we say that a
mild change
of hue family
has occurred. We show below a table corresponding
to a Hamiltonian circuit in the graph of hue families
connected by mild changes, that cycles through the
120 families. In this way, we give a cyclic order to
the hue families, analogous to the trichromatic case,
where you can give a cyclic order to the 6 hue fami-
lies and in fact (but this only in the trichromatic case)
a cyclic order to the hues themselves.
3 ROUNDER COLOUR SPACES
For colour processing, it is better to transform the hy-
percube into a rounder space. Initially, we obtain a
space of the ”hexcone” type, that although not round
is a starting point to obtain the rounder
double-cone
type
space and a round ball type space. Such spaces
are more intuitive and less prone, under colour trans-
formations, to end up at ”forbidden colours” due to
a careless change of coordinates. A useful geometric
technique to obtain rounder spaces is spinning.
The
spin
Sp(N,M,R
3
) in R
3
of a subset N of the
upper half plane of R
2
around
a set M R
2
of dimen-
sion 1 with parametrisation M = {[m
1
(t), m
2
(t)] : t
P R
1
} is given by
Sp(N,M,R
3
) := {[x,y.m
1
(t), y.m
2
(t)] : (x,y) N}
Usually, the coordinates m
i
of M are taken to be in
the interval [-1, 1]; for example, the spin of the trian-
gle {[x,y] R
2
: x [1,1],y = 1 |x|} around a cir-
cle in R
3
, is the double cone {[x,(1 |x|)cos(t),(1
|x|)sin(t)] : x [1,1],t [0,2π)} and the spin of the
same triangle around a hexagon is a hexcone. In a
sense, you are spinning the flag of the upper half of
R
2
, that contains the triangle, with pole the x axis.
Similarly, the
spin
Sp(N,M,R
5
) in R
5
of a subset
N of the upper half plane of R
2
around
a set M R
4
,
usually of dimension 3, with parametrisation M =
{[m
1
(t
1
,t
2
,t
3
),m
2
(t
1
,t
2
,t
3
),m
3
(t
1
,t
2
,t
3
),m
4
(t
1
,t
2
,t
3
)] :
(t
1
,t
2
,t
3
) P R
k
} is given by
Sp(N,M,R
5
) := {[v, w.m
1
(t
1
,t
2
,t
3
),w.m
2
(t
1
,t
2
,t
3
),
w.m
3
(t
1
,t
2
,t
3
),w.m
4
(t
1
,t
2
,t
3
)] : (v,w) N,(t
1
,t
2
,t
3
)
P}.
The equatorial σ sits in R
5
in codimension 2. By
projecting σ onto the 4-hyperplane Π that is orthog-
onal to the achromatic segment φ you get again a PL
S
3
, denoted σ
, this time embedded (in Π) in codi-
mension 1. The equatorial sphere σ has been made
”flat”, and each ray in Π from the origin intersects a
circumscribing, round S
3
, which is key to getting a
round hue space.
In fact, for this projection π : σ Π, π(σ) is an
embedding of σ. To see that it is injective, notice that
π(s) = π([s
0
,s
1
,s
2
,s
3
,s
4
]) = [s
0
k,s
1
k,s
2
k,s
3
k,s
4
k], where k is the average of the s
i
s. Therefore,
if π(
a
s) = π(
b
s) then
a
s
b
s =
b
k
a
k, that is,
a
s and
b
s differ by a constant vector, but since the points of
σ have exactly one 0-valued coordinate and one 1-
valued coordinate, the remaining coordinates being
strictly between 0 and 1, the only possibility is that
such constant vector be the origin.
