diff --git a/main.tex b/main.tex
index 0619df62d963e228d7dee8961bdd2c7ec5f6aca3..7703efa6d99c75d3f583e237c82b9d432c48512a 100644
--- a/main.tex
+++ b/main.tex
@@ -277,7 +277,7 @@ where $C(f)\in\{0,1,1'\}^n$ denotes a configuration with slots on the sites $C$
 	\begin{center}
     	\includegraphics{diagram_gap.pdf}
     \end{center}
-    \caption{\label{fig:diametergap} Illustration of Definition \ref{def:diameter}. A set $C=\{1,2,4,7,9\}\subseteq[n]$ consisting of 5 positions is shown by the red dots. The smallest interval containing $C$ is $[1,9]$, so the diameter is $\diam{C}=9$. The blue squares denote the set $\gaps{C} = \{3,5,6,8\}$. The dotted line at the top depicts the rest of the circle which may be much larger. The largest gap of $C$ is $\maxgap{C}=2$ which is the largest connected component of $\gaps{C}$.}
+    \caption{\label{fig:diametergap} Illustration of Definition \ref{def:diameter}. A set $C=\{1,2,4,7,9\}\subseteq[n]$ consisting of 5 positions is shown by the red dots. The smallest interval containing $C$ is $[1,9]$, so the diameter is $\diam{C}=9$. The blue squares denote the set $\gaps{C} = \{3,5,6,8\}$. The dotted line at the top depicts the rest of the cycle which may be much larger. The largest gap of $C$ is $\maxgap{C}=2$ which is the largest connected component of $\gaps{C}$.}
 \end{figure}
 
 \begin{claim}[Strong cancellation claim] \label{claim:strongcancel}
@@ -366,7 +366,7 @@ Here we prove claim \ref{claim:weakcancel}, the weaker version of the claim. We
 	We say two paths $\xi_i\in\paths{b_i}$ ($i=1,2$) of the Markov Chain are \emph{independent} if $\xi_1$ never resamples a site that was ever zero in $\xi_2$ and the other way around. It is allowed that $\xi_1$ resamples a $1$ to a $1$ that was also resampled from $1$ to $1$ by $\xi_2$ and vice versa. If the paths are not independent then we call the paths \emph{dependent}.
 \end{definition}
 \begin{definition}[Path independence - alternative] \label{def:independence2}
-    Equivalently, on the infinite line $\xi_1$ and $\xi_2$ are independent if there is a site `inbetween' them that was never zero in $\xi_1$ and never zero in $\xi_2$. On the circle $\xi_1$ and $\xi_2$ are independent if there are \emph{two} sites inbetween them that are never zero.
+    Equivalently, on the infinite line $\xi_1$ and $\xi_2$ are independent if there is a site `inbetween' them that was never zero in $\xi_1$ and never zero in $\xi_2$. On the cycle $\xi_1$ and $\xi_2$ are independent if there are \emph{two} sites inbetween them that are never zero.
 \end{definition}
 \begin{claim}[Sum of expectation values] \label{claim:expectationsum}
 When $b=b_1\land b_2\in\{0,1\}^n$ is a state with two groups ($b_1\lor b_2 = 1^n$) of zeroes with $k$ $1$s inbetween the groups, then we have $R_b(p) = R_{b_1}(p) + R_{b_2}(p) + \mathcal{O}(p^{k})$ where $b_1$ and $b_2$ are the configurations where only one of the groups is present and the other group has been replaced by $1$s. To be precise, the sums agree up to and including order $p^{k-1}$.
@@ -463,10 +463,77 @@ It is useful to introduce some new notation. Note that an \emph{event} is a subs
     \end{align*}
     up to any order in $p$.
 \end{lemma}
-The lemma says that conditioned on $j_1$ and $j_2$ not being crossed, the two halves of the circle are independent. 
