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Author Topic: Busy Beaver amount of Coin  (Read 1602 times)

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liopoil

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Re: Busy Beaver amount of Coin
« Reply #25 on: May 12, 2017, 10:50:13 am »
0

Here's a strategy to gain an arbitrarily large amount of actions costing $5 or less. This would make the upper limit depend only on how much you can draw:


Ferry on Hunting Grounds, inherited Catacombs.

Put 2 Quarries in play (Black Market). Play Baron with Watchtower in Hand. You can now do the following:

Trash the gained Estate.
  trigger Catacombs on trash - gain card costing less than Estate (which is now in the trash and thus no longer an action, i.e. it costs $2)
  gain Hunting Grounds (which costs $0)
  trash Hunting Grounds, gaining 3 Estates.
  trash Estate 1 & 2, gaining any actions costing $5 or less
 trash Estate 3, gain Hunting Grounds, repeat.


An issue is that you have to limit actions. A Village added to this would allow you to draw an arbitrarily large amount of cards.
Now this idea seems like it has potential... if we use ~n madmen for our draw then we can have O(n2^n) cards and ~n actions left. Not sure what we would do then though... If there was a way to make this only able to gain unlimited cards costing 4 or less then we could use storyteller.
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faust

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Re: Busy Beaver amount of Coin
« Reply #26 on: May 12, 2017, 01:18:03 pm »
+2

Some ponderings on this.

1) We can include Royal Carriage; it doesn't break anything unless we also include non-terminals costing $5 or less. Royal Carriage means that all our gains can be actions we use. That was my first intiution, but actually:

2) We can substitute Madmen for City Quarter in order to vastly improve our drawing potential:
- play CQ1, drawing ~n Royal carriages.
- play all Royal Carriages
- play CQ2, call n Royal Carriages. This draws O(n2^n) cards.
- repeat. Calling ~n2^n RCs on the second CQ draws O(n2^(n^2n)) overall we should get something like Ω(n*(2↑↑n)^n) cards drawn (this is a lower bound; the computation seems hard).

3) Once the final card is drawn with the final call on the final RC, we can cash in using e.g. Beggar/Miser for that number squared.

This is pretty crazy stuff... can somebody convince me that Royal Carriage doesn't actually allow arbitrary amounts of draw?
« Last Edit: May 12, 2017, 01:22:45 pm by faust »
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Since the number of points is within a constant factor of the number of city quarters, in the long run we can get (4 - ε) ↑↑ n points in n turns for any ε > 0.

faust

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Re: Busy Beaver amount of Coin
« Reply #27 on: May 12, 2017, 01:19:40 pm »
0

(2↑↑n = 2^2^...^2 [n times] for anyone who hasn't seen this yet)
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Since the number of points is within a constant factor of the number of city quarters, in the long run we can get (4 - ε) ↑↑ n points in n turns for any ε > 0.

liopoil

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Re: Busy Beaver amount of Coin
« Reply #28 on: May 12, 2017, 01:32:46 pm »
0

That's pretty insane... I'm concerned that we might be able to get too many actions though. We have to do the baron-watchtower-hunting grounds-catacombs thing for unlimited gains first, right? Yes, it looks like we can only have ~n city quarters total, but I think we might be able to get unlimited +actions and therefore unlimited +cards with hunting grounds. Actually, if we have x royal carriages and hunting grounds then we can call the royal carriages on the last royal carriage for +x actions, then play the hunting grounds and now have 2x royal carriages and hunting grounds. Since we can get unlimited RCs and HGs I think we're stuck.

EDIT: The issue is exactly the "unless we also include non-terminals costing $5 or less". Royal carriage is a non-terminal costing $5 or less.

I totally missed city quarter though when I was looking through cards... I hope we can do something with that.
« Last Edit: May 12, 2017, 01:34:20 pm by liopoil »
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faust

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Re: Busy Beaver amount of Coin
« Reply #29 on: May 12, 2017, 01:36:43 pm »
+1

That's pretty insane... I'm concerned that we might be able to get too many actions though. We have to do the baron-watchtower-hunting grounds-catacombs thing for unlimited gains first, right? Yes, it looks like we can only have ~n city quarters total, but I think we might be able to get unlimited +actions and therefore unlimited +cards with hunting grounds. Actually, if we have x royal carriages and hunting grounds then we can call the royal carriages on the last royal carriage for +x actions, then play the hunting grounds and now have 2x royal carriages and hunting grounds. Since we can get unlimited RCs and HGs I think we're stuck.

EDIT: The issue is exactly the "unless we also include non-terminals costing $5 or less". Royal carriage is a non-terminal costing $5 or less.

