Add a write up of the state delta resolution algorithm

docs/state_delta_resolution.tex

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+\documentclass{article}
+\usepackage[utf8]{inputenc}
+\usepackage{amsfonts}
+\usepackage{amsthm}
+\usepackage{amsmath}
+\usepackage{algorithmic}
+\usepackage{algorithm}
+
+\title{State Delta Resolution Algorithm}
+\author{Erik Johnston}
+\date{April 2018}
+
+\newtheorem{theorem}{Theorem}
+\newtheorem{lemma}[theorem]{Lemma}
+\newtheorem{corollary}[theorem]{Corollary}
+
+\setlength{\parskip}{1em}
+
+\newcommand{\inlinemaketitle}{{\let\newpage\relax\maketitle}}
+
+
+\begin{document}
+	
+\maketitle
+	
+We first define some basic functions that correspond to the synapse code.
+
+Let $K$ be the set of all type/state key tuples and $E$ the set of all events. We can then define arbitrary functions:
+\begin{equation}
+\begin{split}
+f: F \rightarrow E\\
+g: G \rightarrow E\\
+\end{split}
+\end{equation} which we call state maps, for $F, G \subset K$.
+
+We can then compute the set of all ``unconflicted events":
+\begin{equation}
+U_{f,g} = \{x\ |\ \forall x \in F \cap G,\ f(x) = g(x)\} \cup (F \bigtriangleup G)
+\end{equation} i.e. the set of state keys where $f$ and $g$ don't conflict. Similarly, we define:
+\begin{equation}
+\begin{split}
+u_{f,g} : U_{f,g} \longrightarrow &\ E\\
+x \longmapsto &\ \begin{cases}
+	f(x), & \text{if}\ f \in F \\
+	g(x), & \text{otherwise}
+\end{cases}
+\end{split}
+\end{equation} which gets the unconflicted event for a given state key.
+
+
+We can also define a function on $C_{f,g} = F \cup G \setminus U_{f,g}$:
+\begin{equation}
+c_{f,g}: C_{f,g} \rightarrow E
+\end{equation} which is used to resolve conflicts between $f$ and $g$. Note that $c_{f,g}$ is either $f(x)$ or $g(x)$. 
+
+Now we define:
+\begin{equation} 
+\begin{split}
+r_{f,g}: F \cup G \longrightarrow &\ E\\
+x \longmapsto &\ \begin{cases}
+u_{f,g}(x), & \text{if}\ x \in U_{f,g}\\
+c_{f,g}(x), & \text{otherwise}
+\end{cases} 
+\end{split}
+\end{equation} which we call the resolved state of $f$ and $g$.
+
+\begin{lemma}
+	$\forall x \in U_{f,g} \ s.t.\ g(x) = g'(x)$ then $r_{f,g}(x) = r_{f,g'}(x)$
+\end{lemma}
+
+We define 
+\begin{equation}
+\alpha: E \rightarrow \mathbb{P}(K)
+\end{equation} to be the mapping of an event to the type/state keys needed to auth the event, and
+\begin{equation}
+\alpha_{f,g}(x) = \alpha f(x) \cup \alpha g(x)
+\end{equation} which is the set of auth events required for $f(x)$ and $g(x)$. Note that $\alpha_{f,g}(x) \subset F \cup G$.
+
+Further, we can define
+\begin{equation}
+a_{f,g}(x) = \bigcup_{n=0}^\infty (\alpha_{f,g})^n(x)
+\end{equation} to be the auth chain of $f(x)$ and $g(x)$. This is well defined as there are a finite number of elements in $F \cup G$ and $a_{f,g} \rightarrow F \cup G$.
+
+If we consider the implementation of $c_{f,g}$ in Synapse we can see that it depends not only on the values of $x$, but also on the resolved state of their auth events, i.e. $r_{f,g}(\alpha_{f,g}(x))$. By ``depends on" we mean that if those are the same for different values of $f$ and $g$, then the result of $c_{f,g}(x)$ is the same.
+
+\begin{lemma}
+	 $c_{f,g}$ depends only on $a_{f,g}(x)$
+\end{lemma}
+
+\begin{proof}
+	$c_{f,g}(x)$ depends on $x \in a_{f,g}(x)$, and $r_{f,g}(\alpha_{f,g}(x))$. Now:
+	\[
+		\begin{split}
+		r_{f,g}(\alpha_{f,g}(x))\ =\ & u_{f,g}(\alpha_{f,g}(x)) \cup c_{f,g}(\alpha_{f,g}(x))\\
+		\end{split}
+	\] but by definition $u_{f,g}(\alpha_{f,g}(x))$ depends only on $\alpha_{f,g}(x)$, so $r_{f,g}(\alpha_{f,g}(x))$ depends on $a_{f,g}(x)$ and $c_{f,g}\alpha_{f,g}(x)$.
+	
+	By induction, $c_{f,g}\alpha_{f,g}(x)$ depends on $a_{f,g}(x)$ and $c_{f,g}(\alpha_{f,g})^n(x), \forall n$. Since $(\alpha_{f,g})^n(x)$ repeats and we know $c_{f,g}$ is well defined, we can infer that $c_{f,g}(x)$ depends only on $\bigcup_{n=0}^\infty (\alpha_{f,g})^n(x) = a_{f,g}(x)$.
