Dominion Strategy Forum
Dominion => Puzzles and Challenges => Topic started by: Boldot on March 02, 2012, 08:00:15 pm
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So I've been reading up on the Golden Deck (5 card deck where you bishop a Province, and buy a new Province each turn, 5vp chips a turn), and I was wondering if there was anything better?
So the challenge is what's the most number of tokens you can get each turn, every turn. You can only have 5 cards in your entire deck at the start and end of each turn, they must be the same cards, and it's a solo game so no outside assistance. I'd rule out Possession as well.
For hard mode, they must be 5 unique cards. The best I can come up with this is 12vp a turn, works in hard mode as well.
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Hmm, best I can come up with in 10 minutes:
TR-KC-Goons-Goons-Watchtower
Two Tripled Goons = 7 buys granting 2 VP per buy for 14 VP total. Watchtower stays in hand to trash everything you buy.
Doesn't work for your hard mode though.
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I suspect that it's possible to use KC-Bridge-Ironworks tricks to bootstrap yourself from a 5 card hand into a number of VP tokens that turn that is limited only by the number of cards in the supply, then end the game by 3-piling that turn, which maybe technically qualifies. That wouldn't be in the spirit of the puzzle though. ;)
(Edit: I had a different purported easy-mode solution for 12 VP per turn, but it was wrong because I forgot to include Bridge discounting in Bishop value.)
An easy mode solution for 12 VP per turn:
Hand contains KC, KC, Ironworks, Bishop, Monument. KC-KC, playing Ironworks to gain 3x Gardens, which are drawn immediately, then play Bishop to trash all three Gardens for 1+2 VP each, then play Monument for an additional 3 VP. In total, 12 VP.
Hard mode solution for 12 VP per turn:
Hand contains Province, TR, KC, Explorer, Bishop. TR-KC, playing Explorer to gain 3 Golds, then playing Bishop to trash all three Golds for 1+3 VP each or 12 VP total.
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On easy mode:
KC - KC - Ironworks (for 3 x Peddlers) - Smithy (any drawer will do) - Bishop
for 15 VP per turn.
Correction: Didn't work because of the verbiage on Peddler; thanks chesskidnate for pointing it out.
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Hard Mode: 36 VP:
Hand is NV, TR, KC, Bridge, WS.
KC->TR Bridge x2
WS gain two Fairgrounds
NV put two Fairgrounds on NV mat.
NV Mat also has one of each card, including black market's 25 cards.
edit: Oh, unless only VP tokens count.
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On easy mode:
KC - KC - Ironworks (for 3 x Peddlers) - Smithy (any drawer will do) - Bishop
for 15 VP per turn.
During your buy phase, this costs $2 less per action card you have in play, but not less than 0
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blueblimp, your easy mode solution doesn't work since the fairgrounds are devalued by bridge.
edit: That's a clever fix.
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blueblimp, your easy mode solution doesn't work since the fairgrounds are devalued by bridge.
Ah, you are right of course, thanks. =) I've replaced it with a different easy mode solution.
Inspired by the easy mode solution I replaced it with, here is a hard mode solution for 10 VP per turn:
Hand is TR, KC, Ironworks, Bishop, Monument. TR the KC. First play the Ironworks to gain three Islands (or Island+Garden+Silk Road if you don't want to exhaust the piles too fast), which are drawn into your hand--plus you now have 3 actions. Next play the Bishop to trash all three Islands for 1+2 VP each. Then use one of your actions to play the Monument, which gives you 1 VP for a total of 10 VP.
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Easy mode
KC/KC/gov/gov/bishop
Play as follows:
KC -> KC
Gov 1:Gain 2 gold, draw them
Gov 2: Remodel gold -> province, gain gold, draw gold & province
Bishop: trash 2 Golds & a province
Total points: 13
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Easy mode
KC/KC/gov/gov/bishop
Play as follows:
KC -> KC
Gov 1:Gain 2 gold, draw them
Gov 2: Remodel gold -> province, gain gold, draw gold & province
Bishop: trash 2 Golds & a province
Total points: 13
Of course, in a RL game you just gave your opponent 3 silvers, a remodel, and 3 trashes. Although, the silver flood probably screws up THEIR golden deck, so maybe it's a good thing.
