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Mic Qsenoch

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Re: Best Asymptotic Point Scoring
« Reply #50 on: November 11, 2020, 09:38:05 pm »
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I'm saying this on the basis that exchanging a card that is being gained from the Black Market deck shouldn't work, and you'll be forced to take the card you were originally gaining. So in our case, our Fairgrounds would get removed from the Black Market deck and we'd stop being able to buy it.

Though, I'm remembering that gaining Horse can be exchanged with Changeling, and that's not from a supply pile, so maybe it does work after all. I've yet to test this with the Black Market / Trader situation on the client.

Exchanging just requires a pile, it doesn't have to be a Supply pile, so Horses work. Black Market cards do not have a pile, you can't upgrade Page/Peasant from the BM for instance.

The dominionstrategy.com wiki seems to have some contradictions.  The rules given for exchanging are as you have stated, but is the Black Market deck a pile or not?

Their rules clarifications for Black Market include:

"If you buy a card from the Black Market deck and then use Trader to prevent yourself from gaining it, the bought card goes back on top of the Black Market deck."

I wonder what ShuffleIT has implemented?

That particular rule clarification predates the recent Trader errata (note that the current Trader doesn't "prevent yourself from gaining"). You can't Exchange things from the Black Market (see here: http://forum.dominionstrategy.com/index.php?topic=13044.msg483985#msg483985 and some messages below it).
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pitythefool

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Re: Best Asymptotic Point Scoring
« Reply #51 on: December 22, 2020, 07:10:49 pm »
+1

I believe that the latest Trader change kills the triple up arrow solution. It had a good run.  :'(
I tried to make a replacement for bitwise & paulfc's brilliant 3 up-arrow solution, which now seems invalid due to recent rule changes.  I am using Livery & Falconer in place of Black Market & Haggler.  But even with Mandarin thrown in, it does not achieve 3 up-arrow growth.  It is superior to the other 2 up-arrow solutions in this thread, but it does not lend itself to easy analysis.  I am hoping that working together, we may improve upon it.

Kingdom:  Squire, Changeling, Black Market, Scavenger, Patron, Falconer, Livery, Scrying Pool, Golem
Black Market:  Mandarin, Rogue, Vault, Count, City Quarter, Herbalist
Events:  Travelling Fair, Donate
Landmark:  Obelisk on Livery
Projects:  Capitalism, Academy
Ways:  Way of the Seal, Way of the Turtle, Way of the Mouse -> Hermit

Victory points are generated by Obelisk.

Getting started: (in no particular order):
    buy 2 Black Markets and then buy all the cards from the Black Market
    buy the projects Capitalism and Academy
    buy Seaway (using Quarry) and Training for Livery
    buy 2 Scavengers
    buy lots of Scrying Pools (or alternately Squires and trash them with Donate, gain Scrying Pools)
    buy lots of Liveries, Falconers, and Golems
    buy Donate to trash unwanted cards

Starting deck

     1 Mandarin
     1 Herbalist
     1 Vault
     1 Count
     1 Rogue
     1 City Quarter
     2 Black Markets
     2 Scavengers
     S Scrying Pools
     L Liveries
     F Falconers
     G Golems

I have denoted the start and end of the Black Market treasure phases, to indicate where I play action/treasures as treasures.

Turn start:  (Quarry is in hand, Scrying pool has been set aside as the result of the previous turn)

    play Scrying Pool (from Way of the Turtle), draw entire deck (+action)
    play Herbalist as Way of the Seal

<begin outer loop>
    play Black Market
    <begin Black Market treasure phase>
        play all Liveries
    <end Black Market treasure phase>
    <begin Horse gaining loop>
        play Falconer, gain Patron
            Patron is a four coin card so we gain one Horse for every Livery played this turn (not just those currently in play).
            Patron is a card of three types and thus we can react to gaining it by playing another Falconer from our hand.
        repeat this sequence for all Falconers thus playing them all with just one action.
        (each gained action card comes with +action from Academy.  It should be pretty clear by now that we will never run out of actions.)
    <end Horse gaining loop>
    play Black Market
    <begin Black Market treasure phase>
        play Quarry
    <end Black Market treasure phase>
    For all Horses and Patrons we also have in hand,
        play <action> as Way of the Mouse (Hermit), gain Livery
    (Quarry made 5 coin cards affordable to Hermit, but they do not trigger Liveries to produce Horses.)
    (each gained action card comes with +action from Academy.)
    play Scrying Pool, draw Horses, Patrons, and Liveries
    For all Patrons and all Liveries and all-but-one Horse,
        play <action> as Way of the Mouse (Hermit), gain Livery
    play Scrying Pool, draw Liveries
    <begin main Hermit loop>
        For all Liveries,
            play Livery as Way of the Mouse (Hermit), gain Livery
        play Scrying Pool, draw Liveries
        {Exit loop when only 4 Scrying Pools remain, or 5 Scrying Pools if there is only 1 Golem}
    <end main Hermit loop>
    For all Liveries (one final time),
        play Livery as Way of the Mouse (Hermit), gain Falconer
        play Livery as Way of the Mouse (Hermit), gain Squire
    play Scrying Pool, draw Falconers and Squires
    play Vault, draw <two cards>, discard all but Squires, Scavengers, Golem, Mandarin, and <two cards>
    play Scavenger, topdeck Scrying Pool
    play Scavenger, topdeck Count
    play Golem, reveal Count and Scrying Pool
        play Count, discard <two cards>, trash hand which is just Mandarin and a lot of Squires, gain Scrying Pools for trashing Squires
        play Scrying Pool, draw entire deck (only non-action is Quarry which is in play)
    play Rogue, gain Mandarin from trash,
        topdeck treasures (Quarry, Black Markets, Liveries, Patrons, Count, Rogue, Vault, Scavengers, Herbalist)
        put Quarry and a Black Market on top
    play Horse, draw Quarry and a Black Market
    play Scrying Pool, draw entire deck (Quarry already in hand)
    {Exit outer loop when no Golems remain}
<end outer loop>

    play Black Market
        play all Liveries
    For half of your Falconers,
        play Falconer, gain Patron and Horses, convert all to Changelings and discard.
    For the remaining Falconers,
        play Falconer, gain Patron and Horses, topdeck all
    play Scrying Pool.  Draws all of the topdecked Patrons and Horses up to the first Changeling
    play City Quarter.  Draws all of the Changelings.
    play last Scrying Pool as Way of the Turtle.

