b263ba19907ec893d4d51fd1f56e23d98d736cf9
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Ben Fiedler
5 months ago
2082b64

add new draft

1 files changed,117insertions(+),0deletions(-) A content/blog/the-miu-system.md

A content/blog/the-miu-system.md => content/blog/the-miu-system.md +117 -0

@@ 0,0 1,117 @@--- title: "The MIU system" date: 2020-12-12T00:00:00+01:00 tags: math draft: true --- In [Gödel, Escher, Bach](https://en.wikipedia.org/wiki/G%C3%B6del,_Escher,_Bach), 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`. Hofstadter asks the following: > Given the initial string `MI`, is it possible to construct the string `MU` using only the four rules above? Take a few minutes and try for yourself. Many people quickly suspect that it is impossible, but why? Let us add an additional, imaginary rule. <ol start="5"> <li>`xUy -> xIIIy`, where `x` and `y` can be any strings</li> </ol> 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. those with value not divisible by three? The answer turns out to be yes, using a simple 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 the value of `My` is the value of `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 odd, 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) + 3 == value(My) (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.