2
$\begingroup$

I find the following IBE scheme from the videos posted and i don't understand the decryption algorithm, will any one please elaborate the 6th step

scheme

  1. Setup($\lambda$) : $(\mathbb{G}, \mathbb{G}_T, g, P)$ <-- GenBilgroup($\lambda$)

    $\alpha\leftarrow_R\mathbb{F}_p$

  2. Public Parameters : $pp = [ g , y = g^\alpha , g_1, h ] \in \mathbb{G}$

  3. Master secret : $mk = g_1^\alpha $

  4. Keygeneration: $k(mk,id)$: $d_1= mk\cdot(Y^{id\cdot h})^r, d_2 = g^r $ where $r\leftarrow_R\mathbb{F}_p $

  5. Encryption: $E(pp,id,m)$ : $c_1 = g^s, c_2=(Y^{id\cdot h})^s, c_3=m\cdot e(y, g_1)^s$

  6. Decryption: $e(c_1,d_1)/e(c_2,d_2)$ will give $e(y,g_1)^s$ How? and $m$ is obtained by dividing $c_3$ by that value.

$\endgroup$
1
  • 1
    $\begingroup$ You can use MathML / Tex on this site. $\endgroup$
    – Maarten Bodewes
    Commented Feb 16, 2015 at 14:50

1 Answer 1

1
$\begingroup$

I don't know whether it is correct

$$e(c_1,d_1) = e(g^s, mk\cdot(y^{id\cdot h})^r)$$

$$=e(g^s,mk)\cdot e(g^s, (y^{id\cdot h})^r)$$

$$= e(g^s , g_1^\alpha ) \cdot e(g^r , (y^{id\cdot h})^s)$$

$$= e(y , g_1^\alpha ) \cdot e(g^r , (y^{id\cdot h})^s)$$

$$e(c_2,d_2) = e((Y^{id\cdot h})^s , g^r)$$

$$= e((Y^{id\cdot h})^s , g^r)$$

therefore $e(c_1,d_1)/e(c_2,d_2) = e(y , g_1^\alpha )$

$\endgroup$
2
  • 1
    $\begingroup$ Yes, of course it's correct. A straitforward calculation using the bilinearity of e, shows that dividing $c_3$ by $e(y,g_1^\alpha)$ allows to recover m. I confirm. $\endgroup$ Commented Mar 21, 2015 at 17:23
  • $\begingroup$ @manju If the answer fits, you might want to “accept” it… thanks. $\endgroup$
    – e-sushi
    Commented Jul 19, 2015 at 22:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.