From fd66c0b51f6718b0dd734c786920ae9fb33ee87b Mon Sep 17 00:00:00 2001 From: Carlo Zancanaro Date: Mon, 19 Nov 2012 16:58:51 +1100 Subject: Add some more contribution stuff. Proof? --- tex/thesis/contribution/contribution.tex | 250 ++++++++++++++++++++----------- 1 file changed, 166 insertions(+), 84 deletions(-) (limited to 'tex/thesis/contribution/contribution.tex') diff --git a/tex/thesis/contribution/contribution.tex b/tex/thesis/contribution/contribution.tex index e41a39f..acf9cdd 100644 --- a/tex/thesis/contribution/contribution.tex +++ b/tex/thesis/contribution/contribution.tex @@ -18,10 +18,14 @@ \newcommand\all{\mathsf{all}} \newcommand\fix{\mathsf{fix}} -\newcommand\soln{\mathsf{soln}} +\newcommand\solnFP{\mathsf{soln^{FP}}} +\newcommand\RFP{\mathsf{R^{FP}}} +\newcommand\solnSI{\mathsf{soln^{SI}}} +\newcommand\RSI{\mathsf{R^{SI}}} \newcommand\post{\mathsf{post}} \newcommand\pre{\mathsf{pre}} +\newcommand\last{\mathsf{last}} \newcommand\Operators{\mathcal{O}} \newcommand\Systems{\mathbb{E}} @@ -62,7 +66,7 @@ The existing algorithm as presented in Section fixpoint iteration and max-strategy iteration. Each of these operations consists of naively ``evaluating'' the system repeatedly until a further evaluation results in no change. Gawlitza et -al.~proved that these iterations must converge in a finite number of +al.~proved that these iterations converge in a finite number of steps\cite{Gawlitza:2007:PFC:1762174.1762203} , but in practice this naive approach performs a large number of re-calculations of results which are already known. @@ -293,7 +297,7 @@ presented in Listing \ref{algo:wdfs}. \If {$v \ne D[x]$} \State $D = \{ x \mapsto v \} \oplus D$ \State $W = \inflFP[x]$ - \State $\inflFP(x) = \emptyset$ + \State $\inflFP[x] = \emptyset$ \State $\stableFP = \stableFP \backslash W$ \For {$v \in W$} \State $\solve(v)$ @@ -323,7 +327,7 @@ the values of other variables through the $\eval$ function. In the case of mutually-recursive variables this would result in infinite recursion, as each time the variable's value is calculated we -must first calculate the variable's value. In order to prevent this a +first calculate the variable's value. In order to prevent this a $\stable$ set is maintained of variables whose values are either currently being computed or whose values have already been computed. This set provides a ``base'' for the recursion to terminate, @@ -423,9 +427,10 @@ In the particular case of \emph{monotonic}, \emph{expansive} operators it has been shown\cite{Gawlitza:2007:PFC:1762174.1762203} that this process will take a finite number of iterations if we ensure at each iteration that we make a \emph{strict} improvement on our -strategy. This means that each time we improve our strategy we must -make it greater in at least one $\max$-expression, and no worse in the -others. +strategy. This means that each time we improve our strategy we make it +greater in at least one $\max$-expression, and no worse in the +others. That is, we must satisfy $\sigma_{n+1}(\rho) > +\sigma_n(\rho)$, for the $\rho$ in which we are improving. A new function, $\improve: ((E_{\max} \to E) \times (X \to D)) \to (E_{\max} \to E)$, is used below as a \emph{strategy improvement @@ -539,6 +544,8 @@ solve $\max$-strategy iteration problems. $\stableSI$ & $\subseteq E_{\max}$, the set of all $\max$-expressions whose strategies are stable \\ + $\rho$ & $: X \to \CZ$, a variable assignment \\ + $\system$ & $\in \Systems$, an equation system \\ $\improve$ & $: ((E_{\max} \to E) \times (X \to \CZ)) \to (E_{\max} @@ -570,17 +577,24 @@ solve $\max$-strategy iteration problems. \Procedure {solve} {$x \in E_{\max}$} \If {$x \not \in \stableSI$} \State $\stableSI = \stableSI \cup \{x\}$ - \State $\rho = \solve \fixpoint(\sigma(\system))$ \State $v = \improve(\lambda y . \eval(x, y), \rho)[x]$ \If {$v \ne \sigma[x]$} \State $\sigma = \{ x \mapsto v\} \oplus \sigma$ \State $W = \inflSI[x]$ - \State $\inflSI(x) = \emptyset$ + \State $\inflSI[x] = \emptyset$ \State $\stableSI = \stableSI \backslash W$ \For {$v \in W$} - \State $\solve(v)$ + \State $\solve(v)$ \EndFor \EndIf + \State $\rho_{\last} = \rho$ + \State $\rho = \solve \fixpoint(\sigma(\system))$ + \For {$y \in X$} + \If {$\rho_{\last}[y] = \rho[y]$} + \State $\stableSI = \stableSI \backslash \{\system[y]\}$ + \State $\solve(\system[y])$ + \EndIf + \EndFor \EndIf \EndProcedure \end{algorithmic} @@ -594,7 +608,8 @@ fixpoints, but applies them to $\max$-strategy iteration problems. Dependencies between $\max$-expressions are considered and an improvement on the strategy at an expression $e$ is only attempted if some of the $\max$-expressions which influence $e$ themselves -improve strategy. +improve strategy. We also destabilise $\max$-expressions which have +their values changed in the fixpoint iteration step. For this particular algorithm to work, however, we must assume another property of $\improve$. In Listing \ref{algo:adapted-wdfs} we take the @@ -622,7 +637,8 @@ result in the $\max$-strategy improvement at each step being more expensive. The extra work to improve other portions of the strategy will be discarded with only $\improve(...)[x]$ being used. The particular $\improve$ that we are using (see Listing \ref{algo:pmax}) -satisfies this property by way of its anonymous function. +satisfies this property using an anonymous function to delay the +computation. \section{Local Demand-driven Strategy Improvement (LDSI)} @@ -997,6 +1013,13 @@ themselves destabilised as well. \label{algo:ldsi:solve} \end{algorithm} +$\solve \strategy$ is very similar to the other $\solve$ calls we have +already seen. The procedure attempts a strategy improvement for the +$\max$-expression $x$ and, if any improvement is possible, it improves +the strategy $\sigma$ before invalidating and re-evaluating any other +$\max$-expressions which may depend on $x$. We here assume that +$\improve$ has the same properties as those which are assumed in +Section \ref{sec:adapted-wdfs}. @@ -1011,107 +1034,166 @@ and $\max$-strategy iteration. The fixpoint iteration portion of the algorithm is concerned with -finding the greatest fixpoint, $\soln(s) : X \to D$ for the system $s +finding the greatest fixpoint, $\solnFP(s) : X \to D$ for the system $s = \sigma(\system)$. If we consider all valid fixpoints, $\fix(s) = \{ -\rho \mid \rho = s(\rho) \}$, then we find that $\soln(s) \in \fix(s)$ -and $\soln(s) \ge \rho ~ \forall \rho \in \fix(s)$. Our general -approach will be to consider our variable assignment $D \ge \soln(s)$ +\rho \mid \rho = s(\rho) \}$, then we find that $\solnFP(s) \in \fix(s)$ +and $\solnFP(s) \ge \rho ~ \forall \rho \in \fix(s)$. Our general +approach will be to consider our variable assignment $D \ge \solnFP(s)$ and show that it decreases until $D \in \fix(s)$ and thus $D = -\soln(s)$. +\solnFP(s)$. -We begin by stating two invariants of the LDSI algorithm. The set -$\depFP^{+}(x)$ is the set of all variables which $x$ may depend +We begin by stating three invariants of the LDSI algorithm. We use +$\depFP^{+}(x)$ to be the set of all variables which $x$ may depend on. $\depFP^{+}$ is defined as the transitive closure of $\depFP$, -which is defined by $\depFP(x) = \{ y \in X \mid x \in \inflFP(y) -\}$. $R \subseteq X$ is defined as the variables currently on our -recursion