+	
+\end{proof}
+
+By inspecting the actual implementation of $\alpha$ we can define $a_{f,g}^{-1}(x)$ to be a function which $\forall x$, $x \in a_{f,g}^{-1}a_{f,g}(x)$. We can similarly define $\alpha_{f,g}^{-1}(x)$. Note that $\forall x$, $x \in a_{f,g}^{-1}(x)$
+
+We now consider $g': G' \rightarrow E$, where $g(x) = g'(x)$ except for $x \in G_\delta$, i.e. $g'$ is a state map based on $g$.
+
+\begin{lemma}
+	For $f, g, g'$ s.t. $\forall x \notin G_\delta,\ g(x) = g'(x)$, then $\forall x \notin a_{f,g'}^{-1}(G_\delta),\ r_{f,g}(x) = r_{f,g'}(x)$.
+\end{lemma}
+
+\begin{proof}
+	Let $x$ be s.t. $r_{f,g}(x) \neq r_{f,g'}(x)$:
+	\[
+	\begin{split}
+		&\Rightarrow a_{f,g}(x) \neq a_{f,g'}(x)\\
+		&\Rightarrow \exists y \in a_{f,g}(x) \text{ s.t. } g(y) \neq g'(y)\\
+		&\Rightarrow y \in G_\delta\\
+		&\Rightarrow a_{f,g'}^{-1}(y) \subseteq a_{f,g'}^{-1}(G_\delta)\\
+		&\Rightarrow x \in a_{f,g'}^{-1}(G_\delta)
+	\end{split}
+	\]
+\end{proof}
+
+\begin{corollary}
+	For $f, g, g'$ s.t. $\forall x \notin G_\delta$, $g(x) = g'(x)$, then $\forall x \notin C_{f,g} \cap a_{f,g'}^{-1}(G_\delta)$, $r_{f,g}(x) = r_{f,g'}(x)$
+\end{corollary}
+
+\begin{proof}
+	This follows from the previous result and that if $x \in U_{f,g} \setminus G_\delta$ then $r_{f,g}(x) = r_{f,g'}(x)$.
+\end{proof}
+
+This allows us to reuse most of the results of $r_{f,g}$ when calculating $r_{f,g'}$ if $G_\delta$ is small. In particular we can calculate the delta between the two functions without having to inspect $U_{f,g}$, which dramatically cuts down the amount of data used to compute deltas of resolved state of large state maps.
+
+However, we can do better than this. We can note that $r_{f,g}(x)$ only depends on $r_{f,g}(\alpha_{f,g}(x))$ for values of $\alpha_{f,g}(x)$ not in $U_{f,g}$. Concretely, this means for example that if $G_\delta$ includes the membership of the sender of a power level event, but the power level event is in $U_{f,g}$, then we don't need to recalculate all conflicted events---despite the membership event being in every event's auth chain.
+
+\begin{lemma}
+	$\forall x$ s.t. $r_{f,g}(x) \neq r_{f,g'}(x)$ then $\exists y_1, ..., y_n$ s.t. $y_n \in G_\delta$, $y_i \notin U_{f,g}$ and $y_{i+1} \in \alpha_{f,g'}(y_i)$
+\end{lemma}
+
+\begin{proof}
+	If $r_{f,g}(x) \neq r_{f,g'}(x)$ then $\exists y \in G_\delta$ s.t. $y \in a_{f,g'}(x)$. By definition of $a_{f,g'}(x)$, $\exists y_0, ..., y_n$ s.t. $y_0 = x$ and $y_{i+1} \in \alpha_{f,g'}(y_i)$. 
+	
+	We know that $r_{f,g'}(x)$ depends on either $u_{f,g'}(x)$ or $c_{f,g'}(x)$, but if $x \in U_{f,g'}$ then there is no dependency on $x$'s auth events and so $y_n = y_0 = x \in G_\delta$. Otherwise, we have $c_{f,g}(x) \neq c_{f,g'}(x)$, which depends on $f(x)$, $g'(x)$ or $r_{f,g'}(\alpha_{f,g'}(x))$. If $x \notin G_\delta$ then we know $f(x)$ and $g'(x)$ are the same, and so $r_{f,g}(\alpha_{f,g'}(x)) \neq r_{f,g'}(\alpha_{f,g'}(x)) \Rightarrow \exists y_1 \in \alpha_{f,g'}(x)$ s.t. $r_{f,g}(y_1) \neq r_{f,g'}(y_1)$.
+	
+	 Applying the above to $y_1$ then if $y_1 \in U_{f,g'} \Rightarrow y_1 = y_n \in G_\delta$. By induction $y_i \notin U_{f,g'}$ for $i < n$.
+	 
+	 Note that we can assume $y_i \notin G_\delta$ as otherwise we would pick $n = i$,  and so if $y_i \notin U_{f,g} \Leftrightarrow y_i \notin U_{f,g'}$
+\end{proof}
+
+We can use this approach and create an iterative algorithm for computing the set of state keys that need to be recalculated:
+
+\begin{algorithm}
+\caption{Calculate state keys needing to be recalculated}
+\begin{algorithmic}
+	\STATE $to\_recalculate \leftarrow \text{empty set of state keys}$
+	\STATE $pending \leftarrow G_\delta$
+	\WHILE{$pending$ is empty}
+		\STATE $x \leftarrow \text{pop from}\ pending$
+		\IF{$x \notin U_{f,g}$}
+			\STATE $\text{add all in}\ \alpha_{f,g'}^{-1}(x)\ \text{to}\ pending$
+			\STATE $\text{add}\ x\ \text{to}\ to\_recalculate$
+		\ENDIF
+	\ENDWHILE
+	\RETURN $to\_recalculate$
+\end{algorithmic}
+\end{algorithm}
+
+
+\end{document}