+The lemma says that conditioned on $j_1$ and $j_2$ not being crossed, the two halves of the cycle are independent. 
 
 \begin{proof}
-    Note that any path $\xi\in\paths{b} \cap \mathrm{NZ}^{(j_1,j_2)}$ can be split into paths $\xi_1\in\paths{b_1}\cap \mathrm{NZ}^{(j_1,j_2)}$ and $\xi_2\in\paths{b_2}\cap\mathrm{NZ}^{(j_1,j_2)}$. This can be done by taking all resampling positions $r_i$ in $\xi$ and if $r_i$ is ``on the $b_1$ side of $j_1,j_2$'' then add it to $\xi_1$ and if its ``on the $b_2$ side of $j_1,j_2$'' then add it to $\xi_2$. Note that now $\xi_1$ is a path from $b_1$ to $\mathbf{1}$, because in the original path $\xi$, all zeroes ``on the $b_1$ side'' have been resampled by resamplings ``on the $b_1$ side''. Since the sites $j_1,j_2$ inbetween never become zero, there can not be any zero ``on the $b_1$ side'' that was resampled by a resampling ``on the $b_2$ side''.  Vice versa, all paths $\xi_1\in\paths{b_1}\cap \mathrm{NZ}^{(j_1,j_2)}$ and $\xi_2\in\paths{b_2}\cap\mathrm{NZ}^{(j_1,j_2)}$ also induce a path $\xi\in\paths{b} \cap \mathrm{NZ}^{(j_1,j_2)}$ by simply concatenating the resampling positions. Note that $\xi_1,\xi_2$ actually induce $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ paths $\xi$ because of the possible orderings of concatenating the resamplings in $\xi_1$ and $\xi_2$. However, all these paths have smaller weight, and by the same reasoning as in the proof of claim \ref{claim:expectationsum} these weights sum to exactly $1$, so we obtain
+    From any path $\xi\in\paths{b} \cap \mathrm{NZ}^{(j_1,j_2)}$ we can construct paths $\xi_1\in\paths{b_1}\cap \mathrm{NZ}^{(j_1,j_2)}$ and $\xi_2\in\paths{b_2}\cap\mathrm{NZ}^{(j_1,j_2)}$ as follows. Let us write the path $\xi$ as
+    $$\xi=\left( (z_1, s_1, r_1), (z_2, s_2, r_2), ..., (z_{|\xi|}, s_{|\xi|}, r_{|\xi|}) \right)$$
+    where $z_i\in[n]$ denotes the number of zeroes in the state before the $i$th step, $s_i\in [n]$ denotes the site that was resampled and $r_i\in \{0,1\}^3$ is the result of the three resampled bits. We have
+    \begin{align*}
+        \P_b[\xi] &= \P(\text{choose }s_1) \P(r_1) \P(\text{choose }s_2) \P(r_2) \cdots \P(\text{choose }s_{|\xi|}) \P(r_{|\xi|}) \\
+                &= \frac{1}{z_1} \P(r_1) \frac{1}{z_2} \P(r_2) \cdots \frac{1}{z_{|\xi|}} \P(r_{|\xi|}) .
+    \end{align*}
+    To construct $\xi_1$ and $\xi_2$, start with empty sequences $\xi_1,\xi_2$ and for each step $(z_i,s_i,r_i)$ in $\xi$ do the following: if $s_i$ is ``on the $b_1$ side of $j_1,j_2$'' then add $(z^{(1)}_i,s_i,r_i)$ to $\xi_1$ and if its ``on the $b_2$ side of $j_1,j_2$'' then add $(z^{(2)}_i,s_i,r_i)$ to $\xi_2$. Here $z^{(1)}_i$ is the number of zeroes that were on the $b_1$ side and $z^{(2)}_i$ is the number of zeroes on the $b_2$ side so we have $z_i = z^{(1)}_i + z^{(2)}_i$.
