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Messages - heron

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I just like to play different people

General Discussion / Re: Maths thread.
« on: February 24, 2021, 12:14:36 am »
f(x) is the number that you get by applying f to x.

But what is x?

I did point out that you might have this complaint, and also that this complaint makes no sense if you also think that x \in O(x^2) makes perfect sense...

x is any and every real number, the same way it is when you say f(x) = x^2.

General Discussion / Re: Maths thread.
« on: February 22, 2021, 11:48:54 pm »
Idk this f(x) business seems fine.

f(x) \in O(g(x)) means that there exists m,M such that if x > M then |f(x)| ≤ Mg(x).

f(x) is the number that you get by applying f to x. I guess you might complain that this is a statement about f, not a particular number f(x).
But from the sounds of it you don't have a problem with saying x \in O(x^2), which would have the same problem, so this argument does not hold up for me.

tl;dr: how can x be a thing but f(x) not be a thing.

Dominion General Discussion / Re: Boon Power Level Comparison
« on: January 29, 2021, 09:02:43 pm »
none of the Boon-givers are cantrips, what's up with that?

Dominion Articles / Re: Mandarin-Scepter Loops
« on: August 22, 2020, 12:40:35 am »
These combos are not unimaginably rare. Artisan/Scepter/Mandarin/Way of the Mole came up in a league game.

Feedback / Re: Anyway to report negative attitude player?
« on: July 05, 2020, 12:23:33 am »
this is the official report form:

Dominion Articles / Re: Combos/Counters Article list
« on: June 09, 2020, 02:53:54 am »
Am I the only person who thinks that lurker/hunting grounds is easily the most broken combo in all of Dominion?

It's not that broken, because it enables the Lurker+gaining free Hunting Grounds from the trash combo for your opponent and if it's an engine board, that probably beats the Duchy rush.

Actually R-man is right, lurker-hg is quite tough to beat. Idk about most broken combo, but it's pretty high on the list.

Dominion General Discussion / Re: DeepMind for Dominion
« on: April 24, 2020, 09:15:51 pm »
i imagine if they were going to do this for a card game, they'd use MtG since that's been shown to be NP-complete ("does the game end", i believe, was their qualifier problem), whereas dominion (for all non-trivial play examples, ie, player A, player B both draw and discard 5 cards, next turn) should be P, provided none of the truly unbounded vp cards (Monument, Plunder, Chariot Race) are present.

additionally, if they programmed their bot with the goal of "end the game while you have the most points" rather than "get the most points", Dominion should be solvable in P
Huh I think you mean "turing complete" instead of "NP-complete"?

it's that too but given that each "knob" (whether the deck will run out, whether life will run out, etc) is unbounded (Gaea's Blessing, Lifelink, etc), you end up dealing with infinities, so given each deck is doing non-trivial things (draw a card, end your turn being the trivial play) it's impossible given two of all possible 60 card decks to say whether a game will end. ergo NP-complete.

This is not what NP-Complete means.
Perhaps you mean noncomputable? I don't really know anything about magic though.

Solo Challenges / Re: Another infinite loop
« on: April 12, 2020, 07:39:32 pm »
Can do 1 kingdom card a 2 landscapes:

KC-KC (way of the rat, way of the horse to draw new KC + treasure, play new KC, repeat)

Donald says we should avoid using 2 Ways. But sounds nice.

You can replace way of the rat with inventor.

No. It is a mere King's Court, can never gain itself without Way of the Rats.

Instead of playing king's court as way of the rat, you play it as itself and then play inventor 3 times. The first 2 plays are used to gain king's court and inventor, and the third play is way of the horse.

Solo Challenges / Re: Another infinite loop
« on: April 12, 2020, 03:18:22 pm »
Can do 1 kingdom card a 2 landscapes:

KC-KC (way of the rat, way of the horse to draw new KC + treasure, play new KC, repeat)

Donald says we should avoid using 2 Ways. But sounds nice.

You can replace way of the rat with inventor.