The function f : R
5
R
2
with
(p
0
, p
1
, p
2
, p
3
, p
4
) 7→ (µ,ρ), where ρ is the range of
the coordinate values and µ :=
max{p
i
:i5}+min{p
i
:i5}
2
is the midrange, bijects each of the hue triangles
(whose union is [0, 1]
5
) to a canonical, isosceles
µρ-triangle”, also called the
luminance-saturation
triangle
T := f([0,1]
5
). T is an isosceles triangle
with base {(µ,ρ) R
2
: µ [0,1],ρ = 0} and height
{(µ,ρ) R
2
: µ = 0.5,ρ [0,1]}. We say that µ
measures the luminance of a colour on each hue
triangle and that ρ measures its chromatic saturation.
The 5-hypercube is the (union of the) collection
of the hue triangles, each of which has as common
side the achromatic segment φ and oposing vertex a
point in σ; the fact that the hue triangles have diffrent
shapes differentiates this construction of the hyper-
cube [0,1]
5
from a spin.
The spin of the µρ triangle around σ
is (analog
to the the hexcone space) the 5D 120-cone H :=
Sp(T,σ
,R
5
).
The spin of the triangle T around the standard
(Euclidean, round) sphere S
3
R
4
gives the double-
cone type space. This space makes explicit the fact
that, corresponding to maximal 1 and minimal 0
values of luminance (at the vertices of the cone),
there is no hue nor chromatic saturation to be added.
By deforming the triangle T into a semicircle, D
and spinning it around the round S
3
of R
4
, you get
a solid ball B
5
, called Runge space. The process of
spinning adds the hue to the luminance and saturation
of T. In Runge space, the hue and the luminance are
made explicit but not so is the chromatic saturation;
instead, we call the distance from the central middle
gray the
kolorfulness
or vividness of a colour point
VISAPP2015-InternationalConferenceonComputerVisionTheoryandApplications
100
and the corresponding point in the S
4
boundary of the
ball the
kolor
or chromaticity ( defined for achromatic
colours such as s and w as well) of the colour point.
The kolor is the point of the 3-sphere at which the spin
was made and the kolorfulness is the distance to the
central point
middle gray
of the 5-ball.
4 PENTACHROMATIC HUE
In the trichromatic case, the hue is a
cyclic variable
(Restrepo and Estupinan, 2014); in the tetrachromatic
case, the hue is bidimensional and spherical; in the
pentachromatic case the hue is tridimensional, spheri-
cal. In both the tetrachromatic and the pentachromatic
cases, it is not the hues but
families of hue
that can be
cyclically ordered. There are 24 families of hue in the
tetrachromatic case, and 120 families of hue in the
pentachromatic case.
The ordering of the relative contributions of each
of the five primaries giving rise to a coloured light
beam can be seen as a broad property of its colour;
colours are thus characterised as belonging to one of
120 possible
families of hue
. Likewise for the rela-
tive contributions of five photoreceptorswith different
photopigments at a small retinal spot.
In the case of human colour vision, the percep-
tual proper of uniqueness is made explicit (in Marr’s
jargon (Marr, 1982)) at V4 and not at the receptoral
level nor in the retina, the thalamus or V1 (Zeki,
1993). In fact, since the cone responses overlap (as
do the spectral responses of camera filters), it is not
possible to have a stimulus eliciting response from
only one cone type; instead of using the term unique-
ness of hue, to such hypothetical case, we refer to
as (receptoral) the
unariness
of a hue. Further reti-
nal, and cortical processing differentiates further the
hue in these hue families of a receptoral level; in our
trichromatic case, the receptoral family L M S
is subdivided into 2 cortical-hue families: oranges
and citrines (greenish yellows) separated by cortical,
unique yellow ( L = M). The cortical uniques and hue
families are then unique red, reddish oranges, central
orange, yellowish oranges, unique yellow, yellowish
citrines, central citrine, greenish citrines and unique
green.
