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« **on:** February 26, 2021, 08:08:36 pm »
TL;DR: +Ironworks -Forager +Squire should still be finite and boost it up to N -> (2^(2^(2^N))).

Looks neat! I'm a little confused at the end of the Buy phase since Conquest should only give 2 silvers, which of course can be remedied by exchanging instead of trashing on the second-to-last buy and leaving 6 money to buy the Conquest last. (Also, you only need to do this on the last turn anyway.) This shouldn't affect any of the calculations, of course.

I think we can improve this by removing Forager and adding Ironworks (which gets Seawayed), which is a gainer that works with Capitalism but can't gain Mandarins (or Scrying Pools or Universities). Since Ironworks can gain a copy of itself, we have to be careful that this doesn't introduce any infinite loops. I believe this is okay because

1. You cannot gain Scrying Pools or Universities mid-turn (or take them out of play).

2. Without using a University, you cannot gain a Cultist or Mandarin midturn, and cards stay in play.

3. Without playing Scrying Pool or University, the only plays that don't decrease handsize are playing our single Scheme/Fortress/Watchtower, trashing a Cultist, or playing a Cultist, or playing an Ironworks on Estate. We have a finite bound on all these actions besides the Ironworks play. The Ironworks play strictly decreases the number of "available" (in hand + deck + discard) Ironworks. No other play (besides playing University) can increase the number of available Ironworks.

With this, the action phase loop looks like:

play Scrying Pool, drawing entire deck

play Black Market

play all Priests, trashing the Fortress

play all Ironworks, gaining either Priest or Ironworks (I think the best thing is all Ironworks until the second-to-last loop where you get some ratio of Priests)

(The last loop is analogous to pitythefool's solution, as are the remaining phases.)

Analysis:

At the start of the turn, say we have:

N Ironworks

<= N Priests (exact amount doesn't really matter)

U Universities

U+1 Scrying Pools

Every time we play a Pool and University, we double the amount of Ironworks, and on the last step we can give ourselves an equal number of Ironworks and Priests, giving 2^U * N Ironworks and 2^U * N Priests. Let's just set U=N to simplify, so that we have N * 2^N Priest plays then N * 2^N Ironworks plays, giving us on the order of $(N^2 * 4^N) and (N * 2^N) buys.

Every Masterpiece buy/trash multiplies the wealth by N * 2^N, so our total wealth before the end of the buy phase is on the order of

(N^2 * 4^N) * (N * 2^N)^(N* 2^N) ~ (N*2^N)^(N*2^N) < (2^N)^(2^N) < 2^(2^N).

So we can grow from N -> 2^(2^N) every turn. I'm a little confused where this puts us in growth rate at the moment.

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So another thing I think we can do is add Squire and trash it for midturn Scrying Pool gains. Then our finiteness argument should be like

1. You cannot gain Universities mid-turn (or take them out of play).

2. Supposing you do not play any Universities, every card you play does not increase the number of available (Scrying Pool + Cultist + Squire + Priest + Ironworks) that you have.

3. Increasing your handsize strictly decreases this availability count.

4. Besides playing your single Scheme, Fortress, or Watchtower, every action you can do either strictly decreases your hand size or your total availability count (or both).

In this proposed loop, I'm going to ignore +Actions since we aren't using it as a limiting resource and could fix it with Academy or maybe Lost Arts on Ironworks. The loop should look like

Draw deck (with scrying pool)

inner loop start

Requirement: Hand has at least 1 Scrying Pool. If not the last loop before a University, hand has at least 2 Scrying Pools or 1 Scrying Pool, 1 Priest, 1 Squire.

Play all Ironworks, gaining either Ironworks, Priest, or Squire. (In most loops, almost every gain is for Ironworks. Replace with a Priest/Squire gain if needed to make the above condition satisfied.)

If not the last loop before a University:

If there is only 1 Scrying Pool in hand, first Priest to trash a Squire, gaining Scrying Pool. Either way, play a Scrying Pool.

If the last loop before a University (only 1 Scrying Pool in hand, and don't have Priest + Squire):

end inner loop

Outer loop

Do inner loop until it ends (note: in the inner loop for the last iteration of the outer loop (no Universities left), be sure to gain lots of Priests and play them too.)

Play University

Analysis for this:

To simplify, let's just say we have N each of Ironworks and Universities to start, and some negligible amount of Scrying Pools, Priests, and Squires. Once we have 1 Scrying Pool, 1 Priest, 1 Squire left, each inner loop will have to spend two gains on Squire and Priest, and the rest on Ironworks. So if we have X Ironworks to start, we'll end up with X-2 Ironworks on the next iteration. Iterating until the resources are exhausted, we'll have gained/played around X^2/2 Ironworks total. So every full outer loop takes our Ironworks from X->X^2/2. We will do this N times, once for each University.

N iterations of this starting from 1 gives us O(2^(2^N))) at the final iteration. Of course, we're starting at N, not 1, but this will not increase the asymptotics very much.

With 2^(2^N) Priest plays and 2^(2^N) buys, we will wind up with a N -> (2^(2^N))^(2^(2^N)) < 2^(2^(2^N)) growth per turn.