算法补充之KMP作用原理（一）

1. 整体把握

　　一般都可以想到 $O(n^2)$ 的暴力算法，KMP改进暴力算法使其算法复杂度为O(m+n)。本文参考：维基百科。

2.算法学习

2.1 使用局部匹配表

To illustrate the algorithm’s details, consider a (relatively artificial) run of the algorithm, where W = “ABCDABD” and S = “ABC ABCDAB ABCDABCDABDE”. At any given time, the algorithm is in a state determined by two integers:

m, denoting the position within S where the prospective match for W begins,
i, denoting the index of the currently considered character in W.

　　通过例子学习KMP，先从以上定义确定W、S、m、i分别代表的意义，然后看几个规则：

For the moment, we assume the existence of a “partial match” table T, described below, which indicates where we need to look for the start of a new match in the event that the current one ends in a mismatch.局部匹配表的意义

The entries of T are constructed so that if we have a match starting at S[m] that fails when comparing S[m + i] to W[i], then the next possible match will start at index m + i - T[i] in S (that is, T[i] is the amount of “backtracking” we need to do after a mismatch). 局部匹配表如何使用

This has two implications: first, T[0] = -1, which indicates that if W[0] is a mismatch, we cannot backtrack and must simply check the next character;

and second, although the next possible match will begin at index m + i - T[i], as in the example above, we need not actually check any of the T[i] characters after that, so that we continue searching from W[T[i]].

　　假设我们有一个partial match table（如下图所示），重复上面查表的规则和方法，我们可以手动复现下面的KMP算法过程：
这里写图片描述
　　

　　特别强调：每次当前母串中 $m+i$ 位置失配时，查表中对应的i的值，然后将下一次匹配的位置设置为 $m + i - T[i]$ ，但是只需要从T[i]处开始匹配就可以了。

2.2 构造局部匹配表

没有优化的构造方法

这里写图片描述
　　由”ABCDABD”构造其局部匹配表如上图所示，过程如下：

We consider the example of W = “ABCDABD” first. We will see that it follows much the same pattern as the main search, and is efficient for similar reasons. We set T[0] = -1.T[0]默认为-1，注意当前是ｉ位置发生了失配

To find T[1], we must discover a proper suffix of “A” which is also a prefix of pattern W. But there are no proper suffixes of “A”, so we set T[1] = 0.这里需要了解前缀和后缀的定义

To find T[2], we see that the substring W[0] - W[1] (“AB”) has a proper suffix “B”. However “B” is not a prefix of the pattern W. Therefore, we set T[2] = 0.不同于说是AB子串的前缀，而强调了是W的前缀，这里为下面的简易判断方式做了铺垫。

3Continuing to T[3], we first check the proper suffix of length 1, and as in the previous case it fails. Should we also check longer suffixes?

No, we now note that there is a shortcut to checking all suffixes: let us say that we discovered a proper suffix which is a proper prefix (A proper prefix of a string is not equal to the string itself) and ending at W[2] with length 2 (the maximum possible); then its first character is also a proper prefix of W, hence a proper prefix itself, and it ends at W[1], which we already determined did not occur as T[2] = 0 and not T[2] = 1. Hence at each stage, the shortcut rule is that one needs to consider checking suffixes of a given size m+1 only if a valid suffix of size m was found at the previous stage (i.e. T[x] = m) and should not bother to check m+2, m+3, etc.通过反证法，证明了T[i]与T[i-1]的一些关系

　　理解最为困难的就是第二点，在T[3]上我们本应该检查前后缀长度为1和2，但是由于T[2]等于0，所以我们只需要检查长度为1的前后缀。用DP的思想很容易理解，因为T[i]的最长前后缀应该是T[i-1]的基础上再拓展一个字符。当然，并不一定T[3]的最长前后缀长度一定为T[i-1]+1，只是说我们需要检查的长度范围为0到T[i-1]+1。

优化的构造方法

　　有人称之为优化的KMP算法，实际上维基百科直接给的就是优化后的版本，而没有提及上面的优化前的版本，这种优化通过观察在构建局部匹配表上做了改进。维基上的内容不是很好理解，根据这篇，很快理解了优化的本质。如下公式所示：
　　 $if 　p[T[i]]==p[i]，则令T[i]=T[T[i]]$

这里写图片描述
　　　
　　在编程上的实现比思想上更加简单，只要在没有优化的版本增加判断条件，进行优化就可以了。

3. 伪代码

　　学习伪代码，然后尝试自己去实现，在某些字符串的题目中，KMP给我们提供了 $O(m+n)$ 的备用函数，结合深度搜索可以解决很多复杂的问题。
　　

algorithm kmp_search:
    input:
        an array of characters, S (the text to be searched)
        an array of characters, W (the word sought)
    output:
        an array of integers, P (positions in S at which W is found)
        an integer, nP (number of positions)

    define variables:
        an integer, j ← 0 (the position of the current character in S)
        an integer, k ← 0 (the position of the current character in W)
        an array of integers, T (the table, computed elsewhere)