Also in the trichromatic case, the vertices of the
hue hexagon
, an equatorial S
1
in the PL S
2
given
by [0, 1]
3
, called the
chromatic hexagon
in (Re-
strepo, 2011), belong either to {red green, blue} or to
{yellow, violet, cyan} (at the receptoral level, yellow
is a binary colour). With the modification of yellow
to a unique and the corresponding addition of the bi-
nary families orange and cetrine, the hue hexagon be-
comes an octagon with vertices {red, yellow, green,
blue} (the uniques) and segments {orange, cetrine,
cyan, violet} (the binaries). In the trichromatic case,
the
uniqueness
of a hue is the opposite of its
binari-
ness
: the closer you get to a unique hue (red, green,
yellos or blue) the farther you are from either of the
central binary
hues (orange, cetrine, violet or cyan).
Geometrically in σ, define
uniqueness
as the distance
to a closest unique and
binariness
as the distance to a
closest central binary hue.
In the pentachromatic case, a vertex of σ is said to
be a
unary
hue; at a point in the interior of an edge
you have contributions from two primaries and the
hue is said to be
binary
, on a triangle in the complex
σ, you have contributions from 3 primaries and the
hue is said to be
trinary
and inside (i.e. in the interior
of) a tetrahedron of σ, the hue is said to be
quater-
nary
. There are 5 unary hues, e.g. [10000], which are
the vertices of σ having 4 zero-valued coordinates, 10
binary hue families, e.g. [1w000] which are the seg-
ments of σ made of points having 3 0-valued ccor-
diantes; 10 trinary hue families, e.g. [1wx00], which
are the triangles of σ of points having 2 0-valued co-
ordinates and 5 quaternary hue families, e.g. [1wxy0],
which are the tetrahedra of σ having 1 0-valued coor-
dinate. Since a colour point with no zero coordinate
is not a hue (not an element of σ), there are no pentary
hues, such a colour has an achromatic component.
Every colour point can be writtten uniquely either
as an achromatic colour e.g. [0.1,0.1, 0.1,0.1,0.1], or
an achromatic colour plus a colour of a quaternary hue
e.g. [0.1, 0.2,0.2,0.2, 0.2], or an achromatic colour
plus a colour of a quaternary hue plus a colour of a
trinary hue, e.g. [0.1,0.2, 0.4,0.4,0.4], or an achro-
matic colour plus a colour of a quaternary hue plus a
colour of a trinary hue plus a colour of a binary hue
e.g. [0.1,0.2,0.4, 0.7,0.9], or as an achromatic colour
plus colour of a quaternary hue plus a colour of a tri-
nary hue plus a colour of a binary hue plus a colour
of a unique hue as follows: [0.1, 0.2, 0.4, 0.7, 0.9] =
[0.1, 0.1, 0.1, 0.1, 0.1] + [0, 0.1, 0.1, 0.1, 0.1] + [0, 0,
0.2, 0.2, 0.2] + [0, 0, 0, 0.3, 0.3] + [0, 0, 0, 0, 0.2], of
hues, correspondingly,
undefined
, [01111], [00111],
[00011] and [00001].
For a given a 5-tuple q = [q
0
,q
1
,q
2
,q
3
,q
4
]
[0,1]
5
there is at least one (of the 120 possible) per-
mutation p : {0,1, 2, 3, 4} {0,1,2,3,4} such that
a nondecreasing ordering q
p(0)
q
p(1)
q
p(2)
q
p(3)
q
p(4)
results. As already said, in the equatorial
3-sphere σ, the set of 5-tuples so ordered by a given
permutation is a tetrahedron; thus the 120 permuta-
tions p determine a triangulation of the hue sphere
σ into 120 tetrahedra. Correspondingly, the 120 per-
mutations determine a triangulation of [0,1]
5
into 120
RoundColourSpaceforPentachromacy-CircularityofMultidimensionalHue
101
5-simplices where each 4-simplex is the topological
join of the achromatic segment and the corresponding
tetrahedron in the triangulation of σ.