+    Let the resulting paths be
+    \begin{align*}
+        \xi_1 &= \left( (z^{(1)}_{a_1}, s_{a_1}, r_{a_1}), (z^{(1)}_{a_2}, s_{a_2}, r_{a_2}), ..., (z^{(1)}_{a_{|\xi_1|}}, s_{a_{|\xi_1|}}, r_{a_{|\xi_1|}}) \right) \\
+        \xi_2 &= \left( (z^{(2)}_{b_1}, s_{b_1}, r_{b_1}), (z^{(2)}_{b_2}, s_{b_2}, r_{b_2}), ..., (z^{(2)}_{b_{|\xi_1|}}, s_{b_{|\xi_1|}}, r_{b_{|\xi_1|}}) \right)
+    \end{align*}
+    Now $\xi_1$ is a valid (terminating) path from $b_1$ to $\mathbf{1}$, because in the original path $\xi$, all zeroes ``on the $b_1$ side'' have been resampled by resamplings ``on the $b_1$ side''. Since the sites $j_1,j_2$ inbetween never become zero, there can not be any zero ``on the $b_1$ side'' that was resampled by a resampling ``on the $b_2$ side''.
+    The probability of $\xi_1$, started from $b_1$, is given by
+    \begin{align*}
+        \P_{b_1}[\xi_1] &= \P(\text{choose }s_{a_1}) \P(r_{a_1}) \P(\text{choose }s_{a_2}) \P(r_{a_2}) \cdots \P(\text{choose }s_{a_{|\xi|}}) \P(r_{a_{|\xi|}}) \\
+                &= \frac{1}{z^{(1)}_{a_1}} \P(r_{a_1}) \frac{1}{z^{(1)}_{a_2}} \P(r_{a_2}) \cdots \frac{1}{z^{(1)}_{a_{|\xi|}}} \P(r_{a_{|\xi|}})
+    \end{align*}
+    and similar for $\xi_2$.
+    Vice versa, all paths $\xi_1\in\paths{b_1}\cap \mathrm{NZ}^{(j_1,j_2)}$ and $\xi_2\in\paths{b_2}\cap\mathrm{NZ}^{(j_1,j_2)}$ also induce a path $\xi\in\paths{b} \cap \mathrm{NZ}^{(j_1,j_2)}$ by simply interleaving the resampling positions. Note that $\xi_1,\xi_2$ actually induce $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ paths $\xi$ because of the possible orderings of interleaving the resamplings in $\xi_1$ and $\xi_2$. The following diagram illustrates all possible interleavings:
+    \begin{center}
+        \includegraphics{diagram_paths.pdf}
+    \end{center}
+    A particular interleaving is a path through the above grid. So for a fixed $\xi_1,\xi_2$ there are $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ possible interleavings $\xi$, and vice versa there are $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ paths $\xi$ that decompose into the same $\xi_1$ and $\xi_2$. For a fixed $\xi_1,\xi_2$ we now show the following:
+    \begin{align*}
+        \sum_{\substack{\xi\in\paths{b} \cap \mathrm{NZ}^{(j_1,j_2)} \text{ s.t.}\\ \xi \text{ decomposes into } \xi_1,\xi_2 }} \P_b[\xi] &=
+        \sum_{\text{interleavings of }\xi_1,\xi_2} \P(\text{interleaving}) \cdot \P_{b_1}[\xi_1] \cdot \P_{b_2}[\xi_2] \\
+        &= \P_{b_1}[\xi_1] \cdot \P_{b_2}[\xi_2]
+    \end{align*}
+    This is best explained by an example. We have, for an example interleaving:
+    \begin{align*}
+        \xi   &= \left( (z_1, s_1, r_1), (z_2, s_2, r_2), (z_3, s_3, r_3), (z_4, s_4, r_4), \cdots \right) \\
+        \xi_1 &= \left( (z^{(1)}_1, s_1, r_1), \phantom{(z^{(2)}_2, s_2, r_2), (z^{(1)}_3, s_3, r_3),} (z^{(1)}_4, s_4, r_4),\cdots  \right) \\
+        \xi_2 &= \left( \phantom{(z^{(1)}_1, s_1, r_1),} (z^{(2)}_2, s_2, r_2), (z^{(2)}_3, s_3, r_3),\phantom{(z^{(2)}_4, s_4, r_4), } \cdots \right)
+    \end{align*}
+    Remember $z^{(1)}_i + z^{(2)}_i = z_i$.
