a32c9ebf50f8399c34e5712ada692b52926644eb
—
Ben Fiedler
5 months ago
f03adbc

minor improvements

1 files changed,4insertions(+),4deletions(-) 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)`.