The permutations have the structure of a group;
algebraically, they are the elements of the symmetric
group S
5
and, perhaps not an obvious fact, its ele-
ments can be
cyclically sequenced
so that two con-
secutive permutations differ by a transposition of con-
secutive coordinates (Johnson, 1963). From the view-
point of the wavelength domain, two consecutive co-
ordinates of a colour are better related than another
pair of ccordinates. The swapping of two consecu-
tive colour coordinates is called a
mild, hue-family
change
. Also, it so happens that the two tetrahedra of
σ that correspond to such a permutation have exactly
a triangular face in common; also, you can visit each
tetrahedra of σ exactly once going from one tetrahe-
dron to the next through such triangles. Equivalently
stated, there is a Hamiltonian circuit in the Cayley
graph of S
5
(Witte, 1982).
We say that each tetrahedron of σ determines a
family of hues in the sense that in each tetrahedron
the relative contributions of the primaries is fixed.
Also, you have that the 120 hue families are cycli-
cally ordered. It is the families of hue rather than the
hues themsalves that are cyclically ordered, unlike the
trichromatic case where the hue sphere is triangulated
into 6 intervals (Restrepo, 2011), and each interval is
linearly ordered, the points within each tetrahedron of
σ do not have a linear order.
As we said in Section 2, the S
4
boundary Σ of
[0,1]
5
consists of 10 4-cubes; each such 4-cube con-
nects with 8 other 4-cubes via each 3-cube of its
boundary. For example, the 4-cubes {v = 0} and
{z = 1} connect via the 3-cube {v = 0,z = 1}. In the
corresponding connectivity graph, having as nodes
the 4-cubes and as branches the connecting 3-cubes,
there are several Hamiltonian circuits.
Also, the 120 tetrahedra of the S
3
equatorial hue
sphere σ can be grouped into 20 3-cubes; for exam-
ple, the 3-cube {v = 0,w = 0} groups the tetrahedra
{v = 0,w = 0,x y z}, {v = 0,w = 0,y x z},
{v = 0,w = 0,y z x}, {v = 0,w = 0,z y x},
{v = 0,w = 0,z x y} and {v = 0,w = 0,x z y}.
That is, σ can also be
celled
into 20 3-cubes, on
each of which, as in the trichromatic case, the colour
points can be given the trichromatic attributes of hue
saturation and value. Each 3-cube in σ connects with
each of 6 other 3-cubes via a 2-square. For example,
the 3-cubes {v = 1, x = 0} and {x = 0,z = 1} connect
via the square {v = 1, x = 0,z = 1}; there are several
Hamiltonian circuits in the graph corresponding to
this cellular decomposition of σ into 3-cubes. in such
a way that each 3-cube (the union of six tetrahedra)
is visited exactly once and every pair of consecutive
3-cubes have a square (the union of two triangles)
face in common.
We indicate an iterative procedure of obtaining
Hamiltonian circuits in the graph of S
n
. Let e
n
denote
the identity element of S
n
. In Figure 1 the permuta-
tions of two elements are used to get the permutations
of a set of three elements in such a way that you get
a factorisation, into transpositions of consecutive el-
ements, of e
3
from that of e
2
. The symmetric group
S
3
with three elements is generated by the transposi-
tions t
0
:= (0,1) and t
1
:= (1,2). Denote the action
of a transposition t acting on an ordered set, or triple
p = [p
0
, p
1
, p
2
] as t(p), or simply tp; thus, for exam-
ple, (1, 2)[0, 1, 2] = [0, 2, 1]. From Figure 1, the
permutations of [0,1,2] are cyclically ordered as
Figure 1: The lines mustard and red represent the symmet-
ric group S
2
. A new element is positioned at each possible
position for after each transposition in the previous permu-
tation to get S
3
. The iteration of this gives a Hamilton circuit
on each S
n
.