    let nP ← 0

    while j < length(S) do
        if W[k] = S[j] then
            let j ← j + 1
            let k ← k + 1
            //如果k=length w 那么代表一个完整的匹配完成
            if k = length(W) then
                (occurrence found, if only first occurrence is needed, m may be returned here)
                //记录这个匹配信息 模式串开始位置位j-k(模式串长度) np+1，括号里面的话的意思，如果只要找到第一个匹配完成的位置，则这里直接return就可以了。
                let P[nP] ← j - k, nP ← nP + 1
                //重新开始，这里和失配的处理一样，因为即使这个匹配完成了，也存在前面的字符后面匹配的一部分，所以必须trackback
                let k ← T[k] //(T[length(W)] can't be -1)
        else
        //失配发生的时候，直接trackback
            let k ← T[k]
            //如果k<0的时候，直接跳到下一个字符
            if k < 0 then
                let j ← j + 1
                let k ← k + 1

　　仔细观察发现没有 $m$ 了，在上面的算法学习部分是有个 $m$ 标志模式串起始位置的，这里基于两个思考：
　　

m = m + k - T [k] + T [k] = m + k, m = j - k

$m=m+k-T[k]+T[k]=m+k,　m=j-k$
　　注意上面的j是先增加了1，所以这里为 j-k。上面的伪代码实现起来应该很容易，然后看局部匹配表的伪代码如下：

//未优化的版本，伪代码：
algorithm kmp_table:
    input:
        an array of characters, W (the word to be analyzed)
        an array of integers, T (the table to be filled)
    output:
        nothing (but during operation, it populates the table)

    define variables:
        an integer, pos ← 2 (the current position we are computing in T)
        an integer, cnd ← 0 (the zero-based index in W of the next 
character of the current candidate substring)

    (the first few values are fixed but different from what the algorithm 
might suggest)
    let T[0] ← -1, T[1] ← 0
    //pos从3开始，所以刚开始的判断为w[1],w[0],cnd=0 注意cnd代表了当前substring的最长前缀的下一个值。
    while pos < length(W) do
        (first case: the substring continues)
        if W[pos - 1] = W[cnd] then
            let cnd ← cnd + 1, T[pos] ← cnd, pos ← pos + 1
        //如果没有拓展这个最长前缀，按照上面的算法我们应该以--cnd>=0的方式检查
        //但是这样会判断多余的长度，采用cnd=T[cnd]避免检查没有必要的最长前缀长度
        (second case: it does not, but we can fall back)
        else if cnd > 0 then
            let cnd ← T[cnd]
        //第三种情况，从T[pos-1]的cnd值直到0都检查过了，没有符合要求的前后缀，则此时设置T[pos]=0;
        (third case: we have run out of candidates.  Note cnd = 0)
        else
            let T[pos] ← 0, pos ← pos + 1

　　对于cnd=T[cnd]，其实网上很多人都困惑，但它并不是KMP算法的原理部分，而是编程者的技巧，这种编程技巧源自一种观察：
这里写图片描述
观察i=10，此时因为p[pos-1] != p[cnd](此时pos=10 cnd=2)，执行cnd=T[cnd]。我们知道了p[pos-1] != p[cnd]的事实，从KMP原理上此时cnd应该为2，去检查PA == AT?，不过可以多思考一步，如果AT能成为pos=10的W的最长前缀，那么A一定PA的最长前缀。按照这种逻推广到一般情形，遇到上面p[pos-1] != p[cnd]的情况时，下一步应该检查的最长前缀的长度为T[cnd]，而不是cnd-1。这里需要多想想，再次强调这是编程方式的一种优化。再看看优化后的KMP算法的局部匹配表的构造算法：

//经过优化的局部匹配表构造算法
algorithm kmp_table:
    input:
        an array of characters, W (the word to be analyzed)
        an array of integers, T (the table to be filled)
    output:
        nothing (but during operation, it populates the table)

    define variables:
        an integer, pos ← 2 (the current position we are computing in T)
        an integer, cnd ← 0 (the zero-based index in W of the next 
character of the current candidate substring)

    (the first few values are fixed but different from what the algorithm 
might suggest)
    let T[0] ← -1, T[1] ← 0
    while pos < length(W) do
        (first case: the substring continues)
        if W[pos - 1] = W[cnd] then
            let cnd ← cnd + 1, T[pos] ← cnd, pos ← pos + 1
        (second case: it does not, but we can fall back)
        else if cnd > 0 then
            let cnd ← T[cnd]
            //注意这里的continue,不能省略
            continue;
        (third case: we have run out of candidates.  Note cnd = 0)
        else
            let T[pos] ← 0, pos ← pos + 1
        //源自上面的原理，很容易得到修改上面的伪代码。
        if p[T[pos-1]] == p[pos-1] 
            T[pos-1]=T[T[pos-1]]

4. 总结

　　KMP算法记录篇远没有结束，还需要自己转化成C++代码，并且实战一个算法题。对于字符串的算法题目，使用回溯加KMP能解决一些难度较大的算法题。