Ts2

readme.md

@@ -80,10 +80,10 @@ diagram ahh. i aint do Ts.
 
 $$
 \begin{aligned}
-    y_1 &= e_k(x_1 \oplus IV)
-    y_i &= e_k(x_i \oplus y_{i-1})
-    x_1 &= e_k^{-1}(y_1) \oplus IV
-    x_i &= e_k^{-1}(y_i) \oplus y_{i-1}
+    y_1 &= e_k(x_1 \oplus IV) \\
+    y_i &= e_k(x_i \oplus y_{i-1}) \\
+    x_1 &= e_k^{-1}(y_1) \oplus IV \\
+    x_i &= e_k^{-1}(y_i) \oplus y_{i-1} \\
 \end{aligned}
 $$
 
@@ -91,10 +91,10 @@ $$
 
 $$
 \begin{aligned}
-    s_1 &= e_k(IV)
-    s_i &= e_k(s_{i-1})
-    y_i &= s_i \oplus x_i
-    x_i &= s_i \oplus y_i
+    s_1 &= e_k(IV) \\
+    s_i &= e_k(s_{i-1}) \\
+    y_i &= s_i \oplus x_i \\
+    x_i &= s_i \oplus y_i \\
 \end{aligned}
 $$
 
@@ -102,10 +102,10 @@ $$
 
 $$
 \begin{aligned}
-    y_1 &= e_k(IV) \oplus x_1
-    y_i &= e_k(y_{i-1}) \oplus x_i
-    x_1 &= e_k(IV) \oplus y_1
-    x_i &= e_k(y*{i-1}) \oplus y_i
+    y_1 &= e_k(IV) \oplus x_1 \\
+    y_i &= e_k(y_{i-1}) \oplus x_i \\
+    x_1 &= e_k(IV) \oplus y_1 \\
+    x_i &= e_k(y*{i-1}) \oplus y_i \\
 \end{aligned}
 $$
 
@@ -113,7 +113,7 @@ $$
 
 $$
 \begin{aligned}
-    y_i = e_k(IV \|\| CTR_i) \oplus x_i
+    y_i = e_k(IV \| CTR_i) \oplus x_i
 \end{aligned}
 $$
 
@@ -125,20 +125,20 @@ two kinds apparently. idk
 
 1. convert expo to binary. we go left to right. (Ts is buns)
 
-   lets say we have x, e = $\overline{e_1e_2e_3e_4}$ and n. We are trying to do $x^e \pmod n$ or just x^e
+   lets say we have x, e = $\overline{e_1e_2e_3e_4}_2$ and n. We are trying to do $x^e \pmod n$ or just x^e
 
 2. let do Ts
    1. of course
 
-      we have $\overline{e_1} = 1$
+      we have $e_1 = 1$
 
-      do $x^{\overline{e_1}} \pmod n$
+      do $x^{\overline{e_1}_2} \pmod n$
 
    2. sq step
 
       we are trying to expand Ts to $x^{\overline{e_1e_2}}$
 
-      so we first do $x^{\overline{e_10}}$ aka $x^{\overline{e_1} \cdot 2}$
+      so we first do $x^{\overline{e_10}_2}$ aka $x^{\overline{e_1}_2 \cdot 2}$
 
       do $ans^2 \pmod n$
 
@@ -146,7 +146,7 @@ two kinds apparently. idk
 
       if $e_2$ conveniently equals 0, skip Ts.
 
-      so $x^{\overline{e_10}}$ needs to become $x^{\overline{e_11}}$
+      so $x^{\overline{e_10}_2}$ needs to become $x^{\overline{e_11}_2}$
 
       that means you needa multiply by x.
 
@@ -230,14 +230,14 @@ for authentication. is this really you type shit.
 
    calc r = alpha ^ ke mod p
 
-   s = (x - d _ r) _ ke^-1 mod (p-1)
+   $$ s = (x - d \cdot r) k_e^-1 \pmod{p-1}$$
 
    give (r, s) to whoever is gonna verify yo msg
 
 3. verify x
    get ts to be true:
 
-   $\beta^rr^s \equiv \alpha^x \pmod p$
+   $$\beta^rr^s \equiv \alpha^x \pmod p$$
 
    then you are ball.