I keep pointing out that you can do themed trees of combat charms without charmshare, and nobody appears to notice.
Plenty of people have noticed, Jon; the argument is that, if choosing between the alternatives:
(a) Every one of X splats gets Y unique "martial arts" trees, approximately balanced for that splat,
and
(b) There are Y unique "martial arts" trees, shared between the X splats but approximately balanced,
it's not at all clear that (a) is less work. (a) clearly requires creating XY total trees. (b) requires creating X trees, and checking each tree against Y splats. Your thesis seems to be that "checking this tree against a splat" is necessarily more work than creating a tree from scratch, i.e., that the effort for (a) is on the order of XY, while the effort for (b) is some quantity greater than XY - some (edit: thanks,
@grommile) f(X)*Y, where f(X) grows faster than X. The
opposing thesis is that f(X) grows more slowly than X, i.e., that checking a tree against one splat is in general less effort than creating a new tree from scratch.
There's also:
(c) The X splats are meaningfully divided into three tiers, and there are Y trees, each "tweaked" for the tier of character using it.
... which, if the X splats
really are meaningfully divided into three tiers, in principle cuts the labor down to a maximum of 3Y, that is, the number of unique martial arts, times the number of tiers that art needs to be rebuilt and rebalanced for. (In practice, it's argued, the work required is somewhere between Y and 3Y, because some Charms work fine for all three tiers.)
The points of argument seem to be:
-What actually is the growth rate of f(X)?, and
-Is "tiering" actually an accurate way to describe the breakdown of splats?
We can't answer that second question for 3e, yet, because ha ha Ex3 release schedule, but it's not inherently impossible for it to be true. But it's not at all obvious to me that O(f(X)) > O(X) is
necessarily a true statement, and that seems to be the position your argument hangs on; indeed, in the abstract, it seems rather more likely that for many systems O(f(X)) < O(X), in which case (b) or (c) are less net work.