stack. +$\depFP[x] = \{ y \in X \mid x \in \inflFP[y] \}$. $\RFP \subseteq X$ +is defined as the variables currently on our recursion stack. \begin{align} - D &\ge \soln(s) \label{eqn:fp-inv1} \\ + D &\ge \solnFP(s) \label{eqn:fp-inv1} \\ % - D &= \soln(s)[x] & \mbox{if } - x \in \stableFP, - x \not \in R, - \depFP^{+}(x) \cap R = \emptyset - \label{eqn:fp-inv2} \\ + x &\in \inflFP[y] & \forall x \in \stableFP, y \in \depFP[x] + \label{eqn:fp-inv3} \\ % - x \in \inflFP[y] & \forall x \in \stableFP, y \in \depFP(x) - \label{eqn:fp-inv3} + D &= \solnFP(s)[x] & \mbox{if } + x \not \in \RFP, + \depFP^{+}[x] \cap \RFP = \emptyset, + x \in \stableFP + \label{eqn:fp-inv2} \end{align} (\ref{eqn:fp-inv1}) follows from monotonicity. We begin our iteration -at some variable assignment $D \ge \soln(s)$ and monotonically descend -from there. Each evaluation of a variable remains greater than -$\soln(s)[x]$ because our variable assignment $D$ always satisfies $D -\ge soln(s)$. We cannot jump below the greatest fixpoint because to do -so would imply that we could ``turn around'' for at least one -variable, thereby breaking our monotonicity assumption. +at some variable assignment $D \ge \solnFP(s)$ and monotonically descend +from there. We know that $s(D)[x] \ge soln(s)[x] ~\forall D \ge +\solnFP(s), x \in X$, thus each evaluation of $s[x]$ in the context of +$D$ will result in a new value for $D[x]$ which is either closer to or +equal to $\solnFP(s)[x]$. As we only ever evaluate in the context of +variables taken from $D$, and $D[x]$ is set to $\infty$ when $x$ is +invalidated, we know that $D \ge \solnFP(s)$ holds. + +(\ref{eqn:fp-inv3}) follows from the use of the $\eval \fixpoint$ +function. Each variable lookup results in a direct insertion into +$\inflFP$, so during the evaluation of $x$ all relevant $\inflFP$ sets +are updated to reflect the current dependencies. For (\ref{eqn:fp-inv2}) we will examine what has happened when each of the conditions is false: \begin{itemize} \item - If $x \not \in \stableFP$ then a re-evaluation of $x$ will occur - with the next call to $\solve \fixpoint(x)$. This may result in a - change to the value of $D[x]$, or it may leave $D[x]$ stable for the - moment. In either case, the value of $D[x]$ will be decreased, given - the value of the other variables in $X \backslash {x}$. If we denote - the variable assignment before the evaluation of $x$ by $D_{\pre}$, - and the variable assignment after the evaluation of $x$ by - $D_{\post}$ then we find that $D_{\post} \le D_{\pre}$. + If $x \in \RFP$ then we are currently in the context of an earlier + evaluation of $x$, so the value of $D[x]$ will be changed again by + that earlier call. In addition, we conclude that $x \in + \depFP^{+}[x]$ and thus, from the following condition, that $x$ will + be re-evaluated at some point after this one. \item - If $\depFP^{+}(x) \cap R \ne \emptyset$ then a variable which $x$ - depends on is in $R$ at the moment. This will result in the + If $\depFP^{+}[x] \cap \RFP \ne \emptyset$ then a variable which $x$ + depends on is in $\RFP$ at the moment. This will result in the re-evaluation of $x$ if the value of any of the variables in - $\depFP^{+}(x) \cap R$ have changed. + $\depFP^{+}[x] \cap \RFP$ have changed. \item - If $x \in R$ then we are currently in the context of an earlier - evaluation of $x$, so the value of $D[x]$ will be changed again by - that earlier call. In addition, we conclude that $x \in - \depFP^{+}(x)$ and thus, from the previous condition, that $x$ will - be entirely re-evaluated at some point after this one. + If $x \not \in \stableFP$ then a re-evaluation of $x$ will occur + with the next call to $\solve \fixpoint(x)$. This may result in a + change to the value of $D[x]$, or it may leave $D[x]$ stable for the + moment. If we denote the variable assignment before the evaluation + of $x$ by $D_{\pre}$, and the variable assignment after the + evaluation of $x$ by $D_{\post}$ then we find that $D_{\post} \le + D_{\pre}$, due to the monotonicity of $s$. After each change of + $D[x]$, everything which \emph{may} have depended on $D[x]$ is + re-evaluated (see the previous two items), leading to a final point + of stability when $D[x] = \solnFP(s)[x]$, by the definition of + $\solnFP(s)[x]$ and (\ref{eqn:fp-inv1}). \end{itemize} -(\ref{eqn:fp-inv3}) follows from the use of the $\eval \fixpoint$ -function. Each variable lookup results in a direct insertion into -$\inflFP$, so during the evaluation of $x$ all relevant $\inflFP$ sets -are updated. - After an evaluation of $\solve \fixpoint(x)$ we know that $x \in -\solve \fixpoint(x)$. If it is also the case that $R = \emptyset$, as -is the case for top-level calls to $\solve \fixpoint$, then we know -that $D[x] = \soln(s)[x]$. This means that a function $\lambda x +\stableFP$. If it is also the case that $\RFP = \emptyset$, as is the +case for top-level calls to $\solve \fixpoint$, then we know that +$D[x] = \solnFP(s)[x]$. This means that the function $\lambda x . (\solve \fixpoint(x); D[x])$ will act as a variable assignment solving for the greatest fixpoint of $s = \sigma(\system)$, as is required by our $\max$-strategy iteration. As the $\max$-strategy iteration changes $\sigma$ it also induces a change in $s = \sigma(\system)$, so in order to maintain the correctness of this -algorithm we must show that our above are maintained by this process. - -When the $\max$-strategy iteration portion invalidates a part of the -fixpoint iteration it does two things: it removes the variable $x$ -from the $\stableFP$ set, and it sets $D[x] = \infty$. The -invalidation of variables is then propagated to each variable which -transitively depends on $x$: $\{ y \in X \mid x \in \depFP^{+}(y) \}$, -or $\{ y \in X \mid y \in \inflFP^{+}(x) \}$, where $\inflFP^{+}$ is -the transitive closure of $\inflFP$. We know from (\ref{eqn:fp-inv3}) -that this transitive closure of $\inflFP$ will identify the entire -subsystem which may depend on the value of $x$. The invalidation of -transitive dependencies ensures that (\ref{eqn:fp-inv1}) and -(\ref{eqn:fp-inv2}) hold for the entire subsystem which is affected by -the change in strategy, and thus that our invariants hold for our -entire system. - -This is sufficient for us to be confident that the fixpoint iteration -portion of the algorithm is correct. The invariants ensure that $D$ is -approaching $\soln(s)$ and we know that our invariants are maintained -by the $\max$-strategy invalidating the fixpoint iteration. - - +algorithm we must show that our above invariants are maintained by +this process. + +When the $\max$-strategy iteration changes the current $\max$-strategy +$\sigma$ at the variable $x$ it changes the equation system $s = +\sigma(\system)$ that the fixpoint iteration uses. The $\max$-strategy +iteration then invalidates the affected portion of the fixpoint +iteration by doing two things: it removes the variable $x$ from the +$\stableFP$ set, and it sets $D[x] = \infty$. The invalidation of +variables is then propagated to each variable which transitively +depends on $x$: $\{ y \in X \mid x \in \depFP^{+}[y] \} = \{ y \in X +\mid y \in \inflFP^{+}[x] \}$, where $\inflFP^{+}$ is the transitive +closure of $\inflFP$. We know from (\ref{eqn:fp-inv3}) that this +transitive closure of $\inflFP$ will identify the entire subsystem +which may depend on the value of $x$. The invalidation of transitive +dependencies ensures that (\ref{eqn:fp-inv1}) holds for the changed +subsystem, as $\infty \ge z ~ \forall z \in \CZ$. From +(\ref{eqn:fp-inv1}) we can then conclude that (\ref{eqn:fp-inv2}) +holds as well, as the removal of $x$ from $\stableFP$ combined with +(\ref{eqn:fp-inv1}) leads to (\ref{eqn:fp-inv2}). These invariants now +hold for the affected sub-system and are still true for the unaffected +sub-system. Thus our invariants hold for our entire system and our +fixpoint iteration will continue to be correct in the presence of +invalidation by the $\max$-strategy iteration. + + +We move now to the $\max$-strategy iteration. We will use $\rho: X \to +\CZ$ as $\rho = \lambda x . (\solve \fixpoint(x); D[x])$, a variable +assignment which will always calculate the greatest fixpoint of +$\sigma(\system)$ for the current strategy $\sigma$. Each time $\rho$ +is queried for a variable's value it will also record which variables +have had their values changed, whether or not those changed values are +final, in the set $\touchedFP$. + +We can establish similar invariants for the $\max$-strategy iteration +as we had for the fixpoint iteration. We denote the optimal strategy +by $\solnSI(\system)$ and we define $\RSI \subseteq E_{\max}$ to be +the set of $\max$-expressions on our recursion stack. +\begin{align} + e &\in \inflSI[y] & \forall e \in \stableSI, y \in \depSI[e] + \label{eqn:si-inv1} \\ +% + \sigma &= \solnSI(\system)[e] & \mbox{if } + e \not \in \RSI, + \depSI^{+}[e] \cap \RSI = \emptyset, + e \in \stableSI + \label{eqn:si-inv2} +\end{align} -We move now to the $\max$-strategy iteration. -% TODO: finish the proof +(\ref{eqn:si-inv1}) follows from the same logic as (\ref{eqn:fp-inv3}) +for the fixpoint iteration. The $\eval \strategy$ directly updates the +$\inflSI$ sets whenever they are used. This ensures our dependency +information is correct. + +(\ref{eqn:si-inv2}) follows a similar argument to (\ref{eqn:fp-inv2}) +for the fixpoint iteration. At each step we must make a strategy +improvement, such that $\sigma_{n+1}(\system)(\rho) > +\sigma_n(\system)(\rho)$. Every step must be an improvement and, by the +assumptions made in Section \ref{sec:adapted-wdfs}, we know that we +will make an improvement for each individual variable if it is +possible to do so. The invalidation for a changed strategy is +identical to that which is performed for variables in the fixpoint +iteration, so it will not be re-considered here. An additional +complexity is that $\rho$ is not a constant mapping during +$\max$-strategy iteration. In order to maintain (\ref{eqn:si-inv2}) we +ensure that on changes in $\rho$ we invalidate some of our +$\max$-expressions. The set $\touchedFP$ over-approximates the set of +changed variables, so we are certain that all changes will be found by +enumerating $\touchedFP$. We only invalidate those $\max$-expressions +which are the right-hand side of a stable, changed variable. If a +variable has not stabilised then it's value is has not been used +directly (see (\ref{eqn:fp-inv2})) and thus cannot directly influence +the $\max$-strategy. If the variable has stabilised to the same value +as it had previously then it has not changed and thus has no effect on +the $\max$-strategy. + +If both the $\max$-strategy iteration and the fixpoint iteration +correctly solve their respective portions of the problem, and the +communication between them is correct, then we know that the overall +algorithm is correct. \subsection{Implementation} \label{sec:ldsi:implementation} + -- cgit v1.2.3