The MIU System
The first exercise introduced in Gödel, Escher, Bach is the MIU System. This post describes the system and how to solve it.
I’m currently reading “Gödel, Escher, Bach” by Douglas Hofstadter, which is generally about strange loops.
A strange loop is a cyclic structure that goes through several levels in a hierarchical system. It arises when, by moving only upwards or downwards through the system, one finds oneself back where one started. - Wikipedia
One of the strange loops Hofstadter describes is a formal system. The first formal system the author describes is the MIU System. The MIU System has four rules:
- If you possess a string whose last letter is
I, you can add aUat the end. - Suppose you have
Mx. Then you may addMxxto your collection. - If
IIIoccurs in one of the strings in your collection, you may make a new string withUin place ofIII. - If
UUoccurs inside one of your strings, you can drop it.
The goal of the book is to get from MI to MU. I wrote an applet where you can try it out yourself. Part of the blog post is also available there.
So, how to get from I to the U?
The only rule that can reduce the number of Is is the 3rd rule, so we should focus on that. To remove all Is, we need to ensure count(I) mod 3 = 0, which means that the number of Is must be divisible by 3. "count" represents the number of Is, "mod" is the mathematical operation modulo.
With the second rule, we can duplicate the starting I. But doubling a number that is not divisible by 3 will never make it divisible by 3.
Proof sketch:Suppose you have a number
nthat is not divisible by 3, son mod 3 != 0, which means thatn = x * 3 + y, wherexcan be any natural number andyis 1 or 2. When doubling the number, we getn * 2 = x * 6 + y * 2.y * 2is now 2 or 4, both numbers are still not divisible by 3.x * 6is divisible by 3, so adding 2 or 4 will make it not divisible by 3.
Reducing the number of Is by 3 obviously won’t help; we can only use this rule in a useful way when we already have a number that is divisible by 3. All other rules don’t affect the number of Is.
So the first exercise given by Douglas Hofstadter in “Gödel, Escher, Bach” is impossible to solve.
If you don’t believe this proof, simply try it out and check if you’re able to find a solution. You won’t.
But why give the reader an exercise that is not solvable?
Because one encounters a lot of loops in the system. Every time you think, “I might try this,” you’ll get to a previous point, and you notice that the whole system loops itself. You figure out that you can’t escape, and it doesn’t matter how often or long you try, you’ll always go back to where you started.