[0,1,2]
[1,0,2] = (0,1).[0,1,2]
[1,2,0] = (1,2)(0,1).[1,2,0]
[2,1,0] = (0,1)(1,2)(0,1).[0,1,2]
[2,0,1] = (1, 2)(0, 1)(1, 2)(0, 1).[0,1,2]
[0,2,1] = (0,1)(1,2)(0,1)(1,2)(0,1).[0,1,2]
[0,1,2] = (1,2)(0,1)(1,2)(0,1)(1,2)(0,1).[0,1,2]
The last ordering being the initial one. Thus, at
the last line you have a factorization of the identity e
3
of the symmetric group S
3
as
e
3
= Π
3!1
i=0
t
i
= (1,2)(0,1)(1,2)(0,1)(1,2)(0,1)
into 6 transpositions of consecutive elements, either
(0,1) or (1,2). In fact, the 6 permutations
π
j
:= Π
j
i=0
t
i
, j /0,5/
of S
3
are thus cyclically ordered. We simplify further
the notation by writing only the first element of each
transposition in the factorisations, for example e
3
=
1.0.1.0.1.0.
The 24 permutations in S
4
are likewise cyclically
ordered. By interleaving the sequences of transposi-
tions 0:=(0, 1), 1:=(1,2), 2:=(2,3), 2:=(2, 3), and its re-
verse, alternatively in between each of the 6 transpo-
sitions in the factorisation of e
3
S
3
and taking care
of adding a one in the notation of the transpositions in
e
3
, after the end of an inserted sequence 2, 1, 0, but
not after an inserted sequence 0, 1, 2, (extending the
idea in Figure 1) you get
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e
4
= Π
4!1
i=0
t
i
= [0.1.2.(0).2.1.0.(1+1).0.1.2.(0).2.1.0.(1+1).0.1.2.(0)
.2.1.0.(1+1)]
T
where the transpositions in parenthesis are those cor-
responding to e
3
; thus,
e
4
= [0.1.2.0.2.1.0.2.0.1.2.0.2.1.0.2.0.1.2.0.2.1.0.2]
T
where the T in Π
T
indicates a reversing in the order-
ing of the factorization Π. A cyclic ordering of the
elements of S
4
is then {π
j
| j {0,1,2,...23}}, with
π
j
= Π
j
i=0
t
i
.
This procedure can be generalized. A factoriza-
tion of e
5
S
5
into 5 × 24 = 120 transpositions of
consecutive elements is
e
5
= Π
5!1
i=0
t
i
= [α 0 α
T
1’ α 2 α
T
0’ α 2 α
T
1’ α 0 α
T
2’ α 0 α
T
1’
α 2 α
T
0’ α 2 α
T
1’ α 0 α
T
2’ α 0 α
T
1’ α 2 α
T
0’ α 2 α
T
1’ α 0 α
T
2’]
T
where α = 0.1.2.3 and a prime, as in n
, denotes
n+ 1, that is,
= [α 0 α
T
2 α 2 α
T
1 α 2 α
T
2 α 0 α
T
3 α 0 α
T
2 α 2
α
T
1 α 2 α
T
2 α 0 α
T
3 α 0 α
T
2 α 2 α
T
1 α 2 α
T
2 α 0
α
T
3]
T
= [0 1 2 3 0 α
T
2 0 1 2 3 2 α
T
1 0 1 2 3 2 α
T
2 0 1 2 3
0 α
T
3 0 1 2 3 0 α
T
2 0 1 2 3 2 α
T
1 0 1 2 3 2 α
T
2 0 1 2 3
0 α
T
3 0 1 2 3 0 α
T
2 0 1 2 3 2 α
T
1 0 1 2 3 2 α
T
2 0 1 2 3
0 α
T
3]
T
= [0 1 2 3 0 3 2 1 0 2 0 1 2 3 2 3 2 1 0 1 0 1 2 3 2 3 2
1 0 2 0 1 2 3 0 3 2 1 0 3 0 1 2 3 0 3 2 1 0 2 0 1 2 3 2 3 2
1 0 1 0 1 2 3 2 3 2 1 0 2 0 1 2 3 0 3 2 1 0 3 0 1 2 3 0 3 2
1 0 2 0 1 2 3 2 3 2 1 0 1 0 1 2 3 2 3 2 1 0 2 0 1 2 3 0 3 2 1 0 3]
T
and a ciclic ordering of the 120 elements of S
5
is
{π
j
| j {0, 1, 2,...119}}, with
π
j
= Π
j
i=0
t
i
, j /0,119/ (2)
5 CODING AND DECODING THE
HUE
In the 120-cone space, the luminance and the chro-
matic saturation are coded in the µρ triangle while the
hue is coded as a point in the PL hue sphere σ. This
code of σ is the basis for the definition of hue in the
S
3
of the double-cone type space.
