diff --git a/main.tex b/main.tex index efc84dbb86206c78086ffdc2cbde7b352cda1299..a5a24ca064848bb9e28f2e6bf2cd8d27bdba4b88 100644 --- a/main.tex +++ b/main.tex @@ -601,21 +601,25 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i Then \begin{align*} \P_{I}(\Z{j}) &= \P_{I'}(\Z{j}) + \mathcal{O}(p^{d(i_*,j) + 1 - |I|}) \\ - \P_{I}(\Z{j},\NZ{s}) &= \P_{I'}(\Z{j},\NZ{s}) + \mathcal{O}(p^{\dist{s}(i_*,j) + 1 - |I|}) . + \P_{I}(\Z{j},\NZ{s}) &= \P_{I'}(\Z{j},\NZ{s}) + \mathcal{O}(p^{\min\left( \dist{s}(i_*,j), \dist{j}(i_*,s) \right) + 1 - |I|}) . \end{align*} \end{lemma} \begin{proof} - We will prove both statements inductively on $|I|$. For $|I|=1$ we have $I=\{i_*\}$ and $I'=\emptyset$, so $\P_{I'}(\Z{j})=0$ and + Without loss of generality, we can assume that $j=0$ and $0 < i_* < s < n$ (because we can shift $j$ to $0$ and switch the direction to get the correct ordering). Therefore, we have to prove: \begin{align*} - \P_{I}(\Z{j}) &= \mathcal{O}(p^{d(i_*,j)}) \\ - \P_{I}(\Z{j},\NZ{s}) &= \mathcal{O}(p^{\dist{s}(i_*,j)}) + \P_{I}(\Z{0}) &= \P_{I'}(\Z{0}) + \mathcal{O}(p^{d(i_*,0) + 1 - |I|}) \\ + \P_{I}(\Z{0},\NZ{s}) &= \P_{I'}(\Z{0},\NZ{s}) + \mathcal{O}(p^{\min\left( i_*, s-i_* \right) + 1 - |I|}) . \end{align*} - simply because a chain of zeroes has to be formed between $i_*$ and $j$, and in the second case this chain can not go through $s$. Now assume both statements hold up to $|I|-1$, then we prove them both for sets of size $|I|$. + We will prove both statements inductively on $|I|$. For $|I|=1$ we have $I=\{i_*\}$ and $I'=\emptyset$, so $\P_{I'}(\Z{0})=0$ and + \begin{align*} + \P_{I}(\Z{0}) &= \mathcal{O}(p^{d(i_*,0)}) \\ + \P_{I}(\Z{0},\NZ{s}) &= \mathcal{O}(p^{i_*}) = \mathcal{O}(p^{\min\left( i_*, s-i_* \right)}) + \end{align*} + simply because a chain of zeroes has to be formed between $i_*$ and $0$, and in the second case this chain can not go through $s$ so the shortest path has length $i_*$. Now assume both statements hold up to $|I|-1$, then we prove them both for sets of size $|I|$. When we refer to an interval $[a,b]$ on the circle we could be referring to two possible intervals because of the periodicity of the circle. Define $[a,b]_j$ as the interval with vertex $j$ on the \emph{inside}. Furthermore by $-a$ we mean the vertex $n-a$, as one would expect modulo $n$. - %If we refer to only $[a,b]$ then we mean $\{a,a+1,...,b\}$ where numbers are considered modulo $n$. So $[a,b]$ and $[b,a]$ are different intervals that cover the circle together. - Without loss of generality, we can assume that $0=j < i_* < s < n$. We will now consider intervals around vertex 0. + We will now consider intervals around vertex 0. For $l,r\geq 1$ and $l+r\leq n$, define the event ``zeroes patch'' $\mathrm{ZP}^{[-l,r]_0}$ as the event of getting zeroes inside the interval $[-l,r]_0$ but not on the boundary, i.e. $$\mathrm{ZP}^{[-l,r]_0} = \NZ{-l} \cap \Z{-l+1} \cap \cdots \cap \Z{0} \cap \cdots \cap \Z{r-1} \cap \NZ{r}$$ Note that there are $r+l-1$ `zeroes' in this event, so $\P_{J}(\mathrm{ZP}^{[-l,r]_0}) = \mathcal{O}(p^{r+l-1-|J|})$ for $J\subseteq[-l,r]_0$ is a lower bound on the order of $p$.