[opensource-dev] Consensus? was: Client-side scripting in Snowglobe

Colin Kern colin.kern at gmail.com
Mon Feb 22 21:55:58 PST 2010

On Mon, Feb 22, 2010 at 6:51 PM, Carlo Wood <carlo at alinoe.com> wrote:
> On Sun, Feb 21, 2010 at 04:57:18PM -0800, Dzonatas Sol wrote:
>> When LL said "here is a sphere the size of a quarter in diameter...
>> 1 2 3 4 5 6" as one points top, bottom, left, right,  back, front.
>> And says "Stupid" with a superiority look.
>> Obviously the person that was challenged, the one to be hired, said "Odd."
>> If you know if it is "even" or "odd" then you know who gets the last
>> move, and wins.
> This is clearly a way to measure someones spatial insight.
> Now note that if it's a game, and coins are not allowed to be moved
> around once they are placed, then it's very unlikely that you will
> be able to place 6 coins on the surface of that sphere (with diameter
> equal to the coin), because the one who'd put down the fifth coin
> would not put it such that you can put down the sixth coin, but
> somewhere in the middle of the left-over surface, leaving no spot
> for the sixth coin. Calling people stupid over this game surprises
> me however (because I happen to have a extremely large spatial
> insight, officially measured mind you (although, they couldn't
> actually graph it in their graph because I scored not just out of
> the graph but even off the paper that the graph outline was printed
> on)), because I'm having a hardtime to quickly guess if you CAN put
> five coins on the sphere... It seems not unlikely that only four
> coins will make it... which would mean that the one that begins
> will try to leave open as much space as possible when putting
> down the third coin. The person to move second would try to use
> as much space as possible. So, first goes on top say, second on
> grounds of symmetry probably on the bottom, then again forced to
> play without any strategy, the third coin just goes on the side,
> and the fourth coin, wanting to be last, takes the exact
> opposite of that, leaving two places free: Oh hell... that
> way you STILL end up with six coins being placed, even though
> both tried to screw the other with strategy. The only freedom
> that still exists would be the one placing the second coin: by
> not placing it exactly on the opposite side, you'll likely end
> up with only five coins. However, since putting it on the exact
> opposite side caused this player to win, he has little reason
> to play it elsewhere. Hence, due to perfect symmetry, the first
> player has no real choice, ever. And the second one, who wins,
> can control the game completely; hence 6 coins.
> Not THAT simple however.

There's really no way to figure it out without information about
either of the player's strategies.  It seems like what the interview
was really asking is what the maximum number of coins that could be
fit on that surface was, or he should have specified that these two
players were perfect and playing optimally.


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