User:Devanden/scratch

This is just my scratch page.

Lemma 1
The first main lemma of the proof shows that the union of a small finite number of these transforms will contain the entire space of random input strings. Using this fact, a &Sigma;2 sentence and a &Pi;2 sentence can be generated that is true if and only if x&isin;L (see corollary). If $$\frac{|A(x)|}{|R|} > 1 - \frac{1}{2^{|x|}}$$, then $$\exists t_1,t_2,\ldots,t_{|r|}, \mbox{ where } |t_i|=|r| \mbox{ such that } \cup_i A(x) \oplus t_i = R \ $$

Proof. Use the probabilistic method. Pick $$t_1,t_2,\ldots,t_{|r|}$$ at random. Let $$S=\cup_i A(x) \oplus t_i$$. Now, $$\forall r \in R$$, $$\Pr [r \notin S] = \Pr [r \notin A(x) \oplus t_1] \cdot \Pr [r \notin A(x) \oplus t_2] \cdots \Pr [r \notin A(x) \oplus t_{|r|} \le { \frac{1}{2^{|x|}} } ^{|r|}=2^{-|x| |r|}$$ So summing over the elements of $$R$$, $$\Pr [(r_1 \notin S) \vee (r_2 \notin S) \vee \cdots \vee (r_{2^{|r|}} \notin S) ] \le \sum_r \frac{1}{2^{|x| \cdot |r|}} = \frac{1}{2^{|x|}} \ll 1$$ Thus, $$\Pr [S = R] \ge 1 - \frac{1}{2^{|x|}}$$. Thus there exists a choice of $$t_1,t_2,\ldots,t_{|r|}$$ such that $$R = \cup_i A(x) \oplus t_i \ $$. Now when $$x \notin L$$, the set $$A(x)$$ covers and exponentially small fraction of the space, and so there will not be a small set of translations which allow it to cover $$R$$.

Lemma 1
The general idea of lemma one is to prove that if $$A(x)$$ covers a large part of the random space $$R$$ then there exists a set of translations that will cover the entire random space. If $$\frac{|A(x)|}{|R|} > 1 - \frac{1}{2^{|x|}}$$, then $$\exists t_1,t_2,\ldots,t_{|r|}$$, where $$|t_i|=|r| \ $$ such that $$ \cup_i A(x) \oplus t_i = R \ $$ Proof. Randomly pick $$t_1,t_2,\ldots,t_{|r|}$$. Let $$S=\cup_i A(x) \oplus t_i$$. So, $$\forall r \in R$$, $$\Pr [r \notin S] = \Pr [r \notin A(x) \oplus t_1] \cdot \Pr [r \notin A(x) \oplus t_2] \cdots \Pr [r \notin A(x) \oplus t_{|r|}] \le { \frac{1}{2^{|x| \cdot |r|}} } $$

The probability that there will exist one element in R</math not in $$S$$: $$\Pr [(r_1 \notin S) \vee (r_2 \notin S) \vee \cdots \vee (r_{2^{|r|}} \notin S) ] \le \sum_r \frac{1}{2^{|x| \cdot |r|}} = \frac{1}{2^{|x|}} \ll 1$$