This weekend I thought about the flashlight and 8 batteries riddle:

You have a flashlight that works with 2 batteries. You also have 8 batteries, of which 4 are empty and the other 4 are full. There is no way to tell which battery is empty and which one is full, but you can put 2 batteries in the flashlights. If both of them are full, the flashlight will turn on.

What is the minimum number of tries that will guarantee that the flashlight will turn on?

At first, I explored the possible combinations, then I considered playing with probability trees. But then I put this on paper, and soon, things became much clearer. I represented the batteries with dots, and the tries with lines, and instead of playing with abstract concepts, I started to play with lines and dots:

First I found a solution that would try 7 combinations, and would guarantee that the 8th was correct:

But somehow I knew that the solution had to be 7 tries, not 8. So I kept moving the lines and connecting the dots until the bulb lit (pun intended):

No matter where the 8th full battery is, there will be a line connecting it with another full battery. (You can also find a video here).

This shows the importance of visualizing your problem before being able to come up with the answer. I am using a Moleskine notebook and a Baron Fig Squire pen (some say it’s the best pen in the world 🙂 ).