所有原始词集都是主要语言吗？

如果没有单词v且k > 1从而w = v k，则单词$w$称为原始。字母Σ上所有原始词的集合Q是一种众所周知的语言。在WLOG中，我们可以选择Σ = { a ，b }。$v$$k > 1$$w = v^k$$Q$$\Sigma$$\Sigma = \{ a,b \}$

甲语言$L$是素数，如果对于每一种语言$A$和$B$与$L = A \cdot B$我们有$A = \{\epsilon\}$或$B = \{ \epsilon \}$。

Q是素数吗？

随着SAT的帮助下解算器，我可以表明我们要么$\{a,b\} \subseteq A$或$\{a,b\} \subseteq B$另有$\{ ababa, babab \} \subset Q$不能被分解成$A$和$B$，但此后一直停留。

答案是肯定的。假设我们有一个分解$Q = A\cdot B$。

一个简单的观察是，$A$和$B$必须是不相交的（因为对于$w\in A\cap B$我们得到$w^2\in Q$）。特别地，仅一个$A,B$可以包含$\epsilon$。我们可以假设wlog（因为另一种情况下是完全对称的），其$\epsilon\in B$。然后，因为$a$和$b$不能被分解成非空的因素，我们必须有$a,b\in A$。

接下来，我们得到的是$a^mb^n\in A$（并且，类似地完全，$b^ma^n\in A$）的所有$m,n>0$通过感应上$m$：

对于$m=1$，由于$ab^n\in Q$，就必须有$ab^n = uv$与$u\in A, v\in B$。由于$u\neq\epsilon$，$v$必须是$b^k$一些$k\le n$。但是，如果$k>0$，则由于$b\in A$我们得到$b^{1+k}\in Q$，矛盾。所以$v=\epsilon$，和$ab^n\in A$。

用于感应步骤中，由于$a^{m+1}b^n\in Q$我们有$a^{m+1}b^n = uv$与$u\in A, v\in B$。由于再次$u\neq\epsilon$，我们要么$v = a^kb^n$一段$0<k<m+1$，或$v=b^k$一些$k<n$。但是，在前者的情况下，$v$已经是$A$归纳假设，因此$v^2\in Q$，矛盾。在后者的情况下，就必须有$k=0$（即，$v=\epsilon$），因为从$b\in A$我们得到$b^{1+k}\in Q$。所以$u=a^{m+1}b^n\in A$。

现在考虑的与原始词语的一般情况下$r$之间交替$a$和$b$，即$w$是任$a^{m_1}b^{n_1}\ldots a^{m_s}b^{n_s}$，$b^{m_1}a^{n_1}\ldots b^{m_s}a^{n_s}$（对于$r=2s-1$），$a^{m_1}b^{n_1}\ldots a^{m_{s+1}}$, or $b^{m_1}a^{n_1}\ldots b^{m_{s+1}}$ (for $r=2s$); we can show that they are all in $A$ using induction on $r$. What we did so far covered the base cases $r=0$ and $r=1$.

For $r>1$, we use another induction on $m_1$, which works very much the same way as the one for $r=1$ above:

If $m_1=1$, then $w=uv$ with $u\in A, v\in B$, and since $u\neq\epsilon$, $v$ has fewer than $r$ alternations. So $v$ (or its root in case $v$ itself is not primitive) is in $A$ by the induction hypothesis on $r$ for a contradiction as above unless $v=\epsilon$. So $w=u\in A$.

If $m_1>1$, in any factorization $w=uv$ with $u\neq\epsilon$, $v$ either has fewer alternations (and its root is in $A$ unless $v=\epsilon$ by the induction hypothesis on $r$), or a shorter first block (and its root is in A unless $v=\epsilon$ by the induction hypothesis on $m_1$). In either case we get that we must have $v=\epsilon$, i.e. $w=u\in A$.

The case of $Q' := Q\cup\{\epsilon\}$ is rather more complicated. The obvious things to note are that in any decomposition $Q = A\cdot B$, both $A$ and $B$ must be subsets of $Q'$ with $A\cap B = \{\epsilon\}$. Also, $a,b$ must be contained in $A\cup B$.

With a bit of extra work, one can show that $a$ and $b$ must be in the same subset. Otherwise, assume wlog that $a\in A$ and $b\in B$. Let us say that $w\in Q'$ has a proper factorization if $w=uv$ with $u\in A\setminus\{\epsilon\}$ and $v\in B\setminus\{\epsilon\}$. We have two (symmetric) subcases depending on where $ba$ goes (it must be in $A$ or $B$ since it has no proper factorization).

• If $ba\in A$, then $aba$ has no proper factorization since $ba,a\notin B$. Since $aba\in A$ would imply $abab\in A\cdot B$, we get $aba\in B$. As a consequence, $bab$ is neither in $A$ (which would imply $bababa\in A\cdot B$) nor in $B$ (which would imply $abab\in A\cdot B$). Now consider the word $babab$. It has no proper factorization since $bab\notin A\cup B$ and $abab,baba$ are not primitive. If $babab\in A$, then since $aba\in B$ we get $(ba)^4\in A\cdot B$; if $babab\in B$, then since $a\in A$ we get $(ab)^3\in A\cdot B$. So there is no way to have $babab\in A\cdot B$, contradiction.
• The case $ba\in B$ is completely symmetric. In a nutshell: $bab$ has no proper factorization and cannot be in $B$, so it must be in $A$; therefore $aba$ cannot be in $A$ or $B$; therefore $ababa$ has no proper factorization but also cannot be in either $A$ or $B$, contradiction.

I am currently not sure how to proceed beyond this point; it would be interesting to see if the above argument can be systematically generalized.