Restructure Unify
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Andreas Stadelmeier
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unify.tex
332
unify.tex
@ -6,6 +6,102 @@
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% the algorithm only removes wildcards, never adds them
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\section{Unify}\label{sec:unify}
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%\newcommand{\tw}[1]{\tv{#1}_?}
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\begin{description}
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\item[Input:] An environment $\Delta'$
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and a set of constraints $C = \set{ \type{T} \lessdot \type{T}, \type{T} \lessdotCC \type{T}, \type{T} \doteq \type{T} \ldots}$
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\item[Output:]
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Set of unifiers $Uni = \set{\sigma_1, \ldots, \sigma_n}$ and an environment $\Delta$
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\end{description}
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%The transformation steps are not applied all at once but in a specific order:
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\unify{} executes the following steps until a type solution is found:
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\begin{description}
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\item[Step 1:]
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Apply the rules depicted in the figures \ref{fig:normalizing-rules}, \ref{fig:reduce-rules} and \ref{fig:wildcard-rules} exhaustively,
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starting with the \rulename{circle} rule.
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\item[Step 2:]
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%If there are no $(\type{T} \lessdot \tv{a})$ constraints remaining in the constraint set $C$
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%resume with step 4.
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The second step is nondeterministic.
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%\unify{} has to pick the right transformation for each constraint of the form $\type{N} \lessdot \tv{a}$.
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%The rules in figure \ref{fig:step2-rules} offer three possibilities to deal with constraints $\type{N} \lessdot \tv{a}$.
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For every $\type{T} \lessdot \tv{a}$ constraint \unify{} has to pick exactly one transformation from figure \ref{fig:step2-rules}.
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The same principle goes for constraints of the form $\tv{a} \lessdot \type{N}, \tv{a} \lessdot \tv{b}$ and the two transformations in figure \ref{fig:step2-rules2}.
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%They have to be applied until the constraint set holds no constraints of the form $\tv{a} \lessdot \type{N}, \tv{a} \lessdot \tv{b}$.
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If atleast one transformation was applied in this step revert to step 1.
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Otherwise proceed with step 3.
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%This builds a search tree over multiple possible solutions.
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%\unify{} has to try each branch and accumulate the resulting solutions into a solution set.
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%$\type{T} \lessdot \ntv{a}$ constraints have three and $\type{T} \lessdot \wtv{a}$ constraints have five possible transformations.
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\textit{Hint:}
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When implementing this step via backtracking
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the rules \rulename{General} and \rulename{Super} should be tried first.
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% is the Same rule really necessary?
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The \rulename{Settle} and \rulename{Raise} rules should only be used when none of the rules in figure \ref{fig:step2-rules} can be applied.
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\item[Step 3:]
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Apply the rules in figure \ref{fig:cleanup-rules} exhaustively.
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\rulename{Ground} and \rulename{Flatten} deal with constraints containing free variables.
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If a type placeholder is solely used as lower bound \rulename{Ground} can replace it with the bottom type.
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Otherwise the \rulename{Flatten} rule has to remove the wildcards responsible for the free variable.
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\text{Note:} Only one of those rules has to be applied per constraint.
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If the constraint set has been changed by one of these rules the algorithm must return to step 1.
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But if only the \rulename{SubElim} rule is applied or the constraint set is not changed at all,
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the algorithm can proceed with step 4.
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The cleanup step prepares the constraint set for the last step by applying the following concepts:
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%Two transformations are done which also help to remove unnecessary types from the solution set.
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\begin{description}
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\item[Bottom type]
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The bottom type $\bot$ is used to generate \texttt{? extends} wildcard definitions.
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This is the only possible solution when dealing with multiple upper bounds:
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$\tv{a} \lessdot \type{T}, \tv{a} \lessdot \type{S}$ is usually not a correct solution (given $\type{S}$ and $\type{T}$ are no subtypes of eachother).
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But if $\tv{a}$ is a lower bound of a wildcard it can be set to $\bot$.
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Those constraints only stay in the constraint set after the first step if $\type{S}$ and $\type{T}$ do not have a common subtype.
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The \rulename{Ground} rule uses this concept to generate \texttt{extends} Wildcards.
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\end{description}
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\item[Step 4:] \textit{(Generating Result)}
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Apply the rules in figure \ref{fig:generation-rules} until $\wildcardEnv = \emptyset$ and $C = \emptyset$.
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The resulting $\Delta, \sigma$ is a correct solution.
