From: tkwa <kwathomas0@gmail.com>
Date: Thu, 4 Aug 2016 01:37:38 +0000 (-0700)
Subject: Merge branch 'master' of ssh://plrg.eecs.uci.edu/home/git/iotcloud
X-Git-Url: http://demsky.eecs.uci.edu/git/?a=commitdiff_plain;h=1eb744ef67a2f85bd6d59f81a73d3d9529787ecb;p=iotcloud.git

Merge branch 'master' of ssh://plrg.eecs.uci.edu/home/git/iotcloud

Conflicts:
	doc/iotcloud.tex
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1eb744ef67a2f85bd6d59f81a73d3d9529787ecb
diff --cc doc/iotcloud.tex
index 6b812d3,a795fa8..b30c2cd
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+++ b/doc/iotcloud.tex
@@@ -769,15 -854,13 +878,17 @@@ Forward direction: The path of $t$ is 
  Suppose that there is a transitive closure set $\mathscr{S}$ of clients, at index $n$. Then there is some total message sequence $T$ of length $n$ such that every client $C$ in $\mathscr{S}$ sees a partial sequence $P_C$ consistent with $T$. \end{theorem}
  
  \begin{proof}
 -The definition of consistency of $P_C$ with $T$ is that every message $p \in P_C$ with index $i(p) \le n$ is equal to the message in that slot in $T$. Let $C_1$ be some client in the transitive closure set, with partial message sequence $P_{C_1}$, and let $u$ be some message with $i(u) > i$ that $C_1$ shares with another client. Then let $T$ be the portion of the path of $u$ ending at index $i$ and $t$ be the message at that index. Clearly, by Lemma 1, $P_{C_1}$ is consistent with $T$. We will show that, for every other client $D$ with partial sequence $P_D$, $P_D$ has some message whose path includes $t$. Because $D$ is in the transitive closure, there is a sequence of edges from $C_1$ to $D$. Call this $\mathscr{C} = (C_1, C_2, ..., D)$. 
+ 
 -We subsequently prove by induction that $D$ has a message whose path includes $t$.
 +The definition of consistency of $P_C$ with $T$ is that every message $p \in P_C$ with index $i(p) \le n$ is equal to the message in that slot in $T$. Let $C_1$ be some client in the transitive closure set, with partial message sequence $P_{C_1}$, and let $u$ be some message with $i(u) > n$ that $C_1$ shares with another client. Then let $T$ be the portion of the path of $u$ ending at index $n$ and $t$ be the message at that index. Clearly, by Lemma 1, $P_{C_1}$ is consistent with $T$. We will show that, for every other client $D$ with partial sequence $P_D$, $P_D$ has some message whose path includes $t$. Because $D$ is in the transitive closure, there is a sequence of clients $\mathscr{C} = (C_1, C_2, ..., D)$ from $C_1$ to $D$, where each shares an edge with the next.
 +
 +We prove by induction that $P_D$ has a message whose path includes $t$.
  \begin{itemize}
 -\item For the base case, $P_{C_1}$ includes $u$, whose path includes $t$. 
 -\item For the inductive step, suppose $P_{C_k}$ has a message $w$ with a path that includes $t$, and shares message $x$ with $P_{C_{k+1}}$ such that $i(x) > i$. If $i(w) = i(x)$, then $w = x$. If $i(w) < i(x)$, then, by Lemma 1, $w$ is in the path of $x$. If $i(w) > i(x)$, $x$ is in the path of $w$; note again that its index is greater than $i$. In any case, $t$ is in the path of $u_k+1$.
 -\item Let $w$ the message of $D$ whose path includes $t$. By Lemma 1, every message in $P_D$ with index smaller than $i(w)$ is in the path of $w$. Since $t$ is in the path of $w$, every message in $P_D$ with smaller index than $i(t)$ is in $T$. Therefore, $P_D$ is consistent with $T$.
 +\item Base case: $P_{C_1}$ includes $u$, whose path includes $t$.
 +
 +\item Inductive step: Each client in $\mathscr{C}$ has a partial message sequence with a message that includes $t$ if the previous client does. Suppose $P_{C_k}$ has a message $w$ with a path that includes $t$, and shares message $x$ with $P_{C_{k+1}}$ such that $i(x) > n$. By Lemma 1, $w$ or $x$, whichever has the least sequence number, is in the path of the other, and therefore by Lemma 2, $t$ is in the path of $x$.
 +
 +\item Let $z$ be the message of $D$ whose path includes $t$. By Lemma 1, every message in $P_D$ with index smaller than $i(w)$ is in the path of $z$. Since $t$ is in the path of $z$, every message in $P_D$ with smaller index than $i(t)$ is in $T$. Therefore, $P_D$ is consistent with $T$.
++
  \end{itemize}
  \end{proof}