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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.
+
+<br>
+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: $\mathcal{R} = \{ K_1, K_2, \ldots, K_n \}$
+
+```python
+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: $\tilde{K} = k_{\pi} \mathcal{H_p} ( K_{\pi}) \in G$
+    ```python
+    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
+    ```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)
+    ```
+
+3. Compute $c_{\pi + 1} = \mathcal{H}_n ( m, [\alpha G], [\alpha \mathcal{H}_p(K_{\pi})])$
+
+    ```python
+    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):<br>
+    $\alpha \mathcal{H}_p (K_{\pi}) = r_{\pi} \mathcal{H}_p (K_{\pi}) + c_{\pi} \cdot (\tilde{K})$,<br>
+    where $\tilde{K}= k_{\pi} \mathcal{H_p} ( K_{\pi})$, so:<br>
+    $\alpha \mathcal{H}_p (K_{\pi}) = r_{\pi} \mathcal{H}_p (K_{\pi}) + c_{\pi} \cdot (k_{\pi} \mathcal{H}_p(K_{\pi}))$<br>
+    which is equal to,<br>
+    $\alpha \cdot \mathcal{H}_p (K_{\pi}) = (r_{\pi} + c_{\pi} \cdot k_{\pi}) \cdot \mathcal{H}_p(K_{\pi})$<br>
+    From where we can see: $\alpha = r_{\pi} + c_{\pi} \cdot k_{\pi}$<br>
+    which we can rearrange to
+    $r_{\pi} = \alpha - c_{\pi} \cdot k_{\pi}$.<br><br>
+    
+    ```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)
+    ```
+    
+6. Define $r_{\pi} = \alpha - c_{\pi} k_{\pi} \mod{p}$
+
+```python
+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 $\mathcal{R}$.
+- $len(\sigma(m)) = 1+n$
+
+```python
+return [c[0], r]
+```
+
+
+<br><br><br>
+
+#### Step by step (simplified):
+
+<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>
+
+*You can scroll down the images through the step-by-step diagrams.*
+
+<br>
+
+It reminds in some way to the approach to close a box like the one in the picture:
+![](img//posts/ring-sig/box-closed.png)
+
+
+<br><br><br>
+### 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 $c_1 \stackrel{?}{=} c'_i$
+    ```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]
+    ```
+
+<br><br>
+
+## Links
+Toy implementation:
+
+- Sage: https://github.com/arnaucube/math/blob/master/ring-signatures.sage
+- Rust: https://github.com/arnaucube/ring-signatures-rs
+
+Resources:
+
+- *"Zero to Monero"* - https://web.getmonero.org/library/Zero-to-Monero-2-0-0.pdf
+(section *"3.4 Back’s Linkable Spontaneous Anonymous Group (bLSAG) signatures"*)
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