Rules Questions / Re: Cost of Animal Fair revisited
« on: April 11, 2020, 10:59:26 pm »
I think it is clear that when Donald said that Animal Fair has an alternate cost of trashing an action, he isn't using cost as a formal term.
Obviously if there was another card that was like you can trash an action and pay $2 to buy it but it normally cost 25 then you could not remodel animal fair into this hypothetical card.

Solo Challenges / Re: Another infinite loop
« on: April 11, 2020, 10:29:38 pm »
Can do 1 kingdom card a 2 landscapes:

KC-KC (way of the rat, way of the horse to draw new KC + treasure, play new KC, repeat)

General Discussion / Re: Maths thread.
« on: February 04, 2020, 02:38:44 pm »
I found an optimal strategy:

Index the players 1-127.
Each player computes the XOR of the indices of the players with red hats.
If it is 0, they guess that they have a red hat.
If it is their own index, they guess that they have a black hat.
Otherwise, they abstain.

Thoughts on how to come up with this strategy:
Since each guess is correct with 50% probability, in order to maximize the chance of winning we need to make it so that when someone guesses wrong, everyone else does too, and when someone guess right they are the only one to guess.
The easiest example of this sort of behavior is a situation where a player guesses that they have a red hat if all other players have a black hat.
Then everyone guesses wrong if there are all black hats, and exactly one person guesses correctly if there is exactly one red hat.

Basically, we need to tile the space with clusters consisting of a center state (where everyone guesses wrong) and all the adjacent states.
Due to the way that switching a hat color twice doesn't change the state, it seemed natural to use some sort of XOR condition.
So, I picked the states where everyone guesses wrong to be the states where the XOR of the indices of red hats to be 0.
You could pick a different number besides 0 if you want.

That said, here is a similar problem from the most recent putnam:

Let Z^n be the integer lattice in R^n. Two points in Z^n are called neighbors if they differ by exactly 1 in one coordinate and are equal in all other coordinates.
For which integers n ≥ 1 does there exist a subset S of Z^n satisfying the following two conditions?
(1) If p is in S, then none of the neighbors of p is in S.
(2) If p is not in S, then exactly one of the neighbors of p is in S

Dominion General Discussion / Re: Improving play at 3+ players
« on: February 01, 2020, 06:34:52 pm »
I will make a note on this argument.

Until the curses (would have) run out, the rate at which you receive curses is independent of whether or not you buy witch.
Instead, buying witch affects the rate at which the other players receive curses, and does so to the same extent in both 2p and 3p.

General Discussion / Re: Maths thread.
« on: January 28, 2020, 09:06:59 pm »

And, for hat problem number 1:
Decide ahead of time to always guess odd or always even. Then everyone is guessing the same hat distribution.  This produces a 50% chance of victory, because there's an equal number of even and odd distributions.

I think this is the right idea, but there isnít the same number of odd and even distributions. So we would actually have to check what are the odds of having an odd number of white hats, total, and compare with the odds of having an even number. Then everybody agrees to guess the hat color that leads to a total number of white hats that is odd/even, depending on their previous result.

No, see the exchange earlier with MiX and bitwise. bitwise is correct.

General Discussion / Re: Maths thread.
« on: January 24, 2020, 11:02:00 pm »
2. Similar to the previous problem, but there are 127 players. Furthermore, each player simultaneously guesses their hat color OR abstains. The players win if all guesses are correct and there was at least one guess, and otherwise, they all lose. What is the maximum probability of victory that they can achieve?

This one... I think this is the answer.


With the distribution being completely random, there's no way any single person can have more than a 50% chance to guess their own hat colour. No amount of planning a way a person might decide to guess or not guess will give them a better than 50% chance when they actually make their guess. Therefore, one person is designated to guess at random, while everyone else abstains. The person making a guess then has a 50% chance to guess correctly, making everyone win.

This is not correct. Consider the game with 3 people and the strategy of guessing the opposite color if both other players have the same color hat.

It is true that 50% of guesses made are incorrect, but you can load it so that many wrong guesses are made at once, and when a correct guess is made it is the only guess.