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KC-KC-Ironworks-Ironworks-Ironworks, draw 9 Islands and set them aside.
This of course only works once in a normal game, but if you assume infinite pile size..
18VP/Turn
Edit: You can substitute 4/9 of the Ironwork'd cards with Silk Roads or Gardens, potentially getting more VP if you allow for alternate VP values in the puzzle (Silk Roads minimum is 2 anyway since you have 9 VC)
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KC-KC-Ironworks-Ironworks-Ironworks, draw 9 Islands and set them aside.
This of course only works once in a normal game, but if you assume infinite pile size..
18VP/Turn
Edit: You can substitute 4/9 of the Ironwork'd cards with Silk Roads or Gardens, potentially getting more VP if you allow for alternate VP values in the puzzle (Silk Roads minimum is 2 anyway since you have 9 VC)
If you have Islands setting aside other Islands, what does the 9th Island set aside?
But the puzzle says you have to gain VP tokens, not just VP in general. I really can't think of anything decent for the hard mode question, but I like blueblimp's.
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If it can't set aside anything, it'll just set itself aside without another card.
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If it can't set aside anything, it'll just set itself aside without another card.
Ooh. Learn something new everyday. ;D
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Blueblimps first Hard mode of 12 VP is what I first had in mind, but I am loving the other ideas.
And I guess just gaining VP in general works, as long as you maintain the 5 card hand before and after. I'd be interested to see what kind of shenanigans would be possible with things like Island or even Native Village
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Well, maybe someone can make a variation of the Gain "infinitely-many" cards in one turn (http://"http://forum.dominionstrategy.com/index.php?topic=1741.0") puzzle that includes putting VP cards on the Native Village mat, or setting them aside with Island.
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Well, maybe someone can make a variation of the Gain "infinitely-many" cards in one turn (http://"http://forum.dominionstrategy.com/index.php?topic=1741.0") puzzle that includes putting VP cards on the Native Village mat, or setting them aside with Island.
Tricky thing is that there's no way to remove action cards from play, so it's hard to maintain a 5-card hand if you play the new actions you gain.
A way to gain a 18 VP (not tokens) per turn via NV mat, without variable-value VP cards (since otherwise things like Fairgrounds, Gardens, etc. can have ridiculous value if you make assumptions about what's already on your NV mat):
Hand contains KC, KC, Bridge, Feast, NV. KC-KC, playing the Bridge, then the Feast to gain 3x Province and trashing the Feast, then the NV to put all the Provinces on the NV mat. Buy a Feast at $1 using the $3 you got from Bridge.
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BB's plan also works with:
University instead of feast
Has some variable VP (but nothing starting on the mats):
KC/KC/university/IW/golem
KC -> KC -> University (gain three duchies) -> IW (gain three islands & draw three duchies) -> golem (island three duchies).
Next turn gain three dukes instead of duchies
Next turn gain three duchies again.
Net total points after 3 turns of play: 54 (Due to province limits, BB's only nabs 51; of course his can delay the final province haul until he runs out of feasts to buy and play for more turns just gaining & decking duchies).
In a 3 player setup, you can go for four turns yielding 80 points (5 dukes, 7 duchies, and 12 islands). Which is more than all 12 provinces which are worth only 72.
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University can't gain Duchies.
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ack, well there goes those ideas.
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Sorry to bump an old thread, but here is a new easy mode solution for 12 VP tokens per turn:
Hand is KC, KC, Jack-of-all-Trades, Mine, Bishop.
KC-KC, then Jack. Don't trash anything with Jack. This results in 3x Silver, Mine, Bishop in hand.
Now Mine to convert all the Silvers to Golds. Then trash them all with Bishop for 12 VP tokens.
Another one that only gets 10 VP tokens per turn in easy mode but can maybe be tweaked to do better:
Hand is KC, KC, Trusty Steed, Bishop, Bishop.
KC-KC, then Trusty Steed for: +$2, gain 4 Silvers; +2 cards, +$2; +2 cards, +$2. Now you have $6 and your hand contains two Bishops and four Silvers. KC one Bishop, trashing 3 Silvers for 6 VP tokens. KC the other Bishop, trashing one Silver and two of nothing for 4 VP tokens. In total, 10 VP tokens.
