X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=blobdiff_plain;f=article.tex;h=01297a8fb0681c79d7ffc1933fe941f02f77adb5;hp=f8acb2260ffcf7898dd228718269b1efe9f40447;hb=e02ac317b82e521bccdbfadda3ff9c11d6a1533d;hpb=0b6086688e6de7212b4e21c46b3dcaf39e4cf063 diff --git a/article.tex b/article.tex index f8acb22..01297a8 100644 --- a/article.tex +++ b/article.tex @@ -10,6 +10,8 @@ \renewcommand{\ge}{\geqslant} \renewcommand{\le}{\leqslant} +\newcommand{\nge}{\ngeqslant} +\newcommand{\nle}{\nleqslant} \newcommand{\has}{\sqsupseteq} \newcommand{\isin}{\sqsubseteq} @@ -32,6 +34,10 @@ \newcommand{\py}{\pay{P}} \newcommand{\pn}{\pan{P}} +\newcommand{\pr}{\pa{R}} +\newcommand{\pry}{\pay{R}} +\newcommand{\prn}{\pan{R}} + %\newcommand{\hasparents}{\underaccent{1}{>}} %\newcommand{\hasparents}{{% % \declareslashed{}{_{_1}}{0}{-0.8}{>}\slashed{>}}} @@ -49,11 +55,18 @@ \newcommand{\pancsof}[2]{\pancs ( #1 , #2 ) } \newcommand{\pendsof}[2]{\pends ( #1 , #2 ) } -\newcommand{\patchof}[1]{{\mathcal P} ( #1 ) } -\newcommand{\baseof}[1]{{\mathcal B} ( #1 ) } +\newcommand{\merge}{{\mathcal M}} +\newcommand{\mergeof}[4]{\merge(#1,#2,#3,#4)} +%\newcommand{\merge}[4]{{#2 {{\frac{ #1 }{ #3 } #4}}}} + +\newcommand{\patch}{{\mathcal P}} +\newcommand{\base}{{\mathcal B}} +\newcommand{\patchof}[1]{\patch ( #1 ) } +\newcommand{\baseof}[1]{\base ( #1 ) } + +\newcommand{\eqntag}[2]{ #2 \tag*{\mbox{#1}} } \newcommand{\eqn}[2]{ #2 \tag*{\mbox{\bf #1}} } -\newcommand{\corrolary}[1]{ #1 \tag*{\mbox{\it Corrolary.}} } %\newcommand{\bigforall}{\mathop{\hbox{\huge$\forall$}}} \newcommand{\bigforall}{% @@ -68,7 +81,8 @@ \newcommand{\Largenexists}{\mathop{\hbox{\Large$\nexists$}}} \newcommand{\qed}{\square} -\newcommand{\proof}[1]{{\it Proof.} #1 $\qed$} +\newcommand{\proofstarts}{{\it Proof:}} +\newcommand{\proof}[1]{\proofstarts #1 $\qed$} \newcommand{\gathbegin}{\begin{gather} \tag*{}} \newcommand{\gathnext}{\\ \tag*{}} @@ -142,6 +156,17 @@ patch is applied to a non-Topbloke branch and then bubbles back to the Topbloke patch itself, we hope that git's merge algorithm will DTRT or that the user will no longer care about the Topbloke patch. +\item[ $\displaystyle \mergeof{C}{L}{M}{R} $ ] +The contents of a git merge result: + +$\displaystyle D \isin C \equiv + \begin{cases} + (D \isin L \land D \isin R) \lor D = C : & \true \\ + (D \not\isin L \land D \not\isin R) \land D \neq C : & \false \\ + \text{otherwise} : & D \not\isin M + \end{cases} +$ + \end{basedescript} \newpage \section{Invariants} @@ -181,7 +206,7 @@ Ie, the two limbs of the RHS of Tip Contents are mutually exclusive. Let $B = \baseof{C}$ in $D \isin \baseof{C}$. Now $B \in \pn$. So by Base Acyclic $D \isin B \implies D \notin \py$. } -\[ \corrolary{ +\[ \eqntag{{\it Corollary - equivalent to Tip Contents}}{ \bigforall_{C \in \py} D \isin C \equiv \begin{cases} D \in \py : & D \le