General Discussion / Re: Maths thread.
« on: December 08, 2019, 04:30:44 pm »
I found it about the same as usual; I was very disappointed after the test when I realized I misread B5. For some reason I thought it said f(n) = F_{2n + 1} instead of f(2n + 1) = F_{2n + 1}. I had solved a problem that was basically the same as the actual problem before, but the technique could not solve the problem they way I had read it. If I had read more carefully I would almost surely have got it...

I think B6 was unusually easy this year; it was definitely the problem that took me the least time to solve.
A1 and B1 were perhaps more difficult/time consuming than usual.

General Discussion / Re: Maths thread.
« on: December 04, 2019, 03:25:23 pm »
Yea it's my last year for it.

Puzzles and Challenges / Re: Slowest trashing kingdom
« on: September 25, 2019, 06:07:44 pm »
There is also no need for watchtower in the bard kingdom. Just bard + death cart works.

Puzzles and Challenges / Re: Slowest trashing kingdom
« on: September 24, 2019, 11:53:27 pm »
trader + death cart seems pretty slow.

Dominion General Discussion / Re: ThunderDominion Card List Results, 2019
« on: September 06, 2019, 02:54:44 pm »
IMO upgrade should be 3rd. Maybe 2nd. Upgrade >> steward.

General Discussion / Re: Maths thread.
« on: June 28, 2019, 08:51:54 pm »
Sorry this explanation is kind of poorly written.

Note that for x coprime to 2^32, x^(2^30) = 1 mod 2^32.

Suppose that a has maximum order, i.e. a^k ≠ 1 for 0 < k < 2^30.
Let n = c0 + c1*2 + c2*2^2 + ... + c29*2^29.
(a^n)^(2^29) = c^(2^29)
a^(c0 * 2^29) = c^(2^29)
Since a has maximum order, exactly one of c0 = 0 and c0 = 1 will work. And then you can proceed like that to find the other ci.
So where does LTE come in?

Well, since a and c are both 1 mod 4, v2((a^c0)^(2^29) - c^(2^29)) = v2(a^c0 - c) + v2(2^29).
We know the LHS is either 32 or <32 depending on choice of c0. So RHS is either 3 or <3 depending on choice of c0.
Similar argument applies for other ci.

Why v2(a-1) = v2(c-1)?
Well, earlier I had this assumption that a had maximal order. That may not be true. However, if a = 1 mod 4, then o(a) = 32 - v2(a - 1).
So a and c have the same order.
Let that order be 2^h.
Then we can do stuff like
a^(c0*2^(h-1)) = c^(2^(h-1)) for exactly one choice of c0.

General Discussion / Re: Maths thread.
« on: June 24, 2019, 08:49:17 pm »
People were able to compute plenty of derivatives perfectly well far before any of the modern rigorous notions of limits, which was primarily due to Cauchy. They did so mostly with "handwaving."

The key issue here is more subtle, and it's more about being circular than being handwavy. The approximation e^h = 1+h when h is small is only valid for the number e. It's equivalent to the fact that the tangent line to the graph of e^x at x=0 is y=1+x, which can't be established without knowing that the derivative of e^x is e^x in the first place.

Still not quite sure I understand. Doesn't x^h approach 1+h as h approaches 0 for any positive x?

The "handwaving" is just a less formal way of talking about limits, right?

It actually approaches 1 + hlogx. But since log(e) = 1, for e^h it is 1 + h.

Edit: Well, you can say that lim as h-->0 of | x^h - (1 + h) | = 0. But that is only because both of the quantities approach 1. It is more meaningful to consider the difference relative to h, which just gives the limit of the derivative: (x^h - (1 + h)) / h

General Discussion / Re: Maths thread.
« on: June 01, 2019, 02:18:06 pm »
This isn't a great reason, but if you multiply 72 or 288 by 2.5 you get 180 mod 360. It's not a great reason because if you multiply (72 + 360) by 2.5 you get 360 again.

It's also not true: 288*2.5 = 720, while 216 * 2.5 = 240.

Well, fair enough.

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