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Update CSE442T_L12.md

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fix typo
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Zheyuan Wu
2024-11-19 22:42:46 -06:00
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@@ -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: