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      <h1>bLSAG ring signatures overview</h1>

<p><em>2022-07-20</em></p>

<blockquote>
<p>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.</p>
</blockquote>

<p><br>
bLSAG: Back&rsquo;s Linkable Spontaneous Anonymous Group signatures</p>

<ul>
<li>signer ambiguity</li>
<li>linkability</li>
<li>unforgeability</li>
</ul>

<h3>Setup</h3>

<p>Let <span class="math inline">\(G\)</span> be the generator of an EC group.
We use a hash function <span class="math inline">\(\mathcal{H}_p\)</span>, which maps to curve points in EC, and a normal hash <span class="math inline">\(\mathcal{H}_n\)</span>, which maps to <span class="math inline">\(\mathbb{Z}_p\)</span>.
Signer&rsquo;s key pair: <span class="math inline">\(k_{\pi}\)</span>, s.t. <span class="math inline">\(K_{\pi} = k_{\pi} \cdot G \in \mathcal{R}\)</span>, with secret index <span class="math inline">\(\pi\)</span>.
Set of Public Keys: <span class="math inline">\(\mathcal{R} = \{ K_1, K_2, \ldots, K_n \}\)</span></p>

<pre><code class="language-python">def new_key():
    k = F.random_element()
    K = g * k # g is the generator of the EC group
    return K
</code></pre>

<h3>Signature</h3>

<ol>
<li><p>compute key image: <span class="math inline">\(\tilde{K} = k_{\pi} \mathcal{H_p} ( K_{\pi}) \in G\)</span></p>

<pre><code class="language-python">key_image = k * hashToPoint(K)
</code></pre></li>

<li><p>Generate <span class="math inline">\(\alpha \in^R \mathbb{Z}_p\)</span>, and <span class="math inline">\(r_i \in^R \mathbb{Z}_p\)</span>, for <span class="math inline">\(i \in \{1, 2, \ldots, n \}\)</span>, with <span class="math inline">\(i \neq \pi\)</span></p>

<ul>
<li><p><span class="math inline">\(r_i\)</span> is used for the fake responses</p>

<pre><code class="language-python">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)
</code></pre></li>
</ul></li>

<li><p>Compute <span class="math inline">\(c_{\pi + 1} = \mathcal{H}_n ( m, [\alpha G], [\alpha \mathcal{H}_p(K_{\pi})])\)</span></p>

<pre><code class="language-python">c[pi1] = hash(R, m, a * g, hashToPoint(R[pi]) * a, p)
</code></pre></li>

<li><p>for <span class="math inline">\(i=\pi + 1, \pi +2, \ldots, n, 1, 2, \ldots, \pi -1\)</span>, calculate, replacing <span class="math inline">\(n+1 \rightarrow 1\)</span>
$<span class="math inline">\(
c_{i+1} = \mathcal{H}_n (m, [r_i G + c_i K_i], [r_i \mathcal{H}_p (K_i) + c_i \tilde{K}])
\)</span>$</p>

<ul>
<li>Notice that (from step 3 &amp; 4):<br>
<span class="math inline">\(\alpha \mathcal{H}_p (K_{\pi}) = r_{\pi} \mathcal{H}_p (K_{\pi}) + c_{\pi} \cdot (\tilde{K})\)</span>,<br>
where <span class="math inline">\(\tilde{K}= k_{\pi} \mathcal{H_p} ( K_{\pi})\)</span>, so:<br>
<span class="math inline">\(\alpha \mathcal{H}_p (K_{\pi}) = r_{\pi} \mathcal{H}_p (K_{\pi}) + c_{\pi} \cdot (k_{\pi} \mathcal{H}_p(K_{\pi}))\)</span><br>
which is equal to,<br>
<span class="math inline">\(\alpha \cdot \mathcal{H}_p (K_{\pi}) = (r_{\pi} + c_{\pi} \cdot k_{\pi}) \cdot \mathcal{H}_p(K_{\pi})\)</span><br>
From where we can see: <span class="math inline">\(\alpha = r_{\pi} + c_{\pi} \cdot k_{\pi}\)</span><br>
which we can rearrange to
<span class="math inline">\(r_{\pi} = \alpha - c_{\pi} \cdot k_{\pi}\)</span>.<br><br></li>
</ul>

