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In the game of splendor, what is the fewest number of turns you could take to get 15 points?

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  • 5
    I disagree. You can definitely figure out some kind of "perfect setup" and find the minimum number of turns based on that. It will never happen in practice (especially since you would have to assume either a solo game or a really stupid opponent), but it's a valid theoretical question.
    – ConMan
    Jan 31, 2019 at 5:23
  • 2
    BGG has a complete decklist if anyone wants to use it. Jan 31, 2019 at 6:14
  • 2
    On BGG there's also a google sheet with the same data by luless (make sure to scroll down, the one on the top of the page isn't correct) and a group photo of all cards in case anyone wants to double-check.
    – monk-time
    Feb 5, 2019 at 17:32

6 Answers 6

6

I have found 15 points in 15 turns, which works even for 3 player rules:

  1. Take: white, blue, green
  2. Take: white, blue, green
  3. Take: white, blue, red
  4. Claim: white card (3 blue)
  5. Take: green, red, black
  6. Claim: white card (2 red, 1 black)
  7. Take: white, blue, green
  8. Claim: white card (4 green) +1 pt
  9. Claim: white card (6 white) +3 pt
  10. Take: 2 white
  11. Claim: blue card (7 white) +4 pt
  12. Take: white, blue, black
  13. Take: 2 white
  14. Claim: blue card (7 white, 3 blue) +5 pt
  15. Claim: green card (4 white, 2 blue, 1 black) +2 pt

Although, I don't have proof that 14 turns game is not possible.

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  • Given that it's been a year and a half to get a one turn improvement, I don't think you should feel bad that you don't have a 14 turn proof....
    – John
    Aug 16, 2021 at 23:12
  • 14 turns looks really close to possible but I haven't been able to quite get there. This game would be a turn 14 win in 4p except it holds 11 gems one turn.
    – xnor
    Nov 2, 2021 at 7:35
  • Great work on shaving off 1 turn! I've been trying to bruteforce all shorter solutions to find the shortest path but can't quite optimize the code enough.
    – monk-time
    Jan 22 at 9:16
5

16 points in 16 turns:

  1. Take 1 white, 1 green, 1 black
  2. Take 1 white, 1 green, 1 black
  3. Take 1 blue, 1 red, 1 black
  4. Claim Blue 0 pt card (1 white, 2 black)
  5. Claim Blue 0 pt card (1 white, 1 green, 1 red, 1 black); stash: 1 blue, 1 green
  6. Take 2 blue
  7. Take 2 blue
  8. Claim Blue 2 pt card (5 blue); stash: 2 blue, 1 green
  9. Take 1 white, 1 blue, 1 green
  10. Claim Blue 3 pt card (6 blue); stash: 1 white, 2 green
  11. Take 2 blue
  12. Take 2 blue
  13. Claim Green 4 pt card (7 blue); stash: 1 white, 1 blue, 2 green
  14. Take 2 blue
  15. Claim Green 5 pt card (7 blue, 3 green); stash: 1 white
  16. Claim Red 2 pt card (1 white, 4 blue, 2 green); stash: empty

One thing I've noticed is that there's an interesting disbalance among the 2 pt cards for 5+3 gems: only 2/5 cards (blue and green) give the bonus of the same color as one of the requirements. Feels like something that might yield an even better solution.

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  • @John: Yes, I think I have the same version as the ones I have linked in the comment to the main post: i.imgur.com/12DIzjz.jpg
    – monk-time
    Feb 5, 2019 at 20:56
  • I am amazed that there is an imbalance on the 2 pt 5+3 cards. That is very unexpected. Nice catch. Can you exploit it? Your above solution is very clever.
    – John
    Feb 6, 2019 at 3:32
  • I think you have a calc error in there ... I step through it a few times .. but keep getting different #'s at your step #8: #1 (Stash: W,G,B) #2 (Stash: WW,GG,BB) #3 (Stash: WW,GG,BBB,U,R) #4 (Stash: W,GG,B,U,R) (Perm: U) #5 (Stash: G,U) (Perm: U) #6 (Stash: G,UUU) (Perm: U) #7 (Stash: G,UUUUU) (Perm: U) #8 (Stash: G,U) (Perm: UU) (but you claim you have G,UU in stash .. which makes step #10 impossible) :( Did I make a mistake? or are you off there?
    – Ditto
    Feb 6, 2019 at 15:16
  • @Ditto: You missed that I have not one but two blue cards after step #5, which allows me to save 2 blue chips out of 5 that I have on step #8. That leaves me with 2 blue chips.
    – monk-time
    Feb 6, 2019 at 18:31
  • @monk-time: you're right! Gotcha! :)
    – Ditto
    Feb 6, 2019 at 21:06
5