Buy Phase:

    play all Patrons
    play Herbalist
    DO NOT play Quarry

    buy Travelling Fair for extra buys as needed
    then spend everything on Liveries, gaining a lot of Horses in the process

Night Phase:
    exchange the Changelings for Golems

Cleanup:
    topdeck the Quarry (by way of Herbalist)
    one Scrying Pool has been set aside (by Way of the Turtle)

Proof of finiteness

Liveries in large number are a powerful force.  If Horses are allowed to gain more cards that produce Horses unimpeded, unbounded loops are easy to create even without the presence of Mandarin.  Add Mandarin and Capitalism and now you have a lot of exploits to plug.  The kingdom ended up being very complicated because of it, due to the addition of necessary interlocks.  Here's the reasoning why play is bounded.

The case of not purchasing Capitalism.

We have only one Mandarin and one Rogue.  To gain the Mandarin, we first have to trash it and then gain it back from the trash with the Rogue.  We can trash it easily with Hermit.  When we gain the Mandarin back, the only thing going back onto the deck will be the Quarry, so the Rogue cannot be played again and we cannot gain the Mandarin back again.  After that, the playing of all cards will be final.  Unbounded play must therefore be predicated on gaining more cards during play.
The only gainers are Hermit and Falconer.
Without the Quarry in play, the Hermit is limited to gaining three coin cards and Falconers are limited to gaining four coin cards.  The Falconers can therefore produce more Horses.  But the number of Falconers in our deck is finite and will run out.  After which we can play actions as a Hermit and only gain cards that cost up to three coins which will not produce any more Horses.
With the Quarry in play, five coin actions are reduced to a cost of three coins and can be gained with Hermit, but they will not produce more Horses.  We can gain more Falconers, but they are limited to gaining cards which cost less than themselves, and their cost has been reduced to three coins.  Eventually we will not be able to gain any more cards.
There is one final way to gain a card; trashing a Squire.  The Rogue is an attack card, but it is not in the supply, so we can only gain Scrying Pools.
Any action played as a Hermit can gain a Squire and trash a Squire and gain a Scrying Pool.  That seems like a gain, but it isn't.  You are losing two cards and  gaining two cards.  It just makes waiting for the end take a lot longer.

The case of purchasing Capitalism.

Hermit can no longer trash the Mandarin, nor the Squires.  The gaining situation is the same as above.  We can gain Horses only when the Quarry is not in play, but we cannot gain any really useful cards unless the Quarry is in play.  Gaining the Mandarin from the trash will return the Rogue and a lot of other useful action/treasures to the stack to be played again but most importantly, it removes the Quarry from play.  The only card that can trash the Mandarin is the Count and only by trashing the entire hand.  That obviously ends play unless triggered by a Golem.  This is limited because Golems are not treasures and they cannot be returned from play and they cannot be gained, only acquired in the Night phase or bought in the final Buy phase.

Analysis

If we play L Liveries and then F Falconers gaining F Patrons, we will gain L*F Horses.
We then turn those L*F Horses and F Patrons into (L+1)*F more Liveries,
and then turn those Liveries into more Liveries, S-4 more times for O(L*F*S) Liveries.
The final Hermit loop converts O(L*F) Liveries into O(L*F*r) Falconers and O(L*F*(1-r)) Squires.  The Squires are converted to Scrying Pools by the Count.
For the next playing of Falconers, L' = O(L*F*S), F' = O(L*F*r), and S' = O(L*F*(1-r)).
To maximize Liveries gained next iteration, we need to maximize L'*F'*S'= O(L*F*S) * O(L*F*r) * O(L*F*(1-r)) = O(L^3*F^3*S*(1-r)*r).
That at least tells us that we should gain Falconers and Squires in equal number so F = S, and that equates to O(L^3*F^4).
The number of Liveries appears to cube itself each time we play a Golem, but not really since the other components do not grow nearly as fast.
Empirical data suggests that the actual exponent of growth for Liveries is ~2.4 (though my spreadsheet could only do 7 iterations before blowing up).

This is repeated for each Golem, with the number of Golems for the next turn equal to half the number of Horses gnerated in the last iteration.
Empirically, G' = ~(L^(2.4*G)^1.7.  That means that overall growth per turn = L^(2.4*[(L^(2.4*G)^1.7]) or thereabouts not even counting the Liveries we end up buying.
The problem is that the exponents aren't in a favorable place.  I believe, overall, it surpasses 2 up-arrows.

For those that are new to up-arrow notation:

If f(x) = x*2, then f(f(f(2)))) = 2*2*2*2 = 2^4 = 2↑4 = 16  "exponentiation"
If f(x) = x^2, then f(f(f(2)))) = ((2^2)^2)^2 = 2^8 = 256
If f(x) = 2^x, then f(f(f(2)))) = 2^(2^(2^2)) = 2↑↑4 = 65536  "tetration"
If f(x) = 2↑↑x, then f(f(f(2)))) = 2↑↑(2↑↑(2↑↑2) = 2↑↑↑4 = ?  "pentation"

Mathematicians have not blessed that second row with a name that I could find.  Bankers would call it "compounding", and if applied to finance, "amortization".

EDIT:  Forget what I said about amortization.  It's just another form of exponentiation.  I don't have any label for that second row.
« Last Edit: January 27, 2021, 08:19:47 pm by pitythefool »
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bitwise

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Re: Best Asymptotic Point Scoring
« Reply #52 on: December 29, 2020, 05:17:31 pm »
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Exciting stuff :D I'll try to take a closer look at the effect on liveries per golem played since that seems a little bit sketchy to me.

If I'm not mistaken, I believe the second row can be referred to as a double exponential function?
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pitythefool

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Re: Best Asymptotic Point Scoring
« Reply #53 on: January 11, 2021, 07:06:38 pm »
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... seems a little bit sketchy to me.

I've had a chance to reexamine the analysis.  It was indeed a bit sketchy.
I was maximizing the right expression to find the best ratio of Falconers to Squires, but it was looking an iteration further ahead.  For growth, we should simply look at L' relative to L.  L' was correctly stated as O(L*F*S).
Regardless of the starting values, a couple of Golem iterations leaves the cards in very specific ratios;  F = S = H;  L = F^(sqrt(2))
I wrote a program that reimplemented and verified my spreadsheet.  I took it to 23 Golems.  It verified the exponential growth of Liveries to be 1+sqrt(2), which can be calculated from the relationships above, to eight decimal places: 2.41421356.

Liveries after G Golem iterations is L^(2.414*G).  In the final iteration we do not produce more Liveries but convert half the Horses to Golems for the next turn, so we end a turn with G = L^1.707, so growth can be expressed as L^(2.414*L^1.707)) per turn.

