X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=blobdiff_plain;f=article.tex;h=4c5f6751682c83a13e478fdb11d64496abd1bd22;hp=4c487f77c7b1d1c0099f37dc7d70116f5c29e4da;hb=1059b2343f8b2979915db881a067cd3307cdead2;hpb=cc02c1dc4b781d43f63bd036e0d0adbf678f8f2b diff --git a/article.tex b/article.tex index 4c487f7..4c5f675 100644 --- a/article.tex +++ b/article.tex @@ -1,4 +1,5 @@ \documentclass[a4paper,leqno]{strayman} +\errorcontextlines=50 \let\numberwithin=\notdef \usepackage{amsmath} \usepackage{mathabx} @@ -31,6 +32,10 @@ \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{>}}} @@ -48,8 +53,14 @@ \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}[4]{{\mathcal M}(#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{\eqn}[2]{ #2 \tag*{\mbox{\bf #1}} } \newcommand{\corrolary}[1]{ #1 \tag*{\mbox{\it Corrolary.}} } @@ -141,6 +152,17 @@ 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 \merge{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} @@ -226,15 +248,46 @@ 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 $C$, given +\gathbegin + \merge{C}{L}{M}{R} +\gathnext + L \le C +\gathnext + R \le C +\end{gather} +No Replay is preserved. {\it Proof:} + +\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: @@ -333,6 +386,45 @@ $\qed$ 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 + \merge{C}{L}{R^+}{R^-} +\end{gather} + +\subsection{Conditions} + +\[ \eqn{ Unique Tip }{ + \pendsof{L}{\pry} = \{ R^+ \} +}\] +\[ \eqn{ Currently Included }{ + L \haspatch \pry +}\] + +\subsection{Desired Contents} + +xxx need to prove $D \isin C \equiv D \not\in \pry \land D \isin L$. + +\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{Unique Base} + +Need to consider only $C \in \py$, ie $L \in \py$. + +xxx tbd + \section{Merge} Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$): @@ -341,12 +433,7 @@ 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} + \merge{C}{L}{M}{R} \end{gather} \subsection{Conditions} @@ -364,28 +451,7 @@ Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$): \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} @@ -393,9 +459,9 @@ Need to consider only $C \in \py$, ie $L \in \py$, and calculate $\pendsof{C}{\pn}$. So we will consider some putative ancestor $A \in \pn$ and see whether $A \le C$. -$A \le C \equiv A \le L \lor A \le R \lor A = C$. +By Exact Ancestors for C, $A \le C \equiv A \le L \lor A \le R \lor A = C$. But $C \in py$ and $A \in \pn$ so $A \neq C$. -Thus $A \le L \lor A \le R$. +Thus $A \le C \equiv A \le L \lor A \le R$. By Unique Base of L and Transitive Ancestors, $A \le L \equiv A \le \baseof{L}$. @@ -407,32 +473,18 @@ $A \le R \equiv A \le \baseof{R}$. But by Tip Merge condition on $\baseof{R}$, $A \le \baseof{L} \implies A \le \baseof{R}$, so -$A \le \baseof{R} \lor A \le \baseof{R} \equiv A \le \baseof{R}$. -Thus $A \le C \equiv A \le \baseof{R}$. Ie, $\baseof{C} = -\baseof{R}$. - -UP TO HERE +$A \le \baseof{R} \lor A \le \baseof{L} \equiv A \le \baseof{R}$. +Thus $A \le C \equiv A \le \baseof{R}$. +That is, $\baseof{C} = \baseof{R}$. -By Tip Merge, $A \le $ +\subsubsection{For $R \in \pn$:} -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 +By Tip Merge condition on $R$, +$A \le \baseof{L} \implies A \le R$, so +$A \le R \lor A \le \baseof{L} \equiv A \le R$. +Thus $A \le C \equiv A \le R$. +That is, $\baseof{C} = R$. -Let $A \in \pends{C}{\pn}$. -Then by Calculation Of Ends $A \in \pendsof{L,\pn} \lor A \in -\pendsof{R,\pn}$. - - - -%$\pends{C, - -%%\subsubsection{For $R \in \py$:} +$\qed$ \end{document}