3aa9cda6110c3e53813b60533161cfe1c3557bc5
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Ben Fiedler
5 months ago
af91967

publish miu post

1 files changed,33insertions(+),33deletions(-) M content/blog/on-decidability-and-the-mu-puzzle.md

M content/blog/on-decidability-and-the-mu-puzzle.md => content/blog/on-decidability-and-the-mu-puzzle.md +33 -33

@@ 1,8 1,7 @@--- title: "On Decidability and the MU-puzzle" date: 2020-12-12T00:00:00+01:00 tags: [math, decidability] draft: true title: "On Decidability and the MU puzzle" date: 2020-12-13T00:00:00+00:00 tags: [math, decidability, short-and-sweet] --- [Gödel, Escher, Bach](https://en.wikipedia.org/wiki/G%C3%B6del,_Escher,_Bach)@@ 11,16 10,16 @@ In the first few chapters, Hofstadter introduces a formal system called theMIU-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` can be any string 1. `Mx -> Mxx`, where `x` can be any string 1. `xIIIy -> xUy`, where `x` and `y` can be any strings 1. `xUUy -> xy`, where `x` and `y` can be any strings 1. `xI -> xIU`, where `x` matches the rest of the string 1. `Mx -> Mxx`, where `x` matches the rest of the string 1. `xIIIy -> xUy`, where `x` and `y` match the rest of the string 1. `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: 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?@@ 58,37 57,38 @@ value(MU) = 3value(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`. 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 is divisible by 3 iff 3 is one of its (unique) prime factors. When doubling, the only prime factor we potentially add is 2, hence the resulting number also cannot have 3 as a prime factor. Thus it is not divisible by 3. We can express this more succinctly as 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 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, i.e. all strings `Mx` such that `value(Mx) != 0 (mod 3)`? Not only have we shown that `MU` is not constructible, but also any other string with a value divisible by 3, starting from `MI`. 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 (among others) the following algorithm. 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)`.@@ 105,11 105,11 @@ generated by repeatedly applying rule 1 to `MI` has values `1, 2, 4, 8, 16, 32,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 since we can only remove `U`s 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 `UU`s, increases `value(My)` be 3, and everything works out. deleting `UU`s, 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 `III`s to `U`s, in the right positions. Thus we have shown that the@@ 128,7 128,7 @@ 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?". `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@@ 143,8 143,8 @@ 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 type-check a program, the compiler must solve the problem](https://forums.swift.org/t/swift-type-checking-is-undecidable/39024). The author shows that in order to type-check a program, 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