araver

POSTS

T H I N K ROLLO - 0 LOGIC - 0 MAFIA - 0 TRICK - 3 SHRUG - 1 MEMES - 0 BRICK - 2 Nana +5 DRINK - 3 Araver +5 TRUNK - 3 DRINK - 3 Nana +5 THINK - 5 TRUNK - 3 Nana +5 THINK - 5 Marksmanjay +5 +10 Leaderboard: Marksmanjay is up as host

Nope, you need logick. Logic is not enough. J/K But anyone can actually steal your guess with that explanation.

T _ I N _ ROLLO - 0 LOGIC - 0 MAFIA - 0 TRICK - 3 SHRUG - 1 MEMES - 0 BRICK - 2 Nana +5 DRINK - 3 Araver +5 TRUNK - 3 DRINK - 3 Nana +5 THINK - 5 TRUNK - 3 Nana +5

T _ _ N _ ROLLO - 0 LOGIC - 0 MAFIA - 0 TRICK - 3 SHRUG - 1 MEMES - 0 BRICK - 2 Nana +5 DRINK - 3 Araver +5 TRUNK - 3 DRINK - 3 Nana +5 I'm not suing you, I'm getting points

T _ _ _ _ ROLLO - 0 LOGIC - 0 MAFIA - 0 TRICK - 3 SHRUG - 1 MEMES - 0 BRICK - 2 Nana +5 DRINK - 3 Araver +5 TRUNK - 3

_ _ _ _ _ ROLLO - 0 LOGIC - 0 MAFIA - 0 TRICK - 3 SHRUG - 1 MEMES - 0

_ _ _ _ _ ROLLO - 0 LOGIC - 0 MAFIA - 0 P.S. You probably can get rookie to get your old account back. Just drop him a PM and leave some personal details for him to identify it.

@Thalia - oh, but you do _ _ _ _ _ ROLLO - 0

This was a game I spent countless days coming back to, even though it's very simple, it gets addictive Rules (slightly changed from unreality's original thread): 1. The host chooses a 5-letter word and starts the game by posting _ _ _ _ _ 2. Any player can guess a 5-letter word and post it (along with an attempt to logically explain the guess) 3. The host checks the guessed word and returns the number of letters in a correct position. The host gets +1 point per guess. 4A If the player attempted an explanation that was correct, he gets +5 points per explained letter (+10 points for the fifth and final letter) and the host shows the correct letters in the next post e.g. _ _ _ L _ 4B If the player attempted an explanation that was incorrect, the host gets +5 points per incorrect explanation. 5. Conditional explanations are welcome 6. The player that explained the last letter becomes the new host. _ _ _ _ _

One of the problems we all face is indeed time, and yes, it appears a lot of community members have moved on. But as I see it, the forum itself is not dead, so we could get a game started as long as it doesn't involve quick play. I'm open to suggestions as long as it's fun. Maybe something lighter than Mafia at first. There were a lot of smaller interesting games that could be started and allow asynchronous input from players along the time and drop-ins. If one of those games sticks then after a short while we can try a game that requires everybody coming back at least 3 times a day to contribute.

EDIT: nvm, did not see page 2

Well, it looks like an encoding, the code is similar to base64, but it has more than 26x2+10+2=64 chars. Once you determine how many chars it actually has, it should get easier to figure out the base. Little fiddling led me to this (spoiler):

I believe this is

Well

I haven't been able to come up with a better than n=11 alternative. I'm pretty sure (branching my results at n=15) that there's nothing between 12 and 14.

Assuming exponentiation.

Agreeing with Rob. Or kinda disagreeing with both

n=11 in my post above.

I keep forgetting to click "Sort by Date" and when it sorts by rating I get confused. My question was different, let me rephrase: Assume there was no round before (A,B,C) = (4, 2, 5) with B and C active gets at the end of round 1 to (4, 4, 3). 1) Do we stop here even if the active players are not the ones equal? Bonanova's response does imply so i.e. Game ends at the end of the round if ANY 2 players are equal (different wording than OP though). So we count a minimum of 1 round before game with (4,2,5) ends. 2) If we let it go to the next round since the two active players (B and C) are not equal: 2A) assuming A and B could be active AND if we don't check the condition before we subtract, then one of them gets 0, and that may cause infinite rounds. 2B) If we do check the condition A=B since they are the active, then we call the game ending in Round 2. What I meant is that different wording makes either N=1 or N=2. In any case, I'm assuming and counting using rule "Game ends at the end of the round if ANY 2 players are equal". And another submission from me:

Question on wording of the goal: The game ends when the two active players in any round have exactly the same stake. If (A,B,C) go into a round, B and C are active and B = A at the end of the round (e.g. B = A/2, B<C), does that count as stopping the game OR the game goes to the next round where only if C sits out the goal is met?

Sorry, but I was wrong about my "trapezoid" solution. At most two trapezoids would be similar, forcing the third to be similar would make all congruent. I've circled around and am pretty sure now there's no other way to split a triangle into 3 similar yet non-congruent triangles unless it's follows Bonanova's solution. And I think the same hypothesis holds for showing no 3 simlar quadriterals can be obtained either (limit case is 3 congruent trapezoids for an equilateral triangle). With more than 4 sides ... that's even more unlikely, as they won't have the necessary number of sides to begin with (i.e. partition a triangle into 3 pentagons / hexagons).