\\ @@ -633,7 +637,7 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i \end{center} Note that by Claim~\ref{claim:eventindependence} we have \begin{align*} - \P_{I}(\mathrm{ZP}^{[-l,r]_0}) = \P_{I \cap [-l,r]_0}(\mathrm{ZP}^{[-l,r]_0}) \;\cdot\; \P_{I\setminus [-l,r]_0}(\NZ{a},\NZ{b}) + \P_{I}(\mathrm{ZP}^{[-l,r]_0}) = \P_{I \cap [-l,r]_0}(\mathrm{ZP}^{[-l,r]_0}) \;\cdot\; \P_{I\setminus [-l,r]_0}(\NZ{-l},\NZ{r}) \end{align*} We have $i_*\in I \setminus[-l,r]_0$, and $I\cap[-l,r]_0 = I' \cap [-l,r]_0$. Define $J=I\setminus[-l,r]_0$ and $J'=I'\setminus[-l,r]_0$. We have $|J|<|I|$ so we can apply the induction hypothesis to $J$: \begin{align*} @@ -647,7 +651,7 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i 1 - \P_{J'}(\Z{-l},\NZ{r}) - \P_{J'}(\Z{r}) \\ - &\quad + \mathcal{O}(p^{\dist{r}(i_*,-l)+1-|J|}) + &\quad + \mathcal{O}(p^{\min\left( \dist{r}(i_*,-l), \dist{-l}(i_*,r) \right) +1-|J|}) + \mathcal{O}(p^{d(i_*,r)+1-|J|}) \\ &= \P_{J'}(\NZ{-l},\NZ{b}) @@ -666,7 +670,7 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i Where we used Claim~\ref{claim:eventindependence} again. Case separation shows that $$\min\left( \dist{r}(i_*,-l) , d(i_*,r)\right) + l +r \geq d(i_*,0) + 1$$ - which proves the claim. + for $l,r\geq 1$ which proves the claim. The first equality that we have to prove now follows from the fact that the ``zeroes patch'' events are a partition of $\Z{0}$: \begin{align*} @@ -688,31 +692,31 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i \tag{partition of $\Z{0}$}\\ &=\sum_{l=1}^{n-s}\sum_{r=1}^{i_*-1} \P_{I}(\mathrm{ZP}^{[-l,r]_0},\NZ{s}) - +\mathcal{O}(p^{\dist{s}(i_*,0)+1-|I|}) \\ + +\mathcal{O}(p^{i_*+1-|I|}) \\ &=\sum_{l=1}^{n-s}\sum_{r=1}^{i_*-1} \P_{I\cap[s,r]_0}(\mathrm{ZP}^{[-l,r]_0},\NZ{s}) \cdot \P_{I\setminus [s,r]_0}(\NZ{r},\NZ{s}) - +\mathcal{O}(p^{\dist{s}(i_*,0)+1-|I|}) + +\mathcal{O}(p^{i_*+1-|I|}) \tag{Claim~\ref{claim:eventindependence}}\\ - &=\sum_{l=1}^{n-s}\sum_{r=1}^{s} + &=\sum_{l=1}^{n-s}\sum_{r=1}^{i_*-1} \P_{I'\cap[s,r]_0}(\mathrm{ZP}^{[-l,r]_0},\NZ{s}) \cdot \P_{I\setminus [s,r]_0}(\NZ{r},\NZ{s}) - +\mathcal{O}(p^{\dist{s}(i_*,0)+1-|I|}) + +\mathcal{O}(p^{i_*+1-|I|}) \tag{$i_*\in I \setminus[s,r]_0$}\\ - &=\sum_{l=1}^{n-s}\sum_{r=1}^{s} + &=\sum_{l=1}^{n-s}\sum_{r=1}^{i_*-1} \P_{I'\cap[s,r]_0}(\mathrm{ZP}^{[-l,r]_0},\NZ{s}) \cdot \P_{I'\setminus [s,r]_0}(\NZ{r},\NZ{s}) \\ &\qquad +\mathcal{O}(p^{\min\left( \dist{r}(i_*,s) , d(i_*,r)\right)+l+r-|I|}) - +\mathcal{O}(p^{\dist{s}(i_*,0)+1-|I|}) + +\mathcal{O}(p^{i_*+1-|I|}) \tag{same argument as before}\\ - &=\sum_{l=1}^{n-s}\sum_{r=1}^{s} + &=\sum_{l=1}^{n-s}\sum_{r=1}^{i_*-1} \P_{I'\cap[s,r]_0}(\mathrm{ZP}^{[-l,r]_0},\NZ{s}) \cdot \P_{I'\setminus [s,r]_0}(\NZ{r},\NZ{s}) \\ &\qquad - +\mathcal{O}(p^{\dist{s}(i_*,0)+1-|I|}) - \tag{case separation \todo{does not seem to work?}}\\ + +\mathcal{O}(p^{\min\left( i_* , s-i_* \right) +1-|I|}) + \tag{case separation}\\ &= \P_{I'}(\Z{0} , \NZ{s}) - +\mathcal{O}(p^{\dist{s}(i_*,0)+1-|I|}) + +\mathcal{O}(p^{\min\left( i_* , s-i_* \right) +1-|I|}) \end{align*} This finishes the proof. \end{proof}