Even Harder Coin Problem

it is similar but much more difficult.

from my buddy tom:

> ok, here's a tough one (i thought). there are no "aha!" tricks - it
> requires straightforward deductive-reasoning.
>
> you have 12 coins. one of them is counterfeit. all the good coins weigh
> the same, while the counterfeit one weights either more or less than a
> good coin.
>
> your task is to find the counterfeit coin using a balance-scale in 3
> weighs. moreover, you want to say whether the coin weighs more or less
> than is should and, and this is the real kicker, your weighs must be
> non-adaptive.
>
> that is, your choice of what to put on the balance for your
> second weigh cannot depend on the outcome of the first weigh and your
> decision about what to weigh for round 3 cannot depend on what happened on
> either your first or second weigh.
>
> for example, you can't say something like "take coin #1 and coin #2 and
> weigh them. if they balance, then take coins 3,4,5 and weight them
> against 6,7,8...if 1 and 2 don't balance, then weigh #1 vs #12..." you
> have to say something like:
>
> round #1: do this
> round #2: do this
> round #3: do this
>
> if the results are left tilt, balanced, and left tilt, respectively, then
> coin #11 is heavier than it should be.
>
> this problem is solveable...it took me about 1-2 hours of working on it to
> get it. i think even finding the counterfeit using an adaptive solution
> is tough. then non-adaptive constraint makes it quite hard and having to
> find whether it's heavier and lighter is cruel and unusual riddling ;-)
>
> have fun...