SESSION NOVEL ALGORIHMS AND APPLICATIONS + ...

30 Int'l Conf. Foundations of Computer Science | FCS'11 |
secret image. Hence we can finger out the secret when
stacking at least 4 shares. For the following algorithm,
let m = n 2 n n
– 2n = ( n − 3)
C1
+ Cn
−1
=
n
n 2
n −5n
+ 6 n
C2 ( n − 2)
Cn
+ C 2 0
+ .
(4, n) scheme algorithm:
Input: A binary secret image S with size w × h and the
value of n.
Output: n shares R1, R2, …, Rn, each with size w × h.
1. Let C0 = [M(n, 2)||(n – 3) × M(n, n)||((n 2 – 5n + 6)
/ 2) × M(n, 0)] and C1 = [(n – 3) × M(n, 1)||M(n,
n – 1)], the size of C0 and C1 are both n × m.
2. for (1 ≤ i ≤ h ; 1 ≤ j ≤ w)
t = random(1..m);
for (1 ≤ k ≤ n)
if (S(i, j) == 0)
Rk(i, j) = C0(k, t);
else
Rk(i, j) = C1(k, t);
Theorem 1. In the proposed scheme, we stack at least
four shares can reveal the secret, and stack one, two or
three shares cannot.
Proof. We use the light transmission rate to prove that
we cannot recognize the secret if we stack less than
four shares, but we can see the image if at least four
shares stack together. We divided into five cases. The
first three cases are to prove that if we stack less than
four shares, the light transmission rate for the white
and black pixel of the stacked image are in the same.
The last two cases prove the light transmission rate
for the white pixel of the stacked image is larger than
the light transmission rate for the black pixel of the
stacked image when at least four shares stack together.
Therefore we can obey the (4, n)-threshold secret
sharing scheme. For A = C0 or C1, consider the
proposed algorithm, the definition of fi(A) and ℑ = 1 −
(b / p), we have ℑ = 1 − (fi(A) / m) = ℑ(A, k). Note
that Lemma 1 will be used in the following proof.
Case 1. For any one share
ℑ(C0, 1)
1 − ,|− ,| ,
=1 − , + − , +
,
−
= − +
− ,
ℑ(C1, 1)
= 1 − − ,||, −
= 1 − − , + , −
−
= − +
− .
So ℑ(C0, 1) = ℑ(C1, 1) when one get one share,
and one cannot see any information.
Case 2. Stack any two shares
ℑ(C0, 2)
= 1 − ,|− ,| ,
= 1 − , + − ,+
,
−
= − +
− ,
ℑ(C1, 2)
= 1 − − ,||, −
= 1 − − , + ,−
− + −
= − +
− .
So ℑ(C0, 2) = ℑ(C1, 2). That means when
stacking any two shares, we cannot see any
information.
Case 3. Stack any three shares
ℑ(C0, 3)
= 1 − ,||− ,|| ,
=1− , + − , + ,
−
= − +
− ,
ℑ(C1, 3)
= 1 − ,||,
= 1 − − , + ,−
−
− +
− .
Again, ℑ(C0, 3) = ℑ(C1, 3) when stacking any
three shares, so we cannot see any information.
Case 4. Stack any four shares
ℑ(C0, 4)
= 1 − ,– , ,
=1 − , + − , +
,
−
= − +
− ,
ℑ(C1, 4)