The doctor diagnosed me with "wrist pain"; my experience was very close to the old joke:

- Doctor, doctor, it hurts when I do this!

- So don't do that, then.

I'm going to look at voice recognition stuff in Berkeley early next week; I'm trying to make an appointment with a voice recognition technology reseller. (Berkeley is the home of the disability movement, so it has a number of resellers, dealers, and consultants in assistive and adaptive technologies, and a bunch of them have showrooms. Cool!)

I also still haven't gotten around to comment in some of the front-page articles (death of copyright, graydon's technical questions).

Collage

My collage is delayed a lot because of my wrists, but I'll think it's a really bad thing if I can't get it finished this weekend.

12 coins (spoilers)

rillian: you can't search more than 3^n coins, or floor((3^n)/2) coins when you also have to state whether a coin is heavy or light -- this from an information flow argument (each weighing distinguishes one of three possible states, and the possible states described by "coin j is bad" are j in number, the states described by "coin j is bad with error-direction s" are 2j in number).

This doesn't prove that you can search (3^n)/2 coins, just that you can't search more than that.

bwtaylor: In answer to the original problem, we can break it into sub-problems.

1. Given 2 possibly-bad coins known to have the same possible error direction (both heavy or both light), where the shared possible error-direction is known, to find which of the two coins is bad (if either), in one use of the balance: weigh the two coins against each other. (If they balance, neither is bad.)

2. Given 4 possibly-bad coins of unknown error direction (coins of which one is heavy or light, but you don't know which -- call these "U" coins), and at least 1 known-good coin (call this an "N" coin), to find which coin is bad, and its error-direction, in 2 uses of the balance. Weigh two U coins against a third U coin and an N coin. If the two U coins rise: they are suspect-light coins (possibly light, known not to be heavy; call these "L" coins), and the third U coin is suspect-heavy ("H"); then use the solution to problem 1, above, with the two L coins (if neither is bad, then the H coin is bad). If the two U coins sink: they are H coins, and the third U coin is an L coin; use the solution to problem 1, above, with the two H coins (if neither is bad, then the L coin is bad). If the two U coins balance with the third U coin and the N coin: the fourth U coin is bad. Weigh it against the N coin to find out whether it is heavy or light.

3. Given 8 possibly-bad coins of which 4 are H and 4 are L, and at least 4 N coins, to find which coin is bad (its error direction is then automatically known), in 2 uses of the balance. Weigh 3 H coins and 1 L coin against 4 N coins. If the L coins and the H coin rise, then the bad coin is the L. If they sink, then the bad coin is one of the 3 H coins; use the solution to problem 1 with any two of the 3 H coins to find which is bad (if neither, it is the third). If they balance, then the bad coin is one of the 3 remaining L coins or the remaining 1 H coin. Use the solution to problem 2, above, with these 4 coins (we already have more information than we need!) to find which is bad.

4. Given 12 U coins, to find which coin is bad, and its error-direction, in 3 uses of the balance. Weigh 4 of the U coins against 4 others of the U coins. If they balance, one of the remaining 4 U coins is bad; use the solution to problem 2, above, to find which and its error-direction. If they do not balance, the 4 coins on the side which sinks are now H coins, and the 4 coins on the side which rises are now L coins. Use the solution to problem 3, above, to find which is bad.

In order to be able to search floor((3^n)/2) coins, I would need to find a recursive generalization of this somehow. I'm going to talk Ryan's advice and walk around San Francisco for a while today -- it is a beautiful day! -- and see if one comes to me.

Other people's diaries

rakholh: I don't want Lieberman to get elected "just because" he is Jewish. Actually, I don't want him to get elected at all, because he wants to censor the Internet. Actually, I don't want anyone to get elected, because I am an anarchist. But I especially don't want Lieberman to be elected, since he wants to censor the Internet. (All major candidates do, but Lieberman is just more vocal about it.)

I thought it was neat that he was a Jewish major-party nominee, though. In the U.S. it is extremely rare for anyone who doesn't appear to be a church-going white Protestant male to be elected President or Vice President, or even to be nominated by a major party for those offices. The last exception to the "getting elected" bit was John F. Kennedy, a Catholic (still Christian by prevailing standards, still white, still male). If I remember correctly, the Democrats did nominate a woman for Vice President once, but she was not elected. No woman has ever been elected President or Vice President. Neither has an atheist or an adherent of a religion other than Christianity. (Apparently some non-Christian theists who disliked organized religions were, once upon a time.)

Some time I can talk about claims of divine inspiration for scriptures. :-)

lilo: I need to give my wrists a break!