Dominion Strategy Forum
Dominion => Puzzles and Challenges => Topic started by: WrathOfGlod on November 02, 2011, 05:57:37 am
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Before Hinterlands the longest you could go before a reshuffle was 2 turns and the shortest was also 2 turns. However with Hinterlands the shortest you can go is zero turns (Player 4, players 1-3 open with a noble brigand).
The question is: In a solitaire game how many turns can you go before ending the game without triggering a single reshuffle of the discard pile.
(Assume that the order of cards can be chosen).
(I don't have an exact answer to this puzzle).
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Question: Does triggering of Inn count as a reshuffle or is it an independent shuffle? You're not shuffling your entire discard pile per se...
If Inn doesn't count it's obvious you will be buying all Mandarins and 2 random cards, after which you will have to reshuffle.
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Also no real answer, but
it should be possible to start with some Mandarins, get them back to the draw by bying Inns, use Inns+Mandarins to buy KC+Bridges+X, buy ridiculusly many actions, and get them back to the discard by Inns. One could extend this game probably by Watchtowers. Then the whole game is how to time this to get the maximal number of turns. Using Ambassadors to get Inns/Mandarins back to the supply makes it even longer.
With a little luck you can even go infinite with Mandarin/Inn/Ambassador, have to think about it.
Ok, let's try
a) 5/2-> 10xMandarin, Inn(I,10xM) [5xC] [2xC,3xE,10xM,I]
b) IMM(2xM)->I(I,2xM) [5xC,I,2xM] [2xC,3xE,8xM,I]
c) IMM(2xM)->I(2xI,2xM) [5xC,I,2xM] [2xC,3xE,8xM,2xI]
d) IMM(2xM) -> Amb [5xC,2xI,4xM,Amb] [2xC,3xE,6xM,I]
e) IMM(2xM) -> I(3xI,6xM,Amb) [5xC,I,2xM] [2xC,3xE,8xM,2xI]
f) 4xM+Amb(I) -> nothing [5xC,I,6xM,Amb] [2xC,3xE,4xM,1xI]
g) IMM(2xM) -> I(8xM,2xI,Amb) [5xC,I,2xM] [2xC,3xE,8xM,2xI]
h) (g==e), goto f).
I=Inn, M=Mandarin, Amb=Ambassador, in brackets: Inn(discard), Amb(return). first [] discardpile, second [] drawpile.
And just another exampler how everything get's so much easier when passing to infinities. Would have wanted to figure out an optimum if the loop was not possible
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I hadn't considered the ambassador, with it you can easily get to an infinite length with the inn, an interesting thing about this puzzle is that with inn/mandarin/ambassador combo in play getting an infinite loop is almost trivial (if you start 5-2)
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The best I can come up with for what to do after the first round is to pick up KC's (KC mandarin loses 2 cards per hand,Inn mandarin uses 5, plain Mandarin uses 4). However I don't believe it is possible to get a KC on the first inn.
My best so far for the second round is
4 Mandarin+Inn=Market (Draw Discard Estate Mandarin) Return 2 Mandarin
5 Mandarin=3 Action (Arbitrary) Return Mandarin
Mandarin+2 Copper+2Estate=Inn Return Estate
We now have 2 Inns 9 Mandarins and a market in our deck and a mandarin+7copper+2 estate in our discard.
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Is it possible to then trash/ambassador everything away (use counting house to get the discarded coppers) and then island the last card? Then you wouldn't be reshuffling.