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

Carlo Wood carlo at alinoe.com
Mon Feb 22 15:51:22 PST 2010


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.

-- 
Carlo Wood <carlo at alinoe.com>


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