Would be nice if something could be done with the $12 you have to spend after doing this.
Another one for 12 VP tokens per turn in easy mode:
Hand is KC, KC, Bag of Gold, Smithy, Bishop.
KC-KC, then Bag of Gold, then Smithy to draw the 3x Golds, then Bishop to trash them.
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I quite like the Jack solution. Very elegant. But I thought the Jack didn't gain the silver until after it drew, so the last silver is stuck sitting on your deck, no?
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In the description (http://dominionstrategy.com/2011/12/05/hinterlands-jack-of-all-trades/), "Gain a Silver" comes first, so I think I'm safe there.
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Surprised no one mentioned the easiest Easy Mode solution. Sure, it only gets 9VP, but it's soooo straightforward:
KCx2, Monumentx3 - KC->KC, play each Monument 3 times. Continue ad infinitum (not limited on piles!).
The real trick is getting your deck down to that point since you don't have a Bishop to help. You can use your opponent's Bishop or perhaps Islands to get you there.
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In the description (http://dominionstrategy.com/2011/12/05/hinterlands-jack-of-all-trades/), "Gain a Silver" comes first, so I think I'm safe there.
You are right. I guess I was thinking that because I never seem to draw the newly gained Silver. But obviously that would only happen if you had no deck to begin with.
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Surprised no one mentioned the easiest Easy Mode solution. Sure, it only gets 9VP, but it's soooo straightforward:
KCx2, Monumentx3 - KC->KC, play each Monument 3 times. Continue ad infinitum (not limited on piles!).
The real trick is getting your deck down to that point since you don't have a Bishop to help. You can use your opponent's Bishop or perhaps Islands to get you there.
I'll raise you.
You need 4 KC's, 3 Schemes and 7 Monuments.
With KC-KC-KC-KC-Scheme, you can draw the other Schemes and with those KC'd Schemes, you'll draw all the Monuments.
You can KC 4 Monuments for 12 VP and play the other 3 for 3 extra VP.
This gives 15 VP per turn ad infinitum.
You can use the same Island or Bishop trick.
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Surprised no one mentioned the easiest Easy Mode solution. Sure, it only gets 9VP, but it's soooo straightforward:
KCx2, Monumentx3 - KC->KC, play each Monument 3 times. Continue ad infinitum (not limited on piles!).
The real trick is getting your deck down to that point since you don't have a Bishop to help. You can use your opponent's Bishop or perhaps Islands to get you there.
I'll raise you.
You need 4 KC's, 3 Schemes and 7 Monuments.
With KC-KC-KC-KC-Scheme, you can draw the other Schemes and with those KC'd Schemes, you'll draw all the Monuments.
You can KC 4 Monuments for 12 VP and play the other 3 for 3 extra VP.
This gives 15 VP per turn ad infinitum.
You can use the same Island or Bishop trick.
Scrying Pool -> Draw 10X KC 10X Goons 9X Monument 1 Scheme 9 Bishops 1 Watchtower..whatever card gainers you want, etc, too lazy to calculate it out. Buy coin, trash with watchtower, return scrying pool to top of deck.
Orrrr we can assume that this puzzle is without scheme and generally implies a 5 card deck, or a bigger deck still forcibly caused
In which case Tactician, KCX3, MonumentX5, Curse, Tactician is forcing 15 VP turn. Probably can beat that with bishop.
... But I'm pretty sure we want a 5 card starting deck.
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I think the first one gets 21 VP tokens per turn.
Sorry to bump an old thread, but here is a new easy mode solution for 12 VP tokens per turn:
Hand is KC, KC, Jack-of-all-Trades, Mine, Bishop.
KC-KC, then Jack. Don't trash anything with Jack. This results in 3x Silver, Mine, Bishop in hand.
Now Mine to convert all the Silvers to Golds. Then trash them all with Bishop for 12 VP tokens.
Another one that only gets 10 VP tokens per turn in easy mode but can maybe be tweaked to do better:
Hand is KC, KC, Trusty Steed, Bishop, Bishop.
