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 65      <h1>bLSAG ring signatures overview</h1>
 66
 67<p><em>2022-07-20</em></p>
 68
 69<blockquote>
 70<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>
 71</blockquote>
 72
 73<p><br>
 74bLSAG: Back&rsquo;s Linkable Spontaneous Anonymous Group signatures</p>
 75
 76<ul>
 77<li>signer ambiguity</li>
 78<li>linkability</li>
 79<li>unforgeability</li>
 80</ul>
 81
 82<h3>Setup</h3>
 83
 84<p>Let <span class="math inline">\(G\)</span> be the generator of an EC group.
 85We 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>.
 86Signer&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>.
 87Set of Public Keys: <span class="math inline">\(\mathcal{R} = \{ K_1, K_2, \ldots, K_n \}\)</span></p>
 88
 89<pre><code class="language-python">def new_key():
 90    k = F.random_element()
 91    K = g * k # g is the generator of the EC group
 92    return K
 93</code></pre>
 94
 95<h3>Signature</h3>
 96
 97<ol>
 98<li><p>compute key image: <span class="math inline">\(\tilde{K} = k_{\pi} \mathcal{H_p} ( K_{\pi}) \in G\)</span></p>
 99
100<pre><code class="language-python">key_image = k * hashToPoint(K)
101</code></pre></li>
102
103<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>
104
105<ul>
106<li><p><span class="math inline">\(r_i\)</span> is used for the fake responses</p>
107
108<pre><code class="language-python">a = F.random_element()
109r = [None] * len(R)
110for i in range(0, len(R)):
111    if i==pi:
112        continue
113
114
115    r[i] = mod(F.random_element(), p)
116</code></pre></li>
117</ul></li>
118
119<li><p>Compute <span class="math inline">\(c_{\pi + 1} = \mathcal{H}_n ( m, [\alpha G], [\alpha \mathcal{H}_p(K_{\pi})])\)</span></p>
120
121<pre><code class="language-python">c[pi1] = hash(R, m, a * g, hashToPoint(R[pi]) * a, p)
122</code></pre></li>
123
124<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>
125$<span class="math inline">\(
126c_{i+1} = \mathcal{H}_n (m, [r_i G + c_i K_i], [r_i \mathcal{H}_p (K_i) + c_i \tilde{K}])
127\)</span>$</p>
128
129<ul>
130<li>Notice that (from step 3 &amp; 4):<br>
131<span class="math inline">\(\alpha \mathcal{H}_p (K_{\pi}) = r_{\pi} \mathcal{H}_p (K_{\pi}) + c_{\pi} \cdot (\tilde{K})\)</span>,<br>
132where <span class="math inline">\(\tilde{K}= k_{\pi} \mathcal{H_p} ( K_{\pi})\)</span>, so:<br>
133<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>
134which is equal to,<br>
135<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>
136From where we can see: <span class="math inline">\(\alpha = r_{\pi} + c_{\pi} \cdot k_{\pi}\)</span><br>
137which we can rearrange to
138<span class="math inline">\(r_{\pi} = \alpha - c_{\pi} \cdot k_{\pi}\)</span>.<br><br></li>
139</ul>
140
141<pre><code class="language-python">for j in range(0, len(R)-1):
142    i = mod(pi1+j, len(R))
143    i1 = mod(pi1+j +1, len(R))
144
145
146    c[i1] = hash(R, m, r[i] * g + c[i] * R[i],
147            r[i] * hashToPoint(R[i]) + c[i] * key_image, p)
148</code></pre></li>
149
150<li><p>Define <span class="math inline">\(r_{\pi} = \alpha - c_{\pi} k_{\pi} \mod{p}\)</span></p></li>
151</ol>
152
153<pre><code class="language-python">r[pi] = mod(a - c[pi] * k, p)
154</code></pre>
155
156<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>.
157- <span class="math inline">\(len(\sigma(m)) = 1+n\)</span></p>
158
159<pre><code class="language-python">return [c[0], r]
160</code></pre>
161
162<p><br><br><br></p>
163
164<h4>Step by step (simplified):</h4>
165
166<div style="overflow:auto;">
167<div style="width: 60%; float:left; height: 360px; overflow-y:scroll;">
168    
169<img src="img/posts/ring-sig/step00.png" style="width:100%;" />
170<img src="img/posts/ring-sig/step00.png" style="width:100%;" />
171<img src="img/posts/ring-sig/step01.png" style="width:100%;" />
172<img src="img/posts/ring-sig/step02.png" style="width:100%;" />
173<img src="img/posts/ring-sig/step03.png" style="width:100%;" />
174<img src="img/posts/ring-sig/step04.png" style="width:100%;" />
175<img src="img/posts/ring-sig/step05.png" style="width:100%;" />
176<img src="img/posts/ring-sig/step06.png" style="width:100%;" />
177    
178</div>
179
180<div style="width: 40%; float:right; margin-top:80px;">
181    
182    <ul>
183<li>Generate $r_i \in^R \mathbb{Z_p}$</li>
184<li>Compute $c_{i+1}$ from $r_i$</li>
185<li>Link $r_{\pi}$ with $c_{\pi}$</li>
186</ul>
187
188</div>
189</div>
190
191<p><em>You can scroll down the images through the step-by-step diagrams.</em></p>
192
193<p><br></p>
194
195<p>It reminds in some way to the approach to close a box like the one in the picture:
196<img src="img/posts/ring-sig/box-closed.png" alt="" /></p>
197
198<p><br><br><br></p>
199
200<h3>Verification</h3>
201
202<ol>
203<li><p>check <span class="math inline">\(p \tilde{K} \stackrel{?}{=} 0\)</span></p>
204
205<ul>
206<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>
207</ul></li>
208
209<li><p>for <span class="math inline">\(i = 1, 2, \ldots, n\)</span>, replacing <span class="math inline">\(n+1 \rightarrow 1\)</span>
210$<span class="math inline">\(
211c'_{i+1} = \mathcal{H}_n (m, [r_i G + c_i K_i], [r_i \mathcal{H}_p (K_i) + c_i \tilde{K}])
212\)</span>$</p></li>
213
214<li><p>check <span class="math inline">\(c_1 \stackrel{?}{=} c'_i\)</span></p>
215
216<pre><code class="language-python">c[0] = c1
217for j in range(0, len(R)):
218    i = mod(j, len(R))
219    i1 = mod(j+1, len(R))
220    c[i1] = hash(R, m, r[i] * g + c[i] * R[i],
221                 r[i] * hashToPoint(R[i]) + c[i] * key_image, p)
222
223
224assert c1 == c[0]
225</code></pre></li>
226</ol>
227
228<p><br><br></p>
229
230<h2>Links</h2>
231
232<p>Toy implementation:</p>
233
234<ul>
235<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>
236<li>Rust: <a href="https://github.com/arnaucube/ring-signatures-rs">https://github.com/arnaucube/ring-signatures-rs</a></li>
237</ul>
238
239<p>Resources:</p>
240
241<ul>
242<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>
243(section <em>&ldquo;3.4 Back’s Linkable Spontaneous Anonymous Group (bLSAG) signatures&rdquo;</em>)</li>
244</ul>
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