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1
Puzzles and Challenges / Re: Easy Puzzles
« on: July 12, 2019, 09:03:48 pm »
Not sure if either of these actually work, but
  • Innovation + Experiment
  • Changeling (gaining a different card)

2
You could have Mandarin as its own supply pile, Haggler in play, Trader in hand, and a $6+ cost in the Black Market pile. Then you can buy the 6 cost, revealing trader to not actually take it from the deck, but still gain a Mandarin from the supply.

3
Variants and Fan Cards / Re: Really bad card ideas
« on: June 06, 2019, 02:59:30 pm »
Aspiring Scholar
Action - $5
+7 Cards.
Discard your hand.

4
Puzzles and Challenges / Re: Asymptotic analysis of kingdoms puzzle
« on: April 20, 2019, 10:21:15 pm »
Inheritance(...): Looks good, yay!
<= 3$: Looks good, yay!

5
Puzzles and Challenges / Re: Asymptotic analysis of kingdoms puzzle
« on: April 19, 2019, 02:15:09 pm »
Hint for Inheritance + ...
Silver seems to be better than Gold... maybe Copper is somehow better than Silver? ???

Treasures looks like roughly the right engine, but with several things that can be improved upon.
  • Idols - You can get a lot more bang for your buck from your idols.
  • Buys - You can improve this too. It's important here that Crowns are "free" to buy.
  • Boons - Two other Boons are useful, and they will help reduce the need to trash.

6
Puzzles and Challenges / Re: Asymptotic analysis of kingdoms puzzle
« on: April 18, 2019, 09:02:54 pm »
Oh yeah, I've taken a look at the answers: _DDKI_G_C_URT oh god whyyyyyy
https://giphy.com/gifs/2zelCiUo5KJyN8MgMr/html5

Inheritance
I wasn't able to get wolfram alpha to take in all those inequalities, so I couldn't check super well. The total actions constraint seems like it's supposed to be W+T >= V+2V+4V+F instead of the current right hand side, but that shouldn't change the answer.
I think you need an additional draw constraint as well-- maybe you can draw WVs, Estates, and SMs for free with a small number of Scrying Pools, but the non-actions have to be drawn with your WVs/Estates. Need something like W + T >= U+V+2V+8V. At a glance, it looks like the values you have wouldn't meet this bound.

As for the actual strategy, it's getting closer but you can do something a bit better.

7
Puzzles and Challenges / Re: Asymptotic analysis of kingdoms puzzle
« on: April 18, 2019, 01:46:33 am »
Reactions:
That's right too, nice!

You have a lot of the kingdoms solved now (9/13 I think) and I would strongly encourage looking at the letters you have and guessing at what it says. Hint hint. :P

8
Puzzles and Challenges / Re: Asymptotic analysis of kingdoms puzzle
« on: April 18, 2019, 01:03:54 am »
Debt:
Yup, nice!

Inheritance, ...
Procession is indeed useless here, sorry about that. You can do better than that solution. Hint: / You don't need to buy any stonemasons or worker's villages. (Well, besides at the beginning.

9
Puzzles and Challenges / Re: Asymptotic analysis of kingdoms puzzle
« on: April 12, 2019, 12:37:16 pm »
Debt:
ghostofmars is almost right, but keep in mind that that deck can only support 3R^2/2 * T^2 estates.

10
Puzzles and Challenges / Re: Asymptotic analysis of kingdoms puzzle
« on: April 11, 2019, 04:10:11 pm »
6+:
As you can probably tell from the possible answers, this is valid but you can do better. KC Hunting Grounds Expand is good, but you need to do give it some help to make it even better.

Artisan, ...
Yup, that's right. I was thinking of posting a challenge for the best polynomial possible, either with a valid kingdom (10 cards + some number of sideways cards). It turns out you can get a lot higher with Renaissance.

You have probably solved enough to start guessing at the message. :)

11
Puzzles and Challenges / Re: Asymptotic analysis of kingdoms puzzle
« on: April 10, 2019, 02:01:09 pm »
I verified your solution and checked whether adding Weddings would bring a benefit
Normal turn: A * TF + Ex, B * TF + Qu, C * TF + Cu
Mission turn: X * TF + Ex, Y * TF + Qu, Z * TF + We
A ratio R of my total cards are Gold the rest are Curses

I find the optimal values A = 0.526, B = 0.044, C = 0.009, X = 0.551, Y = 0.048, Z = 0.002, R = 0.912. This leads to a growth rate of (1 + C/(1-R))^T = 1.103^T. Including the cubic cashing-out, we get 1.341^T.

Here are the equations
Code: [Select]
5 A + 2 B + 2 C = 3 R
2 B = (1 - R)
2 A = 1 + B + C
5 X + 2 Y + 9 Z = 3 (R + B)
2 Y = (1 - R) + C
2 X = 1 + B + C + Y + Z
(1 - R) * (B + Y + Z) = R * C

Yup, that's right! I would put a smiley here but it shows through the spoiler tags, heh.

12
Puzzles and Challenges / Re: Asymptotic analysis of kingdoms puzzle
« on: April 08, 2019, 04:45:36 pm »
Thanks for trying the puzzle out.  :)
Comments per kingdom, if you're interested.

Re: Durations
Correct! I like how nice this one turned out. / anti-spoiler dummy text

Re: Debt
These are the right cards to use, but you can do better with a different greening strategy. More spoilers : / Even though getting estates on turns you buy triumph is twice as efficient, you don't actually want to maximize the proportion of turns you do this, just due to the way that it increases the numerator by a linear amount but the denominator by a quadratic amount.

Re: Events
If you have Y=X/11, then shouldn't the base growth rate be 12/11 -> 1.0909^T, not 1.111? Anyway, you can improve this strategy. More spoilers : / You can improve both the building up and the cashing out. I hadn't considered not always drawing all your gold (besides possible endgame shenanigans) but I don't think it is helpful. There's also another card that helps you build up.

13
Variants and Fan Cards / Re: Really bad card ideas
« on: April 03, 2019, 06:05:07 pm »
I feel like this has to have been done before, but:

Counterspell
Reaction - $5

When another player plays a card, you may reveal and discard this card from your hand. If you do, they get no effect rather than following the card's instructions.

No effect rather than following the card's instructions? Isn't that just the normal effect from playing a card? Did you mean to say "rather" there?

I have seen serious suggestions for cards that react to "when an opponent plays a card"; at least one that would cancel the card.
The intention was that this completely negates the effect of the card (much like Enchantress negates the effects of the card, replacing it with +1 Card/+1 Action). It's different from Enchantress since you get to pick which card to negate, and has the mess of only affecting one other player in a 3+ player game.

I guess to make it not a completely awful idea, it needs to give the person who played it +1 Action, otherwise you could just use it on the first action played.

14
Variants and Fan Cards / Re: Really bad card ideas
« on: April 03, 2019, 05:51:19 pm »
I feel like this has to have been done before, but:

Counterspell
Reaction - $5

When another player plays a card, you may reveal and discard this card from your hand. If you do, they get no effect rather than following the card's instructions.

15
Puzzles and Challenges / Re: Asymptotic analysis of kingdoms puzzle
« on: April 03, 2019, 02:44:52 pm »
Nice! Thanks for trying the puzzle out. :)

I will say that the strategy is allowed to depend on T, meaning that you know how many turns you have, so you can choose to delay buying points until the very end if that would benefit your strategy. Doing that should improve some of your kingdom's scores.

If you'd like to know, the ones you have correct are: Potion, Reserve, Hamlet/Watchtower, Travellers .
Also, I would say that you have the right strategy for Durations but seem to have analyzed it incorrectly?

16
Puzzles and Challenges / Re: Asymptotic analysis of kingdoms puzzle
« on: March 29, 2019, 12:50:21 am »
What are the challenge's rules concerning possession? Can we assume a cooperative opponent?
The opponent can't buy or play cards, even when possessed, so you can't do much with them. You can assume that they are cooperative and you can assume perfect luck for how they draw cards though, like if you needed to Ambassador them cards so that you could Jester them, or something.