Also, a couple tweaks to the procedure.  For the main Hermit loop, I stated "Exit loop when only 4 Scrying Pools remain, or 5 Scrying Pools if there is only 1 Golem".  However, you can stay in this loop until you are down to 2 Scrying Pools.  The 2nd Scrying Pool will draw the entire deck which includes more Scrying Pools created by the Count when it trashed Squires.
Also, it left a very large number of Horses in our hand at the beginning of the buy phase.  I did not want to play the Quarry since I wanted to gain Horses in the buy phase.  But I could have played the Quarry, played the Horses as Hermits and gained Liveries.  We would not gain Horses for our purchases, but I think having more Liveries for the first iteration is a win.
But I thought of an even better strategy.  Do not play Quarry, but still play the Horses as Hermits and gain Squires.  Then in the buy phase, Donate and trash the Squires to gain Scrying Pools.  While at it, pre-trash the Mandarin as well.  Just a slight catch:  the Quarry which we topdecked gets shuffled again.  We can work around that by setting two Scrying Pools aside with Way of the Turtle and simply take Herbalist out of the kingdom.
This is all insignificant but fun to think about.

EDIT:  Corrected the number of Horses and Golems again.
« Last Edit: January 16, 2021, 08:31:28 am by pitythefool »
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pitythefool

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Re: Best Asymptotic Point Scoring
« Reply #54 on: January 15, 2021, 06:43:41 pm »
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This is all insignificant but fun to think about.
Another thing my model has let me discover is the optimum ratio of Falconers-to-Squires to gain.  Earlier analysis had pegged it at 0.5 (equal amounts of each).  This seemed reasonable since they contributed to the increase of Liveries nearly equally and neither Falconers nor Scrying Pools could be returned from play.  It turns out that the algorithm is not sensitive to this value at all.  If you place the ratio at the extremes (close to 0.0 or 1.0), the exponential growth drops off  to about 2.0, but anywhere in between and the growth tends towards 2.414.  But a ratio of 0.689655 gets there just a bit quicker.  It's because Falconers also produce Patrons and Patrons do get returned from play by gaining a Mandarin, hence they repeatedly contribute to gaining more Liveries.

If you start with just 5 Liveries, 5 Falconers, 5 Scrying Pools, and 7 Golems, along with the minimum number of other components, and acquire Falconers/Squires at the 0.5 ratio, you will end up with 3.967*10^310 Liveries.  At a ratio of 0.689655, you end up with 5.742*10^317 Liveries; more than 10 million times as many.  The fact that this isn't really significant is a testimony to how large these numbers are.

If you continue both for a total of 23 Golems the difference in Liveries is 10^(4.132*10^8) versus 10^(4.229*10^8).
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majiponi

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Re: Best Asymptotic Point Scoring
« Reply #55 on: January 26, 2021, 10:49:35 am »
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How about this? infinite?

Kingdom: Page, Stonemason, Scrying Pool, Scheme, Watchtower, Philosopher's Stone, Fortress
Landmark: Tomb

Turn:
play Schemed Scrying Pool to draw all
play n Stonemasons to gain 2n Stonemasons (topdeck)
play Hero to gain Philosopher's Stone (topdeck)
play Scrying Pool to discard Philosopher's Stone to draw all
play Stonemason to gain 2 Scrying Pool
play 2n-1 Stonemasons to gain 4n-2 Stonemasons
repeat this until you have no Hero in hand
play all Pages, Treasure Hunters, Warriors to draw all
play Stonemasons to gain Pages
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pitythefool

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Re: Best Asymptotic Point Scoring
« Reply #56 on: January 27, 2021, 08:21:22 pm »
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play Hero to gain Philosopher's Stone (topdeck)

There have been a lot of Scrying Pool/Stonemason engines posted here.  I really like the usage of Hero, though.  That's novel.  It takes 3 turns for gained Pages to be used as Heroes though.  That really dampens the growth.  Imagine that you could buy Heroes directly and use them the very next turn.  And also, imagine that they generate 3.5 Scrying Pools instead of just two.  Then the growth would be on par with an engine I posted in this thread earlier [Reply #24 on: November 10, 2018, 11:12:10 pm].

But having said that, I like when people post new ideas and yours has potential.  It has two things going for it that I see.
One:  Warriors produce a lot of coin.  Try calculating how much (and remember that Scrying Pools are also attack cards).  Once again, though, you would be looking at a two turn delay in Warrior growth;  training on Stonemason may generate more.
(EDIT:  I confused Warrior with Soldier)
Two:  With Capitalism purchased, Hero is considered a treasure and may be returned to your deck by gaining a Mandarin.  Unfortumately, I can not come up with any way to limit Mandarin gains, in this kingdom, off the top of my head.
« Last Edit: January 28, 2021, 12:28:45 pm by pitythefool »
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Mic Qsenoch

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Re: Best Asymptotic Point Scoring
« Reply #57 on: January 27, 2021, 08:31:38 pm »
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play Hero to gain Philosopher's Stone (topdeck)

There have been a lot of Scrying Pool/Stonemason engines posted here.  I really like the usage of Hero, though.  That's novel.  It takes 3 turns for gained Pages to be used as Heroes though.  That really dampens the growth.  Imagine that you could buy Heroes directly and use them the very next turn.  And also, imagine that they generate 3.5 Scrying Pools instead of just two.  Then the growth would be on par with an engine I posted in this thread earlier [Reply #24 on: November 10, 2018, 11:12:10 pm].

But having said that, I like when people post new ideas and yours has potential.  It has two things going for it that I see.
One:  Warriors produce a lot of coin.  Try calculating how much (and remember that Scrying Pools are also attack cards).  Once again, though, you would be looking at a two turn delay in Warrior growth;  training on Stonemason may generate more.
Two:  With Capitalism purchased, Hero is considered a treasure and may be returned to your deck by gaining a Mandarin.  Unfortumately, I can not come up with any way to limit Mandarin gains, in this kingdom, off the top of my head.

Warrior doesn't make money, that's Soldier.
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majiponi

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Re: Best Asymptotic Point Scoring
« Reply #58 on: January 28, 2021, 11:21:37 am »
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play Hero to gain Philosopher's Stone (topdeck)

There have been a lot of Scrying Pool/Stonemason engines posted here.  I really like the usage of Hero, though.  That's novel.  It takes 3 turns for gained Pages to be used as Heroes though.  That really dampens the growth.  Imagine that you could buy Heroes directly and use them the very next turn.  And also, imagine that they generate 3.5 Scrying Pools instead of just two.  Then the growth would be on par with an engine I posted in this thread earlier [Reply #24 on: November 10, 2018, 11:12:10 pm].