+    The probability associated to this interleaving is
+    \begin{align*}
+        \P_b[\xi] &= \frac{1}{z_1} \P(r_1) \frac{1}{z_2} \P(r_2) \frac{1}{z_3} \P(r_3) \frac{1}{z_4} \P(r_4) \cdots \\
+        &=
+        \frac{z^{(1)}_1}{z_1} \frac{1}{z^{(1)}_1} \P(r_1) \;
+        \frac{z^{(2)}_2}{z_2} \frac{1}{z^{(2)}_2} \P(r_2) \;
+        \frac{z^{(2)}_3}{z_3} \frac{1}{z^{(2)}_3} \P(r_3) \;
+        \frac{z^{(1)}_4}{z_4} \frac{1}{z^{(1)}_4} \P(r_4)
+        \cdots \\
+        &=
+        \frac{z^{(1)}_1}{z_1}
+        \frac{z^{(2)}_2}{z_2}
+        \frac{z^{(2)}_3}{z_3}
+        \frac{z^{(1)}_4}{z_4}
+        \cdots
+        \P_{b_1}[\xi_1] \P_{b_2}[\xi_2]
+    \end{align*}
+    \todo{write more}
+    \begin{align*}
+        \P(\text{interleaving}) = \P(\text{choose step of }\xi_1) \P(\text{choose step of }\xi_2) \P(\text{choose step of }\xi_2) \P(\text{choose step of }\xi_1) \cdots
+    \end{align*}
+    These choices depend on the number of zeroes present in the state:
+    \begin{align*}
+        \P(\text{choose step of }\xi_1) &=
+        \frac{z^{(1)}_1}{z^{(1)}_1 + z^{(2)}_1} \\
+        \P(\text{choose step of }\xi_2) &=
+        \frac{z^{(2)}_1}{z^{(1)}_1 + z^{(2)}_1}
+    \end{align*}
+    These are the $p_i$ and $1-p_i$ shown in the grid diagram. In $\P_{b_1}[\xi_1]$ we have a factor $\frac{1}{z^{(1)}_1}$, so together with the probability above this gives $\frac{z^{(1)}_1}{z^{(1)}_1 + z^{(2)}_1} \frac{1}{z^{(1)}_1} = \frac{1}{z_1}$, as in the original expression for $\P_b[\xi]$. In the grid we see that at every point the probabilities sum to 1, and we always reach the end, so we know the sum of all paths in the grid is 1. This means the sum of all $\P(\text{interleaving})$ is equal to 1.
+
+    We obtain
     \begin{align*}
         \mathbb{P}_b(\mathrm{NZ}^{(j_1,j_2)},A_1,A_2)
         &= \sum_{\substack{\xi\in\paths{b} \cap \\ \mathrm{NZ}^{(j_1,j_2)}\cap A_1\cap A_2}} \mathbb{P}[\xi] \\
@@ -479,10 +546,10 @@ The lemma says that conditioned on $j_1$ and $j_2$ not being crossed, the two ha
         \mathbb{P}_{b_2}(\mathrm{NZ}^{(j_1,j_2)},A_2).
     \end{align*}
     The second equality follows directly from $\mathbb{P}(A\mid B)=\mathbb{P}(A,B)/\mathbb{P}(B)$ and setting $A_1,A_2$ to the always-true event.