I totally missed city quarter though when I was looking through cards... I hope we can do something with that.
You cannot call RC on RC since the played RC is no longer in play when you would call the RC from the tavern mat. That is the beauty of why we can include it in this.
« Last Edit: May 12, 2017, 01:42:04 pm by faust »
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Since the number of points is within a constant factor of the number of city quarters, in the long run we can get (4 - ε) ↑↑ n points in n turns for any ε > 0.

liopoil

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Re: Busy Beaver amount of Coin
« Reply #30 on: May 12, 2017, 01:58:30 pm »
0

Oh wow! Ok, it might work...

The kingdom has to be bounded no matter how you play it. So it still has to work if you played 4 quarries... but since city quarter costs I think we still can't gain it from the watchtower trick?

I think everything still works even if we play hunting grounds in between the city quarters... for example, after CQ2 we have n2^n royal carriages... and a hunting grounds, and n actions. That means we can play a hunting grounds n2^n times to draw 4n2^n cards, and repeat that n times for 4nn2^n cards. Hmm, this might just change it to 8↑↑n or something? But yeah, it does seem that actions are limited. I'm impressed
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faust

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Re: Busy Beaver amount of Coin
« Reply #31 on: May 12, 2017, 02:11:41 pm »
0

We can also take advantage of the oodles of actions were getting:

After fully resolving 1 City Quarter and call RCs, let's draw a hand of RCs and a single Hunting Grounds. We can use that to quadruple our handsize. Then repeat until there is only 1 actions left. This gives us a factor of 4^x for each step, where x is the number of actions (which is double the number of Royal Carriages called on the City Quarter).

This means that directly before playing Qity Quarter 3, we would have O(8^n*n*2^n)=O(n*16^n) cards in hand.

Iterating (and finally squaring, this would yield


PPE: 1
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Since the number of points is within a constant factor of the number of city quarters, in the long run we can get (4 - ε) ↑↑ n points in n turns for any ε > 0.

faust

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Re: Busy Beaver amount of Coin
« Reply #32 on: May 12, 2017, 02:17:47 pm »
0

You can also probably Disciple every City Quarter that you have.

EDIT: Nah, Teacher must not be available.   :(
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Since the number of points is within a constant factor of the number of city quarters, in the long run we can get (4 - ε) ↑↑ n points in n turns for any ε > 0.

liopoil

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Re: Busy Beaver amount of Coin
« Reply #33 on: May 12, 2017, 02:26:23 pm »
+1

I'm thinking that since we can quadruple ~2n times before each city quarter we'll end up with something like 32 ↑↑ n actually. I had the same thought with disciple, but nope.

I'm assuming you meant to use big-O notation there instead of a 0... but that's not really appropriate anymore either. Big-O just swallows up constant factors, and the fact that we have more like n - 5 city quarters rather than n city quarters already decreases the answer by a quintuply-exponential factor.

There's also really no reason to square it at the end. Just draw coppers the last time that we draw.
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faust

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Re: Busy Beaver amount of Coin
« Reply #34 on: May 12, 2017, 02:40:00 pm »
0

That's true. We need better notation!

I'm thinking of how to improve performance further. Patrol can draw 7 cards per play, but then only 3 of those are usable.

Actually, what about Patrol + Crossroads? Patrol draws 3 good cards and 4 green cards, Crossroads draws for all the green cards in our hand.
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Since the number of points is within a constant factor of the number of city quarters, in the long run we can get (4 - ε) ↑↑ n points in n turns for any ε > 0.

faust

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Re: Busy Beaver amount of Coin
« Reply #35 on: May 12, 2017, 02:44:22 pm »
0

Of course kingdom space might run out. We have:

City Quarter, Hunting Grounds, Catacombs, Royal Carriage, Quarry, Black Market, Watchtower.

We can trigger the trashing stuff by buying a Sprawling Castle off of Black Market, so no Baron needed.

Thus we can handily include Patrol+Crossroads+Island.

EDIT: Patrol is pretty unnecessary as soon as we have > 7 green cards in hand.
« Last Edit: May 12, 2017, 02:50:42 pm by faust »
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Since the number of points is within a constant factor of the number of city quarters, in the long run we can get (4 - ε) ↑↑ n points in n turns for any ε > 0.

liopoil

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Re: Busy Beaver amount of Coin
« Reply #36 on: May 12, 2017, 02:52:38 pm »
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That's true. We need better notation!

I'm thinking of how to improve performance further. Patrol can draw 7 cards per play, but then only 3 of those are usable.

Actually, what about Patrol + Crossroads? Patrol draws 3 good cards and 4 green cards, Crossroads draws for all the green cards in our hand.
I was just to post about crossroads... but rather than use patrol, I was thinking that we should use crossroads + island in-between city quarters instead of hunting grounds. That should make a huge difference. At this point city quarter is actually mainly serving to give us a large bounded number of actions, and I think it is actually the only card that can do that for us. I thought about trying to incorporate golem but I think it goes unbounded with royal carriage.