But having said that, I like when people post new ideas and yours has potential.  It has two things going for it that I see.
One:  Warriors produce a lot of coin.  Try calculating how much (and remember that Scrying Pools are also attack cards).  Once again, though, you would be looking at a two turn delay in Warrior growth;  training on Stonemason may generate more.
Two:  With Capitalism purchased, Hero is considered a treasure and may be returned to your deck by gaining a Mandarin.  Unfortumately, I can not come up with any way to limit Mandarin gains, in this kingdom, off the top of my head.

The idea is non-Kingdom cards avoids infinity.  Your Expanding post is nice, but using Platinum/Colony will spoil it (Stonemason Expand to Gold, Expand Gold to Platinum, Stonemason Platinum to 2 Provinces, Stonemason them to 4 Expands). I first tried Tragic Hero, went inf. Next Treasurer, inf. Then, Mint, inf. Then I realized that depending Kingdom cards creates infinite loops too often. That's why I posted Heroes idea.
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pitythefool

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Re: Best Asymptotic Point Scoring
« Reply #59 on: January 28, 2021, 12:27:13 pm »
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play Hero to gain Philosopher's Stone (topdeck)

There have been a lot of Scrying Pool/Stonemason engines posted here.  I really like the usage of Hero, though.  That's novel.  It takes 3 turns for gained Pages to be used as Heroes though.  That really dampens the growth.  Imagine that you could buy Heroes directly and use them the very next turn.  And also, imagine that they generate 3.5 Scrying Pools instead of just two.  Then the growth would be on par with an engine I posted in this thread earlier [Reply #24 on: November 10, 2018, 11:12:10 pm].

But having said that, I like when people post new ideas and yours has potential.  It has two things going for it that I see.
One:  Warriors produce a lot of coin.  Try calculating how much (and remember that Scrying Pools are also attack cards).  Once again, though, you would be looking at a two turn delay in Warrior growth;  training on Stonemason may generate more.
Two:  With Capitalism purchased, Hero is considered a treasure and may be returned to your deck by gaining a Mandarin.  Unfortumately, I can not come up with any way to limit Mandarin gains, in this kingdom, off the top of my head.

The idea is non-Kingdom cards avoids infinity.  Your Expanding post is nice, but using Platinum/Colony will spoil it (Stonemason Expand to Gold, Expand Gold to Platinum, Stonemason Platinum to 2 Provinces, Stonemason them to 4 Expands). I first tried Tragic Hero, went inf. Next Treasurer, inf. Then, Mint, inf. Then I realized that depending Kingdom cards creates infinite loops too often. That's why I posted Heroes idea.

Yes, I do like the Hero.  It does plug a lot of loopholes not being in the Kingdom.
It turns out that the Expand approach is flawed, and you don't even need Platinum/Colony.  Province will suffice as I pointed out way back here in a post about an improved kingdom.
  Alas, Expand Fortress to Gold, Expand Gold to Province, Stonemason Province to 2 Golds, Stonemason 2 Golds to 4 Expands yields a gain of 2 Expands, hence another unbounded engine.

And yes, I was totally wrong about Warriors.  (Thanks Mic Qsenoch)  There's so many cards now I have trouble remembering them all.  Shame on me for not double checking.

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majiponi

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Re: Best Asymptotic Point Scoring
« Reply #60 on: January 28, 2021, 09:38:02 pm »
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It turns out that the Expand approach is flawed, and you don't even need Platinum/Colony.  Province will suffice as I pointed out way back here in a post about an improved kingdom.
  Alas, Expand Fortress to Gold, Expand Gold to Province, Stonemason Province to 2 Golds, Stonemason 2 Golds to 4 Expands yields a gain of 2 Expands, hence another unbounded engine.

Stonemason Golds to Expands?
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pitythefool

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Re: Best Asymptotic Point Scoring
« Reply #61 on: January 29, 2021, 07:14:34 am »
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It turns out that the Expand approach is flawed, and you don't even need Platinum/Colony.  Province will suffice as I pointed out way back here in a post about an improved kingdom.
  Alas, Expand Fortress to Gold, Expand Gold to Province, Stonemason Province to 2 Golds, Stonemason 2 Golds to 4 Expands yields a gain of 2 Expands, hence another unbounded engine.

Stonemason Golds to Expands?
Some days, I can't seem to get my foot out of my mouth.  The quote was from a kingdom that considered using Ferry.  Without the price reduction, it would be Expand Fortress to Gold, Expand Gold to Province, Stonemason Province to 2 Expands.  And that would yield no gain.  So the kingdom in the original thread was valid after all, since it did not contain Platinum/Colony.  Now I have to go edit it again.
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majiponi

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Re: Best Asymptotic Point Scoring
« Reply #62 on: January 29, 2021, 09:53:18 am »
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It turns out that the Expand approach is flawed, and you don't even need Platinum/Colony.  Province will suffice as I pointed out way back here in a post about an improved kingdom.
  Alas, Expand Fortress to Gold, Expand Gold to Province, Stonemason Province to 2 Golds, Stonemason 2 Golds to 4 Expands yields a gain of 2 Expands, hence another unbounded engine.

Stonemason Golds to Expands?
Some days, I can't seem to get my foot out of my mouth.  The quote was from a kingdom that considered using Ferry.  Without the price reduction, it would be Expand Fortress to Gold, Expand Gold to Province, Stonemason Province to 2 Expands.  And that would yield no gain.  So the kingdom in the original thread was valid after all, since it did not contain Platinum/Colony.  Now I have to go edit it again.

Sorry, but I've noticed that your Kingdom goes infinite even without Ferry.

Expand Alchemist to Possession
Stonemason Possession to Golem and Gold
Stonemason Golem to 2 Alchemists
Expand Gold to Province
Stonemason Province to 2 Expands

Removing Alchemist, no infinity, but no interest either.
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pitythefool

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Re: Best Asymptotic Point Scoring
« Reply #63 on: January 29, 2021, 05:24:30 pm »
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It turns out that the Expand approach is flawed, and you don't even need Platinum/Colony.  Province will suffice as I pointed out way back here in a post about an improved kingdom.
  Alas, Expand Fortress to Gold, Expand Gold to Province, Stonemason Province to 2 Golds, Stonemason 2 Golds to 4 Expands yields a gain of 2 Expands, hence another unbounded engine.

