Gödel, Escher, Bach takes the reader on a journey through mind, music, machines and self-reference. In the first few chapters, Hofstadter introduces a formal system called the MIU-system. The MIU-system consists of four simple rules for manipulating strings consisting of the characters M, I and U.

1. xI -> xIU, where x matches the rest of the string
2. Mx -> Mxx, where x matches the rest of the string
3. xIIIy -> xUy, where x and y match the rest of the string
4. xUUy -> xy, where x and y match the rest of the string

Note that the placeholders x and y must always match the entire string, i.e. the application MII -> MIII, choosing x = I, is not valid. The correct application is MII -> MIIII.

Then Hofstadter asks the reader to answer the MU puzzle:

Given the initial string MI, is it possible to construct the string MU using only the four above rules?

Take a few minutes and try for yourself. Many people quickly suspect that it is impossible, but why?

The solution

Let us add an additional, imaginary rule.

Each string Mx in the new system now has a form using only an M followed by Is, constructed by expanding all Us.

   MUIU
=> MIIIIIII

   MU
=> MIII

   MI
=> MI

Define the value of a string to be the number of Is in it, after transforming it to its MIIIIII...III form.

value(MUIU) = 7
value(MU) = 3
value(MI) = 1

The value of our target MU is 3, which is divisible by 3, while the value of our starting string MI is 1, which is not divisible by 3. If we can show that, starting with a string of value not divisible by 3, every rule application cannot create a string with value divisible 3, then it is also impossible to get MU by starting with MI.

So take a string Mx whose value is not divisible by 3. Rules 1, 3 and 4 preserve the value of Mx modulo 3, so by assumption the resulting string also has value not divisible by 3.

Rule 2 doubles the value of a string. However, by doubling any number which is not a multiple of 3 we can never create a number divisible by 3: a number n is divisible by 3 iff 3 is one of its prime factors. When doubling n the only prime factor we add is 2, hence the resulting number also cannot have 3 as a prime factor. We can express this more succinctly as

∀x. x ≠ 0 (mod 3) --> 2x ≠ 0 (mod 3)

Thus 2*n is not divisible by 3, and we can never construct the string MU, starting from MI.

Characterizing all generatable strings

Not only have we shown that MU is not constructible, starting from MI, but also any other string with a value divisible by 3. The question remains: which strings can we generate? Is it possible to generate all other strings, i.e. all strings Mx such that value(Mx) != 0 (mod 3)?

The answer turns out to be yes, using the following algorithm.

1. Generate My = MIIIIII...III by applying rule 1 to MI, such that the following holds: the value of My is larger than Mx and value(My) = value(Mx) (mod 3).
2. Append U if value(My) != value(Mx) (mod 6).
3. Merge IIIIII to UU and delete until value(My) == value(Mx).
4. Replace the MIIII...III with Mx by applications of rules 2 and 3.

It is always possible to apply step 1: the infinite sequence of strings generated by repeatedly applying rule 1 to MI has values 1, 2, 4, 8, 16, 32, ..., generating all powers of 2. Taking these values modulo 3 we get 1, 2, 1, 2, 1, 2, 1, 2, ..., i.e. 2^i (mod 3) is 1 if i is even, and 2 otherwise. Since value(Mx) != 0 by assumption, there always exists a longer string My such that value(My) = value(Mx) (mod 3).

In step 3 we need to delete U pairs until we have that value(My) = value(Mx). Unfortunately, we can only decrease value(My) in steps of six, since we can only remove Us in pairs. This is where rule 2 comes into play: if value(My) != value(Mx) (mod 6), then there would always be one U left over. (Note: since these values are congruent modulo 3, the only possible case is that value(Mx) == value(My) + 3 (modulo 6)). Appending an additional U before deleting UUs, increases value(My) by 3, and everything works out.

Step 4 is simple: Mx has the same value as My and we can use rule 2 to convert IIIs to Us, in the right positions. Thus we have shown that the MIU system lets us generate precisely the strings which have a value not divisible by 3.

The MIU-system and decidability

The MIU-system isn’t just a neat puzzle to solve: Hofstadter shows the reader that some questions about formal systems cannot be answered solely from within. Rather, we had to step outside the restrictions placed upon us by the four rules to successfully answer the question.

Given infinite time we could have concluded this ourselves, by generating all possible strings: however, in this case there exists a solution which is finite. We have constructed a decision procedure which solves not only the MU-problem, but any decision problem of the form “Does candidate string Mx belong to the MIU-system, starting from MI?”.

It is not always possible to find a finite decision procedure. Take for example all strings which are valid C programs (or choose any other sufficiently powerful language, it doesn’t matter). The decision problem “Does a given C program terminate at some point?” is not solvable in finite time, as shown by Alan Turing (1936). This problem is also known as the halting problem, and is one of the most famous undecidable problems.

These abstract problems can even have real world consequences: this year it was shown that type-checking a Swift program is also an undecidable problem. The author shows that in order to type-check a program, the compiler must solve the word problem for finitely generated groups. I like how this example shows us that some abstract problems pop up in unexpected places, and why seemingly purely theoretical knowledge matters, even for applied problems such as building compilers.

If you have any questions or comments feel free to reach out to me via my public inbox. If you are interested in undecidability in other programming languages, you might also like this website I made: typing-is-hard.ch.