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For this step to succeed there should only be four kinds of constraints left.
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\begin{enumerate}
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%\item\label{item:3} $\tv{a} \lessdot \tv{b}$ %, with $a$ and $b$ both isolated type variables
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\item $\tv{a} \doteq \tv{b}$
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%\item $\wtv{a} \doteq \type{G}$
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\item\label{item:1} $\tv{a} \lessdot \wctype{\ol{\wtype{W}}}{C}{\ol{\type{T}}}$, with $\text{fv}(\wctype{\ol{\wtype{W}}}{C}{\ol{\type{T}}}) \subseteq \Delta_in$
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\item\label{item:2} $\tv{a} \doteq \wctype{\ol{\wtype{W}}}{C}{\ol{\type{T}}}$, with $\tv{a} \notin \ol{\type{T}}$ % and $\text{fv}(\wctype{\ol{\wtype{W}}}{C}{\ol{\type{T}}}) = \emptyset$
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\item\label{item:3} $\tv{a} \doteq \rwildcard{X}$
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\end{enumerate}
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%Each type placeholder $\tv{a}$ must solely appear on the left side of a constraint.
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\unify{} fails if there is any $\tv{a} \doteq \type{T}$ such that $\tv{a}$ occurs in $\type{T}$.
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For the cases \ref{item:1}, \ref{item:2}, and \ref{item:3} the placeholder $\tv{a}$
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cannot appear anywhere else in the constraint set.
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Otherwise the generation rules \rulename{GenSigma} and \rulename{GenDelta} will not be able to process every constraint.
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% \begin{displaymath}
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% \deduction{
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% \wildcardEnv \cup \set{\wildcard{B}{\type{G}}{\type{G'}}} \vdash C \implies \Delta, \sigma
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% }{
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% \wildcardEnv \vdash C \implies \Delta \cup \set{\wildcard{B}{\type{G}}{\type{G'}}}, \sigma
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% }\quad \text{tph}(\type{G}) = \emptyset, \text{tph}(\type{G'}) = \emptyset,
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% \rwildcard{B} \notin \text{dom}(\Delta)
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% \quad \rulename{AddDelta}
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% \end{displaymath}
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\end{description}
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\subsection{Transformation Rules (Step 1)}
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Our \unify{} process uses a similar concept as the standard unification by Martelli and Montanari \cite{MM82},
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consisting of terms, relations and variables.
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Instead of terms we have types of the form $\exptype{C}{\ol{T}}$ and
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@ -462,7 +558,75 @@ and every added wildcard gets a fresh unique name.
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This ensures the wildcard environment $\wildcardEnv{}$ never
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gets the same wildcard twice.
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\subsection{Branching Step}
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\begin{figure}
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\begin{center}
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\leavevmode
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\fbox{
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\begin{tabular}[t]{l@{~}l}
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\rulename{Subst} &
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$\begin{array}[c]{l}
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\wildcardEnv \vdash C \cup \set{\ntv{a} \doteq \type{T}}\\
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\hline
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[\type{T}/\tv{a}]\wildcardEnv \vdash [\type{T}/\tv{a}]
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C \cup \set{\ntv{a} \doteq \type{T}}
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\end{array}
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\quad \begin{array}{c}
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\ntv{a} \notin \type{T} \\
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\text{fv}(\type{T}) \subseteq \Delta', \, \text{wtv}(\type{T}) = \emptyset
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\end{array}$\\
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\\
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\rulename{Subst-WC} &$
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\begin{array}[c]{l}
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\wildcardEnv \vdash C \cup \set{\wtv{a} \doteq \rwildcard{T}}\\
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\hline
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[\type{T}/\wtv{a}]\wildcardEnv \vdash [\type{T}/\wtv{a}]C
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\end{array} \quad \wtv{a} \notin \type{T}
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$\\
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\\
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\rulename{Normalize} &
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$\begin{array}[c]{l}
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\wildcardEnv \vdash C \cup \set{\ntv{a} \doteq \type{T}}\\
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\hline
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[\ntv{b}/\wtv{b}]\wildcardEnv \vdash [\ntv{b}/\wtv{b}]
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C \cup \set{\ntv{a} \doteq [\ntv{b}/\wtv{b}]\type{T}}
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\end{array}
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\quad \begin{array}{c}
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\wtv{b} \in \text{wtv}(\type{T})\\
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\ntv{b} \ \text{fresh}
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\end{array}$\\
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\\
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\rulename{Contract} &
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$\begin{array}[c]{l}
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\wildcardEnv \cup \set{\wildcard{A}{\type{U}}{\type{L}}} \vdash C \cup \set{\ntv{a} \doteq \type{T}}\\
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\hline
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[\type{U}/\type{A}]\wildcardEnv \vdash [\type{U}/\type{A}]
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C \cup [\type{U}/\type{A}]\set{\ntv{a} \doteq \type{T}, \type{L} \doteq \type{U}}
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\end{array}
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\quad \begin{array}{c}
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\rwildcard{A} \in \text{fv}(\type{T})\\
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\end{array}$\\
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\end{tabular}}
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\end{center}
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\caption{Substitution rules}\label{fig:subst-rules}
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\end{figure}
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\unify{} must not replace normal type placeholders with free variables
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except variables initially passed by $\Delta_{in}$.
