The MU Puzzle Solution
My very first blog post was about the MU Puzzle from Gödel, Escher, Bach by Douglas Hofstader. I promised that I would post a follow up with the solution. That was a year and a half ago.
I had decided against writing one, until I got an email from an actual person who read my blog:
Hi Mitchel. I saw your blog post about getting mu from mi. This problem has been driving me crazy. Everywhere online says that no number not divisible by three can be doubled and be divisible by three. Is this even true? What about 2^54. I believe is divisible by three. Hope you can help. - Andrew
Well now I have to write a post, don’t I? (Also there’s two L’s in my name but that’s fine. Spelling is made up anyway.)
So, as Andrew’s figured out, the key to the puzzle has something to do with the number of I’s in our word. We’re trying to get from MI to MU, but somehow we can’t get rid of all those pesky I’s! When I started this puzzle, I just wrote down some attempts:
MIU (Rule 1)
MIUIU (Rule 2)
MIUIUIUIU (Rule 2)
MII (Rule 1)
MIIII (Rule 1)
MIU (Rule 3)
(We’ve seen this before!)
MII (Rule 1)
MIIII (Rule 1)
MIIIIIIII (Rule 1)
MIIIIIU (Rule 3)
MIIUU (Rule 3)
MII (Rule 4)
And so on. A couple things quickly become apparent. First of all, every word will always start with an M. That’s because we start with MI, and none of our rules change the first letter. Second, no other letters beside the first one will ever be an M.
Third, and most importantly, it seems really hard to get rid of all the I’s. If you play around with the rules long enough, you might start to suspect that it’s impossible. And indeed, that is what we’ll show.
We want to show that
No matter how you apply the 4 production rules to MI, the result will always have an I in it.
This is the same as saying you can’t get from MI to MU, because MU doesn’t have an I in it.
Now, if you look at the examples above you’ll notice a pattern emerge. The number of I’s either doubles (Rule 1), or decreases by 3 (Rule 3). Rules 2 and 4 don’t affect the number of I’s. Since Rule 3 is the only way to get rid of I’s, we’ll need to figure out whether it’s possible to make the number of I’s a multiple of 3.
I’ll claim this isn’t possible. Why? Because if you start with one I, then you can never get a multiple of three by any combination of doubling or subtracting 3. First, notice that subtracting 3 from a number will not change whether it is a multiple of 3. Second, notice that doubling a number will not change whether it is a multiple of 3. This is easier to see pictorally, so I made a short animation below. (You can refresh to restart.)
There are 3 disjoint cases. Either the number is divisible by 3, in which case doubling it gives us twice as many groups of 3. If the number is not divisible by 3, then it has either 1 or 2 left over. If it has 1 left over, then doubling that 1 gives us 2, which is not divisible by 3. If we have 2 left over, then doubling those 2 gives us 4, which is also not divisible by 3.
More mathematically, say we have some number \(m\). Then
If \(m = 3n\) then \(2m = 2(3n) = 3(2n)\)
If \(m = 3n + 1\) then \(2m = 2(3n + 1) = 3(2n) + 2\)
If \(m = 3n + 2\) then \(2m = 2(3n + 2) = 3(2n+1) + 1\)
So if a number is divisible by 3, it will still be divisible by 3 after doubling. If a number is not divisble by 3, it will still not be divisible by 3 after doubling. This means no matter how many times we apply Rule 1, we’ll never have the right number of I’s to get rid of all of them with Rule 3.
Andrew, however, seems to have some doubts about this proof. Specifically, what about 2^54? Sure, the proof seems unassailable. But most proofs do until you prove them wrong. Can we be sure that it’s the same even if we multiply a huge number of times? You’ve never seen someone calculate 2^54 have you?
Well, fortunately we have computers to crunch these big calculations for us. In fact, 2^54 isn’t that big; you only need a 54 bit integer to represent it in binary. Most computers use 64 bits by default, but obviously if you’re clever you can handle much bigger numbers.
Anyway, if we use the proof above, we would predict that 2^54 should have exactly one remainder when divided by 3. This is because each time you multiply by two, you flip flop between having one remainder and having two remainders.
And indeed, if you use this big decimal calculator you can see that 2^54 = (6004799503160661 * 3) + 1. If you don’t believe me, feel free to write it out.