The potion kingdom would be more interesting if you could possess them and actually buy cards; I regret not having it that way.

17
Puzzles and Challenges / Re: Asymptotic analysis of kingdoms puzzle
« on: March 28, 2019, 05:00:28 pm »
Fair enough, I edited the first post to make it more clear.
I also added some practice kingdoms, a list of the ones in the puzzle I think are the most approachable, and some fun bonus kingdoms. :)

Fun kingdoms to try out:
  • Bishop, Cursed Village, Seaway
  • Bridge, Hunting Grounds, Donate, Canal, Inheritance

18
Puzzles and Challenges / Re: Asymptotic analysis of kingdoms puzzle
« on: March 25, 2019, 03:31:20 pm »
This is a mystery hunt-style puzzle, so not all the instructions are not explicitly given. If you want to just think about the dominion and math, then (minor spoiler)  For each kingdom, find its optimal score under the rules given. Each kingdom's answer is in the list of answers to the right. If you find all the answers in order, the corresponding letters will spell out a message. The eventual goal is to have a final answer (english word or phrase) for the entire puzzle.

19
Puzzles and Challenges / Asymptotic analysis of kingdoms puzzle
« on: March 23, 2019, 08:44:26 pm »
Hi f.ds, sharing a puzzle that a friend and I made (well, mostly my friend) for another friend of ours. You probably won't enjoy the puzzle unless you like both dominion and math, but hopefully everyone here likes at least one of the two. :)

The rules of our puzzle's asymptotic dominion are similar to those in http://forum.dominionstrategy.com/index.php?topic=17378.0, but the main difference is that in our puzzle, you can assume you will have perfect luck, as it asks about the best possible score. Also, this puzzle was made before Renaissance came out, so no Renaissance cards are in any of the kingdoms.

The puzzle is at https://walledvillage.github.io/puzzle.pdf. Each of the 13 bullet points on the left is a separate kingdom to consider. Most of these kingdoms might not have exactly 10 kingdom cards (in fact, one has none), but should be considered as if that were the kingdom anyway. Each of them matches with an answer on the right, and taking the corresponding letters will spell out a message. The eventual goal is to have a final answer, which is a word or phrase.

Feel free to ask for any hints or clarifications here, and I hope you enjoy!

Note: Based off a sample size of 1, this puzzle may take quite a while to complete.
---
Edit: Adding some extra info here.

Some basic kingdoms and their answers, to get the idea across.
  • Market, Chapel. Answer: 1.2^T
  • Market. Answer: 1.2^(T/2) = 1.095^T

A list of sets that I think are easier, since completing the whole puzzle is probably unreasonable.
In approximately increasing difficulty:
  • Potion cost
  • Durations
  • Reactions
  • Artisan, ...
  • Hamlet, Watchtower
  • Debt cost

A kingdom I would say is pretty fun and decently approachable is Artisan, ....
Here's some other kingdoms not included in the puzzle that are fun, and not toooo crazy to analyze:
  • Bishop, Cursed Village, Seaway
  • Bridge, Hunting Grounds, Donate, Canal, Inheritance

20
Puzzles and Challenges / Re: Best Asymptotic Point Scoring
« on: November 21, 2018, 04:03:59 pm »
Really great work. As far as I understand the growth analysis, it seems fine, but I cannot follow two steps, which are more related to Dominion rules.
If we started with 2A BMs and 2A Hs, doing one step of this loop gains us (2A)^2=4A^2 cards. We can choose this to be 2A^2 BMs and 2A^2 Hs. We can repeat the loop by playing all the Hs then all the BMs again. Even ignoring the Hagglers played in previous steps, one step of this loop changes A to A^2 (i.e. A -> A^2 -> A^4 -> A^8 -> ... ), so with B SPs, we can end up with 4*A^(2^B) actions in our hand costing less than 6.

It seems to me that this needs additional support, because every pair of BM and Haggler gives you 4 coins, but 6 are necessary for Fairgrounds. Asymptotically, this would not change a lot I guess, since two-thirds of the cards could still be gained.
Good catch. It doesn't really matter though, we can do 1 BM per 2 Hagglers for example. Now if we start with 1/3 A BM and 2/3 A H, we get (2/3)*(1/3) A^2 cards to work with. My original solution starts with A and gets 1/4 A^2; this starts with A and gets 2/9 A^2. As you suggested, it could also work to do half BM and half H, but only buy with 2/3 of the BM that we gain, which would end up as 1/6 A^2. Any constant factor for the A^2 is fine and will lead to a A^(c*2^A) after all A steps of the inner loop, for some constant c. We only need the inner loop to be 2^A, so there is a lot of room to work with.

In fact, this loop would work even if we couldn't gain Hagglers midturn-- with something like 12 Hagglers, we can gain BMs only, even with only being able to use one in three of them to buy something. That would reduce one outer loop from X -> 2 ↑↑ 2X to x -> O(1) ↑↑ X, but that would be fine.
 
Now, we can repeat this second loop for every Golem we started with. Since we started with X BM, H, SP, and got to repeat this loop X times, we went from A to 2^(2^A) X times, starting at X, which is (2↑↑(2X))^X > 8 * 2↑↑(2X). Letting Y = 2↑↑(2X), we can let this be 4Y SPs in hand and 4Y changelings on the top of our deck, and play all the SPs, drawing the 4Y changelings. Then, in our Night phase, we use the Changelings to gain Y each of BM, H, SP, G, topdecking them (putting the scrying pools on top for next turn).

Where do the changelings come from? I could imagine that these could be exchanged tradered silvers, but you mention drawing some silvers and trashing them with Count.
In the last step of the outer loop, we don't have to gain BMs and Hs--we can gain any non-victory card costing less than 6 that we want. Just pick some of those to be Changelings.

21
Puzzles and Challenges / Re: Best Asymptotic Point Scoring
« on: November 21, 2018, 12:00:27 pm »
Wow!
I need to carefully go through this again, but I guess, there should also be a gardens to gain some VP. Or am I missing something?

Yes, normally you can convert your engine to points on the last 1 or 2 turns easily but since all the gains are from Haggler and Changeling, this isn't actually possible as is. We can just add one Gardens or Obelisk on any pile.

If we were to add Capitalism and Mandarin to the kingdom, we could gain a Mandarin here and then not only get the 2A^2 BM plays, but instead could play all BMs we had already again. I am too lazy right now to work out whether this significantly increases the overall growth rate.
I don't think it really helps, since you can do this only once per loop, and on every step of a loop you square the number of BMs + Hs you have, which is a lot more than the sum all of your previous steps. Example: If you've done X, X^2, X^4, X^8, your next step would be X^16 things, and the Mandarin trick would let you get X+X^2+X^4+X^8 extra BM plays, which is a lot less than X^16.

I thought about my analysis some more and realized that applying X -> 2↑↑(2X) n times doesn't give 2↑↑↑2n. You would need to apply X -> 2↑↑(2↑↑X) n times to do that. The actual value is hard to reason about, but it is definitely greater than 2↑↑↑n (and is also greater than c↑↑↑n for any constant c > 0). However, I don't think it's bigger than any 2↑↑↑(1+c)n, for any c is a constant > 0.

I will edit my post with these fixes.

With the help of paulfc, I think we were able to get to 3 up arrows!

That was some truly inspired sh*t!  Congrats.
Thanks! Actually, it was your latest post that gave me the idea.  :D

22
Puzzles and Challenges / Re: Best Asymptotic Point Scoring
« on: November 21, 2018, 01:41:28 am »
With the help of paulfc, I think we were able to get to 3 up arrows!

EDIT: The original analysis was slightly wrong (2 ↑↑↑ n instead of 2 ↑↑↑ 2n). I also forgot to add a way to score.