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The \rulename{Subst} rule checks if a type $\type{T}$ contains any
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free variables or wildcard placeholders before replacing a normal type placeholder with it.
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%This ensures that free variables are never substituted for normal type placeholders.
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This ensures that a normal type placeholder is never replaced by a type containing free variables.
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A type solution for a normal type placeholder will never contain free variables.
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This is needed to guarantee well-formed type solutions and also keep free variables inside their scope
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(see challenge \ref{challenge3}).
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\rulename{Subst-WC} does not need to do that and can freely replace wildcard placehodlers with types despite their free variables.
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We do not keep replacements for wildcard placeholders and they will not show up in the final type solution.
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If the \rulename{Subst} rule is not applicable then either the \rulename{Normalize}
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or \rulename{Contract} transformation has to be used to remove wildcard placeholders and
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wildcards.
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\subsection{Branching Step (Step 2)}
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\unify{} is described as a nondeterministic algorithm.
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Some constraints allow for multiple transformations from which
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the algorithm has to pick the right one.
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@ -474,7 +638,9 @@ and gathers all possible type solutions.
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We skip the definition of this practice, because it is already described in \cite{TIforFGJ}
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and only needed for a proof of completeness.
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\subsection{Generate Result (Step 4)}
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The generation rules defined in figure \ref{fig:generation-rules} are similar to the other transformation rules but contain an additional part,
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the result output consisting of a wildcard environment and a set of unifier $\sigma$.
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\subsection{Adding Wildcards to the mix}
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%\unify{} is able to create wildcard solutions even when the input set of constraints do not contain any wildcard variables.
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@ -622,101 +788,6 @@ The \texttt{concat} method takes two lists of the same generic type.
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\subsection{Algorithm}
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%\newcommand{\tw}[1]{\tv{#1}_?}
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\begin{description}
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\item[Input:] An environment $\Delta'$
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and a set of constraints $C = \set{ \type{T} \lessdot \type{T}, \type{T} \lessdotCC \type{T}, \type{T} \doteq \type{T} \ldots}$
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\item[Output:]
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Set of unifiers $Uni = \set{\sigma_1, \ldots, \sigma_n}$ and an environment $\Delta$
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\end{description}
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%The transformation steps are not applied all at once but in a specific order:
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\unify{} executes the following steps until a type solution is found:
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\begin{description}
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\item[Step 1:]
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Apply the rules depicted in the figures \ref{fig:normalizing-rules}, \ref{fig:reduce-rules} and \ref{fig:wildcard-rules} exhaustively,
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starting with the \rulename{circle} rule.
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\item[Step 2:]
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%If there are no $(\type{T} \lessdot \tv{a})$ constraints remaining in the constraint set $C$
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%resume with step 4.
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The second step is nondeterministic.
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%\unify{} has to pick the right transformation for each constraint of the form $\type{N} \lessdot \tv{a}$.
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%The rules in figure \ref{fig:step2-rules} offer three possibilities to deal with constraints $\type{N} \lessdot \tv{a}$.
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For every $\type{T} \lessdot \tv{a}$ constraint \unify{} has to pick exactly one transformation from figure \ref{fig:step2-rules}.
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The same principle goes for constraints of the form $\tv{a} \lessdot \type{N}, \tv{a} \lessdot \tv{b}$ and the two transformations in figure \ref{fig:step2-rules2}.
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%They have to be applied until the constraint set holds no constraints of the form $\tv{a} \lessdot \type{N}, \tv{a} \lessdot \tv{b}$.
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If atleast one transformation was applied in this step revert to step 1.
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Otherwise proceed with step 3.