C \\ @@ -227,15 +252,46 @@ by the LHS. And $A \le A''$. \[ \eqn{Calculation Of Ends:}{ \bigforall_{C \hasparents \set A} \pendsof{C}{\set P} = - \Bigl\{ E \Big| + \left\{ E \Big| \Bigl[ \Largeexists_{A \in \set A} E \in \pendsof{A}{\set P} \Bigr] \land \Bigl[ \Largenexists_{B \in \set A} E \neq B \land E \le B \Bigr] - \Bigr\} + \right\} }\] XXX proof TBD. +\subsection{No Replay for Merge Results} + +If we are constructing $C$, with, +\gathbegin + \mergeof{C}{L}{M}{R} +\gathnext + L \le C +\gathnext + R \le C +\end{gather} +No Replay is preserved. \proofstarts + +\subsubsection{For $D=C$:} $D \isin C, D \le C$. OK. + +\subsubsection{For $D \isin L \land D \isin R$:} +$D \isin C$. And $D \isin L \implies D \le L \implies D \le C$. OK. + +\subsubsection{For $D \neq C \land D \not\isin L \land D \not\isin R$:} +$D \not\isin C$. OK. + +\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R) + \land D \not\isin M$:} +$D \isin C$. Also $D \isin L \lor D \isin R$ so $D \le L \lor D \le +R$ so $D \le C$. OK. + +\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R) + \land D \isin M$:} +$D \not\isin C$. OK. + +$\qed$ + \section{Commit annotation} We annotate each Topbloke commit $C$ with: @@ -334,6 +390,80 @@ $\qed$ If $D = C$, trivial. For $D \neq C$: $D \isin C \equiv D \isin A \equiv D \le A \equiv D \le C$. $\qed$ +\section{Anticommit} + +Given $L, R^+, R^-$ where +$R^+ \in \pry, R^- = \baseof{R^+}$. +Construct $C$ which has $\pr$ removed. +Used for removing a branch dependency. +\gathbegin + C \hasparents \{ L \} +\gathnext + \patchof{C} = \patchof{L} +\gathnext + \mergeof{C}{L}{R^+}{R^-} +\end{gather} + +\subsection{Conditions} + +\[ \eqn{ Unique Tip }{ + \pendsof{L}{\pry} = \{ R^+ \} +}\] +\[ \eqn{ Currently Included }{ + L \haspatch \pry +}\] +\[ \eqn{ Not Self }{ + L \not\in \{ R^+ \} +}\] + +\subsection{No Replay} + +By Unique Tip, $R^+ \le L$. By definition of $\base$, $R^- \le R^+$ +so $R^- \le L$. So $R^+ \le C$ and $R^- \le C$ and No Replay for +Merge Results applies. $\qed$ + +\subsection{Desired Contents} + +\[ D \isin C \equiv [ D \notin \pry \land D \isin L ] \lor D = C \] +\proofstarts + +\subsubsection{For $D = C$:} + +Trivially $D \isin C$. OK. + +\subsubsection{For $D \neq C, D \not\le L$:} + +By No Replay $D \not\isin L$. Also $D \not\le R^-$ hence +$D \not\isin R^-$. Thus $D \not\isin C$. OK. + +\subsubsection{For $D \neq C, D \le L, D \in \pry$:} + +By Currently Included, $D \isin L$. + +By Tip Self Inpatch, $D \isin R^+ \equiv D \le R^+$, but by +by Unique Tip, $D \le R^+ \equiv D \le L$. +So $D \isin R^+$. + +By Base Acyclic, $D \not\isin R^-$. + +Apply $\merge$: $D \not\isin C$. OK. + +\subsubsection{For $D \neq C, D \le L, D \notin \pry$:} + +By Tip Contents for $R^+$, $D \isin R^+ \equiv D \isin R^-$. + +Apply $\merge$: $D \isin C \equiv D \isin L$. OK. + +$\qed$ + +\subsection{Unique