5e3a009a4d1c8cb055211e034b40f2fc47a16d17
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Ben Fiedler
3 months ago
2d1865c

Another update

1 files changed,21insertions(+),22deletions(-) R content/blog/{the-mu-puzzle.md => on-decidability-and-the-mu-puzzle.md}

R content/blog/the-mu-puzzle.md => content/blog/on-decidability-and-the-mu-puzzle.md +21 -22

@@ 1,7 1,7 @@--- title: "The MU puzzle" title: "On Decidability and the MU-puzzle" date: 2020-12-12T00:00:00+01:00 tags: math tags: math, decidability draft: true ---@@ 11,10 11,10 @@ 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 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 1. `xIIIy -> xUy`, where `x` and `y` can be any strings 1. `xUUy -> xy`, where `x` and `y` can be any strings 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@@ 58,25 58,22 @@ value(MU) = 3value(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. 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 2, 3 and 4 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 1 doubles the value of a string. However, by doubling any number which is 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. 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. 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@@ 88,10 85,10 @@ We can express this more succinctly asWe 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. 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 for example the following algorithm. The answer turns out to be yes, using (among others) 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)`.@@ 144,17 141,19 @@ problem](https://en.wikipedia.org/wiki/Halting_problem), and is one of the mostfamous [undecidable problems](https://en.wikipedia.org/wiki/List_of_undecidable_problems). These abstract problems can even have real world consequences: This year it was 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 by showing 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 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). interested in undecidability in other programming languages, you might also like this website I made: [typing-is-hard.ch](https://typing-is-hard.ch).