KC-KC, then Trusty Steed for: +$2, gain 4 Silvers; +2 cards, +$2; +2 cards, +$2. Now you have $6 and your hand contains two Bishops and four Silvers. KC one Bishop, trashing 3 Silvers for 6 VP tokens. KC the other Bishop, trashing one Silver and two of nothing for 4 VP tokens. In total, 10 VP tokens.
Would be nice if something could be done with the $12 you have to spend after doing this.
Another one for 12 VP tokens per turn in easy mode:
Hand is KC, KC, Bag of Gold, Smithy, Bishop.
KC-KC, then Bag of Gold, then Smithy to draw the 3x Golds, then Bishop to trash them.
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Disregard.
I think the first one gets 21 VP tokens per turn.
Sorry to bump an old thread, but here is a new easy mode solution for 12 VP tokens per turn:
Hand is KC, KC, Jack-of-all-Trades, Mine, Bishop.
KC-KC, then Jack. Don't trash anything with Jack. This results in 3x Silver, Mine, Bishop in hand.
Now Mine to convert all the Silvers to Golds. Then trash them all with Bishop for 12 VP tokens.
Another one that only gets 10 VP tokens per turn in easy mode but can maybe be tweaked to do better:
Hand is KC, KC, Trusty Steed, Bishop, Bishop.
KC-KC, then Trusty Steed for: +$2, gain 4 Silvers; +2 cards, +$2; +2 cards, +$2. Now you have $6 and your hand contains two Bishops and four Silvers. KC one Bishop, trashing 3 Silvers for 6 VP tokens. KC the other Bishop, trashing one Silver and two of nothing for 4 VP tokens. In total, 10 VP tokens.
Would be nice if something could be done with the $12 you have to spend after doing this.
Another one for 12 VP tokens per turn in easy mode:
Hand is KC, KC, Bag of Gold, Smithy, Bishop.
KC-KC, then Bag of Gold, then Smithy to draw the 3x Golds, then Bishop to trash them.
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This may stretch the definition of "5 card deck" a bit, but I think it works for easy mode (15 VP chips/turn):
Hand: KC/KC/Mint/Bishop/haven
Turn begins KC-> Haven (from last turn) returns 3 plats to hand
Play hand as follows: KC -> KC -> Mint (gain 3 plats to discard) -> bishop (trash 3 plats in hand for 15 points) -> haven (set aside 3 plats from discard)
Clean up - KC/Haven goes into the discard (regenerating the starting deck); KC/Haven/3 Plats are on the mat for next turn
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This may stretch the definition of "5 card deck" a bit, but I think it works for easy mode (15 VP chips/turn):
Hand: KC/KC/Mint/Bishop/haven
Turn begins KC-> Haven (from last turn) returns 3 plats to hand
Play hand as follows: KC -> KC -> Mint (gain 3 plats to discard) -> bishop (trash 3 plats in hand for 15 points) -> haven (set aside 3 plats from discard)
Clean up - KC/Haven goes into the discard (regenerating the starting deck); KC/Haven/3 Plats are on the mat for next turn
Seems legit to me since it's not vulnerable to deck attacks like Swindler, Saboteur, etc.
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Surprised no one mentioned the easiest Easy Mode solution. Sure, it only gets 9VP, but it's soooo straightforward:
KCx2, Monumentx3 - KC->KC, play each Monument 3 times. Continue ad infinitum (not limited on piles!).
The real trick is getting your deck down to that point since you don't have a Bishop to help. You can use your opponent's Bishop or perhaps Islands to get you there.
You can replace the third Monument with a Bishop for the same number of points.
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Anybody who considers jomini's latest solution borderline is definitely not going to like this one. :P It scores 76 guaranteed tokens each go and has an empty draw deck & discard during your opponent's turn (so it's a little more legitimate than KC-Scheme, in my opinion).
The cards you possess: 5 King's Courts, 4 Goons, 3 Havens, 2 Wharfs, an Outpost, and a Watchtower. In total, 16 cards.
