Quote:
Originally Posted by Wyman
Though, if you knew that the "odd" ball was heavier, you could identify it in 2 weighings -- so you have to be careful with these kind of arguments.
I figured it out I think. I'm getting 3 weighings, which is consistent with log base 2 of 8.
So, if the balls are numbered 1-8, take 1-3 and weigh it against 4-6.
If they are equal, weigh 7v8. If these are equal, we're done. If they are different, weigh one against ball 1. Then you can figure out which one is the odd one out, and whether or not its heavier or lighter.
If the 3-3 weighing is different, without loss of generality, assume 123 is the heavier group of balls.
Now, weigh 1 and 2 against 3 and 7 (since we know 7 is a normal ball). If they are different, (say 1 and 2 are heavier), we can weigh them against each other to find the wrong ball.
If they are the same, weigh 4 and 5 (two of the lighter balls). If they're the same, 6 is the light ball. Otherwise, you will know which ball is lighter.
If we generalize this to n balls, with n-1 identical, would the number of weighings be the smallest integer greater than or equal to log base 2 of n?