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I am trying to work through an example of Baek and Zheng’s “threshold identity-based decryption scheme” and I tried a lot to solve this, but somewhere I’m doing something wrong.

I hope someone can help me rectify my error, telling me where I might be making a mistake.

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Example:

Let $q=11,$
$\mathbb{G}=\mathbb{Z}_{11}=\{0,1,2,3,4,5,6,7,8,9,10\},\\ P=1,$
$\mathbb{G}_T=\text{Quadratic residues of }\mathbb{Z}_{23}=\mathbb{Z}^*/23=\{1,2,3,4,6,8,9,12,13,16,18\}$,

$e(x,y)=3^{xy} \bmod{23}$
$H_1(x)= x \bmod{11}\\ H_2(x) = x \text{ in binary}\\ H_3(x,y)=xy \bmod{11}$

Let $s=4$, then $PK=sP=4$.

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Example :

Let $n=7$,

$SG=\{P_1,P_2,P_3,\cdots,P_7\}$,

$t´=3$,

$ID_{SG}=14$

Now, $SK_{SG}=sH_1(ID_{SG}) = 4*3 \equiv 1 \mod 11 $

$$R(z)=1+2z+3z^2$$

Let $i$ be the public identity for $P_i$.

Now, $$[SK_{SG}]_1 = R(1) =6,\\ [SK_{SG}]_2 = R(2) =6,\\ [SK_{SG}]_3 = R(3) =1,\\ [SK_{SG}]_4 = R(4) =2,\\ [SK_{SG}]_5 = R(5) =9,\\ [SK_{SG}]_6 = R(6) =0,\\ [SK_{SG}]_7 = R(7) =8$$

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Example:

Let $\ell=5,$

$ m =(11101)_2,$

$ r=4$

Now, $$k=e(4,3)^4 = 5$$

$$U=rP=4\\ V=H_2(k) \oplus m = 00101 \oplus 111101 = 11000\\ W=4*H_3(4,24)=4*8=32\\ C=(U,V,W)=(4,24,32)$$

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Example:

Now, $e(P,W)=e(1,32)=e(4,8)=e(U,H_3(U,V))$

So, $$k_1=e(U,[SK_{SG}]_1)=e(4,6)=4\\ k_2=e(U,[SK_{SG}]_2)=e(4,6)=4\\ k_3=e(U,[SK_{SG}]_3)=e(4,1)=12\\ k_4=e(U,[SK_{SG}]_4)=e(4,2)=5\\ k_5=e(U,[SK_{SG}]_5)=e(4,9)=3\\ k_6=e(U,[SK_{SG}]_6)=e(4,0)=1\\ k_7=e(U,[SK_{SG}]_7)=e(4,8)=9$$

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Let $$A=\{P_1,P_2,P_3\}$$

Now,

$$\lambda^{A}_{01}=3,\lambda^{A}_{02}=-3,\lambda^{A}_{03}=1\\ k_1=4,k_2=4,k_3=12\\ k=k_1^{\lambda^{A}_{01}}*k_2^{\lambda^{A}_{02}}*k_3^{\lambda^{A}_{03}}=4^3*4^{-3}*12^1=12$$

… which is not equal to actual $k$ (which is $5$)!?!

Please tell me where I went wrong, or maybe even provide me with a numerical example (in integers) as a reference/guidance for me to follow.

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