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bLSAG ring signatures overview

2022-07-20

Note: I’m not a mathematician, I’m just an amateur on math. These notes are just an attempt to try to sort the notes that I took while learning abut bLSAG.


bLSAG: Back's Linkable Spontaneous Anonymous Group signatures
  • signer ambiguity
  • linkability
  • unforgeability

Setup

Let $G$ be the generator of an EC group. We use a hash function $\mathcal{H}p$, which maps to curve points in EC, and a normal hash $\mathcal{H}n$, which maps to $\mathbb{Z}p$. Signer's key pair: $k{\pi}$, s.t. $K{\pi} = k{\pi} \cdot G \in \mathcal{R}$, with secret index $\pi$. Set of Public Keys: $\mathcal{R} = { K_1, K_2, \ldots, K_n }$

def new_key():
    k = F.random_element()
    K = g * k # g is the generator of the EC group
    return K

Signature

  1. compute key image: $\tilde{K} = k_{\pi} \mathcal{H_p} ( K_{\pi}) \in G$

    key_image = k * hashToPoint(K)
    
  2. Generate $\alpha \in^R \mathbb{Z}_p$, and $r_i \in^R \mathbb{Z}_p$, for $i \in {1, 2, \ldots, n }$, with $i \neq \pi$

    • $r_i$ is used for the fake responses
        a = F.random_element()
        r = [None] * len(R)
        for i in range(0, len(R)):
            if i==pi:
                continue
    
            r[i] = mod(F.random_element(), p)
    
  3. Compute $c_{\pi + 1} = \mathcal{H}_n ( m, [\alpha G], [\alpha \mathcal{H}_p(K_{\pi})])$

    c[pi1] = hash(R, m, a * g, hashToPoint(R[pi]) * a, p)
    
  4. for $i=\pi + 1, \pi +2, \ldots, n, 1, 2, \ldots, \pi -1$, calculate, replacing $n+1 \rightarrow 1$ $$ c_{i+1} = \mathcal{H}_n (m, [r_i G + c_i K_i], [r_i \mathcal{H}_p (K_i) + c_i \tilde{K}]) $$

    • Notice that (from step 3 & 4):
      $\alpha \mathcal{H}p (K{\pi}) = r_{\pi} \mathcal{H}_p (K_{\pi}) + c_{\pi} \cdot (\tilde{K})$,
      where $\tilde{K}= k_{\pi} \mathcal{H_p} ( K_{\pi})$, so:
      $\alpha \mathcal{H}_p (K_{\pi}) = r_{\pi} \mathcal{H}_p (K_{\pi}) + c_{\pi} \cdot (k_{\pi} \mathcal{H}_p(K_{\pi}))$
      which is equal to,
      $\alpha \cdot \mathcal{H}_p (K_{\pi}) = (r_{\pi} + c_{\pi} \cdot k_{\pi}) \cdot \mathcal{H}_p(K_{\pi})$
      From where we can see: $\alpha = r_{\pi} + c_{\pi} \cdot k_{\pi}$
      which we can rearrange to $r_{\pi} = \alpha - c_{\pi} \cdot k_{\pi}$.

    for j in range(0, len(R)-1):
        i = mod(pi1+j, len(R))
        i1 = mod(pi1+j +1, len(R))
    
        c[i1] = hash(R, m, r[i] * g + c[i] * R[i],
                r[i] * hashToPoint(R[i]) + c[i] * key_image, p)
    
  5. Define $r_{\pi} = \alpha - c_{\pi} k_{\pi} \mod{p}$

r[pi] = mod(a - c[pi] * k, p)

Signature: $\sigma(m) = (c_1, r_1, \ldots, r_n)$, with key image $\tilde{K}$ and ring $\mathcal{R}$.

  • $len(\sigma(m)) = 1+n$
return [c[0], r]




Step by step (simplified):

<ul>
  • Generate $r_i \in^R \mathbb{Z_p}$
  • Compute $c_{i+1}$ from $r_i$
  • Link $r_{\pi}$ with $c_{\pi}$
  • You can scroll down the images through the step-by-step diagrams.


    It reminds in some way to the approach to close a box like the one in the picture:




    Verification

    1. check $p \tilde{K} \stackrel{?}{=} 0$
      • to ensure that $\tilde{K} \in G$ (and not in a cofactor group of $G$)
    2. for $i = 1, 2, \ldots, n$, replacing $n+1 \rightarrow 1$ $$ c'_{i+1} = \mathcal{H}_n (m, [r_i G + c_i K_i], [r_i \mathcal{H}_p (K_i) + c_i \tilde{K}]) $$
    3. check $c_1 \stackrel{?}{=} c'_i$
      c[0] = c1
      for j in range(0, len(R)):
          i = mod(j, len(R))
          i1 = mod(j+1, len(R))
          c[i1] = hash(R, m, r[i] * g + c[i] * R[i],
                       r[i] * hashToPoint(R[i]) + c[i] * key_image, p)
      
      assert c1 == c[0]
      



    Toy implementation:

    Resources: