关于\(LCP\)有如下两个公式:

  • \(LCP~Lemma:\) 对任意 \(1\le i<j<k\le n\) ,存在 \(LCP(i,k)=min\{LCP(i,j),LCP(j,k)\}\) 成立。
  • \(LCP~Theorem:\) 对任意 \(i<j\),存在 \(LCP(i,j)=^{~~~~~min}_{i+1 \le k \le j}\{LCP(k-1,k)\}\) 成立。

\(LCP~Lemma\) 的证明:

\[p=min\{LCP(i,j),LCP(j,k)\}\]
则有 \[LCP(i,j) \ge p,LCP(j,k) \ge p\]
可得 \[LCP(i,k) \ge p\]
又设 \[LCP(i,k)=q>p\]
\(Suffix_i\)\(Suffix_k\)\(q\)个字符相同。即:\[Suffix_{i,1}=Suffix_{k,1}\] \[Suffix_{i,2}=Suffix_{k,2}\] \[…\] \[Suffix_{i,q}=Suffix_{k,q}\]
\[min{LCP(i,j),LCP(j,k)}=p\]
说明 \[Suffix_{i,p+1}!=Suffix_{j,p+1}~~\text{或}~~Suffix_{j,p+1}!=Suffix_{k,p+1}\]
那么一定有如下式子成立 \[Suffix_{i,p+1}!=Suffix_{k,p+1}\]
于是,\(q>p\)不成立,即\[LCP(i,k) \le p\]
于是\[LCP(i,k)=p=min\{LCP(i,j),LCP(j,k)\}~~\text{得证。}\]

\(LCP~Theorem\) 的证明:

\(LCP~Lemma\)得:\[LCP(i,j)=min\{LCP(i,i+1),LCP(i+1,j)\}\]
又:\[LCP(i+1,j)=min\{LCP(i+1,i+2),LCP(i+2,j)\}\]
经归纳得:\[LCP(i,j)=^{min}_{i<k \le j}\{LCP(k-1,k)\}\text{得证。}\]

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