-    For the third equality, note that again by the same reasoning as in the proof of claim \ref{claim:expectationsum} we have
+    For the third equality, by the same reasoning we can decompose the paths
     \begin{align*}
         \mathbb{P}_b(\mathrm{NZ}^{(j_1,j_2)},A_1,A_2) R_{b,\mathrm{NZ}^{(j_1,j_2)},A_1,A_2}
-        &:= \sum_{\substack{\xi\in\paths{b}\\\xi \in \mathrm{NZ}^{(j_1,j_2)}\cap A_1\cap A_2}} \mathbb{P}[\xi] |\xi| \\
+        &\equiv \sum_{\substack{\xi\in\paths{b}\\\xi \in \mathrm{NZ}^{(j_1,j_2)}\cap A_1\cap A_2}} \mathbb{P}[\xi] |\xi| \\
         &= \sum_{\substack{\xi_1\in\paths{b_1}\\\xi_1 \in \mathrm{NZ}^{(j_1,j_2)}\cap A_1}}
           \sum_{\substack{\xi_2\in\paths{b_2}\\\xi_2 \in \mathrm{NZ}^{(j_1,j_2)}\cap A_2}}
         \mathbb{P}[\xi_1]\mathbb{P}[\xi_2] (|\xi_1| + |\xi_2|) \\
@@ -591,13 +658,13 @@ The intuition of the following lemma is that the far right can only affect the z
 
 	The main insight that Lemma~\ref{lemma:probIndep} gives is that if we separate the slots to two halves, in order to see the cancellation of the contribution of the expected resamples on the right, we can simply pair up the left configurations by the particle filling the leftmost slot. And similarly for cancelling the left expectations we pair up right configurations based on the rightmost filling. 
 	
-	Also this claim finally ``sees'' how many empty places are between slots. These properties make it possible to use this lemma to prove the sought linear bound. We show it for the infinite chain, but with a little care it should also translate to the circle.
+	Also this claim finally ``sees'' how many empty places are between slots. These properties make it possible to use this lemma to prove the sought linear bound. We show it for the infinite chain, but with a little care it should also translate to the cycle.
 
 ~
 
-Here, I (Tom) tried to set do the same Lemma but for the circle instead of the infinite chain.
-\begin{lemma}[Startingstate dependence] \label{lemma:probIndepCircle}
-    Let $d(a,b)$ be the distance between $a,b\in[n]$ on the circle, so $d(a,b)=\min(|a-b| , n-|a-b|)$. Let $\dist{s}(a,b)$ be the distance between $a,b$ when taking the path that does \emph{not} cross $s$. Let $I\subseteq [n]$ be a non-empty set of vertices. Let $i_* \in I$ and define $I' = I \setminus \{i_*\}$. Let $j,s\notin I$, with $j\neq s$ be any vertices not in $I$.
+Here, I (Tom) tried to set do the same Lemma but for the cycle instead of the infinite chain.
+\begin{lemma}[Startingstate dependence] \label{lemma:probIndepCycle}
+    Let $d(a,b)$ be the distance between $a,b\in[n]$ on the cycle, so $d(a,b)=\min(|a-b| , n-|a-b|)$. Let $\dist{s}(a,b)$ be the distance between $a,b$ when taking the path that does \emph{not} cross $s$. Let $I\subseteq [n]$ be a non-empty set of vertices. Let $i_* \in I$ and define $I' = I \setminus \{i_*\}$. Let $j,s\notin I$, with $j\neq s$ be any vertices not in $I$.
     Then
     \begin{align*}
         \P_{I}(\Z{j})        &= \P_{I'}(\Z{j})        + \mathcal{O}(p^{d(i_*,j) + 1 - |I|}) \\
@@ -617,7 +684,7 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i
     \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 $n$.
+    When we refer to an interval $[a,b]$ on the cycle we could be referring to two possible intervals because of the periodicity of the cycle. 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 $n$.
 
  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.
