September 2022

S M T W T F S
    123
45678910
11121314151617
181920 21222324
2526 27282930 

Style Credit

Expand Cut Tags

No cut tags

June 29th, 2008

gfish: (Default)
Sunday, June 29th, 2008 09:15 am
If you want a breakdown of points for the 4 placing teams, they've put up a rather large PDF of the workshop presentation from Friday. It includes the actual images returned, so you can see what kind of quality we're capable of achieving. (UBC's bounding boxes are normally a bit better, but there was a bug in that part of the code.)
gfish: (Default)
Sunday, June 29th, 2008 11:11 pm
Friday night, after the SRVC results were announced, most of us went out drinking. We ended up at a pretentious little place that sold crazy Belgian beers. It was rather more expensive than I'm used to, but worth it. Each bottle was a meal's worth of flavor and substance. And as we sat there getting rather drunk, we fell into a (logic) riddle contest. It was a lot of fun, and not just because I had a lot of luck figuring them out.

One of them completely stumped me, so I spent a good amount of the flight home working on it. I didn't solve it until I had generalized it into a set of equations, but I can see how one could figure it out directly. Since I had some much fun working on it, I thought I'd share.

You're given n coins under a towel and told that m of them are heads up. (The numbers don't matter, beyond the obvious restriction that 0 < m < n.) You can stick your hands under the towel to manipulate them, but you can't in any way sense which ones are heads up. How do you separate them into two piles (possibly of different sizes) such that each pile has the same number of heads? We're looking for a real solution, not something "clever" like 'stand them all on their edges'.

I'll post the solution in the comments.