From a32c9ebf50f8399c34e5712ada692b52926644eb Mon Sep 17 00:00:00 2001 From: Ben Fiedler Date: Sun, 13 Dec 2020 00:08:06 +0100 Subject: [PATCH] minor improvements --- content/blog/on-decidability-and-the-mu-puzzle.md | 8 ++++---- 1 file changed, 4 insertions(+), 4 deletions(-) diff --git a/content/blog/on-decidability-and-the-mu-puzzle.md b/content/blog/on-decidability-and-the-mu-puzzle.md index cf60ffb..82b203a 100644 --- a/content/blog/on-decidability-and-the-mu-puzzle.md +++ b/content/blog/on-decidability-and-the-mu-puzzle.md @@ -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)`. -- 2.30.1