Stonemason Golds to Expands?
Some days, I can't seem to get my foot out of my mouth.  The quote was from a kingdom that considered using Ferry.  Without the price reduction, it would be Expand Fortress to Gold, Expand Gold to Province, Stonemason Province to 2 Expands.  And that would yield no gain.  So the kingdom in the original thread was valid after all, since it did not contain Platinum/Colony.  Now I have to go edit it again.

Sorry, but I've noticed that your Kingdom goes infinite even without Ferry.

Expand Alchemist to Possession
Stonemason Possession to Golem and Gold
Stonemason Golem to 2 Alchemists
Expand Gold to Province
Stonemason Province to 2 Expands

Removing Alchemist, no infinity, but no interest either.

You're absolutely right.  That kingdom is busted.  I'll go edit it for a third time.
I clearly had that problem in mind while working on the followup kingdom, but I somehow didn't connect it to the earlier kingdom.

Hero is looking better all of the time!
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pitythefool

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Re: Best Asymptotic Point Scoring
« Reply #64 on: February 22, 2021, 08:14:24 pm »
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TLDR:  A simpler Mandarin/Scrying Pool engine that overpays for Masterpieces to make Conquest points with increases of N^(4*N^2) per turn.


Kingdom:  Black Market, Forager, Masterpiece, Priest, Changeling, University, Scrying Pool, Cultist, Mandarin
Black Market:  Scheme, Fortress, Watchtower
Events:  Seaway, Conquest
Projects:  Capitalism

Victory points are generated by Conquest.

Getting started: (in no particular order):
    buy 1 Black Market and all the cards from the Black Market
    buy the project Capitalism
    buy Seaway for Forager
    trash all Copper and Estates and Potions

Starting deck

     1   Black Market, Cultist, Scheme, Fortress, and Watchtower
     N   Foragers
     N   Priests
     N   Universities
     N+1 Scrying Pools

loop start
    play Scrying Pool, draw entire deck (nothing but actions)
    play Black Market (so we can play Priests and Foragers without requiring actions)
    play all Priests, trashing the Fortress
    play all Foragers, trashing the Fortress
    play University, gain Mandarin (the Black Market and all Priests and Foragers are returned from play)
repeat until one University remaining
    do the loop a final time, but exchange one Silver the Mandarin for a Changeling
    play the last Scrying Pool, drawing all the cards on the stack and then the Changeling
    play Black Market and all Priests and Foragers as before
    play Scheme (drawing nothing)
    play Cultist (drawing nothing)

buy phase
    buy Masterpiece, overpaying by all of the coins you have
        reveal Watchtower, trash Masterpiece and all Silvers for coin benefit.
    repeat until only two buys left
    buy Masterpiece, overpaying by all of the coins you have except six, exchange Masterpiece and all Silvers for Changelings
    buy Conquest, exchange Silvers for Changelings

night phase
    trash the Changeling in hand, gain a Cultist, reveal Watchtower and trash, draw 3 Changelings
    repeat, thus trashing a third of the Changelings, but drawing the other two thirds to the hand
    trash all Changelings, gaining equal numbers of Universities, Scrying Pools, Priests, and Foragers.*

*(Gaining equal numbers is not best but it exhibits uniform growth and simplifies the math. More Foragers are preferred but the optimal ratio is not a constant.)

cleanup
    put a Scrying Pool on top of your deck (due to Scheme)

Proof of finiteness.

    All phases are limited.
    Action phase:  The only gainer is University which cannot gain itself and cannot be returned from play.  When the last one is played, all other cards can only be played one more time.
    Buy phase:  Additional buys cannot be bought in the buy phase.
    Night phase:  Changelings cannot gain any card that yields more than one card costing more than three coins, when gained or trashed.

Analysis:

    N Priests are played N times each generating O(N^4) coins.  The benefit for trashing a card reaches 2*N^2 coins.  The last play of the Foragers thus adds O(N^3) coins, but this is not significant.  Spending all of it on a Masterpiece and then trashing all gained cards effectively multiplies the wealth by 2*N^2.  That is repeated for every buy up until the last.
    N Foragers played N times each yields 2*N^2 buys.  All Silvers trashed or exchanged for Changelings are still considered "gained".
So the purchase of Conquest yields O[(N^4)*(2*N^2)^(2*N^2)] = O(N^(4*N^2)).  Universities, Scrying Pools, Priests, and Foragers increase by the same order.

« Last Edit: February 27, 2021, 07:29:20 am by pitythefool »
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bitwise

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Re: Best Asymptotic Point Scoring
« Reply #65 on: February 26, 2021, 08:08:36 pm »
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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.

---

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.
« Last Edit: February 26, 2021, 09:28:03 pm by bitwise »
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pitythefool

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Re: Best Asymptotic Point Scoring
« Reply #66 on: February 27, 2021, 07:42:17 am »
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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.
Yes, that was written poorly by me.  But you managed to interpret it correctly.  I have edited the post to correct that and one other mistaken remnant from an earlier version.

Quote
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).
I had considered Squire, but not in combination with another gainer.  Awesome work.
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bitwise

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Re: Best Asymptotic Point Scoring
« Reply #67 on: February 27, 2021, 02:29:20 pm »
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A N -> (2^(2^(2^N))) growth should amount to f(n) = 2 ↑↑ 3n growth, which seems like the best so far if I'm understanding correctly. The triple up arrow seems elusive again though :(

Great work on your part as well! I think it's a lot harder to build out the solution from nothing than to try to improve it. :P
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pitythefool

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Re: Best Asymptotic Point Scoring
« Reply #68 on: May 16, 2021, 10:37:56 am »
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This is not like my usual posts.  I will be posting a new engine next, but it will use some extreme mechanisms that many of you may be unaware of and I thought it would be best to explain them and other findings first.


Ways and means

I am fond of Capitalism.  It transforms a lot of useful actions into action/treasures which allows you to play them in treasure phases.  Many people misunderstand how this interacts with Ways.  If I play an action/treasure as a treasure, I am still playing a card that is an action.  As such, I can elect to play it as a Way.  The Way does not change the type of the card, it only changes the action taken when playing it.  A discussion of this can be found here; https://www.reddit.com/r/dominion/comments/g16wdr/capitalism_with_ways, where Donald X. himself weighs in with: "Ways do not change card text; they do not affect what cards Capitalism applies to."


There's a fine line dividing an action and its effect

Playing an action as a Way does not change any special effect listed on the card below a dividing line, only the action above it.  If I play an Herbalist as Way of the Ox, I can still topdeck a treasure when I discard the Herbalist.  With Capitalism in effect, I can even elect to topdeck the Herbalist itself.  On the other hand, if I play a Scheme as Way of the Ox, I lose the ability to topdeck an action card from play at cleanup since that is granted by the action of the card, not as an effect (e.g above the line versus below the line).