The cyclic ordering of S
5
given by Equation 2, to-
gether with the initial ordering v w x y z,
denoted shortly as vwxyz, determine a cyclic ordering
of the orderings of the 5-tuples [v,w,x,y,z] of coordi-
nates. Thus, an ordering of [v,w,x,y,z] determines a
number j given by the position in the list (see Table 1)
of the cyclic ordering of S
5
, and viceversa. Equation
Table 1: A cyclic ordering of the 120 orderings of the 5-
tuple [vwxyz].
# ordering # ordering # ordering
0 vwxyz 40 vwzxy 80 vwyzx
1 wvxyz 41 wvzxy 81 wvyzx
2 wxvyz 42 wzvxy 82 wyvzx
3 wxyvz 43 wzxvy 83 wyzvx
4 wxyzv 44 wzxyv 84 wyzxv
5 xwyzv 45 zwxyv 85 ywzxv
6 xwyvz 46 zwxvy 86 ywzvx
7 xwvyz 47 zwvxy 87 ywvzx
8 xvwyz 48 zvwxy 88 yvwzx
9 vxwyz 49 vzwxy 89 vywzx
10 vxywz 50 vzxwy 90 vyzwx
11 xvywz 51 zvxwy 91 yvzwx
12 xyvwz 52 zxvwy 92 yzvwx
13 xywvz 53 zxwvy 93 yzwvx
14 xywzv 54 zxwyv 94 yzwxv
15 xyzwv 55 zxywv 95 yzxwv
16 xyzvw 56 zxyvw 96 yzxvw
17 xyvzw 57 zxvyw 97 yzvxw
18 xvyzw 58 zvxyw 98 yvzxw
19 vxyzw 59 vzxyw 99 vyzxw
20 vxzyw 60 vzyxw 100 vyxzw
21 xvzyw 61 zvyxw 101 yvxzw
22 xzvyw 62 zyvxw 102 yxvzw
23 xzyvw 63 zyxvw 103 yxzvw
24 xzywv 64 zyxwv 104 yxzwv
25 xzwyv 65 zywxv 105 yxwzv
26 xzwvy 66 zywvx 106 yxwvz
27 xzvwy 67 zyvwx 107 yxvwz
28 xvzwy 68 zvywx 108 yvxwz
29 vxzwy 69 vzywx 109 vyxwz
30 vxwzy 70 vzwyx 110 vywxz
31 xvwzy 71 zvwyx 111 yvwxz
32 xwvzy 72 zwvyx 112 ywvxz
33 xwzvy 73 zwyvx 113 ywxvz
34 xwzyv 74 zwyxv 114 ywxzv
35 wxzyv 75 wzyxv 115 wyxzv
36 wxzvy 76 wzyvx 116 wyxvz
37 wxvzy 77 wzvyx 117 wyvxz
38 wvxzy 78 wvzyx 118 wvyxz
39 vwxzy 79 vwzyx 119 vwyxz
1 gives the j-th ordering π
j
[vwxyz] of the 5-tuple; on
the other hand, to recover the number j from a per-
mutation of [vwxyz], we proceed as follows.
