--- title: "The MU puzzle" date: 2020-12-12T00:00:00+01:00 tags: math draft: true --- [Gödel, Escher, Bach](https://en.wikipedia.org/wiki/G%C3%B6del,_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. `Mx -> Mxx`, where `x` can be any string 2. `xIIIy -> xUy`, where `x` and `y` can be any strings 3. `xUUy -> xy`, where `x` and `y` can be any strings 4. `xI -> xIU`, where `x` can be any 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.
1. `xUy -> xIIIy`, where `x` and `y` can be any strings
Each string `Mx` in the new system now has a form using only an `M` followed by `I`s, constructed by expanding all `U`s. ``` MUIU => MIIIIIII MU => MIII MI => MI ``` Define the _value_ of a string to be the number of `I`s in it, after transforming it to its `MIIIIII...III` form. ``` value(MUIU) = 7 value(MU) = 3 value(MI) = 1 ``` The rules 2, 3, and 4 preserve value of a string modulo 3. The only operation which changes the number of `I`s is rule 1: it duplicates `x`, hence it doubles the value of a string. 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. Our goal is to show that, starting with a string of value not divisible by 3, every rule application cannot create a string with value divisible 3. So take a string `Mx` whose value is not divisible by 3. Rules 2, 3 and 4 preserve the value of `Mx` modulo 3, so by assumption the resulting string also has value not divisible by 3. Rule 1 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 is divisible by 3 iff 3 is one of its (unique) prime factors. By doubling a number the only prime factor we potentially add is 2, hence the resulting number also cannot have 3 as a prime factor, and is not divisible by 3. We can express this more succinctly as ``` ∀x. x ≠ 0 (mod 3) --> 2x ≠ 0 (mod 3) ``` # Characterizing all generatable strings We have just shown that any string with value divisible by three cannot be generated in the MIU system if starting from `MI`. The question remains: Which strings can we generate? Is it possible to generate all other strings `Mx`, i.e. all strings `Mx` such that `value(Mx) != 0 (mod 3)`? The answer turns out to be yes, using for example 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 `U`s in pairs. This is where rule 2 comes into play: if `value(My) != value(Mx) (mod 6)`, then there would 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 `UU`s, increases `value(My)` be 3, and everything works out. Step 4 is simple: `Mx` has the same value as `My` and we can use rule 2 to convert `III`s to `U`s, 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*](https://en.wikipedia.org/wiki/Decision_problem) which solves not only the MU-problem, but any decision problem of the form "Does candidate string `Mx` belong to the MIU-system?". 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](https://en.wikipedia.org/wiki/Turing_completeness) 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)](https://www.cs.virginia.edu/~robins/Turing_Paper_1936.pdf). This problem is also known as the [halting problem](https://en.wikipedia.org/wiki/Halting_problem), and is one of the most famous [undecidable problems](https://en.wikipedia.org/wiki/List_of_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](https://forums.swift.org/t/swift-type-checking-is-undecidable/39024), by showing that in order to solve type-checking, the compiler must solve the [word problem for finitely generated groups](https://en.wikipedia.org/wiki/Word_problem_for_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](https://lists.sr.ht/~bfiedler/public-inbox). If you are interested in undecidability in other programming languages, you might like [typing-is-hard.ch](https://typing-is-hard.ch).