Update 2024-01-31-Sumcheck.md

Change = to \in

_posts/2024-01-31-Sumcheck.md

@@ -394,11 +394,11 @@ $f(x,y)\cdot f(y,z)\cdot f(z,x) = 1$
 
 Even if one of the two vertices out of the three doesn't have an edge between them that $f$ will evaluate to $0$ i.e. the above multiplication will result $0$ & it will not form a triangle. So the above equation can be used to count the number of triangles in a graph. However, the triangle formed by $3$ edges $(xy)(yz)(zx)$ is the same as that formed by $(yx)(zy)(xz)$ - each triangle can be represented in 6 different ways by $f$. So to get the count of the total number of triangles in the graph, we have to divide by $6$.
 
-$\frac {1}{6} \times \sum_{x=\lbrace 0, 1 \rbrace^{2}} \sum_{y=\lbrace 0, 1 \rbrace^{2}} \sum_{z=\lbrace 0, 1 \rbrace^{2}} f(x,y)\cdot f(y,z) \cdot f(z,x)$
+$\frac {1}{6} \times \sum_{x \in \lbrace 0, 1 \rbrace^{2}} \sum_{y \in \lbrace 0, 1 \rbrace^{2}} \sum_{z \in \lbrace 0, 1 \rbrace^{2}} f(x,y)\cdot f(y,z) \cdot f(z,x)$
 
 We can also write this as 
 
-$\frac {1}{6} \times \sum_{x_1=\lbrace 0, 1 \rbrace}\sum_{x_2=\lbrace 0, 1 \rbrace} \sum_{y_1=\lbrace 0, 1 \rbrace}\sum_{y_2=\lbrace 0, 1 \rbrace} \sum_{z_1=\lbrace 0, 1 \rbrace}\sum_{z_2=\lbrace 0, 1 \rbrace} f(x_1,x_2,y_1,y_2)\cdot f(y_1,y_2,z_1,z_2) \cdot f(z_1,z_2,x_1,x_2)$
+$\frac {1}{6} \times \sum_{x_1 \in \lbrace 0, 1 \rbrace}\sum_{x_2 \in \lbrace 0, 1 \rbrace} \sum_{y_1 \in \lbrace 0, 1 \rbrace}\sum_{y_2 \in \lbrace 0, 1 \rbrace} \sum_{z_1 \in \lbrace 0, 1 \rbrace}\sum_{z_2 \in \lbrace 0, 1 \rbrace} f(x_1,x_2,y_1,y_2)\cdot f(y_1,y_2,z_1,z_2) \cdot f(z_1,z_2,x_1,x_2)$
 
 So we need 6 different variables & not just 4. We can take our original $f$ & create 3 MLEs from it. Instead of $x_i, y_i, z_i$ we will use  6 variables $x_1, x_2, x_3, x_4, x_5 , x_6$ & then create 3 symmetrical $f_i$'s like below
 
@@ -419,7 +419,7 @@ $g(x_1,x_2,x_3,x_4,x_5,x_6) = f_1 \cdot f_2 \cdot f_3$
 
 When $\mathcal P$ computes the below,
 
-$\sum_{x_1=\lbrace 0, 1 \rbrace}\sum_{x_2=\lbrace 0, 1 \rbrace} \sum_{x_3=\lbrace 0, 1 \rbrace}\sum_{x_4=\lbrace 0, 1 \rbrace} \sum_{x_5=\lbrace 0, 1 \rbrace}\sum_{x_6=\lbrace 0, 1 \rbrace} g(x_1,x_2,x_3,x_4,x_5,x_6)$
+$\sum_{x_1 \in \lbrace 0, 1 \rbrace}\sum_{x_2 \in \lbrace 0, 1 \rbrace} \sum_{x_3 \in \lbrace 0, 1 \rbrace}\sum_{x_4 \in \lbrace 0, 1 \rbrace} \sum_{x_5 \in \lbrace 0, 1 \rbrace}\sum_{x_6 \in \lbrace 0, 1 \rbrace} g(x_1,x_2,x_3,x_4,x_5,x_6)$
 
 she will get the answer $24$. And $\frac {1}{6}\times 24 = 4$