As a first step,
reduce
a given ordering of
[v, w,x,y,z] to an ordering of [y, z] by deleting the let-
ters v, w and x from the given ordering; for exam-
ple, zyxwv 7→ zy; then, notice if this corresponds to
an even or an odd permutation (odd in the example)
of yz; next, reduce the initial permutation by deleting
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only the letters v and w, obtaining a permutation of
[x,y,z]; in the previous example, you get zyxwv 7→ zyx
and notice if the reduced permutation (xyz 7→ zyx in
the example) is even or odd (again odd, in the exam-
ple) next notice the parity of the reduced-by-v permu-
tation (wxyz 7→ zyxw, even, in the example); finally,
check the parity of the original (complete) permuta-
tion (vwxyz 7→ zyxwv, even, in the example.) Notice
also, as if going from a reduced ordering to a less re-
duced ordering, the
position
at which the added ele-
ment is inserted; this position is sometimes measured
from left to right and sometimes from right to left. If
you are at an even ordering (e.g. 5432) the new ele-
ment (e.g. 1) is assumed to be inserted from left to
right (thus, in 54321, the 1 has been inserted at the 4-
th position) while, if you are based at an odd ordering,
the inserted element is assumed to have been inserted
from right to left (thus, 54 being an odd permutation
tells us that in 543, the 3 has been inserted at the 0-th
position). These positions are the weight factors t
k
in
the computation of the number j of the permutation,
that is,
j =
n
k=2
n!
k!
t
k
(3)
where n = 5, and t
k
/0, k 1/ is calculated as
described above.
Given a colour point, for example [0.5, 0.4, 0.3,
0.2, 0.1], with Equation 3, we find that we are in the
tetrahedron number 64 of σ; also, its luminance and
its chromatic saturation are given by µ =
0.1+0.5
2
= 0.3
and ρ = 0.5 0.1 = 0.4. Next, we find where in tetra-
hedron 64, the point [1, 0.75, 0.5, 0.25, 0] is (see
Equation 1); that is, we compute the barycentric co-
ordinates of point [1, 0.75, 0.5, 0.25, 0] in the tetra-
hedron with vertices [10000], [11000], [11100] and
[11110] which, in a sense, are the barycentric coordi-
nates of [0.75, 0.5, 0.25] in the tetrahedron with ver-
tices [1000], [1100], [1110] and [1111] of R
4
. Since
[1,0.75, 0.5,0.25,0] = α[1000] + β[1100] + γ[1110] +
δ[1111] and α+ β+ γ + δ = 1, you can get α,β,γ and
δ.
6 HUE PROCESSING
A simple yet useful way of processing pentachromatic
images is to process separately the hue. The remain-
ing pair of colour attributes of the luminance and the
saturation can likewise be processed separately using
e.g. exponential laws (A. Restrepo, 2009). The lu-
minance and saturation can be jointly processed with
flows
of the points in the luminance-saturation trian-
gle. Likewise, the kolor can be processed as a rotation
or a homeomorphism of S
4
and the kolorfulness with
an exponential law, separately. In the remaining of
this section we concentrate on the separate process-
ing of the hue, by ”PL rotations” of σ, a tool for the
transformation of pentachromatic hue.
By a
PL rotation
of σ we mean the following. Ini-
tially, σ is projected on the 4-subspace Π of R
5
that is
orthogonal to φ. Call σ
the embedding π(σ). Then,
Π is rotated; then, the rotated version of σ
(a PL S
3
)
is ”fitted back (see below) on σ
. Then, σ
is back
projected to σ.
The projection π is linear and each tetrahedron of
σ projects to a tetrahedron of σ
. A point of σ such
as h = [0,w, x, y,1], with average η =
v+x+y
5
, has an
image π(h) in σ
given by h
= [η,w η, x η,y
η,1 η] which has zero average. Notice that the or-
dering of the coordinates of h is the same as that of h
.
h
defines the ray on R
5
given by {th
:t [0,)}. To
each ray in Π R
5
there corresponds a unique point
of σ
(and viceversa).
