Probabilistic Visual Cryptography Schemes - Visual Cryptography and Secret Image Sharing - page 144

choices of ` 0 and m 0 , for the case of n. Since for n = 3 scheme S D has m = 3
clearly for m 0 = 3 we get a deterministic scheme; such a scheme is obtained
for ` 0 = 1, h 0 = 2. It is worth to notice that for other choices of ` 0 the
probabilistic factor of the scheme is 0 meaning that no probabilistic scheme
with pixel expansion m 0 = 3 can be constructed with Construction 1. For the
case of m 0 = 2 ` 0 = 1 and ` 0 = 2 are valid choices and they both yield a
1

3 -probabilistic scheme.

TABLE 5.1
Values of for scheme S 0 for
n = 3. The max over each
row is in boldface.
m 0 n` 0 ;h 0 0,1 1,2 2,3

1 1=3 { {

2 1=3 1=3 {

3 0 1 0

TABLE 5.2
Values of for n = 4. The max over each row is in
boldface.
m 0 n` 0 ;h 0 0,1 1,2 2,3 3,4 4,5 5,6

1 1=3 { { { {

0:333

2 1=5 7=15 { { { {

0:200 0:467

3 1=20 1=2 9=20 { { {

0:050 0:500 0:450

4 0 1=5 4=5 1=3 { {

0:050 0:800 0:333

5 0 0 1=2 1 1=6 {

0:167 0.500

6 0 0 0 1 1 0

Table 5.2 gives the resulting values of the probabilistic factor of S 0 over all
possible choices of ` 0 and m 0 , for the case of n = 4. Notice that for n = 4 scheme
S D has m = 6, hence, for m 0 = 6 deterministic schemes can be obtained (in
this case both ` 0 = 3 and ` 0 = 4 yield a deterministic scheme). As reported in
the table, also for m 0 = 5 a deterministic scheme can be obtained by choosing
` 0 = 3. For the other possible choices of m 0 , the table shows in boldface the
biggest probabilistic factor found.
Finally, Table 5.3 gives the resulting values of the probabilistic factor
of S 0 over all possible choices of ` 0 and m 0 , for the case of n = 5. For n = 5
scheme S D has m = 10, hence, m 0 can range from 2 to 10. As reported in