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%This builds a search tree over multiple possible solutions.
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%\unify{} has to try each branch and accumulate the resulting solutions into a solution set.
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%$\type{T} \lessdot \ntv{a}$ constraints have three and $\type{T} \lessdot \wtv{a}$ constraints have five possible transformations.
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\textit{Hint:}
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When implementing this step via backtracking
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the rules \rulename{General} and \rulename{Super} should be tried first.
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% is the Same rule really necessary?
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The \rulename{Settle} and \rulename{Raise} rules should only be used when none of the rules in figure \ref{fig:step2-rules} can be applied.
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\item[Step 3:]
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Apply the rules in figure \ref{fig:cleanup-rules} exhaustively.
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\rulename{Ground} and \rulename{Flatten} deal with constraints containing free variables.
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If a type placeholder is solely used as lower bound \rulename{Ground} can replace it with the bottom type.
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Otherwise the \rulename{Flatten} rule has to remove the wildcards responsible for the free variable.
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\text{Note:} Only one of those rules has to be applied per constraint.
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If the constraint set has been changed by one of these rules the algorithm must return to step 1.
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But if only the \rulename{SubElim} rule is applied or the constraint set is not changed at all,
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the algorithm can proceed with step 4.
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The cleanup step prepares the constraint set for the last step by applying the following concepts:
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%Two transformations are done which also help to remove unnecessary types from the solution set.
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\begin{description}
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\item[Bottom type]
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The bottom type $\bot$ is used to generate \texttt{? extends} wildcard definitions.
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This is the only possible solution when dealing with multiple upper bounds:
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$\tv{a} \lessdot \type{T}, \tv{a} \lessdot \type{S}$ is usually not a correct solution (given $\type{S}$ and $\type{T}$ are no subtypes of eachother).
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But if $\tv{a}$ is a lower bound of a wildcard it can be set to $\bot$.
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Those constraints only stay in the constraint set after the first step if $\type{S}$ and $\type{T}$ do not have a common subtype.
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The \rulename{Ground} rule uses this concept to generate \texttt{extends} Wildcards.
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\end{description}
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\item[Step 4:] \textit{(Generating Result)}
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Apply the rules in figure \ref{fig:generation-rules} until $\wildcardEnv = \emptyset$ and $C = \emptyset$.
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The resulting $\Delta, \sigma$ is a correct solution.
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For this step to succeed there should only be four kinds of constraints left.
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\begin{enumerate}
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%\item\label{item:3} $\tv{a} \lessdot \tv{b}$ %, with $a$ and $b$ both isolated type variables
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\item $\tv{a} \doteq \tv{b}$
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%\item $\wtv{a} \doteq \type{G}$
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\item\label{item:1} $\tv{a} \lessdot \wctype{\ol{\wtype{W}}}{C}{\ol{\type{T}}}$, with $\text{fv}(\wctype{\ol{\wtype{W}}}{C}{\ol{\type{T}}}) \subseteq \Delta_in$
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\item\label{item:2} $\tv{a} \doteq \wctype{\ol{\wtype{W}}}{C}{\ol{\type{T}}}$, with $\tv{a} \notin \ol{\type{T}}$ % and $\text{fv}(\wctype{\ol{\wtype{W}}}{C}{\ol{\type{T}}}) = \emptyset$
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\item\label{item:3} $\tv{a} \doteq \rwildcard{X}$
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\end{enumerate}
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%Each type placeholder $\tv{a}$ must solely appear on the left side of a constraint.
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\unify{} fails if there is any $\tv{a} \doteq \type{T}$ such that $\tv{a}$ occurs in $\type{T}$.
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For the cases \ref{item:1}, \ref{item:2}, and \ref{item:3} the placeholder $\tv{a}$
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cannot appear anywhere else in the constraint set.
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Otherwise the generation rules \rulename{GenSigma} and \rulename{GenDelta} will not be able to process every constraint.
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% \begin{displaymath}
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% \deduction{
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% \wildcardEnv \cup \set{\wildcard{B}{\type{G}}{\type{G'}}} \vdash C \implies \Delta, \sigma
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% }{
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% \wildcardEnv \vdash C \implies \Delta \cup \set{\wildcard{B}{\type{G}}{\type{G'}}}, \sigma
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% }\quad \text{tph}(\type{G}) = \emptyset, \text{tph}(\type{G'}) = \emptyset,
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% \rwildcard{B} \notin \text{dom}(\Delta)
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% \quad \rulename{AddDelta}
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% \end{displaymath}
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\end{description}
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With \texttt{C} being class names and \texttt{A} being wildcard names.
