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176

Puzzles and Challenges / Re: Easy Puzzles

« on: May 13, 2014, 05:06:29 pm »

Inspired by the "empty the supply in 3 turns" challenge:

Gain a Province on your second turn.

Hard mode: gain a Colony on your second turn.

To clarify, you may control all players' actions, but one player must gain a Province (resp. Colony) on her second turn.

177

General Discussion / Re: Maths thread.

« on: May 09, 2014, 05:51:39 pm »

I think that proof works. Nice. Also, yes, I probably should have specified that no three points are collinear.

Here is mine:


Consider 4 points which are the endpoints of two intersecting line segments. Those points form a convex quadrilateral, and the segments are the diagonals of that quadrilateral. We can re-pair those 4 points so that those two segments are two opposite sides of the quadrilateral and do not intersect. From the triangle inequalities, we know that the sum of the lengths of two opposite sides of a convex quadrilateral is strictly less than the sum of the lengths of the diagonals.

Now, consider the pairing of points which leads to the least total length of the line segments. We know that such a pairing exists since there is a finite number of points. Assume that two of the segments in this pairing intersect. Then, we can re-pair them as described above, and the total length of the line segments will decrease. But that is a contradiction, since we started with the pairing with the least total length.

Therefore the pairing with the least total length of line segments will not have any intersecting segments.

Oh, very nice.

178

General Discussion / Re: Maths thread.

« on: May 08, 2014, 08:00:07 pm »

Here is a fun maths problem: Let n be a natural number. 2n distinct points are plotted in the plane. Half of them are colored red, and the other half are colored orange. The points are then divided into n pairs, each with 1 red and 1 orange point. Then, a line segment is drawn between the 2 points in each pair. Prove that there is a way to pair the points such that after the line segments are drawn, none of the line segments intersect each other.

Assume that no three points lie on the same straight line, since otherwise it fails for e.g. four points on a straight line coloured RROO in that order.

Call the set of points A. We will do this by induction on n. Clearly, when n=1 it's possible.

So for n>1, consider the convex hull of A, which is a convex polygon with vertices at points of A such that no point of A is outside the polygon. There are two cases: all the vertices of the convex hull are the same colour, or some are different. If some are different, then two adjacent vertices F and G will be different, and we can safely draw a line between them; no line between two points will ever leave the convex hull, and F and G are both outside the convex hull of the remaining points. Then the number of remaining points is less than n, so by induction we're done.

If all the points on the convex hull have the same colour - red, without loss of generality - then pick two, call them X and Y, and consider a straight line L perpendicular to the line between X and Y. If the line L is very close to X, then there are more red points than orange points on the side of L closer to X. If the line L is very close to Y, then there are more orange points than red points on the side of L closer to X. The aim is to move L from close to X to close to Y, and at some point, there will be equal numbers of orange points and red points on the side of L closer to X.

The only way this can fail to happen is if there are two orange points P and Q which lie on the same line K which is perpendicular to the line between X and Y, and there is one more red point than orange point on the side of K closer to X. (So if L is just closer to X than K, there is one more red than orange on X's side, and if L is just closer to Y than K, there is one more orange than red on X's side). In this situation, however, putting L in the same place as K and then twisting it very slightly around the midpoint of the line segment PQ will give us a line L with equal numbers of orange points and red points on the side of L closer to X, as wanted.

Then we can apply induction to the points on the side of L closer to X, and separately to the points on the side of L closer to Y, to get a set of lines that works for all the points.

Remark: I think this proof only actually requires no four points to lie on the same line, but it was easier to write up with three.

179

General Discussion / Re: Moral quandry....about travelling to work

« on: May 08, 2014, 02:22:16 pm »

Go by train, and use the fact that when you're on the train you don't need to think about driving to do something else. Time in a car is way deader than time on a train, even a commuter train.

180

Puzzles and Challenges / Re: What Card Am I?

« on: May 06, 2014, 07:00:36 pm »

Or do you think I have nothing to give?

Too hard, apparently. Let's see whether I can drop some useful hints.
What word describes land that gives nothing?

Baron (barren)?

181

Innovation Game Reports / Re: Possible bug with Yunus

« on: April 28, 2014, 12:34:47 pm »

Yunus's karma effect: "If any player would take a Dogma action, first you may return a card. If you do, you have the sole majority in its featured icon until the end of the action."

The "its" refers to the card you returned. So returning Oars gave you castle majority, which isn't any use for a factory card.

182

Dominion Articles / Re: Combo: Hermit/Madman + Market Square

« on: April 26, 2014, 03:07:02 am »

How are you planning to gain Madmen mid-turn?

183

Dominion General Discussion / Re: Homage to the Best Card

« on: April 25, 2014, 11:02:03 am »

Is it bad that I've actually opened Scout/develop?

...did you win?

184

Variants and Fan Cards / Re: Dominion: Enterprise (Beta)

« on: April 24, 2014, 12:46:43 pm »

My random thought idea for Conscripts:

"+1 action, +$2; Each other player with 5 or more cards in his hand reveals his hand and discards a card that you choose, then draws a card."

- the idea being, replace a good card with a random one.

It stacks, which I think is important for a non-terminal attack, yet even if you play lots there's still the chance that the top card gives your opponent something to do. The turn-killing potential might still be too high, though.

[also, there's a minor theme-related point that in general cursers have witch-names whereas discard attacks have soldier-names]

185

Variants and Fan Cards / Re: Really bad card ideas

« on: April 17, 2014, 07:54:06 pm »

Inspired by a tangent in the Interview with Donald X. thread:

Student Loan $1 Action-Duration
Gain a University from the University pile.
---
Remains in play forever. At the start of the Buy phase each turn this is in play, if you have at least $5, -1$.

Setup: If University is not in the Supply, add an extra University pile which is not in the Supply.

186

Dominion General Discussion / Re: Interview with Donald X.

« on: April 17, 2014, 08:33:06 am »

Is there a specific reason why "+3 Actions" and "+2 Buys" were only used on a single card each (Crossroads and Squire), or was it a coincidence?

Sir Martin?

187

Variants and Fan Cards / Re: Penny Pincher

« on: April 16, 2014, 03:47:16 pm »

Also, welcome to the forums! (Both navical and gambit05.)

Thankyou

188

Variants and Fan Cards / Re: Penny Pincher

« on: April 15, 2014, 01:56:18 pm »

Would it be silly to start with 2 each for 2 players, and 1 each for more players?