The core kingdom cards we need are:
Black Market (BM), Haggler (H), Scrying Pool (SP), Squire (S), Golem (G), Count.
We also need a single Trader from the Black Market deck, and Trader cannot be a Kingdom pile.
We also need enough +Actions (Page ->Champion is fine). We also need (just one) Royal Seal for topdecking.
We also need a source of buying Golems at the end of our turn. I'll use Changeling for that.
Finally, we need some source of points, since Haggler and Changeling are not very good at getting points. Obelisk on any pile, say BM, works.

The Black Market deck will contain only Fairgrounds, Trader, Royal Seal. To set up our deck, we buy the Trader and Royal Seal, and also obtain and play Champion.

At the beginning of our turn, our deck will have 1 Trader, 1 Royal Seal, X BMs, X Hs, X SPs, and X Gs. (We don't need exactly the same number of each, just within some constant factor is good enough.) We may also have some number of Counts but don't particularly care how many. We can guarantee that 2 Scrying Pools will be in our starting hand, letting us draw all the actions in our deck.

The sequence of our turn will basically be a nested loop.

The basic step of first loop is to play all the Hagglers in our hand, then all the Black Markets in our hand. (The first time we do this loop, we also play the Royal Seal with Black Market.) With each Black Market, we buy the Fairgrounds, but use Trader to gain a Silver instead of instead of the Fairgrounds. If we have A Hagglers out, then each time we buy Fairgrounds, we gain A cards costing less than 6. We can topdeck all actions gained this way, but none of the Silvers. After playing all the Hagglers and Black Markets in our hand, we play a Scrying Pool to get all the actions gained this way (and a single Silver).

If we started with 2A BMs and 2A Hs, doing one step of this loop gains us (2A)^2=4A^2 cards. We can choose this to be 2A^2 BMs and 2A^2 Hs. We can repeat the loop by playing all the Hs then all the BMs again. Even ignoring the Hagglers played in previous steps, one step of this loop changes A to A^2 (i.e. A -> A^2 -> A^4 -> A^8 -> ... ), so with B SPs, we can end up with 4*A^(2^B) actions in our hand costing less than 6.

We will repeat this loop for every SP in our hand, except leaving one extra in our hand. On the last step of the loop, we will gain some constant fraction each of BM, H, Squire. We will also divert some of these gains to be Count, and we need one Count per Golem in our hand, plus 3 extra Counts. This number will always be way way smaller than the number of cards we can gain (at least X^(2^X) cards we can gain, and at most X Golems in our hand), so we can basically ignore the number in our analysis.

With all the actions in our hand, we can play the BMs and Hs to gain some amount of BMs and Hs on the top of our deck. Then, we play Count for topdeck and money to topdeck all of our Golems except for one. With our remaining 4 Counts, we use 3 of them for topdeck and money, topdecking Trader, SP, and our last Count. Now our hand is a bunch of Squires and one Golem. The top of our deck is Count, SP, Trader, the rest of our Golems, and a bunch of BMs and Hs, then some Silvers. We can now use Golem, which will hit Count and SP. We pick Count then SP as our order. With Count, we pick topdeck and trash, topdecking a Squire, then trashing our hand full of Squires. With each Squire, we gain a SP and topdeck it. Then, our Golem plays the SP, letting us draw all the actions we have left (a bunch of SPs, the Trader, the Golems, a bunch of BMs and Hs (and 1 Squire that we don't really care about)).

Doing this is a single step of our second loop. We perform the first loop for all the SPs in our hand, then expend one Golem to gain a bunch more SPs. If we end the steps of the first loop with C BMs+Hs and D Ss, we will have O(C^2) BMs, O(C^2) Hs, and O(D) SPs after all these operations. If we started with A each of BM, H, SP, we will have 4*A^(2^A) actions to allocate between C and D. We can allocate a square root amount to C and the rest to D and end up with D = O(A^(2^A)), which is at least 2^(2^A). So, one step of this loop turns A into 2^(2^A).

Now, we can repeat this second loop for every Golem we started with. Since we started with X BM, H, SP, and got to repeat this loop X times, we went from A to 2^(2^A) X times, starting at X, which is (2↑↑(2X))^X > 8 * 2↑↑(2X). Letting Y = 2↑↑(2X), we can let this be 4Y SPs in hand and 4Y changelings on the top of our deck, and play all the SPs, drawing the 4Y changelings. Then, in our Night phase, we use the Changelings to gain Y each of BM, H, SP, G, topdecking them (putting the scrying pools on top for next turn).

