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---
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.