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About araver

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  1. 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):
  2. I believe this is
  3. 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.
  4. Assuming exponentiation.
  5. Agreeing with Rob. Or kinda disagreeing with both
  6. n=11 in my post above.
  7. 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:
  8. 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?
  9. 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).
  10. I don't have spoilers on mobile interface, sorry. Does the problem state ANY triangle? I can do this for right triangles (which are not isosceles) with 3 similar non congruent triangles and I can do it for equilateral triangles with 3 trapezoids similar non-congruent. But not for any triangle. EDIT: just saw Bonanova's that is my solution as well, but it does not work for A=B.
  11. I still pop into here from time to time, but I rarely have time to host a game. However, if you see this, message/PM me back - I'm willing to host a quick one (6-8) in August / September 2015 if you're interested.