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Author Topic: The exponential card  (Read 5339 times)

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timchen

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The exponential card
« on: April 04, 2012, 05:45:11 pm »
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There is only one card, in the entire Dominion universe, as I know, to be truly exponential. What is the card and what is the exponent under ideal circumstances?
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rrenaud

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Re: The exponential card
« Reply #1 on: April 04, 2012, 06:08:33 pm »
+1

It really depends on what the exponent is a function of.

market + quarry + quarry buys an exponential number of markets with regards to turn (and unbounded supply).

But my best answer is crossroads.  It's exponential in the average case, not just best case
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Insomniac

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Re: The exponential card
« Reply #2 on: April 04, 2012, 06:11:52 pm »
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King's Court? It exponentiates the cards it plays KC=1 Action 3 times  KC+KC=3 actions 3 times  KC+KC+KC=3 actions 3 times + 2 actions 3 times                  Hence KC=3   2KC=9   3KC=15+3(KC was played 3 times)=18     
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O

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Re: The exponential card
« Reply #3 on: April 04, 2012, 06:13:25 pm »
+1

King's Court? It exponentiates the cards it plays KC=1 Action 3 times  KC+KC=3 actions 3 times  KC+KC+KC=3 actions 3 times + 2 actions 3 times                  Hence KC=3   2KC=9   3KC=15+3(KC was played 3 times)=18     

Each additional KC of a KC "adds" (9-1=8) actions, so it's actually linear
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timchen

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Re: The exponential card
« Reply #4 on: April 04, 2012, 06:22:32 pm »
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It really depends on what the exponent is a function of.

market + quarry + quarry buys an exponential number of markets with regards to turn (and unbounded supply).
In the number of copies that cards being played in a single turn I guess.

As for your market example, I think it is very nice. It actually illustrates that the whole game is in some sense exponential when you are not limited by your hand size and +buy; i.e., if the quality and number of cards you can buy is proportional to the quality and number of cards you have played, your deck power will grow exponentially. And that is one essential problem of the game. :P
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Insomniac

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Re: The exponential card
« Reply #5 on: April 04, 2012, 06:25:53 pm »
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King's Court? It exponentiates the cards it plays KC=1 Action 3 times  KC+KC=3 actions 3 times  KC+KC+KC=3 actions 3 times + 2 actions 3 times                  Hence KC=3   2KC=9   3KC=15+3(KC was played 3 times)=18     

Each additional KC of a KC "adds" (9-1=8) actions, so it's actually linear

Not if you count the times kings court is played. Im not talking about "added" actions, im talking sheer number of actions played and since KCs play KCs which isnt counted in 9-1=8.

KC=3+1
KC+KC=3KC plays+3*3=9+3=12+1
KC+KC+KC=6 KC plays=6+3*3=15+2*3=21+1
skip
KC+KC+KC+KC+KC= KC plays KC plays other 3 KCS each of the KCs is played 3x except the first =12. 3 KCs have full actions free = 3*3*3=27+12=39+1=40

or more simply
The first KC is played (1)
It plays a KC 3 times (4)
that kc plays 3 KC 3 times (13)
You now have 3 KCs KC'd to resolve each one resolving the chosen action thrice allowing for 9 actions to be played thrice (9*3=27+13=40)

1+8 !=4     4+8!=13   13+8+8!=40  Thusly KC does not add 8

looking at the series

1, 4, 13, x, 40


Note the initial KC is left out of the calculations and that is the one added at the end


« Last Edit: April 04, 2012, 06:36:41 pm by Insomniac »
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timchen

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Re: The exponential card
« Reply #6 on: April 04, 2012, 06:32:43 pm »
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Not if you count the times kings court is played. Im not talking about "added" actions, im talking sheer number of actions played and since KCs play KCs which isnt counted in 9-1=8.

KC=3+1
KC+KC=3KC plays+3*3=9+3=12+1
KC+KC+KC=6 KC plays=6+3*3=15+2*3=21+1
skip
KC+KC+KC+KC+KC= KC plays KC plays other 3 KCS each of the KCs is played 3x except the first =12. 3 KCs have full actions free = 3*3*3=27+12=39+1=40

Note the initial KC is left out of the calculations and that is the one added at the end




Still linear. the formula is (2n-1)*4+1=8n-3 except for n=1
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Re: The exponential card
« Reply #7 on: April 04, 2012, 06:34:57 pm »
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Your maths makes no sense. 3*3 doesn't equal 9+3, and 9+3 doesn't equal 12+1. Don't use mathematical symbols in the context of maths if you aren't going to use them correctly.

And even then you're still wrong. The formula for how many actions are played, using the way you're counting, is 9n-5, where n is the number of KCs. That's not exponential. Exponential means it multiplies by itself each time multiplies by a constant each time, and linear is nowhere near that. Even something like 10^(10^10)x is slower than even the slowest exponential in the long run.
« Last Edit: April 04, 2012, 06:43:52 pm by Tables »
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...spin-offs are still better for all of the previously cited reasons.
But not strictly better, because the spinoff can have a different cost than the expansion.

