~bfiedler/website

a32c9ebf50f8399c34e5712ada692b52926644eb — Ben Fiedler 2 months ago f03adbc
minor improvements
1 files changed, 4 insertions(+), 4 deletions(-)

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 +4 -4
@@ 77,14 77,14 @@ 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`
Thus `2*n` is not divisible by 3, and we can never construct the string `MU`,
starting from `MI`.


# Characterizing all generatable strings

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
Not only have we shown that `MU` is not constructible, starting from `MI`, but
also any other string with a value divisible by 3. 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)`?



@@ 99,7 99,7 @@ The answer turns out to be yes, using the following algorithm.
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 even, and 2 otherwise. Since
2, 1, 2, 1, 2, ...`, i.e. `2^i (mod 3)` is `1` if `i` is even, and `2` otherwise. Since
`value(Mx) != 0` by assumption, there always exists a longer string `My` such that
`value(My) = value(Mx) (mod 3)`.