Science Puzzle
Two-Pan Boudin Balance
You have nine sausages that look identical. One is slightly heavier than the rest. You have a two-pan balance scale but no weights.
What is the minimum number of weighings needed to guarantee finding the heavier sausage, and how do you do it?
The Answer
Two weighings. Divide the nine sausages into three groups of three. Put group A on the left pan and group B on the right. Set group C aside.
If one side is heavier, the odd sausage is in that group. If the pans balance, it is in group C. Either way, after one weighing you have narrowed the search to three sausages.
For the second weighing, take two sausages from the known heavy group and place one on each pan. If one side is heavier, that is the odd sausage. If they balance, the third sausage (set aside) is the heavy one.
This is a ternary search: each weighing divides the remaining candidates into three equal groups and eliminates two of them, which is the most information a balance scale can give per weighing.
The principle: Ternary search. A balance scale gives three outcomes (left heavy, right heavy, equal), so the optimal strategy divides candidates into three equal groups each time, not two.