seadalton
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hello,
I had a job interview today with the ole 12 snooker ball question (a similar question is hosted on this PF site: physicsforums.com/archive/t-40405_12_red_snooker_balls.html ). The interviewer started with an 8 ball scenerio, then we went to 12 balls. In each case I had to sweat out the answer on a white board b/c it wasn't all that obvious to me.
Anywho, the last question from the interviewer for this topic was to create an algorithm for any number of balls to determine the minimum number of weight measurements needed to distingush the heavier ball. My simpleton answer was x/4 because 2 is the minimum measurement for 8 balls and 3 is the minimum for 12 balls. The interviewer thought my formula was pretty funny b/c the answer should have been an algorithm and not linier function. I thought he was pretty funny b/c I'm not working in a pool hall, but that's a different matter.
If you're familiar with the problem and you know the algorithm, could you please post it or send it to me (email address is in my profile).
Gratefully yours,
seadalton
I had a job interview today with the ole 12 snooker ball question (a similar question is hosted on this PF site: physicsforums.com/archive/t-40405_12_red_snooker_balls.html ). The interviewer started with an 8 ball scenerio, then we went to 12 balls. In each case I had to sweat out the answer on a white board b/c it wasn't all that obvious to me.
Anywho, the last question from the interviewer for this topic was to create an algorithm for any number of balls to determine the minimum number of weight measurements needed to distingush the heavier ball. My simpleton answer was x/4 because 2 is the minimum measurement for 8 balls and 3 is the minimum for 12 balls. The interviewer thought my formula was pretty funny b/c the answer should have been an algorithm and not linier function. I thought he was pretty funny b/c I'm not working in a pool hall, but that's a different matter.
If you're familiar with the problem and you know the algorithm, could you please post it or send it to me (email address is in my profile).
Gratefully yours,
seadalton