## Wednesday, August 11, 2010

### The Ferry Crossing Problem

I first encountered this logic problem in a book, probably sci-fi, probably in 4th or 5th grade, which is why it has made such a lasting impression. The long and short of it is there are 6 people, 3 good guys and 3 bad guys, and a boat that seats 2 people. Bad guys can never outnumber good guys and all the good guys need to get to the other side without the bad guys. How do you do it. I've always liked this problem even when, like this morning, I frantically IM devoted partner asking him what it's called (doesn't actually have a name), and if you don't want to take the time to figure it out, at the end of the post and after some carriage returns, I'll print the answer.

Now I like math and I like logic and I like to be smug about my skills in both. Which is why the personal ferry crossing problem that I am struck by is so horribly humiliating. And it comes up a lot.

I cannot cannot cannot wrap my mind around the math associated with two cars and two drivers and catch myself almost weekly abandoning logic in my mind for reasons I cannot explain. For some reason I find myself thinking: if devoted partner and I go to a bar in separate cars and I get too drunk to drive, devoted partner can drive us both home and then go back to get my car. I think this, in one way or another, ALL THE TIME. Take our upcoming solo airplane trips. I have thought: ok, I'll drive my car to devoted partner's office, then he can drive me to the airport in his car, and then take my car back home. I simply cannot comprehend that devoted partner is unable to drive two cars simultaneously while also being unable to understand why the idea begins in my head as a good one.

And frankly it's starting to drive me a little crazy. I mean even the dullest crayon in the box figures out AFTER A YEAR that two cars require two drivers. But then there's me, happily plotting impossible itineraries.

Solution
G=good guy

First crossing: G + B
Return trip: G
Second crossing: B + B
Return: B
Third: G + G
Return: G + B
Fourth: G + G
Return: B