Finding an LCS is one way of computing how similar two sequences are: the longer the LCS is, the more similar they are.
查找LCS是计算两个序列相似程度的一种方法:LCS越长,两个序列越相似。
So, the length of an LCS for these two sequences is 5.
所以,这两个序列的LCS长度是5。
But after a decade of politics and design-by-committee, the LCS has turned out to be anything but small, cheap and numerous.
但在为期十年的政争和委员会设计之后,LCS除了体积小、造价低、数目众多外,便什么也不是了。
Hence, the number in the lower, right-most cell is the length of an LCS of the two strings S1 and S2 - GCCCTAGCG and GCGCAATG in this case.
所以,表格右下角的数字就是字符串s1和S2(在本例中是GCCCTAGCG和GCGCAATG)的LCS的长度。
In this case, the LCS of S1 and S2 is clearly a zero-length string.
在这种情况下,S1和S2的LCS显然是长度为零的字符串。
This cell will eventually contain a number that is the length of an LCS of GCGC and GCCCT.
这个单元格最终包含的数字就是GCGC和GCCCT的LCS的长度。
You「ll use these arrows later in 「tracing back」 to construct an actual LCS (as opposed to just discovering the length of one).
后面的「回溯」一节将用这些箭头建立实际的LCS(与仅仅发现LCS长度相反)。
Nor could the LCS take on the Type 022 in direct combat and count on winning.
LCS在战斗中也无法直接同022型较量,并指望获得胜利。
To compute the LCS efficiently using dynamic programming, you start by constructing a table in which you build up partial results.
为了用动态编程有效地计算lcs,首先需要构建一个表格,用它保存部分结果。
So, the way you construct an LCS is by starting in the lower-right corner cell and then following the pointer arrows backward.
所以,构建LCS的途径就是从右下角的单元格开始,沿着箭头一路返回。
The score in the bottom-right cell contains the maximum alignment score for S1 and S2, just as it contains the length of an LCS in the LCS algorithm.
右下角单元格中的得分包含S1和S2的最大比对得分,就像在LCS算法中包含LCS的长度一样。
The idea is similar to the LCS algorithm.
它的思路与LCS算法相似。
It turns out that an LCS of these two sequences is GCCAG.
这两个序列的LCS是GCCAG。
Recall that the number in any cell is the length of an LCS of the string prefixes above and below that end in the column and row of that cell.
回想一下,任何单元格中的数字都是该单元格所在行之上和列之前的字符串的LCS长度。
Finally, you construct an actual LCS using the traceback.
最后,用回溯的方式构建实际的lcs。
First, think about how you might compute an LCS recursively. Let.
首先,考虑如何递归地计算lcs。
And, similarly to the LCS algorithm, to obtain S1 」and S2「, you trace back from this bottom-right cell, following the Pointers, and build up S1 」and S2「 in reverse.
而且,与LCS算法类似,要获得S1」和S2「,要从右下角单元格开始沿着指针回溯,反向构建S1」和S2'。
As with the LCS algorithm, for each cell you have three choices and pick the maximum one.
同lcs算法一样,对于每个单元格,都有三个选择,要从中选择最大的。