Base} + +Need to consider only $C \in \py$, ie $L \in \py$. + +xxx tbd + +xxx need to finish anticommit + \section{Merge} Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$): @@ -342,13 +472,9 @@ Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$): \gathnext \patchof{C} = \patchof{L} \gathnext - D \isin C \equiv - \begin{cases} - (D \isin L \land D \isin R) \lor D = C : & \true \\ - (D \not\isin L \land D \not\isin R) \land D \neq C : & \false \\ - \text{otherwise} : & D \not\isin M - \end{cases} + \mergeof{C}{L}{M}{R} \end{gather} +We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$. \subsection{Conditions} @@ -362,31 +488,25 @@ Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$): \text{otherwise} : & \false \end{cases} }\] +\[ \eqn{ Removal Merge Ends }{ + X \not\haspatch \p \land + Y \haspatch \p \land + M \haspatch \p + \implies + \pendsof{Y}{\py} = \pendsof{M}{\py} +}\] +\[ \eqn{ Addition Merge Ends }{ + X \not\haspatch \p \land + Y \haspatch \p \land + M \nothaspatch \p + \implies \left[ + \bigforall_{E \in \pendsof{X}{\py}} E \le Y + \right] +}\] \subsection{No Replay} -\subsubsection{For $D=C$:} $D \isin C, D \le C$. OK. - -\subsubsection{For $D \isin L \land D \isin R$:} -$D \isin C$. And $D \isin L \implies D \le L \implies D \le C$. OK. - -\subsubsection{For $D \neq C \land D \not\isin L \land D \not\isin R$:} -$D \not\isin C$. OK. - -\subsubsection{For $D \neq C \land D \not\isin L \land D \not\isin R$:} -$D \not\isin C$. OK. - -\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R) - \land D \not\isin M$:} -$D \isin C$. Also $D \isin L \lor D \isin R$ so $D \le L \lor D \le -R$ so $D \le C$. OK. - -\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R) - \land D \isin M$:} -$D \not\isin C$. Also $D \isin L \lor D \isin R$ so $D \le L \lor D \le -R$ so $D \le C$. OK. - -$\qed$ +See No Replay for Merge Results. \subsection{Unique Base} @@ -409,34 +529,91 @@ $A \le R \equiv A \le \baseof{R}$. But by Tip Merge condition on $\baseof{R}$, $A \le \baseof{L} \implies A \le \baseof{R}$, so $A \le \baseof{R} \lor A \le \baseof{L} \equiv A \le \baseof{R}$. -Thus $A \le C \equiv A \le \baseof{R}$. Ie, $\baseof{C} = -\baseof{R}$. +Thus $A \le C \equiv A \le \baseof{R}$. +That is, $\baseof{C} = \baseof{R}$. \subsubsection{For $R \in \pn$:} By Tip Merge condition on $R$, -$A \le \baseof{L} \implies A \le R$ +$A \le \baseof{L} \implies A \le R$, so +$A \le R \lor A \le \baseof{L} \equiv A \le R$. +Thus $A \le C \equiv A \le R$. +That is, $\baseof{C} = R$. -UP TO HERE +$\qed$ + +\subsection{Coherence and patch inclusion} -Let $S = - \begin{cases} - R \in \py : & \baseof{R} \\ - R \in \pn : & R - \end{cases}$. -Then by Tip Merge $S \ge \baseof{L}$, and $R \ge S$ so $C \ge S$. - -Consider some $A \in \pn$. If $A \le S$ then $A \le C$. -If $A \not\le S$ then +Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$. +This