Score:-2

Is it working?? (Asymmetric bilinear pairing)

fr flag

Let we have $G_1 \times G_2 \to G_t$

let $g_1 \gets G_1, \\g_{2.1} \gets G_2, \\g_{2.2} \gets G_2$

$$e(g_1^a,g_{2.1}^b) = e(g_1, g_{2.1})^{ab}$$

$$e(g_1^a,g_{2.2}^b) = e(g_1, g_{2.2})^{ab}$$

Is it working?

kelalaka avatar
in flag
Welcome to [cryptography.se]. Could you check the edits? and verify the answer is on the correct path.
Score:0
my flag

Is it work?

Well, it's a bit hard to figure out what you mean by your notation, but in general:

$$e(g_1^a,g_{2.1}^b) = e(g_1, g_{2.2})^{ab}$$

will be true only if $g_{2.1} = g_{2.2}$ (or $a=0$ or $b=0$). As you appear to say that these will both be selected independently, this is not likely.

Or, do you mean that both these hold simultaneously:

$$e(g_1^a,g_{2.1}^b) = e(g_1, g_{2.1})^{ab}$$ $$e(g_1^a,g_{2.2}^b) = e(g_1, g_{2.2})^{ab}$$

If that's what you're asking, well, that's true.

user104620 avatar
fr flag
Oh yes!! Second one is my question!! Thank you very much!!
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