计算一个二进制串的Lempel-谢夫（LZ）复杂性（又名序列复杂性） [英] calculating Lempel-Ziv (LZ) complexity (aka sequence complexity) of a binary string

问题描述

我需要计算一个二进制字符串的LZ复杂性。 LZ的复杂度是遇到作为流differencet子的数量从开始时到结束时被观看。作为一个例子：

S = 1001111011000010

在不同的子标记序列的复杂性C（S）= 6： S = 1/0/01 /一千一分之一千一百一十/ 0010 /

有人可以指导我找到一个简单的解决方案？我相信应该有一些很直接的实现这一众所周知的问题，但我也很难找到他们。能不能做到简单地构建后缀树或类似的东西做的。如果是的话，究竟怎么了？和我应该怎么办？

有人知道的任何C / C ++源$ C ​​$ C来完成任务？

在此先感谢。

要明确的答案提出的树的构建。请问树是这个样子？

  0
       / \
      ○○
     / \ / \
    ○○○○
       / /
      ○○
 

解决方案

1 0 01 11 10 110 00 010
序列的复杂性是8，因为分区8不是6 - 1/0/01/11/10/110/00/010

I need to calculate the LZ-complexity of a binary string. The LZ-complexity is the number of differencet substrings encountered as the stream is viewed from begining to the end. As an example:

s = 1001111011000010

Marking in the different substrings the sequence complexity c(s) = 6: s = 1 / 0 / 01 / 1110 / 1100 / 0010 /

can someone guide me to find a simple solution for that? I am sure there should be some very straight-forward implementations for this well-known problem, but I have difficulty finding them. Can it be done simply done with constructing a suffix tree or something similar. If yes, exactly how? and what should I do?

anyone knows of any c/c++ source code to accomplish the task?

thanks in advance.

to clarify the construction of the tree suggested in the answers. Does the tree looks like this?

         o
       /   \
      o     o
     / \   / \
    o   o o   o
       /     /
      o     o

解决方案

1 0 01 11 10 110 00 010
Complexity of sequence is 8 because the partitions are 8 not 6 - 1/0/01/11/10/110/00/010

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