Update 2023-02-07-R1CSQAP.md

Author: RisenCrypto

_posts/2023-02-07-R1CSQAP.md

@@ -272,11 +272,13 @@ This has 6 left polynomials, one in each row.
 
 Our solution vector $S = [1, out, x, var1, var2, var3]$ also has 6 elements. So we can do a dot product of $S\cdot Lm$ whose output again will be a polynomial. Likewise for the other 2 polynomial matrices.
 
+
 ~~~ruby
 # We define the solution vector also in the field
 S = vector(Fp,[1, 35, 3, 9, 27, 30])
 
 # Create the Lx, Rx & Ox polynomial
+# In the program, we use Fx to denote F(x)
 Lx = Rp(list(S*PolyM[0]))
 Rx = Rp(list(S*PolyM[1]))
 Ox = Rp(list(S*PolyM[2]))
@@ -290,9 +292,9 @@ Ox = 537*x^3 + 296*x^2 + 499*x + 600
 
 ~~~
 
-So by taking a dot product of $S$ with each of $Lm$, $Rm$ and $Om$, we created 3 polynomials $Lx$, $Rx$ and $Ox$ 
+So by taking a dot product of $S$ with each of $Lm$, $Rm$ and $Om$, we created 3 polynomials $L(x)$, $R(x)$ and $O(x)$ 
 
-We can define a new polynomial $T = Lx *  Rx - Ox$.
+We can define a new polynomial $T(x) = L(x) *  R(x) - O(x)$.
 
 ~~~ruby
 ....: # New Polynomial T
@@ -322,7 +324,7 @@ T(4) = 0
 However, the Verifier doesn't know the polynomial $T$, nor can he compute it since he doesn't know the solution vector. So the Prover has to prove it to the Verifier that $T$ is zero at $x=1, x=2, x=3,x=4$. 
 If $T$ is zero at $x=1, x=2, x=3,x=4$, then it means these are all roots of $T$. So we create a new polynomial $Z$ known by both the Prover & the Verifier.
 
-$Z = (x-1)(x-2)(x-3)(x-4)$
+$Z(x) = (x-1)(x-2)(x-3)(x-4)$
 
 If $T$ is divided by $Z$, then it will be perfectly divisible & will leave no remainder i.e. there must exist a Polynomial $H$ such that $T(x) = H(x) \cdot Z(x)$