## Friday, August 13, 2010

### P ≠ NP (part 1)

I'm sure most people who read this blog know a bit about what this means, but I want to try to explain it in a way my mom could understand. I won't go into excruciating detail, and I'll probably gloss over interesting points.

Let's start with an analogy. Suppose you go to the store to buy a present for your nephew. He likes jigsaw puzzles, so you want to get him one. He's going to be twelve years old, so you need to find one that has the appropriate challenge. A simple one with four pieces might be great for a two year-old, but your nephew would find that boring. A big one with ten thousand pieces is probably too much for him. You know he just finished one with five hundred pieces, so you pick one with six hundred pieces because it won't be too easy, but not too frustrating. What you are doing is estimating the complexity of the puzzle.

When you get to the store you find that they are sold out of jigsaw puzzles. They have a variety of other puzzles, though. They have a Rubik's cube, a Sudoku book, some cryptograms, a Tower of Hanoi, etc. They have models you can put together as well. But these things aren't jigsaw puzzles. You can't estimate the complexity the same way. A model with five hundred pieces would be quite a bit more difficult to assemble than a jigsaw puzzle with the same number of pieces.

Perhaps you look at the age ratings for the various puzzles. The eight-piece Tower of Hanoi puzzle is rated for ages ten and up, so maybe a twelve piece? The standard 3x3x3 Rubik's cube is rated for ages eight and up, but would the 4x4x4 be too hard or too easy? What about the 5x5x5?

The kind of rating system you use to describe the difficulty of a puzzle is analogous to the “complexity class” of a problem. The complexity class roughly describes how hard a problem will be to solve, and more importantly, how much harder ‘big’ problems are compared to the ‘small’ ones. Here's what I mean: if I take a jigsaw puzzle, any jigsaw puzzle, and double the number of pieces in it, I'll make it roughly twice as hard (more or less). If I take a cryptogram and double the number of words in it, I'll actually make it a lot easier. If I take the Tower of Hanoi and double the number of discs, I'll make it incredibly harder, maybe even impossible. These puzzles are in different complexity classes.

As it turns out, a lot of puzzles are in the same complexity class as the jigsaw puzzle: the difficulty is reasonably proportional to the number of pieces. Adding a piece or two doesn't change things that much, and it is easy to figure out how long it will take to solve if you've done a few of them. A lot of puzzles are more like the Tower of Hanoi. Adding a single piece will double the amount of time it takes to solve them.

P and NP are complexity classes for problems you might want to solve on a computer. There are all sorts of rating systems, but everyone is familiar with the MPAA movie ratings. There are all sorts of complexity classes, but most computer scientists are familiar with P and NP.

I'll pause here because the analogy is wearing thin. The main point is that “P ≠ NP” is a statement about complexity classes. In particular, it talks about two of the most popular complexity classes and that is part of the reason it is interesting.

John Cowan said...

My wife's the jigsaw puzzlist in the family, and she's not sure (too many confounding factors), but my intuition says that the difficulty of jigsaw puzzles grows as O(N^2), because at least at the beginning you have to find out which of N pieces matches with which of N-1 other pieces.

But of course quadratic growth is not really much worse than linear in this regime.

Brian (Binder) Makin said...

The brute force approach could indeed be N^2 but luckily we don't have to solve them that way. A 10k piece puzzle takes hours (for my mother and I anyway).

If you could do a 1000 piece puzzle in one hour and it grew at N^2 then the 10k piece puzzle should take 100 hours.

Unknown said...
This comment has been removed by the author.
Unknown said...

The Towers of Hanoi has at least one simple algorithm, so you can do it with any disks.

It is never impossible.

Joe Marshall said...

Towers of Hanoi is simple enough, but it is impractical. Although a tower of, say, thirty disks is possible to solve in some theoretic sense, it is impractical for a human being to carry out.