The the rigid motions of S
3
R
4
(the rotations of
R
4
) can coded with the help of unit quaternions, as
in rot(s) = asb, with |a| = |b| = 1. The correspond-
ing group is known as SO(4). After Π is rotated,
to each point h
σ
there corresponds a new point
h
∗∗
Π, not necessarily in σ
. We find an appro-
priately corresponding in σ
as follows. The order-
ing of the coordinates of h
∗∗
determines a tetrahedron
of σ
that contains the point intersection of the ray
through h
∗∗
and σ
; this intersection point is the a
”rotated” version of h
. We compute the barycen-
tric coordinates of this intersection point with respect
to the vertices of the tetrahedron. Then we com-
pute in the correspoding tetrahedron of σ, the cor-
responding barycentric combination, with the com-
puted coordinates gives the ”PL rotated” version of
h. The embedding of σ
= π(σ) Π of σ has an
inverse given by the addition of the constant vector
such that the resulting largest coordinate has value 1
and the smallest one has value 0. For example, con-
sider the hue h = [0,0.4,0.5, 0.6,1] of average 0.5 and
projection h
= [0.5,0.1, 0, 0.1,0.5] that is rotated
to g
= [0.5,0.1, 0, 0.1,0.5] via multiplication by
the orthogonal matrix [00001; 00010; 00100; 01000;
10000]; the ray through h
∗∗
is in the terahedron of σ
with vertices 11110, 11100, 11000, 10000 which it
intersects at the same point h
∗∗
= g
and that is back
projected to point [.5 .1 0 -.1 -.5] + [.5 .5 .5 .5 .5] = [1
.6 .5 .4 0]]
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7 CONCLUDING REMARKS
A model for colour vision in the case of 5 photore-
ceptor types as well as a colour space corresponding
to the additive combination of five primary lights are
presented here. The approach allows to do pentachro-
matic colour image processing and the study of pen-
tachromatic metamerism.
Pentachromacy is relevant in the study of the vi-
sion systems of many animals, e.g. pigeons (David
M. Hunt and Davies, 2009) and dragonflies and flies
(Kelber, 2006). Most mammals are dichromatic (dol-
phins are monochromatic)and old-world monkeysare
trichromatic; the reduction in cone type of mammals
is probably related to the fact that they evolved as
nocturnal animals. Among vertebrates, only some
rodents and marsupials take advantage of ultraviolet
light. Light of short wavelengths is less absorbed by
water and penetrates deeper in water.
By feeding three of the five channels of a pen-
tachromatic image to the RGB inputs of a visualis-
ing device, useful information can be made explicit.
This, combined with pentachroamtic colour process-
ing should provide with a useful tool for the search
of objects with given spectral surface reflectance in
pentachromatic images.
The visualisation of multispectral images can be
done in time; for this, continuous changes of hue
(therefore of SO(4)) are useful (Lenz and Homma,
1996).
The colour vision of animals such as pigeons and
dragonflies can be tested according to these colour
components of luminance, saturation, 3D hue, kolor
and kolorfulness with the models provided and a five-
primary illuminating system that include two types of
UV, in addition to RGB. Pentachromatic metamerism
can also be studied along the lines of (Restrepo,
2014b), (Restrepo, 2014a). This has applications in
illumination, photography and animal vision as well.
In the screen ilumination industry, it is also use-
ful to have models for more than three primary lights
(Shmuel Roth, 2010), (Roger P. A. Delnoij, 2012) in
the visible spectrum. It is a fact that three lights can-
not reproduce all colours, mainly when seen isolated;
nevertheless when colours are seen in context, chro-
matic contrast gives the illusion of many more colours
that those that can be projected in isolated form.
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