Anyway, what we do now is:

CQ1, draw ~n royal carriages, crossroads, and an island.
Crossroads, call the royal carriages, draw O(2^n) islands and O(2^n) royal carriages.
CQ2, call the royal carriages, draw on the order of (2^n)*2^(2^n) royal carriages and a crossroads
Crossroads, call the royal carriages, draw on the order of 2^2^2^n islands and royal carriages, also a crossroads. That only took one action, and we still have 2^n actions left. So do it again and again, and draw 2 ↑↑ (2^n) cards total by my count.
CQ3, call 2 ↑↑ (2^n) royal carriages, ...

I'm not sure how many up-arrows this is anymore. It's at least 3.

EDIT: In case it isn't clear what's happening with the crossroads, each call of crossroads draws only more islands, except for the last one which draws royal carriages. Therefore it always gets just as many royal carriages as it does islands.
« Last Edit: May 12, 2017, 03:19:02 pm by liopoil »
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faust

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Re: Busy Beaver amount of Coin
« Reply #37 on: May 12, 2017, 02:55:27 pm »
+1

Island is actually unnecessary since we can always gain Estates from trashing as many Hunting Grounds as we like, so we'd still have 2 free spaces in our kingdom. Feels like there should be still something we can use them for.
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Since the number of points is within a constant factor of the number of city quarters, in the long run we can get (4 - ε) ↑↑ n points in n turns for any ε > 0.

liopoil

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Re: Busy Beaver amount of Coin
« Reply #38 on: May 12, 2017, 03:05:52 pm »
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Island is actually unnecessary since we can always gain Estates from trashing as many Hunting Grounds as we like, so we'd still have 2 free spaces in our kingdom. Feels like there should be still something we can use them for.
Ah, I forgot that estate counts as an action card. We have three spaces if we put young witch in the black market deck. I think we're at three up-arrows now, but I'm not totally sure. I don't totally remember how debt costs work... are we allowed to stonemason fortunes into city quarters?

EDIT: I think we are. So if we include both then we are down to one spot left, and I think we can use ~2n city quarters. Oh, but I forgot that it's a split pile.
« Last Edit: May 12, 2017, 03:10:58 pm by liopoil »
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faust

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Re: Busy Beaver amount of Coin
« Reply #39 on: May 12, 2017, 03:07:45 pm »
0

Island is actually unnecessary since we can always gain Estates from trashing as many Hunting Grounds as we like, so we'd still have 2 free spaces in our kingdom. Feels like there should be still something we can use them for.
Ah, I forgot that estate counts as an action card. We have three spaces if we put young witch in the black market deck. I think we're at three up-arrows now, but I'm not totally sure. I don't totally remember how debt costs work... are we allowed to stonemason fortunes into city quarters?
Yes, but it's not very clear how split piles work with infinite pile sizes. Maybe the first Fortune only waits at position omega+1, then we're out of luck.
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Since the number of points is within a constant factor of the number of city quarters, in the long run we can get (4 - ε) ↑↑ n points in n turns for any ε > 0.

liopoil

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Re: Busy Beaver amount of Coin
« Reply #40 on: May 12, 2017, 03:36:49 pm »
+1

A lower bound on the end-result, just to check that it really is three arrows:

Let's assume that every time we play city quarter or crossroads called as many times as we have royal carriages that our hand has just one estate in it to start, even though it will actually be huge. Then if we have x royal carriages, we draw about 2^x royal carriages each time we play crossroads or city quarter. Each time we play a city quarter we get +2x actions (O(x)), and each time we play a crossroads we get -1 action.

CQ1: 1 action, ~n RCs
Crossroads (O(1) times): 2^n RCs, 1 action
CQ2: O(2^n) actions, 2↑(2↑n) RCs
Crossroads (O(2^n) times): 2 ↑↑ (2 ↑ n) RCs
CQ3: 2 ↑↑ (2 ↑ n) actions and RCs
Crossroads (2 ↑↑ (2 ↑ n) times): 2 ↑↑ (2 ↑↑ (2 ↑ n)) RCs

And so on. We end up with ~n layers of "2 ↑↑ (". That's the definition of 2 ↑↑↑ n. In the end just draw coppers and we get 2 ↑↑↑ n coins. I don't think there's much point in trying to be more precise than that. Maybe we should say it's more like 2 ↑↑↑ (n - 5), or if we are allowed to start with fortunes and stonemason them, then maybe 2 ↑↑↑ (2n - 11) or something. If we want to do better than that we need to use the last few kingdom spots to put a whole new layer into the strategy.

From the "How high can you go?" thread:
I'm just watching this thread to see when you guys start using up-arrows.
Also, lol:
A number with more digits than atoms in the universe would require a tetration sort of operation, which I have not seen in dominion.
« Last Edit: May 12, 2017, 03:51:11 pm by liopoil »
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