strategy: wip

[topbloke-formulae.git] / strategy.tex

diff --git a/strategy.tex b/strategy.tex
index e0b126657816909da6179b1a86cb118cb1c9c63f..26907129b2ab120c5e473fb24a4959a509f011e8 100644 (file)
--- a/strategy.tex
+++ b/strategy.tex
@@ -14,7 +14,7 @@ The set of direct dependencies (in the form $\py$)
 requested in the commit $K$ ($K \in \pn$) for the patch $\p$.
 
 \item[ $\pc \hasdirdep \p$ ]
-The Topbloke commit set $\pc$ has as a direct contributors the
+The Topbloke commit set $\pc$ has as a direct contributor the
 commit set $\p$.  This is an acyclic relation.
 
 \item[ $\p \hasdep \pq$ ]
@@ -96,16 +96,19 @@ $\pqn$ involved in the cycle.
 For each such $\p$, after updating $\hasdep$, we recursively make a plan
 for $\pc' = \p$.
 
+
+
 \section{Execution phase}
 
 We process commit sets from the bottom up according to the relation
 $\hasdep$.  For each commit set $\pc$ we construct $\tipfc$ from
 $\tipzc$, as planned.  By construction, $\hasdep$ has $\patchof{L}$
 as its maximum, so this operation will finish by updating
-$\tipfa{\patchof{L}}$.
+$\tipca{\patchof{L}}$ with $\tipfa{\patchof{L}}$.
 
-After we are done, the result has the following properties:
-\[ \eqn{Tip Inputs}{
+After we are done with each commit set $\pc$, the
+new tip $\tipfc$ has the following properties:
+\[ \eqn{Tip Sources}{
   \bigforall_{E_i \in \set E_{\pc}} \tipfc \ge E_i
 }\]
 \[ \eqn{Tip Dependencies}{
@@ -127,16 +130,16 @@ it is, are we fast forward to $E_i$
 and drop $E_i$ from the planned ordering.
 
 Then we will merge the direct contributors and the sources' ends.
-
 This generates more commits $\tipuc \in \pc$, but none in any other
-commit set.  We maintain XXX FIXME IS THIS THE BEST THING?
+commit set.  We maintain
 $$
  \bigforall_{\p \isdep \pc}
- \pancsof{\tipcc}{\p} \subset \left[
-   \tipfa \p \cup
-   \bigcup_{E \in \set E_{\pc}} \pancsof{E}{\p}
-  \right]
+ \pancsof{\tipcc}{\p} \subset
+   \pancsof{\tipfa \p}{\p}
 $$
+\proof{
+ For $\tipcc = \tipzc$, $T$ ...WRONG WE NEED $\tipfa \p$ TO BE IN $\set E$ SOMEHOW
+}
 
 \subsection{Merge Contributors for $\pcy$}
 
@@ -144,16 +147,18 @@ Merge $\pcn$ into $\tipc$.  That is, merge with
 $L = \tipc, R = \tipfa{\pcn}, M = \baseof{\tipc}$.
 to construct $\tipu$.
 
-Merge conditions: Ingredients satisfied by construction.
+Merge conditions:
+
+Ingredients satisfied by construction.
 Tip Merge satisfied by construction.  Merge Acyclic follows
 from Perfect Contents and $\hasdep$ being acyclic.
 
-Removal Merge Ends: For $\p = \pc$, $M \nothaspatch \p$.
-For $p \neq \pc$, by Tip Contents,
+Removal Merge Ends: For $\p = \pc$, $M \nothaspatch \p$; OK.
+For $\p \neq \pc$, by Tip Contents,
 $M \haspatch \p \equiv L \haspatch \p$, so we need only
 worry about $X = R, Y = L$; ie $L \haspatch \p$,
 $M = \baseof{L} \haspatch \p$.
-By Tip Contents for $L$, $D \le L \equiv D \le M$.  $\qed$
+By Tip Contents for $L$, $D \le L \equiv D \le M$.  OK.~~$\qed$
 
 WIP UP TO HERE
 
@@ -162,7 +167,11 @@ done the execution phase for $\pcn$ and $\py$.  By
 Perfect Contents for $\pcn$, $\tipfa \pcn \haspatch \p$ i.e.
 $R \haspatch \p$.  So we only need to worry about $Y = R = \tipfa \pcn$.
 By Tip Dependencies $\tipfa \pcn \ge \tipfa \py$.
-And by Tip Inputs $\tipfa \py \ge $
+And by Tip Sources $\tipfa \py \ge $
+
+want to prove $E \le \tipfc$ where $E \in \pendsof{\tipcc}{\py}$
+
+$\pancsof{\tipcc}{\py} = $
 
 
 computed $\tipfa \py$, and by Perfect Contents for $\py$