A dog is a man's best friend

Specifically a Sheepdog.  Whenever you gain a card, you can elect to play a Sheepdog from your hand.  Drawing two cards can sometimes be useful in an engine, but why is it so great?  It isn't so much the effect of the card, but the fact that you can play it whenever you gain another card.  There are probably a lot more useful action cards that you wish had the ability to be played like this.  Well, you can achieve it if the kingdom has Way of the Mouse with Vassal as the action card set aside.  Now when you gain, say, a Scrying Pool, topdeck it, reveal Sheepdog from your hand, elect to play it, play it as Way of the Mouse, Vassal then discards the Scrying Pool, and you can elect to play it.  Sheepdog partnered with Way of the Mouse/Vassal can thus enable practically any engine in any turn phase.


Temporal association of Buy & Haggle

Capitalism is strongest in combination with Mandarin.  I like to use a lot of Hagglers to gain a lot of action/treasure cards including a Mandarin and watch a lot of cards get returned to the top of my draw pile.  I used to think that the Mandarin had to be gained last, since it would remove the Hagglers from play and the Haggle effect is while-in-play, and then I wouldn't be able to gain more cards, but this is not so.  As soon as a buy is made, all Hagglers in play are activated to gain a card and you can choose the cards to be gained in any order.
Even more surprising is that the Hagglers do not even need to be played before a card is bought.  If you can manage to play a Haggler while the when-buy is still being processed, you can haggle a card after the buy.  This is similar to being attacked and reacting with a Caravan Guard.  If you draw another Caravan Guard, you can play that too even though it was not in your hand when the attack card was played.
I've already revealed Way of the Mouse (Vassal) and this is the easiest way to achieve the effect.

With a hand of four Hagglers and four Sheepdogs:

    play all Hagglers
    buy Travelling Fair (for topdecking)
    buy a Gold
        from Hagglers
            gain a Haggler and topdeck it
                reveal Sheepdog and play it as Way of the Mouse (Vassal)
                    Vassal plays Haggler
            gain a Haggler and topdeck it
                reveal Sheepdog and play it as Way of the Mouse (Vassal)
                    Vassal plays Haggler
            gain a Haggler and topdeck it
                reveal Sheepdog and play it as Way of the Mouse (Vassal)
                    Vassal plays Haggler
            gain a Haggler and topdeck it
                reveal Sheepdog and play it as Way of the Mouse (Vassal)
                    Vassal plays Haggler
            gain a Haggler and topdeck it
            gain a Haggler and topdeck it
            gain a Haggler and topdeck it
            gain a Haggler and topdeck it
        gain the Gold

From four Hagglers, we have gained eight more.  Could we have done better?  Yes, quite a bit if Capitalism and Mandarin are in the kingdom.  The Haggle effect of Haggler is "below the line".  We could play the Haggler as a Way and the Haggle effect would still happen.

    play all Hagglers
    buy Travelling Fair (for topdecking)
    buy a Gold
        from Hagglers
            gain a Haggler and topdeck it
            gain a Haggler and topdeck it
            gain a Haggler and topdeck it
            gain a Haggler and topdeck it
                reveal Sheepdog and play it as Way of the Mouse (Vassal)
                    Vassal plays Haggler
                        play Haggler as Way of the Mouse (Vassal)
                            Vassal plays Haggler
                                play Haggler as Way of the Mouse (Vassal)
                                    Vassal plays Haggler
                                        play Haggler as Way of the Mouse (Vassal)
                                            Vassal plays Haggler
                                                play Haggler as an actual Haggler
                                                4th Haggler finished
                                        3rd Haggler finished
                                2nd Haggler finished
                        1st Haggler finished
                This concludes the Sheepdog-Vassal chain.
        <return to when-buy processing>
            from Hagglers just played
                gain a Mandarin
                    all Hagglers are returned to the top of the deck
                gain a Haggler and topdeck it
                gain a Haggler and topdeck it
                gain a Haggler and topdeck it

    We now have 7 Hagglers on our deck.  Reveal another Sheepdog and do it again, this time ending with 13 Hagglers.
    The 3rd Sheepdog ends with 25 Hagglers and the 4th with 49.

    Surprisingly, most of the cards were gained with Hagglers that were still in the supply when the purchase was made and some were not even in play when the haggles were exercised.  This scenario has been verified on ShuffleIT (as with everything mentioned here) and can be achieved, or at least could be if the kingdom contained more than 10 Hagglers.

    Can we do better yet?  Yes.  It is possible to gain an unbounded number of Hagglers.  I leave it for you to ponder.  Hint: gain more Sheepdogs.


Other ways to trigger actions

    Falconer can also be played when a card is gained.  But the card must be of at least two types, so it is more restrictive.  Though both Sheepdog and Falconer can be triggered by gaining themselves.
    A Noble Brigand does not get played when gained.  The new wording is simply "When you buy this, do its attack".
    Village Green may be played, not when gained, but when discarded other than at clean up.  Doctor overpay and Exile mechanics are a good way to limit that.


Delayed gratification

    Why didn't the Hagglers played above immediately grant the gaining of a card?  When a when-buy effect is being processed, no other when-buy effects can be processed until the first one is completed.  The Hagglers are registered as being played during the buy, but do not take effect until when-buy processing is resumed.  This is especially disheartening for Doctor overpay.  Overpaying for a Doctor is considered a when-buy effect.  If you have other when-buy effects to process (such as Hagglers in play), you will be granted the option of choosing the order in which to process them.  When you choose overpay, you must use up all of the overpay credit before processing any other when-buy effects.


No two-for-one specials

    If you Throne Room a Haggler after the buy, you do not get two cards, only one.  It is the act of putting the Haggler in play, not playing it, that causes the effect.


The art of the deal

    Where does the art of haggling come in?  It seems you have no say in determining the price of a haggled card.  Or do you?  When you play a Haggler during a when-buy period, the price point of the gained card is set at the time the Haggler is processed.  This is not necessarily the cost of the card at the time the original buy was made, nor the price of the card when the Haggler was put in play.  For example, if I buy a Farmland (which has its own when-buy effect), then manage to play a Bridge, then a Haggler, and then another Bridge, and then return to when-buy processing, I'll be asked to gain a card cheaper than four coins.  Not the six coins I paid for Farmland, nor the five coins it would have cost when I played the Haggler.
    The price adjustments are applied to the card we bought and the card we can gain equally.  So it appears we do not gain any advantage for all of our effort.
You can also go the other way and it behaves the same; play a Quarry and two Hagglers, then buy a Grand Market for four coins.  For the first Haggler, we can gain a card cheaper than four coins.  Choose a Mandarin and the Quarry is removed from play.  For the second Haggler, we are asked to gain a card cheaper than six coins.  It does not change our selection of action cards that can be gained.  But maybe you wanted a Venture, in which case you at least saved two coins when buying the Grand Market.
    There are four cards whose price CAN be manipulated when using Hagglers.  See http://forum.dominionstrategy.com/index.php?topic=7339.msg866270#msg866270 for a fun puzzle.


