\section{Planning phase}
The results of the planning phase consist of:
\begin{itemize*}
\item{ The relation $\hasdirdep$ and hence the partial order $\hasdep$. }
\item{ For each commit set $\pc$, a confirmed set of sources $\set S_{\pc}$. }
\item{ For each commit set $\pc$, the order in which to merge the sources
$E_{\pc,j} \in \set E_{\pc}$. }
\item{ For each $E_{\pc,j}$ an intended merge base $M_{\pc,j}$. }
\end{itemize*}
We use a recursive planning algorith, recursing over Topbloke commit
sets (ie, sets $\py$ or $\pn$). We'll call the commit set we're
processing at each step $\pc$.
At each recursive step
we make a plan to merge all $\set E_{\pc} = \{ E_{\pc,j \ldots} \}$
and all the direct contributors of $\pc$ (as determined below)
into $\tipzc$, to make $\tipfc$.
We start with $\pc = \pl$ where $\pl = \patchof{L}$.
\subsection{Direct contributors for $\pc = \pcn$}
The direct contributors of $\pcn$ are the commit sets corresponding to
the tip branches for the direct dependencies of the patch $\pc$. We
need to calculate what the direct dependencies are going to be.
Choose an (arbitrary, but ideally somehow optimal in
a way not discussed here) ordering of $\set E_{\pc}$, $E_{\pc,j}$
($j = 1 \ldots m$).
For brevity we will write $E_j$ for $E_{\pc,j}$.
Remove from that set (and ordering) any $E_j$ which
are $\le$ and $\neq$ some other $E_k$.
Initially let $\set D_0 = \depsreqof{\tipzc}$.
For each $E_j$ starting with $j=1$ choose a corresponding intended
merge base $M_j$ such that $M_j \le E_j \land M_j \le T_{\pc,j-1}$.
Calculate $\set D_j$ as the 3-way merge of the sets $\set D_{j-1}$ and
$\depsreqof{E_j}$ using as a base $\depsreqof{M_j}$. This will
generate $D_m$ as the putative direct contributors of $\pcn$.
However, the invocation may give instructions that certain direct
dependencies are definitely to be included, or excluded. As a result
the set of actual direct contributors is some arbitrary set of patches
(strictly, some arbitrary set of Topbloke tip commit sets).
\subsection{Direct contributors for $\pc = \pcy$}
The sole direct contributor of $\pcy$ is $\pcn$.
\subsection{Recursive step}
For each direct contributor $\p$, we add the edge $\pc \hasdirdep \p$
and augment the ordering $\hasdep$ accordingly.
If this would make a cycle in $\hasdep$, we abort . The operation must
then be retried by the user, if desired, but with different or
additional instructions for modifying the direct contributors of some
$\pqn$ 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
$\tipca{\patchof{L}}$ with $\tipfa{\patchof{L}}$.
After we are done with each commit set $\pc$, the
new tip $\tipfc$ has the following properties:
\[ \eqn{Tip Sources}{
\bigforall_{E_i \in \set E_{\pc}} \tipfc \ge E_i
}\]
\[ \eqn{Tip Dependencies}{
\bigforall_{\pc \hasdep \p} \tipfc \ge \tipfa \p
}\]
\[ \eqn{Perfect Contents}{
\tipfc \haspatch \p \equiv \pc \hasdep \py
}\]
For brevity we will sometimes write $\tipu$ for $\tipuc$, etc. We will start
out with $\tipc = \tipz$, and at each step of the way construct some
$\tipu$ from $\tipc$. The final $\tipu$ becomes $\tipf$.
\subsection{Preparation}
Firstly, we will check each $E_i$ for being $\ge \tipc$. If
it is, are we fast forward to $E_i$
--- formally, $\tipu = \text{max}(\tipc, E_i)$ ---
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
$$
\bigforall_{\p \isdep \pc}
\pancsof{\tipcc}{\p} \subset
\pancsof{\tipfa \p}{\p}
$$
\proof{
For $\tipcc = \tipzc$, $T$ ...WRONG WE NEED $\tipfa \p$ TO BE IN $\set E$ SOMEHOW
}
\subsection{Merge Contributors for $\pcy$}
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.
Tip Merge satisfied by construction. Merge Acyclic follows
from Perfect Contents and $\hasdep$ being acyclic.
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 $L$, $D \le L \equiv D \le M$. OK.~~$\qed$
WIP UP TO HERE
Addition Merge Ends: If $\py \isdep \pcn$, we have already
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 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 $\py$
with $M=M_j, L=T_{\pc,j-1}, R=E_j$,
and calculate what the resulting desired direct dependencies file
(ie, the set of patches $\set D_j$)
would be. Eventually we
So, formally, we select somehow an order of sources $S_i$. For each
Make use of the following recursive algorithm, Plan
recursively make a plan to merge all $E = \pends$
Specifically, in