involves considering $D \in \py$. -Let $A \in \pends{C}{\pn}$. -Then by Calculation Of Ends $A \in \pendsof{L,\pn} \lor A \in -\pendsof{R,\pn}$. +\subsubsection{For $L \nothaspatch \p, R \nothaspatch \p$:} +$D \not\isin L \land D \not\isin R$. $C \not\in \py$ (otherwise $L +\in \py$ ie $L \haspatch \p$ by Tip Self Inpatch). So $D \neq C$. +Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$. +\subsubsection{For $L \haspatch \p, R \haspatch \p$:} +$D \isin L \equiv D \le L$ and $D \isin R \equiv D \le R$. +(Likewise $D \isin X \equiv D \le X$ and $D \isin Y \equiv D \le Y$.) +Consider $D = C$: $D \isin C$, $D \le C$, OK for $C \haspatch \p$. -%$\pends{C, +For $D \neq C$: $D \le C \equiv D \le L \lor D \le R + \equiv D \isin L \lor D \isin R$. +(Likewise $D \le C \equiv D \le X \lor D \le Y$.) -%%\subsubsection{For $R \in \py$:} +Consider $D \neq C, D \isin X \land D \isin Y$: +By $\merge$, $D \isin C$. Also $D \le X$ +so $D \le C$. OK for $C \haspatch \p$. + +Consider $D \neq C, D \not\isin X \land D \not\isin Y$: +By $\merge$, $D \not\isin C$. +And $D \not\le X \land D \not\le Y$ so $D \not\le C$. +OK for $C \haspatch \p$. + +Remaining case, wlog, is $D \not\isin X \land D \isin Y$. +$D \not\le X$ so $D \not\le M$ so $D \not\isin M$. +Thus by $\merge$, $D \isin C$. And $D \le Y$ so $D \le C$. +OK for $C \haspatch \p$. + +So indeed $L \haspatch \p \land R \haspatch \p \implies C \haspatch \p$. + +\subsubsection{For (wlog) $X \not\haspatch \p, Y \haspatch \p$:} + +$C \haspatch \p \equiv M \nothaspatch \p$. + +\proofstarts + +One of the Merge Ends conditions applies. +Recall that we are considering $D \in \py$. +$D \isin Y \equiv D \le Y$. $D \not\isin X$. +We will show for each of +various cases that $D \isin C \equiv M \nothaspatch \p \land D \le C$ +(which suffices by definition of $\haspatch$ and $\nothaspatch$). + +Consider $D = C$: Thus $C \in \py, L \in \py$, and by Tip +Self Inpatch $L \haspatch \p$, so $L=Y, R=X$. By Tip Merge, +$M=\baseof{L}$. So by Base Acyclic $D \not\isin M$, i.e. +$M \nothaspatch \p$. And indeed $D \isin C$ and $D \le C$. OK. + +Consider $D \neq C, M \nothaspatch P, D \isin Y$: +$D \le Y$ so $D \le C$. +$D \not\isin M$ so by $\merge$, $D \isin C$. OK. + +Consider $D \neq C, M \nothaspatch P, D \not\isin Y$: +$D \not\le Y$. If $D \le X$ then +$D \in \pancsof{X}{\py}$, so by Addition Merge Ends and +Transitive Ancestors $D \le Y$ --- a contradiction, so $D \not\le X$. +Thus $D \not\le C$. By $\merge$, $D \not\isin C$. OK. + +Consider $D \neq C, M \haspatch P, D \isin Y$: +$D \le Y$ so $D \in \pancsof{Y}{\py}$ so by Removal Merge Ends +and Transitive Ancestors $D \in \pancsof{M}{\py}$ so $D \le M$. +Thus $D \isin M$. By $\merge$, $D \not\isin C$. OK. + +Consider $D \neq C, M \haspatch P, D \not\isin Y$: +By $\merge$, $D \not\isin C$. OK. + +$\qed$ \end{document}