Cryptography Reference
In-Depth Information
through only some light. The color (0; 0; 0), which we will denote also with
the symbol "," is the black color: indeed all lters are 0, meaning that there
is no light left. The color (100; 100; 100), which we will denote also with the
symbol "," is white because no light is absorbed by the lters. The colors
red, green, and blue are represented, respectively, by (100; 0; 0); (0; 100; 0),
and (0; 0; 100); we will refer to these colors also as
R
,
G
, and
B
, respec-
tively. The colors cyan, magenta, and yellow are represented, respectively,
by (0; 100; 100); (100; 0; 100), and (100; 100; 0); we will refer to these colors
also as
C
,
M
, and
Y
, respectively. The color (50; 0; 0) is also a red, because that
is the only component present, but it is darker since some red light has been
absorbed. The higher the value of the component, the lighter is the color. If all
components are equal, i.e., (x;x;x), then the resulting color is a gray whose
intensity depends on x: the smaller is x, the darker is the gray.
Recall that this representation works fine both for the additive model and
for the subtractive model. In the additive model we start from (0; 0; 0) and add
light while in the subtractive model we start from (100; 100; 000) and subtract
light.
In the context of visual cryptography we can think of the transparencies as
lters; starting from a white light we "subtract" some light applying lters (the
transparencies). The remaining light determines the color that we see when
superposing several transparencies. At this point it is worth emphasizing that
"white" on a transparency is actually "transparent." We assume to start with
a pure white light; if the transparency does not have any ink on it, then the
white light just passes through the transparency and we see white.
What is the color of the pixel resulting from the superposition of one or
more transparencies?
When we drop some ink on the transparency and hold the transparency
to the light we see the color that the ink lets pass through. When more trans-
parencies get stacked together, the color of the resulting pixel depends on the
absorption properties of the inks on all of the transparencies.
Let
1
= (x
1
;y
1
;z
1
) and
2
= (x
2
;y
2
;z
2
) be two colors and assume that
two pixels of color
1
and of color
2
are printed on two different transparen-
cies.
The following operator
add
describes the color superposition operation:
add
(
1
;
2
) =
x
1
x
2
L
;
int
y
1
y
2
L
;
int
z
1
z
2
L
:
int
Notice that taking into account only the inks that we have used for each
transparency is a simplification: the perception of the final color depends also
on the material of the transparencies and the aberrations that the stack of
transparencies produces. Moreover, it is likely that the initial light we start
with is not a pure white light and that there are also other sources of light in
the environment. However, the
add
operator is a quite good approximation.
The
add
operation is commutative and thus the order in which we su-
perpose the colors is irrelevant. As expected, it results that
add
(
Y
;
M
) =
R
,
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