Curses, foiled again

    One when-buy event that cannot be triggered after the buy is Embargo.  If you buy a card then manage to play an Embargo on its supply pile during the when-buy processing, you do not gain a curse.  This seems inconsistent with other when-buy processes but it relates to Embargo's wording "For the rest of the game, when a player buys a card, ..."


Doctor, Doctor, give me the news

    While playing with Doctor, I came to realize that the overpay effects are greatly underrated.  An example is a deck that contains a single Tunnel.  Buy a Doctor and overpay by 1000 coins.  Discard the Tunnel, reveal it, gain a Gold.  Next we will reshuffle and encounter either the Tunnel again or the Gold.  Keep trashing Gold and discarding the Tunnel and this sequence can continue until the overpay runs out.
   With topdecking and good strategy, you can prepare a sequence of cards for the Doctor to encounter before buying it.  Squire, Cultist, Catacombs, and Hunting Grounds all have useful effects when trashed and there are others.  Another powerful card to consider here is Village Green.  When discarded, you can elect to play it.  And of course, that invites playing it as a Way.  There are also over two dozen cards that have a side effect when gained and other while-in-play effects you can establish beforehand.
    It is possible to create an engine that runs during a Doctor buy.  It has have the advantage of being bounded by its nature.  You must overpay by a finite amount up front and cannot extend it while the engine is running.  If you play Hagglers in the engine, you are restricted to gaining cards that cost less than a Doctor.  These are very handy restrictions when trying to prevent an engine from running unbounded.


Vassal and Village Green; a powerful combo

    When a Village Green is discarded, you can elect to play it.  When you play a Vassal, it first discards the card on top or your deck, and then if it is an action, you can play it.  So when a Vassal discards a Village Green, you can play the Village Green twice.
    When you pair this effect with Way of the Mouse/Vassal, powerful sequences can be built up.  If you have N Village Greens on your deck and you discard the first one with a Vassal and elect to play it also as a Vassal, then you get to play the second Village Green twice before playing the first one a second time.  If you thus chain through all N Village Greens, you will have N+1 Village Green actions to resolve.  Let's call them VG credits.  And for each VG credit you can either play the Village Green as normal to draw the top card of your deck into your hand and gain two actions, or play the Village Green as a Vassal to either discard the top card of your deck or play it if it is an action.  If you can somehow gain and topdeck more Village Greens, you can increase your Village Green credits and keep the engine running.
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pitythefool

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Re: Best Asymptotic Point Scoring
« Reply #69 on: May 16, 2021, 04:51:26 pm »
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If you are having trouble following some of the play, it may help to read my earlier post.

Kingdom:  Black Market, Village Green, Mandarin, Forum, Royal Carriage, Haggler, Border Village
Black Market:  Druid, Watchtower, Scheme, Treasurer
Boons for Druid are Forest's Gift (+coin, +buy), Flame's Gift (trash a card), and Field's Gift (+action, +coin)

Event:  Training
Landmark:  Obelisk on Haggler (for Victory points)
Project:  Capitalism
Way of the Mouse: Vassal

getting started

    buy all Black Market cards
    buy the project Capitalism
    buy Training for Haggler
    trash all unwanted cards and the Mandarin with Flame's Gift

starting deck

    1 Druid, Watchtower, Scheme, Scrying Pool, Treasurer.  (Mandarin in trash)
    many Hagglers
    many Royal Carriages

action phase

    play Scrying Pool, draw deck
    play Scheme, draw nothing
    play all Royal Carriages
    play Druid, select Forest's Gift (always)
        call all Royal Carriages, replaying Druid

treasure phase

    play Treasurer, gain Copper from trash, reveal Watchtower, trash Copper
    play all Hagglers

buy phase

    buy Transport
        exile a Village Green

    buy Border Village
        from N Hagglers,
            gain Mandarin,
                topdeck Treasurer and all Hagglers
            gain N-1 Village Greens (topdeck)
                after gaining the last Village Green, discard the Village Green from exile
                    play discarded Village Green as Way of the Mouse: Vassal
                        play topdecked Village Green as Way of the Mouse: Vassal
                            repeat for all topdecked Village Greens
                                ...
<start of Haggler chain>
                                play Haggler as Way of the Mouse: Vassal
                                    play Haggler as Way of the Mouse: Vassal
                                        repeat for all topdecked Hagglers
                                            ...

                                            play Treasurer, gain Mandarin from trash and trash it again
                                                topdeck Treasurer and all Hagglers
<repeat Haggler chain N times>
    The Nth time through the loop, we play the Haggler chain up to the Treasurer, then discard the Treasurer without playing it.
    The Village Green chain is thus completed, returning us to when-buy processing.
    Post-buy Hagglers now kick in.
        We have played N Hagglers N times each, so we stand to gain N^2 cards.
            gain N^2 Hagglers and topdeck them
        gain the Border Village
            gain a Haggler (topdeck)

    buy Transport
        exile a Village Green

    buy Forum
        from Hagglers,
            gain all Village Greens and topdeck them

Forums grant an extra buy when buying them, so we can keep buying them.  Reserve 6 coins, but keep buying Forums and loading the deck with Village Greens till the money runs out.

    buy Border Village
        from M Hagglers,
            gain M-2 Hagglers (topdeck)
            gain 1 Mandarin,
                topdeck all Hagglers
            gain 1 Village Green (topdeck)
                after gaining the Village Green, discard the Village Green from exile
                    play discarded Village Green as Way of the Mouse: Vassal
                        play topdecked Village Green as Way of the Mouse: Vassal
                            play Haggler as Way of the Mouse: Vassal
                                play Haggler as Way of the Mouse: Vassal
                                    repeat for all topdecked Hagglers
                                        ...
                            We will eventually run into all of the Village Greens, play them as Vassals too, continuing the chain.
                            If we encounter Forums or Border Villages, play them as Vassals to continue the chain.
                            The last Vassal will draw the Treasurer from the discard pile, play it and gain and trash the Mandarin.
                            Then all the treasures in play get restacked onto the deck and we do it all again.
                            Keep repeating this sequence, discarding the Treasurer with the last Village Green credit.

        with post-buy Hagglers,
            gain Hagglers and topdeck them
        gain the Border Village
            gain a Haggler (topdeck)

    buy another Transport, more Forums, and then repeat by buying another Border Village

We eventually run out of buys.
When the last Border Village is bought, use the post-buy Hagglers to gain Royal Carriages instead of Hagglers.
At cleanup, topdeck the Scrying Pool (from playing Scheme)

Proof of finiteness

    The two possibilities for unbounded play are gaining an unbounded number of cards or repeatedly playing the same cards an unbounded number of times.
The only gainers are a single Treasurer which can only regain a card from the trash, Black Markets, and Hagglers.  Black Markets can only gain cards from the Black Market deck which is finite, so gaining from there cannot be sustained.  Hagglers can only gain cards when you buy something, so they cannot contribute to unbounded play in the action phase.  If the Treasurer is played and does not gain the Mandarin from the trash, it cannot be replayed in the action phase, so that also does not help.  There are other ways to gain Mandarins in the buy phase, but not in the action phase.  It is possible to gain multiple treasure cards from the trash by combining the play of the Treasurer with replays via Royal Carriages.  But it only gains treasures from the trash, so it is limited to what you can put into the trash.  Also, the Royal Carriages cannot be replayed, hence this is also not sustainable.
    The only cards that can draw cards are Border Village, Forum, Village Greens, the single Scheme, and even Watchtower.  None of these are considered treasures under Capitalism and hence cannot be replayed.  But we have introduced the Way of the Mouse: Vassal as a way of playing cards without drawing them.  Indeed, it is now more convenient to play the cards right off our stack.  Doing so in the action phase is of no benefit, since we could play them normally.  Doing so in the buy phase requires discarding a Village Green.  There are only two ways to do this; playing a Forum or discarding one from exile.  Playing a Forum doesn't help since we need to first discard a Village Green to allow us to play a Forum so we can discard a Village Green.  Simple chicken and egg problem.  That only leaves discarding a Village Green from exile.  You must first exile a Village Green and every time you do that requires buying a Transport.
    The buy phase is limited in part because the number of buys are limited.  The single Druid is the only way to obtain buys.  We can gain more Royal Carriages but  cannot play them on the Druid except immediately after.  There is one exception to the limited buys and that is buying Forums.  We can buy as many Forums as we can afford and that is the clincher.  We always run out of money.  There is no way to gain coin in the buy phase without playing actions (which would require buying another Transport).
    A Forum is five coins.  A Haggler is five coins.  We cannot gain more Hagglers when buying Forums.  We can gain more Village Greens, which is mainly the point.  These can trigger the playing of actions, including the replaying of Hagglers.  But the action chain must stop before the Hagglers take effect.  In other words, you can only gain more cards after the ability to play them is gone.  To start another action chain requires buying another Transport.  Or we could trigger actions just to play them as Vassals and earn more coins.  More coins allows us to buy more Forums, but to repeat such a sequence for another Forum buy would require another Transport buy first.  So no engine can be sustained indefinitely.


Analysis

    We start a turn with N number of Hagglers and N^3 number of Royal Carriages.  Each Royal Carriage earns us 2 buys; a Transport and a Border Village (Forums don't count), so we will increase our Haggler count N^3 times.
    Whether we play Hagglers as Hagglers or Vassals, we get the +1 coin token from Training for 3 coins per Haggler.  To simplify the math, I'm reducing that to half the price of a Forum.
When we buy a Border Village (after the first), the pre-buy Hagglers first double the number of Hagglers and then all Hagglers are played multiple times for massive post-buy gains.

The first Border Village increases Hagglers from N to O(N^2).
Then we will buy O(1/2*N^2) Forums, gaining O(1/2*N^4) Village Greens, allowing us to play O(2*N^2) Hagglers O(1/2*N^4) times for O(N^6) Hagglers.
Then we will buy O(1/2*N^6) Forums, gaining O(1/2*N^12) Village Greens, allowing us to play O(2*N^6) Hagglers O(1/2*N^12) times for O(N^18) Hagglers.
Then we will buy O(1/2*N^18) Forums, gaining O(1/2*N^36) Village Greens, allowing us to play O(2*N^18) Hagglers O(1/2*N^36) times for O(N^54) Hagglers.
... and we repeat this N^3 in a turn, yielding O(N^(3^(N^3))) Hagglers.

Sometimes the strength of an engine is stated as growth-per-turn.  Here we turn N Hagglers into N^(3^(N^3)) which can be stated as g(N) = N^(3^(N^3)).  Obelisk grants us two victory points per Haggler, so we actually multiply this two to get VP.  The factor of two is insignificant so I'll just drop it.

It is often gratifying to talk about the Victory points at some turn n as a function of n, usually after a small constant number of turns required to get the engine up and running.  For reasons that will become clear later, I'll drop the constant and assume we start the first turn with 3 Hagglers.

f(0) = 3
f(1) = g(3) = 3^(3^(3^3)) = 3↑↑4 = 37,625,597,484,987
f(2) = g(g(3)) = [3^(3^(3^3))]^3^(3^([3^(3^(3^3))]^3)) which is much greater than 3^(3^(3^(3^(3^(3^3))))) = 3↑↑7 = ??!
f(n) = gn(3) >> 3↑↑(3*n+1)

The actual value of f(n) will far outpace the double up-arrow expression given.  For the evaluation of f(2), we replaced [3^(3^(3^3))] with just 3 at the base of the exponents.  This was necessary since those exponents were being evaluated bottom-up and we can only consider evaluating top-down to continue the chain.  Consider for a moment how much larger the value really would be without that substitution.  The increased value would be inserted into f(3) at the base and also at the exponent 2nd from the top and that's where things really start to get larger.  It would easily outpace the double up-arrow expression even if we were to assume a larger number of Hagglers at the start.  And that's where things get interesting.  What if we started with 27 Hagglers?  27 = 3^3.  That does not work out very well.  But if we started with 3^(3^3), that clearly allows us to add more exponents of 3 to our chain.  Leading me to conclude that
for sufficiently large n, f(n) = 3↑↑(k*n) for arbitrary values of k.
Okay, maybe that's a bit of a stretch, but this math is so strange who can tell.  I think the analysis of the engine is more of an art than creating the engine.

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silverspawn

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Re: Best Asymptotic Point Scoring
« Reply #70 on: January 08, 2022, 07:31:50 am »
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I don't know if you would be interested in this but it seems similar to the solutions in this thread (which I haven't read in detail).
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