<pre><code class="language-python">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)
</code></pre></li>

<li><p>Define <span class="math inline">\(r_{\pi} = \alpha - c_{\pi} k_{\pi} \mod{p}\)</span></p></li>
</ol>

<pre><code class="language-python">r[pi] = mod(a - c[pi] * k, p)
</code></pre>

<p>Signature: <span class="math inline">\(\sigma(m) = (c_1, r_1, \ldots, r_n)\)</span>, with key image <span class="math inline">\(\tilde{K}\)</span> and ring <span class="math inline">\(\mathcal{R}\)</span>.
- <span class="math inline">\(len(\sigma(m)) = 1+n\)</span></p>

<pre><code class="language-python">return [c[0], r]
</code></pre>

<p><br><br><br></p>

<h4>Step by step (simplified):</h4>

<div style="overflow:auto;">
<div style="width: 60%; float:left; height: 360px; overflow-y:scroll;">
    
<img src="img/posts/ring-sig/step00.png" style="width:100%;" />
<img src="img/posts/ring-sig/step00.png" style="width:100%;" />
<img src="img/posts/ring-sig/step01.png" style="width:100%;" />
<img src="img/posts/ring-sig/step02.png" style="width:100%;" />
<img src="img/posts/ring-sig/step03.png" style="width:100%;" />
<img src="img/posts/ring-sig/step04.png" style="width:100%;" />
<img src="img/posts/ring-sig/step05.png" style="width:100%;" />
<img src="img/posts/ring-sig/step06.png" style="width:100%;" />
    
</div>

<div style="width: 40%; float:right; margin-top:80px;">
    
    <ul>
<li>Generate $r_i \in^R \mathbb{Z_p}$</li>
<li>Compute $c_{i+1}$ from $r_i$</li>
<li>Link $r_{\pi}$ with $c_{\pi}$</li>
</ul>

</div>
</div>

<p><em>You can scroll down the images through the step-by-step diagrams.</em></p>

<p><br></p>

<p>It reminds in some way to the approach to close a box like the one in the picture:
<img src="img//posts/ring-sig/box-closed.png" alt="" /></p>

<p><br><br><br></p>

<h3>Verification</h3>

<ol>
<li><p>check <span class="math inline">\(p \tilde{K} \stackrel{?}{=} 0\)</span></p>

<ul>
<li>to ensure that <span class="math inline">\(\tilde{K} \in G\)</span> (and not in a cofactor group of <span class="math inline">\(G\)</span>)</li>
</ul></li>

<li><p>for <span class="math inline">\(i = 1, 2, \ldots, n\)</span>, replacing <span class="math inline">\(n+1 \rightarrow 1\)</span>
$<span class="math inline">\(
c'_{i+1} = \mathcal{H}_n (m, [r_i G + c_i K_i], [r_i \mathcal{H}_p (K_i) + c_i \tilde{K}])
\)</span>$</p></li>

<li><p>check <span class="math inline">\(c_1 \stackrel{?}{=} c'_i\)</span></p>

<pre><code class="language-python">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]
</code></pre></li>
</ol>

<p><br><br></p>

<h2>Links</h2>

<p>Toy implementation:</p>

<ul>
<li>Sage: <a href="https://github.com/arnaucube/math/blob/master/ring-signatures.sage">https://github.com/arnaucube/math/blob/master/ring-signatures.sage</a></li>
<li>Rust: <a href="https://github.com/arnaucube/ring-signatures-rs">https://github.com/arnaucube/ring-signatures-rs</a></li>
</ul>

<p>Resources:</p>

<ul>
<li><em>&ldquo;Zero to Monero&rdquo;</em> - <a href="https://web.getmonero.org/library/Zero-to-Monero-2-0-0.pdf">https://web.getmonero.org/library/Zero-to-Monero-2-0-0.pdf</a>
(section <em>&ldquo;3.4 Back’s Linkable Spontaneous Anonymous Group (bLSAG) signatures&rdquo;</em>)</li>
</ul>

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