コルモゴロフ複雑性定義の同等性

Kolmogorov-Complexityを定義するには多くの方法があり、通常、これらすべての定義は加法定数まで同等です。場合であることを$K_1$及び$K_2$コルモゴロフ複雑機能である（別の言語又はモデルを介して定義された）、その後一定存在$c$ようにそのすべての文字列のための$x$、$|K_1(x) - K_2(x)| < c$。これは、すべてのコルモゴロフ複雑度関数$K$およびxごとに$x$、K （x ）≤ | x | + c（定数 cの場合）。$K(x) \le |x| +c$$c$

チューリングマシンに基づいたKの次の定義に興味があります$K$

1. 状態の数：q状態のTMが空の文字列でxを出力するように、K 1（x ）を最小数qとして定義します。$K_1(x)$$q$$q$$x$
2. プログラムの長さ：定義：K 2（xは）出力がその最短「プログラム」されるようにxは。つまり、TMをバイナリ文字列にエンコードする方法を修正します。マシンのMは、のようにコードの（バイナリ）を示す⟨ M ⟩。 K 2（x ）= 分| ⟨ M ⟩ | ここで、最小値は、空の入力でxを出力するすべてのMのものです。$K_2(x)$$x$$M$$\langle M \rangle$$K_2(x) = \min |\langle M \rangle|$$M$$x$

あるK 1とK 2と同等か？それらの間の関係は何ですか？また、コルモゴロフの複雑さの概念をよりよく把握しているのは、それらが同等でない場合です。$K_1$$K_2$

特に私を悩ませているのは、xでのK 2の増加率で、これは超線形ではないように思われます（または| x | + cではなく、K 2 < C | x |のような定数C > 1で少なくとも線形）。xを出力する最も単純なTMを考えてみましょう- 状態と遷移関数の一部としてxを単にエンコードするTMです。それは見てすぐにあるという K 1（X ）≤ | x | + $K_2$$x$$C>1$$K_2 < C|x|$$|x|+c$$x$$x$$K_1(x) \le |x|+1$。しかし、同じマシンのエンコーディングははるかに大きく、私が得る自明な限界はK 2（x ）≤ | x | ログ| x | 。$K_2(x) \le |x|\log |x|$

I apologize in advance for that I give way too many details, but I'm about to contradict people.

About $K(x)≤K'(x)+c$

The fact that $K_1(x)≤K_2(x)+c$ usually comes from an interpreter of the description language #2 into the description language #1 and not from a translation from programs of #2 into programs of #1.

For example $K_\mathtt{C}(x)≤K_\mathtt{Python}(x)+c_\mathtt{py2c}$ and you get this inequality as simply as this:

void py_run(char * s) {
    // code of your Python interpreter
}

int main(void) {
    py_run("Put here your Python program of size Kpython(x)");
}

Then your constant $c_\mathtt{py2c}$ will be something like $528+490240688$ where $528$ is the number of bits for this code and $490240688$ bits is the size the official Python interpreter written in C. Of course you only need to interpret what is possible in your description language for Python so you can do better than 69 MB :-)

What is important is that you can write your Python program linearly in your C code. For example, a language where you need to put "BANANA" between every character is not a very good description program and the property is then false. (But if the description language authorizes you to write data in a separate file or in a block, this problem disappear)

Why your $K_1(x)=q$ is flawed

The problem with your definition of $K_1$ is that you may need more than $q$ bits to describe a Turing machine with $q$ states because you need to encode transitions.

So no $K_1$ and $K_2$ are probably not equivalent, but that's mainly $K_1$'s fault. I think that we can prove that for all $a>0$ there is a $c_a$ such that $K_1(x)≤a|x|+c_a$. Of course any $a<1$ is enough to disprove the fact that $K_1$ is not a valid function, since it would mean that we can encode more all $2^n$ possible strings of length $n$ into $an+c_a$ bits.

But the size is an incredibly tight bound when building Turing machines. The idea is that in a block of $b$ states there are $b^{2b}$ ways to find transitions for each state and that's better than the usual $2^b$ ways you can fill $b$ bits. Then you can store in each block $\log_2 b$ bits of information. (not $2\log_2 b$ because you have to go in and out the block one way or another)

So yeah... With blocks of size $2^{1/a}$ you could probably prove $K_1(x)≤a|x|+c_a$. But I already written way too much about why the number of states is not a valid Kolmogorov complexity function. If you want me to elaborate, I will.

Now about $K_2$

The naive descriptive language corresponds roughly to $K_2(x)=q\cdot 2 \cdot (\log_2 q + 2)$ (i.e. $\log_2q$ for each next state and details about write and termination).

As you seem to be, I'm convinced that a better/cheater way would be to authorize to encode "data" into Turing machine, maybe by adding a binary tag in the description language that says whether if a state is a data state (that just writes a bit and go to the next state) or if it does something else. That way you could store one bit of your $x$ in one bit of your descriptive language.

However if you keep the same $K_2$ you could use the same technique I used in the previous part to save a few bits, but I seem to be stuck at $K_2(x)≤a|x|\log|x|+c$ (for any $a>0$) .. maybe less than $\log|x|$, even, but obtaining $O(|x|)$ seems hard. (And I expect it should be $|x|$, not even $O(|x|)$.)