# p1gd0g.github.io

MRL’s paper is confusing, it’s better to read the source code.

There is a new output we want to make a “range proof”.

$C=aG+10H$

$10$ is the amount, $a$ is the secret key. G and H are different base point.

We split it in four, we get:

$C_0=a_0G+0 \times 1H$

$C_1=a_1G+1 \times 2H$

$C_2=a_2G+0 \times 4H$

$C_3=a_3G+1 \times 8H$

because $2+8=10$. $a_i$ is random.

For the first line, we get $(C_0,C_0 - 1 \times 1H)$ these two points.

We know:

1. The first point’s secret key is $a_0$.
2. We can’t compute the second point’s secret key.
3. The difference between the tow points is $1H$.

We sign a ring signature on these two points, a ring contains only two points.

$L_0= \alpha G$

$\alpha$ is random.

$q_1=H(L_0)$

$H()$ is a hash function to covert a point to scalar.

$L_1=s_1G+q_1P_1$

$s_1$ is random, $P_1$ is the second point.

$q_0=H(L_1)$

$s_0= \alpha -q_0a_0$ since $L_0= \alpha G=s_0G+q_0P_0=( \alpha G -q_0a_0)G+q_0P_0$

It’s easy to verify this signature because:

$L_0+L_1=(s_0+s_1)G+(q_0+q_1)H$

The second line is similar but we should change the order of $(P_0,P_1)$ because we only know the second point’s secret key.

At last we make four range proof.

In practice, the code is a little different for space-saving.