EDIT: fixed violation of limit on taking 2 of same gems (Forbidden when <4 gems remaining).

First of all, @Arcanist Lupus has provided good answer with simple yet efficient approach and also provided link to cards table in the comments, so I want to thank them. But looking at the table I noticed that 4 pts. cards have better conversion rates than 5 pts. ones, and a lot of gems are lost when going straight for 5 pts. cards.

So, assuming all required cards are available, no nobles and no interference from other players, here is my solution:

Cards I'm going for: White 4 for 7 blacks; Blue 4 for 7 whites; Green 4 for 7 blues; Green 3 for 6 greens.

Steps:


  1. +2 black
  2. +2 black
  3. +1 black, 1 green, 1 {any}
  4. +1 black, 2 {any}
  5. +1 black, 1 white, 1 green. Return 3 {any}
  6. Claim White 4 card for 7 blacks. (1 white and 2 green remain in the stash)
  7. +2 white
  8. +2 white
  9. +1 white, 1 blue, 1 green
  10. Claim Blue 4 card for 6 whites + card. (1 blue and 3 green remain in the stash)
  11. +2 blue
  12. +2 blue
  13. +1 blue, 1 green, 1 {any}. Return 1 {any}
  14. Claim Green 4 card for 6 blues + card. (4 green remain in the stash)
  15. +1 green, 2 {any}
  16. Claim Green 3 card for 5 greens + card

Done. 15 points in 16 turns.

  • steps 3 and 4 are just about getting 2 blacks, since it's forbidden to get 2 at once at that point. Alternatively you can take golden tokens and secure couple of cards for yourself, in that case in step 15 just take 2 greens (you will have 1 less, so it would be legal move).
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    This is good, but you've forgotten that you can only take 2 gems of the same color if there are 4 or more left, so taking 2 of the same 3 times in a row is impossible Feb 1, 2019 at 18:27
  • @ArcanistLupus Gotcha, forgot that rule. Only played the game once. Looking at the steps, I think only step 3 violates the rule, so it can be split into 2 "take 1 of each" to get same 2 blacks, which bumps total steps to 16. I'll edit after more thorough check.
    – Deo
    Feb 2, 2019 at 18:23
1

Each color has a single card worth 5 points, which can be taken by 7 gems of one color and 3 of another. Conveniently, the maximum number of gems you can hold at once is 10, and in a 4 player game there are 7 gems of each color available, which means that you can go after these cards immediately.

For example:

  1. 2 red
  2. 2 red
  3. 1 red, 1 black, 1 {any}
  4. 1 red, 1 black, 1 {any}
  5. 1 red, 1 black, 1 {any}, return 3 {any}
  6. Claim the 5 pt black card
  7. 2 black
  8. 2 black
  9. 1 black, 1 white, 1 {any}
  10. 1 black, 1 white, 1 {any}
  11. 1 white, 1 blue, 1 {any}, return 3 {any}
  12. Claim the 5 pt white card (keep the blue)
  13. 2 blue
  14. 2 blue
  15. 1 blue, 1 white, 1 {any}
  16. 1 blue, 1 white, 1 {any}
  17. Claim the 5 pt blue card.

Congratulations! You've won!

It's possible that there's a faster Nobles strategy, but I'm not sure. 3 Nobles requires at least 12 turns of taking cards, which means you have a total of 4 spare turns for gem collecting (5 to tie my score, 4 to win outright), and it takes three gem turns to get your first 3 cards. Plus you have to stay in cards that produce only 3 different colors.

