-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