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.

enter image description here


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 $PK=sP=4$.

enter image description here

Example :

Let $n=7$,




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


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

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

enter image description here


Let $\ell=5,$

$ m =(11101)_2,$

$ r=4$

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

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

enter image description here


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

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

enter image description here

enter image description here

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


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

… 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.