1 files changed, 33 insertions(+), 33 deletions(-)

M content/blog/on-decidability-and-the-mu-puzzle.md

@@ 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 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` 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) = 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`.
+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