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The wildcard type $\wildcard{X}{U}{L}$ consist of an upper bound $\type{U}$, a lower bound $\type{L}$
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and a name $\mathtt{X}$.
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@ -820,73 +891,6 @@ which are used for the upper and lower bounds.
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% if there are a <. List<x?> constraints remaining in the end, then this can be a sign of a irregular input constraint set.
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\begin{figure}
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\begin{center}
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\leavevmode
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\fbox{
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\begin{tabular}[t]{l@{~}l}
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\rulename{Subst} &
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$\begin{array}[c]{l}
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\wildcardEnv \vdash C \cup \set{\ntv{a} \doteq \type{T}}\\
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\hline
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[\type{T}/\tv{a}]\wildcardEnv \vdash [\type{T}/\tv{a}]
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C \cup \set{\ntv{a} \doteq \type{T}}
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\end{array}
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\quad \begin{array}{c}
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\ntv{a} \notin \type{T} \\
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\text{fv}(\type{T}) \subseteq \Delta', \, \text{wtv}(\type{T}) = \emptyset
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\end{array}$\\
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\\
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\rulename{Subst-WC} &$
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\begin{array}[c]{l}
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\wildcardEnv \vdash C \cup \set{\wtv{a} \doteq \rwildcard{T}}\\
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\hline
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[\type{T}/\wtv{a}]\wildcardEnv \vdash [\type{T}/\wtv{a}]C
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\end{array} \quad \wtv{a} \notin \type{T}
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$\\
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\\
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\rulename{Normalize} &
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$\begin{array}[c]{l}
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\wildcardEnv \vdash C \cup \set{\ntv{a} \doteq \type{T}}\\
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\hline
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[\ntv{b}/\wtv{b}]\wildcardEnv \vdash [\ntv{b}/\wtv{b}]
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C \cup \set{\ntv{a} \doteq [\ntv{b}/\wtv{b}]\type{T}}
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\end{array}
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\quad \begin{array}{c}
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\wtv{b} \in \text{wtv}(\type{T})\\
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\ntv{b} \ \text{fresh}
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\end{array}$\\
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\\
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\rulename{Contract} &
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$\begin{array}[c]{l}
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\wildcardEnv \cup \set{\wildcard{A}{\type{U}}{\type{L}}} \vdash C \cup \set{\ntv{a} \doteq \type{T}}\\
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\hline
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||||
[\type{U}/\type{A}]\wildcardEnv \vdash [\type{U}/\type{A}]
|
||||
C \cup [\type{U}/\type{A}]\set{\ntv{a} \doteq \type{T}, \type{L} \doteq \type{U}}
|
||||
\end{array}
|
||||
\quad \begin{array}{c}
|
||||
\rwildcard{A} \in \text{fv}(\type{T})\\
|
||||
\end{array}$\\
|
||||
\end{tabular}}
|
||||
\end{center}
|
||||
\caption{Substitution rules}\label{fig:subst-rules}
|
||||
\end{figure}
|
||||
|
||||
\unify{} must not replace normal type placeholders with free variables
|
||||
except variables initially passed by $\Delta_{in}$.
|
||||
The \rulename{Subst} rule checks if a type $\type{T}$ contains any
|
||||
free variables or wildcard placeholders before replacing a normal type placeholder with it.
|
||||
%This ensures that free variables are never substituted for normal type placeholders.
|
||||
This ensures that a normal type placeholder is never replaced by a type containing free variables.
|
||||
A type solution for a normal type placeholder will never contain free variables.
|
||||
This is needed to guarantee well-formed type solutions and also keep free variables inside their scope
|
||||
(see challenge \ref{challenge3}).
|
||||
\rulename{Subst-WC} does not need to do that and can freely replace wildcard placehodlers with types despite their free variables.
|
||||
We do not keep replacements for wildcard placeholders and they will not show up in the final type solution.
|
||||
If the \rulename{Subst} rule is not applicable then either the \rulename{Normalize}
|
||||
or \rulename{Contract} transformation has to be used to remove wildcard placeholders and
|
||||
wildcards.
|
||||
|
||||
% \begin{example}{Free variables must not leave the scope of the surrounding \texttt{let} statement}
|
||||
|
||||
% \noindent
|
||||
|
||||
Loading…
Reference in New Issue
Block a user