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  • Actually, I think you can shave off a turn by saving a gem in the second round, but it's too late tonight to lay it out completely. Jan 31, 2019 at 7:05
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    nice upper bound, I think you can do better than 21
    – Cohensius
    Jan 31, 2019 at 13:04
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    @Cohensius You are indeed correct! Especially since I couldn't math, and it turns out that 6*3 is 18, not 21 Jan 31, 2019 at 15:07
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    @Cophensius 4 Green + 4 Blue + 4 Red card gems, is 12 cards. What am I missing?
    – John
    Feb 4, 2019 at 16:52
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    must been very late when I wrote the comment, indeed 4 Green + 4 Blue + 4 Red card gems, is 12 cards.
    – Cohensius
    Feb 8, 2019 at 20:14
1

The top answer by Marcin_smu does not work in two player games. On move 10, he picks up 2 white chips but has one in his hand, in 2 player this is not possible. In 3 player or 4 player, it would be possible.

I have 15 points in 15 turns for 2 player:

  1. 1 blue + 1 black + 1 red (0 points, 1u1b1r, 0 cards)
  2. 1 blue + 1 black + 1 red (0 points, 2u2b2r, 0 cards)
  3. 1 blue + 1 black + 1 red (0 points, 3u3b3r, 0 cards)
  4. Claim the 3 black for a blue card (0 points, 3u3r, 1 blue)
  5. 1 blue + 1 green + 1 red (0 points, 4u1g4r, 1 blue)
  6. Claim the 4 red for a blue card plus 1 point (1 points, 4u1g, 2 blue)
  7. Claim the 6 blue for a blue card plus 3 points card (4 points, 1g, 3 blue)
  8. 1 blue + 1 blue (4 points, 2u1g, 3 blue)
  9. Claim the 5 blue for a blue card plus 2 points card (6 points, 1g, 4 blue)
  10. 1 blue + 1 blue (6 points, 2u1g, 4 blue)
  11. 1 blue + 1 green + 1 white (6 points, 3u1g1w, 4 blue)
  12. Claim 7 blue for a green card plus 4 points (10 points, 1g1w, 4 blue 1 green)
  13. 1 blue + 1 blue (10 points, 2u1g1w, 4 blue 1 green)
  14. 1 blue + 1 green + 1 white (10 points, 3u2g2w, 4 blue 1 green)
  15. Claim 7 blue 3 green for a green card plus 5 points (15 points, 2w, 4 blue 2 green)

note: a duplicate version of this could be done with green but not red, black or white because the 2 point 5 chip card is not monochrome

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  • As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.
    – Community Bot
    Mar 1 at 15:22
-2

As a partial solution, here are some of the qualities I suspect will need to be part of such a "perfect game":

  1. All Noble tiles claimed (since they represent a large source of points for no additional action taken).

  2. Very few turns where chips are taken, and a turning point somewhere in the middle of the game after which no chips are taken (since every turn taking chips is a turn you aren't taking cards which gain you points).

Additionally, I suspect that the structure of the game will be something like (assuming a 2-player setup, so there are 9 points available in nobles):

  1. Spend a few turns taking chips and acquiring cheap level 1 cards.

  2. Get enough level 1 cards that all future purchases are free, probably earning the first noble near the end of this step.

  3. Get 1 or 2 level 2 cards worth 2-3 points, earning the second noble.

  4. Get a level 3 card and the third noble.

It might be possible to shortcut step 3 if you get a couple of the 1 point level 1 cards in step 2, but I don't think that there's a lot to be gained going that way (the 1 point cards are not particularly cost-effective).

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    Maybe I am missing this but this does not answer the questions of how fast can the game be won but is just a general strategy of how to win faster
    – Joe W
    Jan 31, 2019 at 12:46
  • It's me trying to theory-craft what the fastest game is likely to look like, since I haven't had the chance to actually sit down and turn the above "shape" of a game into actual turns. The fact that it might also be decent advice on winning faster is a neat corollary.
    – ConMan
    Jan 31, 2019 at 22:32

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