Since every turn changes X from 2↑↑(2X), this kingdom has growth rate f(n) > 2↑↑↑(2n-c) for some constant c f(n) > 2↑↑↑n. The actual amount is bigger than c↑↑↑n for any constant c, but I'm not sure how to express it. At the very least, we can correctly remove the -c from n and still have a correct lower bound, so there's that.

This kingdom is not an infinite kingdom because:
  • Golems cannot be gained midturn, and there is only one Trader.
  • The only draw that can be gained midturn is Scrying Pool, and no cards are gained to hand.
  • To gain Scrying Pools midturn that can still be played on the same turn, a Golem or Trader must be used.

In summary:
  • Kingdom: Black Market, Haggler, Scrying Pool, Squire, Golem, Count, Changeling, Page. Sideways cards: Obelisk on BM.
    (BM deck: Fairgrounds, Trader, Royal Seal).
  • Growth rate: f(n) = 2↑↑↑n
  • No Renaissance cards needed!

23
Puzzles and Challenges / Re: Best Asymptotic Point Scoring
« on: November 20, 2018, 09:17:50 pm »
Glad that this thread has continued to generate interest :)

A question for tim17.  I'd like a ruling on how split piles work in an infinite kingdom.  Would it be possible to gain a Fortune or would they be buried under an infinity of Gladiators?  Or is it 5 Gladiators, 5 Fortunes, 5 Gladiators, 5 Fortunes, ....   While we're on the subject, how about Knights and Ruins and other mixed supply piles?


I forget if I had thought about this when I had initially posed the question. I don't have strong feelings one way or another; I tend to prefer to let people assume whatever they want and then see what they can come up with. Accordingly, let's say that any pile with more than one card name has infinitely many of each, and you get to choose the order of the pile during setup. If for some reason this causes problems, we can try some other interpretation, but let's go with this for now.
I propose the following interpretation: instead of infinite piles, have the following rule: whenever a pile would be emptied, replenish the pile with cards in the original state. (And if you were somehow gaining 1000 cards at the same time, you replenish the pile multiple times in between gains as necessary.) This lets us include knights and split piles without too much confusion. For example, the Gladiator/Fortune pile would effectively keep being 5 Gladiators then 5 Fortunes alternating.

24
Puzzles and Challenges / Re: Best Asymptotic Point Scoring
« on: November 14, 2018, 08:50:19 pm »
"Each turn d' = (31/88)d, so we end up with around 31/88 ↑↑ n points."  I'm claiming d' = 256^d.
Your post looks right to me (though I didn't check super carefully), but since 4↑↑(n+2) > 256↑↑(n), it's only 2 turns better than the 4↑↑n kingdom (supposing the setup took the same amount of time) despite seeming way better.
Ha.  My setup would definitely take a lot longer.  I wanted to show the usefulness of Hagglers though.  When used to best advantage, they can double your tetration base.  Unfortunately, City Quarters cannot be haggled.  I am also thinking that using Mines to gain actions directly into your hand may be very useful.  But I guess it's three arrows now or bust, eh?
If you put City Quarter in the kingdom and Gladiator/Fortune, you could Haggle for City Quarter that way, but since you can't gain them mid-turn it doesn't seem like that would really help.

One other little thing, you should be able to play your Hagglers when you Black Market, so you don't need quite as many Fishing Villages.

25
Puzzles and Challenges / Re: Best Asymptotic Point Scoring
« on: November 14, 2018, 08:10:05 pm »
"Each turn d' = (31/88)d, so we end up with around 31/88 ↑↑ n points."  I'm claiming d' = 256^d.
Your post looks right to me (though I didn't check super carefully), but since 4↑↑(n+2) > 256↑↑(n), it's only 2 turns better than the 4↑↑n kingdom (supposing the setup took the same amount of time) despite seeming way better.

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