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From the paper Predicate Encryption Supporting Disjunctions, Polynomial Equations, and Inner Products, the vector of the secret key is generated from a polynomial $p(x)$. I understand the univariate polynomial, but what does the vector of a bivariate polynomial look like? For example:

univariate polynomial

$$p(x) = (x - a) \cdot (x - b) = x^2 -(a+b)x + ab$$

  • Secret vector $p = (1, -(a+b), ab)$
  • Vector $w$ (with attribute $a$) = $(a^2, a, 1)$ (I am not sure if I understand it correctly)
  • Inner product: $\left<p,w\right> = a^2 - a^2 - ab + ab = 0$

bivariate polynomial

$$p(x) = (x_1 - a) \cdot (x_2 - b)$$

  • Secret vector $p = ?$
  • Vector $w$ with attribute $a = ?$

Many thanks.

Hilder Vitor Lima Pereira
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Tri Vo Hoang
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1 Answers1

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In general, to encode a polynomial $p(x) = a_n x^n + a_{n-1} x^{n-1} + ... a_1 x + a_0$, $$\mathbb{p} = (a_n, a_{n-1}, ... a_1, a_0)$$ and $$\mathbb{w} = (x^n, x^{n-1}, ... x, 1)$$

So for your univariate case, $\mathbb{w} = (x^2, x, 1)$. You can see that $\left<\mathbb{p}, \mathbb{w} \right> = p(x)$, and you substitute the attribute as you need. So I think your univariate case is correct.

For the bivariate case, it's the same, but with mixed powers. You'll have terms like $x_1^i x_2^j$, where $i+j \leq n$. So in the general case where $n=2$ with two variables / attributes, you have:

$$ p(x_1, x_2) = a_{20}x_1^2 + a_{11}x_1 x_2 + a_{02} x_2^2 + a_{10}x_1 + a_{01}x_2 + a_{00}$$

In your case, you have:

$$ p(x_1, x_2) = x_1 x_2 - x_1 b - x_2 a + ab$$

So $a_{20} = a_{02} = 0$, $a_{11} = 1$, $a_{10} = -b$, $a_{01} = -a$, and $a_{00} = ab$. Finally,

$$\mathbb{p} = (0,1,0,-b,-a,ab)$$

and

$$\mathbb{w} = (x_1^2, x_1 x_2, x_2^2, x_1, x_2, 1)$$

Again, you can see that $\left<\mathbb{p}, \mathbb{w} \right> = p(x_1, x_2)$.