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Re: The exponential card
« Reply #8 on: April 04, 2012, 06:36:41 pm »
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I think there are at least two, depending upon what definition of exponential you want to use. The less meaningful a definition you use the more exponential cards there will be.

Goons have exponential scoring based on number of buys x number of goons in play which is approximately goons x goons.

Silk Roads have a peculiar non-continuous exponential increase : silk road score = [other vp cards + silk road cards]/4 x silk roads.
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Insomniac

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Re: The exponential card
« Reply #9 on: April 04, 2012, 06:38:16 pm »
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Your maths makes no sense. 3*3 doesn't equal 9+3, and 9+3 doesn't equal 12+1. Don't use mathematical symbols in the context of maths if you aren't going to use them correctly.

And even then you're still wrong. The formula for how many actions are played, using the way you're counting, is 9n-5, where n is the number of KCs. That's not exponential. Exponential means it multiplies by itself each time, and linear is nowhere near that. Even something like 10^(10^10)x is slower than even the slowest exponential in the long run.

My math is fine im adding back in the stuff from before. I point out that its 3 for example and that they each play an action thrice which is 3*3=9+3
Im adding the words back in. Saving time rather than having a new line for each word to add back in. Having taken numerous college level math courses and the putnam exam and doing well in all of them I am well aware that the form im using is ok in SOME contexts provided it is clear why you are adding things after an equals sign IE described before hand. Since my math wasnt the point of the post but the final number was AND I described why I was doing it there shouldn't be a problem.

I was still incorrect because I hadn't put enough thought into it to see the formula. I could see from my own sequence of numbers in the second post it wasnt exponential. I just disagreed with add 8

And I know what exponential is, I just didnt put much thought into it.
« Last Edit: April 04, 2012, 06:54:47 pm by Insomniac »
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Re: The exponential card
« Reply #10 on: April 04, 2012, 06:43:23 pm »
+1

I think there are at least two, depending upon what definition of exponential you want to use. The less meaningful a definition you use the more exponential cards there will be.

Goons have exponential scoring based on number of buys x number of goons in play which is approximately goons x goons.

Silk Roads have a peculiar non-continuous exponential increase : silk road score = [other vp cards + silk road cards]/4 x silk roads.

Neither of these are Exponential, though. Both of them are just Quadratic. Wait... whoops, I defined Quadratic instead of Exponential in my earlier post, didn't I :-/? Uhh... actually, looks like I defined something... else. Exponentials should, each time you increase the number of cards by 1, multiply the effect through by a constant. So if e.g. 1 CardX gave you 2Vps and 2 CardX's gave you 4VPs, then 3 would give you 8VPs, 4 would be 16VPs... and 10 would be 1024VPs!

Compare this to Goons... the 1st is 2, the 2nd is 6... continuing that, the 3rd has to be 18, but it's not, it's 12. The 4th should be 54, but it's actually 20
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...spin-offs are still better for all of the previously cited reasons.
But not strictly better, because the spinoff can have a different cost than the expansion.

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Re: The exponential card
« Reply #11 on: April 04, 2012, 06:45:55 pm »
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True my maths is rusty.
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Insomniac

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Re: The exponential card
« Reply #12 on: April 04, 2012, 06:47:50 pm »
+1

I think there are at least two, depending upon what definition of exponential you want to use. The less meaningful a definition you use the more exponential cards there will be.

Goons have exponential scoring based on number of buys x number of goons in play which is approximately goons x goons.

Silk Roads have a peculiar non-continuous exponential increase : silk road score = [other vp cards + silk road cards]/4 x silk roads.

Neither of these are Exponential, though. Both of them are just Quadratic. Wait... whoops, I defined Quadratic instead of Exponential in my earlier post, didn't I :-/? Uhh... actually, looks like I defined something... else. Exponentials should, each time you increase the number of cards by 1, multiply the effect through by a constant. So if e.g. 1 CardX gave you 2Vps and 2 CardX's gave you 4VPs, then 3 would give you 8VPs, 4 would be 16VPs... and 10 would be 1024VPs!

Compare this to Goons... the 1st is 2, the 2nd is 6... continuing that, the 3rd has to be 18, but it's not, it's 12. The 4th should be 54, but it's actually 20

If its truly exponential 0 of the card should give the effect once
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AJD

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Re: The exponential card
« Reply #13 on: April 04, 2012, 06:53:32 pm »
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My vote's with rrenaud; exponential and quadratic cards were discussed in <a href="http://forum.dominionstrategy.com/index.php?topic=1269.msg20678#msg20678">this previous thread</a>.
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Re: The exponential card
« Reply #14 on: April 04, 2012, 07:00:14 pm »
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Heh, true... I think we can make an exception for that (or maybe even... not).

I think the answer is Crossroads though, as rrenaud suggested. Suppose you have a density of p victory cards in deck. After playing Crossroads once, you draw 4p cards, giving you a hand of 4+4p=4(p+1) cards. After a second play, you've played one card giving you 4p+3 cards in hand and so draw (4p+3)p, giving you a hand of 4p^2+7p+3=(p+1)(4p+3) cards. A third play gives you (4p^2+7p+2)p+4p^2+7p+2=(p+1)(4p^2+7p+2). And this pattern unsurprisingly continues, subtract 1, multiply by (p+1). So, it's... more or less exponential in p. Not perfectly, because of the discrete subtract 1 bit, but more or less. Fun fact: It still works for the 0 case.
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...spin-offs are still better for all of the previously cited reasons.
But not strictly better, because the spinoff can have a different cost than the expansion.