Before your normal turn begins, you have KC, KC, Outpost, Wharf, Wharf in hand. 2 King's Courts, all Goons, and the Watchtower are returned to your hand by Havens; that's 7 cards, which was accomplished by the King's Court and 3 Havens that are in play.
So, your hand is 4 KC, Outpost, 2 Wharf, 4 Goons, Watchtower for this turn. KC-KC-KC-KC, which allows you to apply KC to 7 cards: everything except the Watchtower. From the KC'd Goons and Wharves, you have 19 buys, so buy copper (trashing with Watchtower) to score 76 points.
At the end of your normal turn, everything gets cleaned up except one KC, the Outpost, and both Wharves, which all stay out.
In your Outpost turn, the KC'd Wharves kick in and draw 12 cards in addition to the 3 you drew at the end of your normal turn's clean-up phase. Actually, you only end up with 12 cards in hand (because that's all you have). Play KC-KC-Haven-Haven-Haven to set aside 2 KC, all Goons, and the Watchtower (so your hand is now empty).
At the end of your Outpost turn's clean-up phase, one KC and 3 Havens remain in play, and you draw KC, KC, Outpost, Wharf, Wharf.
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Haha, oh wow that's glorious, abusing haven. I wonder if it's possible to take it any further than that, or have we reached the theoretical maximum ridiculousness with haven?
The biggest thing is having your deck be 0 cards at the end of your turn every turn in order for it to count.
Makes me wonder if there's a way to abuse Native village and Outpost similar to how you're working haven.
As an aside, what's the best you can do to make a card that's completely immune to all attacks as well? I suppose you could just include moat or lighthouse, probably lighthouse, but what if those aren't available?
Time to put my own thinkin cap on.
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Haha, oh wow that's glorious, abusing haven. I wonder if it's possible to take it any further than that, or have we reached the theoretical maximum ridiculousness with haven?
As far as my post goes, there's definitely extra that can be squeezed out. For example, there's room to throw in an extra King's Court and Goons without needing to change anything else. That would increase the point count to 110.
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In order to have a 5 card deck, native village is mostly a no go for insane level abuse. Sans outpost, you cannot play the cards on the mat. You might be able to come up with some sort of pull cards off, generate new copies, trash the originals, and put the new copies back on the mat, but the 5 card deck cap is a bit harsh.
Sans outpost, this means you need a NV at the beginning of the turn, and a NV at the end of the turn. That leaves 3 cards to trash and regenerate multiple cards - not so easy to do.
Using outpost/nv is cute, but still has practical problems. This means that outpost takes up one of your slots (leaving you with 4 slots). Unlike haven, KC/NV takes up two more slots. This means that during your outpost turn (having played your payout cards during the first turn), you have only 4 cards you can play. KC/KC/NV/NV means that at most you could have six cards that are actually played on the mat.
I think we can modify BB's solution into something like the following nightmare for hard mode (I don't have time to work through this for optimal setup/play at the moment):
Starting hand:KC/goons/watchtower/outpost/scyring pool
On the mat: KC/KC/KC/KC -> 9 havens -> holding the following 27 cards: 5 KC, 9 goons, 9 throne rooms, scheme, and 4 markets?
So my playable hand begins with playing out the 9 TR in an efficient tree. Off of that I will play a KC tree. At the end of the KC tree I will play a whatever combination lets me get down all 10 goons with as many buys as possible with the following two caveats:
1. I have to have an unplayed watchtower to trash my buys of mass copper.
2. I have to have a spare action to play scheme -> scrying pool at turn end (these can be played off KC or TR branches if needed).
I then buy to trash how ever many dozen coppers I'm allowed and play the outpost. I place Scrying pool on deck top and draw it to start the outpost turn.
My outpost turn consists of playing the pool to draw everything -> KCx4 -> haven to regenerate the deck and then KC/goons for 4 more points with watchtower trashing.
I'm not sure if it is optimal to use all the TR's, but at least a few should be good to KC more buys, so the optimal mix of cards may have more markets and fewer TR. Likewise, the optimal mix of KC and TR used during the normal turn vs used to bulk up card storage (and hence buys)I leave as an exercise for the reader.
I haven't calced it, but I believe this should break 150 VP chips/turn without too much trouble.