Unraveling the Strategy of an Unnamed Game

[T@P]

two people are playing the amazingly complex and strategy filled game that has no name.

the way you play is like this: you take an ideal and round table, and a lot of ideal circular coins (normal coins, little cylinders, nothing crazy). each player then (by turn) puts a coin onto the table. the loser is the one that can't put another coin on to the table so that its not on top of any other coins. also, you can't move the coins once they are placed. you can't put the coins on the table sideways. nothing funny.

question is, is there a strategy (to win) for the first player or the second player?

major hint. there is. :rofl: otherwise it would be a boring question

 

[BicycleTree]

Player 1 puts a coin in the center of the table, and then he mirrors every move by player 2 by a move the same distance from the center but exactly opposite. This is just like the winning strategy for eot-cat-cit!

 

[T@P]

too fast! lol exactly right. what i like about the puzzle is it complete lack of details - and the totally general solution. any way i didnt read the eot posts, so ill trust you on that. maybe i should...

 

[hemmul]

great :) really cool puzzle!

 

[moose]

player one could pick a table the size of his coin :P

 

[T@P]

hehe actually a table that was even a little less than twice the radius of his coin would work :)

 

[BicycleTree]

What if the table is an equilateral triangle?

 

[T@P]

hmmm. triangles don't have and "even" radial symmetry...

this means you can't find "the" spot to put your coin and copy his/her moves. unless I am wrong.

you could attempt going for the height and trying to fill it up, and then follow symmetry along that axis, but it would then depend on who puts the last coin in the height. (depends on the height)

well that's all applying the same basic idea as in the round one. there maybe a new idea to use here. did you have anything in mind bicycletree?

the equilateral triangle idea would be a good one for three people. that way you kick out the loser (the one who goes second) and then the last two face off in the round table game... but its predetermined. and the other two may "team up" against the third (this is why more than 2 player games are hard to analyze this way).

anyone else have some brilliant ideas?

 

[BicycleTree]

No, I just tossed it out there. I don't know how to solve it. Maybe if you looked at small triangles and considered all possible types of moves, some strategy might appear.

 

[BicycleTree]

If the triangle is just large enough to fit a central coin and three other coins in the corners, then putting the first coin in the center loses, so a perfect strategy can't always be based around putting a coin in the center.

 

[T@P]

very true. only if one were to have a winning strategy, it must depend on the first persons ability to go first, other wise you have the old white wins in double chess trick. therefore, if the first player does not make a *unique* move, or one that the second player can copy and end up where they started, you can say that the strategy does not exist for that player. (if you know the double chess trick, this should make sense. to those who dont, in a nutshell, it is like this: assume you play chess in where a player goes twice in one turn. you must go twice. the question is, who wins? and the answer is like this: assume black has a winning strategy. then white moves the knight in and out an suddenly black plays first (white did nothing on the first turn) and you can think of white as black and the other way too. that means that white should win, but we assumed black does. contradiction, therefore white *has a winning strategy*. I am not sure anyone knows what it is though...

 

[Jimmy Snyder]

therefore white *has a winning strategy*.

No, therefore black does not have a winning strategy.

 

[Uno Lee]

Who gets to choose the table? Player 1 or 2 (or 3?)

 

[T@P]

haha i guess player one would always choose the round table, 2 the triangular and 3 the four dimensional torus table.

and jimmysnyder, we proved that black can't have a winning strategy, which more or less entails that it can't win (if played correctly etc.) because if it did have a strategy, white would simply *become* black and then win with the smae strategy.

 

[Uno Lee]

If I was player 2 and could choose the table, I would choose a size, of any shape table, that was too small for any "ideal." Then player 1 (or 3) would be unable to play, thus, forfeiting the game. Player 2 wins. yeh!

 

[T@P]

i think the 19 dimensional torus would be better, since the coint would keep falling off...

 

[Jimmy Snyder]

T@P,

Your proof is a good one. It starts out with "assume black has a winning strategy" and by "reductio ad absurdum" proves that statement false. In other words "Black does not have a winning strategy."

However, this is not the same thing as "White has a winning strategy." It may be that neither has a winning strategy. In any case, you have not proved that White has a winning strategy, only that Black doesn't. If you don't see this, perhaps you have not accounted for the fact that Black might be able to force a draw.

Again, your proof that Black has no winning strategy is a good one. My only issue is with your statement that I quoted before "therefore white *has a winning strategy*."

 

[T@P]

you raise a good point. however, i read the proof in some bigwig book and I am sure its not just a figment of my imagination. so there's that justification ;)

however, i think i mis-spoke. the idea is that you assume *black will WIN* not black has a strategy. you prove that black *cannot* win (if everyone plays their best). clearly white wins then, which proves the existence of a strategy. i hope i explained it better now :)

 

[Jimmy Snyder]

the idea is that you assume *black will WIN* not black has a strategy.

Statements like "Black will win" or "White will win" are almost certainly not true. Either side can probably lose deliberately even if there is a winning strategy.

 

[Bartholomew]

The game might end in a draw.

 

[T@P]

i mean they will win given they play their best (theoretically) game. really, maybe I am not explaining it good, but I am not making it up.

 

[Uno Lee]

If there are only 2 players, turns alternate and the object is to get the last ideal on the table; then why do you need more than one colour of "ideal?"