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Re: The exponential card
« Reply #15 on: April 04, 2012, 07:05:50 pm »
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Assume you have 2 cards in hand, crossroads, and a vp card. Assume that the cards you've played previously have afforded you unlimited actions and there are unlimited vp cards. Furthermore assume perfect shuffle luck.

Play Crossroads Draw 1 Card. This does not allow exponential growth

Assume then you have 3 cards in hand 2 vps and a crossroads

Play Crossroads Draw 2 cards, 1 VP and 1 Crossroads
Play Crossroads Draw 3 cards,  2 VP and 1 Crossroads
Play Crossroads Draw 5 cards, 4 VP and 1 Crossroads
Play Crossroads Draw 9 cards, 8Vp and 1 Crossroads
Play Crossroads Draw 17 Cards, 16 VP and 1 Crossroads
Play Crossroads Draw 23 Cards, 22 VP and 1 Crossroads
Play Crossroads Draw 45 Cards, 44 VP and 1 Crossroads

So without much consideration I can say that in these 2 cases, crossroads is not SIMPLY exponential, though thats not to say it isn't exponential at some other hand sizes, or that there isnt some formula that with subtraction.

Growth is exponential up till 22 and 24. and I cant see a correlation without looking strictly at growth. I suppose you could argue though you have ALL 10 crossroads in hand 1st in which case


Play Crossroads Draw 2
Play crossroads Draw 4
play crossroads draw 8
play crossroads draw 16
play crossroads draw 32
play crossroads draw 64
play crossroads draw 128
play crossroads draw 256

so IN the ideal case it is exponential (where you have a ridiculous number of vp cards you can draw and all crossroads in your hand and at least 7 actions. and your deck is only VP left (until cr 10))
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Re: The exponential card
« Reply #16 on: April 04, 2012, 07:33:42 pm »
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You, uh, made an arithmetic error. Drawing 17 cards should have left your deck with 33 VP cards, but you said you draw 23. It's not hard to see under your assumption it should double every time.

And, you don't need the ideal case for it to be exponential. It averages out as being exponential, certainly, but you only need the crossroads in hand to start with for it to be a true exponential average, no need to assume your deck is entirely green.

Finally, the case with just 1 card is technically exponential. The best kind of exponential, the trivial kind, where you multiply by 1 every time.
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...spin-offs are still better for all of the previously cited reasons.
But not strictly better, because the spinoff can have a different cost than the expansion.

timchen

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Re: The exponential card
« Reply #17 on: April 04, 2012, 08:09:58 pm »
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Yes, Crossroad is the card I am thinking about.
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jimjam

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Re: The exponential card
« Reply #18 on: April 05, 2012, 01:17:17 am »
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Scrying Pool is unbounded.
« Last Edit: April 05, 2012, 01:23:51 am by jimjam »
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jonts26

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Re: The exponential card
« Reply #19 on: April 05, 2012, 01:42:20 am »
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Only if you deck is infinite and purely actions. If there is an action density less than 100%, then its average draw is actually a constant with actual draw having some mean and standard deviation that I don't care enough to figure out right now.
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DStu

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Re: The exponential card
« Reply #20 on: April 05, 2012, 01:59:53 am »
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Only if you deck is infinite and purely actions. If there is an action density less than 100%, then its average draw is actually a constant with actual draw having some mean and standard deviation that I don't care enough to figure out right now.

Only if the action density is constant. If it increases ... I have to leave...
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jimjam

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Re: The exponential card
« Reply #21 on: April 05, 2012, 06:01:02 am »
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Wouldn't any draw in any finite deck be bounded linearly by the size of the deck (unless you're executing a KC-gain combo)? Under this criterion, Crossroads wouldn't be exponential.

The way I would do this rigorously is to define the draw growth of a card c as follows: c(n,k)=max (over all finite decks D without any c's, multisets of cards K with size k, none of which are c): cards in the hand after starting with n c's and K in hand, and playing only c's.
Then,c(n) is the max over k the growth rate of c(n,k) using traditional asymptotic analysis. Crossroads is exponential in this circumstance.
Now for something to be unbounded, it would mean that for any function f:N->Z, there exists a k where c(n,k)>f(n) for all n (which means c(n) has to be undefined).
of course, c(n,k) is undefined for Scrying Pool for all n in Z+ and k, so it would be unbounded.
Counting House is also unbounded in this criterion.
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timchen

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Re: The exponential card
« Reply #22 on: April 05, 2012, 02:40:10 pm »
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Physically I will define that we work in the regime where D>>exp(n)>>1. Then the card draw c(n,k) is certainly exponential in n for crossroad, but constant for Counting House and Scrying Pool (albeit at the order of D.) c(n,k) is proportional to k^-1 for the scrying pool though. If you think about scaling in k that is even faster than exponential at small k.
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