From 466dee3331a3e9fdad38cdf1caeaa7ffb9aa644d Mon Sep 17 00:00:00 2001
From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com>
Date: Tue, 19 Nov 2024 22:42:46 -0600
Subject: [PATCH] Update CSE442T_L12.md

 fix typo
---
 pages/CSE442T/CSE442T_L12.md | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/pages/CSE442T/CSE442T_L12.md b/pages/CSE442T/CSE442T_L12.md
index 13d6451..8eaeb07 100644
--- a/pages/CSE442T/CSE442T_L12.md
+++ b/pages/CSE442T/CSE442T_L12.md
@@ -23,7 +23,7 @@ If $D$ distinguishes $M(X_n)$ and $M(Y_n)$ by $\mu(n)$ then $D(M(\cdot))$ is als
 
 ### Hybrid Lemma
 
-Let $X^0_n,X^1_n$ are ensembles indexed from $1,..,m$
+Let $X^0_n,X^1_n,\dots,X^m_n$ are ensembles indexed from $1,..,m$
 
 If $D$ distinguishes $X_n^0$ and $X_n^m$ by $\mu(n)$, then $\exists i,1\leq i\leq m$ where $X_{n}^{i-1}$ and $X_n^i$ are distinguished by $D$ by $\frac{\mu(n)}{m}$
 
@@ -42,7 +42,7 @@ If all $|p_{i-1}-p_i|<\frac{\mu(n)}{m},|p_0-p_m|<\mu_n$ contradiction.
 
 In applications, only useful if $m\leq q(n)$ polynomial
 
-If $X_0$ and $X^m$ are distinguishable by $\frac{1}{p(n)}$, then $2$ inner "hybrids" are distinguishable $\frac{1}{p(n)q(n)}=\frac{1}{poly(n)}$
+If $X^0_n$ and $X^m_n$ are distinguishable by $\frac{1}{p(n)}$, then $2$ inner "hybrids" are distinguishable $\frac{1}{p(n)q(n)}=\frac{1}{poly(n)}$
 
 Example: