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[heron]

How few King's Court's and Throne Rooms do you need to buy all of the colonies in one turn if all of the other cards in your deck are ruins?
Rules: You are allowed to use more than 10 KC's, ruined libraries, etc. You are not limited by the number of cards there actually are in the game.
Solutions are scored by how many KC's are used, with TR's as a tiebreaker. I will probably post a solution* later today.

*It's not a very good solution.

[eHalcyon]

It should be possible to create a solution using only TR and no KC at all, but that would be tedious.


TR-TR-TR-TR will let you play 4 actions twice (I think... I always have trouble mapping these out).  If 3 are Ruined Library, that draws a total of 6 cards and there is still one action left to go. 

Thus you can get a slightly bigger hand and still have the ability to play more actions.

Therefore, without restrictions on number of TRs and Ruins, you CAN get a solution with 0 KCs.  It would just be really, really long and annoying.


I suppose "all the Colonies" means 8 of them, and I suppose we can assume perfect shuffle luck.

This isn't a solution, but it's a start.


To buy 8 Colonies, you need +7 Buy, +$88.

You could attempt to KC enough Ruined Villages to have actions to play 7 Ruined Markets and 88 Abandoned Mines, but you would also need to KC enough Ruined Libraries to draw all those cards.  Better to skip the Villages and go straight for the others.

To get 7 plays of Ruined Market, you can KC-KC-RM-RM-RM for 9 plays.  I think this is the most efficient way, even though you get more plays than necessary.  So you need to KC 3 of them.

To get 88 plays of Abandoned Mine, you need to KC 30 of them.

To start, you need to KC Ruined Library to draw all the other Ruins and KCs.

I don't really want to work through how many of each are necessary...

[Powerman]

It should be possible to create a solution using only TR and no KC at all, but that would be tedious.


TR-TR-TR-TR will let you play 4 actions twice (I think... I always have trouble mapping these out).  If 3 are Ruined Library, that draws a total of 6 cards and there is still one action left to go. 

Thus you can get a slightly bigger hand and still have the ability to play more actions.

Therefore, without restrictions on number of TRs and Ruins, you CAN get a solution with 0 KCs.  It would just be really, really long and annoying.


I suppose "all the Colonies" means 8 of them, and I suppose we can assume perfect shuffle luck.

This isn't a solution, but it's a start.

Nope.  Using Throne Rooms and +1 Cards, you cannot get a hand of over 5 cards.

[heron]

The simple KC-Library-Mine-Market solution uses 44 KC's.

[eHalcyon]

It should be possible to create a solution using only TR and no KC at all, but that would be tedious.


TR-TR-TR-TR will let you play 4 actions twice (I think... I always have trouble mapping these out).  If 3 are Ruined Library, that draws a total of 6 cards and there is still one action left to go. 

Thus you can get a slightly bigger hand and still have the ability to play more actions.

Therefore, without restrictions on number of TRs and Ruins, you CAN get a solution with 0 KCs.  It would just be really, really long and annoying.


I suppose "all the Colonies" means 8 of them, and I suppose we can assume perfect shuffle luck.

This isn't a solution, but it's a start.

Nope.  Using Throne Rooms and +1 Cards, you cannot get a hand of over 5 cards.

Ah, I forgot to count that you would need those extra RVs to throne. 

But still, you could probably cut down a lot of KCs by using mass TRs instead.

[qmech]

The simple KC-Library-Mine-Market solution uses 44 KC's.

I get 33 for the direct approach.

Only slightly tweaked hits 32.

1 Throne Room
32 King's Court
31 Ruined Library
30 Ruined Mine
3 Ruined Market
for a total of 97 cards.  You get 64 useful KC plays, and can draw 93 cards, so you draw your deck with a card to spare.  The Mines and Markets leave you with $90 and 10 Buys, more than enough to clear the Colonies.

There's slack in the number of cards drawn, the money you have to spend, and the buys you have.  I haven't managed to exploit it to get the number of KC down yet though.  I'd be amazed if it goes below 30.

[heron]

You're right, I wonder why I counted wrong.
I agree that there is very little room for optimization.

[pacovf]

Actually, you need 88 plays of abandoned mine, and 7 plays of ruined market, and enough king's court/throne room/ruined library to support them.

Chaining King Courts provides 2*n-1 effective plays of King Courts for one action. King courting a throne room provides 6 plays of the card in question, but requires three copies of the card and the throne room, instead of one; so, its only benefit is allowing a division of the multiplied actions, but it's just as good/bad from a card draw perspective. Maybe you can dig this to find a solution with more throne rooms and less king courts?*

Anyway, we need one king courted ruined market and two throne roomed; 28 king courted abandoned mines, and two throne roomed. That means two king courted throne rooms (and we are left with two uses of throne room to spare), so 48 (16 of which are king courts) cards to draw. Two king courted ruined libraries provide six cards, and costs three cards (the king court, with one use left for the next one; and the two ruined libraries), for a net gain of three cards. Since we have to draw 43 cards (48, minus the 5 you start with), and we had two spare throne rooms (the first applied on a ruined library, the second on a king court applied on two libraries), the end result is the need to draw 39 cards, so that means another 13 (+1) king courts, and 26 (+3) ruined libraries, and no card draw left.

End result:
-30 king courts
-2 throne rooms
-29 ruined libraries
-28 abandoned mines
-3 ruined markets

And no spare buy/coin/card draw.
*Fast thinking: normally, once your king court chain is going, you would do king court (ruined library; ruined library; king court (abandoned mine; abandoned mine; REPEAT), which draws itself (6 cards) and provides 6 coin for two king courts.
Instead, you could do king court (ruined library; ruined library; king court (ruined library; ruined library; king court (ruined library; ruined library; king court (throne room (abandoned mine; abandoned mine; abandoned mine); throne room (abandone mine; abandoned mine; abandoned mine); REPEAT))); which draws itself (18 cards), and provides 12 coins for four king courts and two throne rooms, so you're worse than in the first case;

So I'm pretty sure I got the best answer.

Man, I sure love procrastinating

[qmech]

Unfortunately, I think you missed a couple of Abandoned Mines.  You're almost there, and the patched version still beats mine.

31 King's Courts - 61 useful KC plays
30 Ruined Libraries - 90 cards
28 Abandoned Mines - $84
2 Ruined Markets - 6 Buys

1 Throne Room
2 Abandoned Mine - $4
1 Ruined Market - 2 Buys

95 cards total