The Longest Common Subsequence LCS is to extract all the possible subsequences from the given two sequences X and Y. The possible extra characters are arranged in the order in which they are arranged in the original sequence. The algorithm for LCS problems has a wide range of uses. For example, in the management of different versions of software, the LCS algorithm is used to find the similarities and differences between the old and new versions; in software testing, the LCS algorithm is used to compare recorded and played back sequences. In the field of genetic engineering, the LCS algorithm is used The algorithm checks the similarities and differences between the patient's DNA strand and the bond's DNA strand; in the anti-plagiarism system, the LCS algorithm is used to check the plagiarism rate of the paper. The LCS algorithm can also be used for program code similarity measurement, human running sequence retrieval, video segment matching, etc., so research on the LCS algorithm has high application value.
Basic concept
Subsequence: A subsequence of a specific sequence is a subsequence of a given sequence. The result obtained by removing zero or more elements (without changing the relative order between elements). For example, the subsequences of the sequence are: , , Common subsequence: Given sequences X and Y, sequence Z is a subsequence of X and a subsequence of Y, then Z is the common subsequence of X and Y. For example, X=[A,B,C,B,D,A,B], Y=[B,D,C,A,B,A[, then the sequence Z=[B,C,A] is X and Y The common subsequence of , its length is 3. But Z is not the longest common subsequence of X and Y, and the sequences [B, C, B, A] and [B, D, A, B] are also the longest common subsequences of X and Y, with a length of 4 , and X and Y do not have a common subsequence with a length greater than or equal to 5. For the common subsequences of the sequence [A, B, C] and the sequence [E, F, G], there is only the empty sequence []. Longest common subsequence: Given the sequences X and Y, select the one or several with the longest length from all their common subsequences.
We can see The subsequence is not necessarily continuous, what is continuous is the substring.
Problem Analysis
We still start the analysis from a matrix and derive the state transition equation ourselves. First of all, we convert the problem into a concept that is familiar enough to the front end. Instead of calling it serially, it can be thought of as an array or a string. To keep things simple, let's just assume that two strings are being compared. We focus on the concept of "subsequence", which can delete multiple or zero ones, or all of them. At this time our first subsequence is an empty string (if our sequence is not a string, we can still)! This is something you really need to pay attention to! Many people just can't understand the chart in "Introduction to Algorithms", and there are also many bloggers who pretend to understand. We always compare from left to right, and of course the first string, because it is the height of the matrix, is placed vertically.
x
""
B
D
C
A
B
A
""
##A
B
C
D
A
B
Suppose The solution of the LCS equation is a number, so this table can only be filled with numbers. The length of the common area of two empty strings is 0.
x
""
B
D
C
A
B
A
##""
0
A
B
C
D
A
B
Then we don’t move X and continue to let it go The string comes out of the array, and Y makes "B" come out of the array. Obviously, the length of their common area is 0. Y is replaced by other characters, D, C, or their continuous combinations DC and DDC. The situation remains unchanged. is 0. Therefore, the first row is all 0. Then we do not move Y, and Y only outputs the empty word string, then it is the same as the above analysis, both are 0, and the first column is all 0.
x
""
B
D
C
A
B
A
#""0
0
0
0
0
0
0
##A
0
B
0
C
0
D
0
A
0
B
0
The LCS problem is a little different from the knapsack problem. The knapsack problem can also be set to -1 row, and the longest common subsequence has an empty subsequence. The appearance of , fixed the left side and the top side from the beginning.
Then we enlarge the problem a little further. This time both sides produce a character. Obviously, only when both are the same, can there be a common subsequence that is not an empty string, and the length can also be understood as 1.
A is "X", Y is any subsequence of "BDCA"##x""
B
D
C
A
B
A
##""
0
0
0
0
0
0
0
##A
0
0
0
0
1
##B
0
C
0
D
0
##A
0
B
0
Continue to fill in the blanks to the right, how to fill in the blanks? Obviously, LCS cannot be greater than the length of X. How can the subsequence of Y starting from the A string be equal to 1 compared with the A sequence of B.
x
""
BD
C
A
B
A
#""
0
0
0
0
0
0
0
##A
0
0
0
0
1
1
1
##B
0
C
0
D
0
A
0
B
0
If Then let's look at B first, ${X_1} == ${Y_0}, we get a new public substring, and we should add 1. why? Because our matrix is a state table, describing the state migration process from left to right and top to bottom, and these states are accumulated based on existing states. What we need to confirm now is the relationship between the value of the grid we want to fill in and the values of the grids around it that have already been filled in. At present, there is too little information and it is just an isolated point, so just fill in 1.
x
""
B
D
C
A
B
A
##""
0
0
0
0
0
0
0
##A
0
0
0
0
1
1
1
##B
0
1
##C
0
D
0
A
0
B
0
Then we let Y have an extra D as a helper, {"",A,B,AB} vs {"",B,D,BD}, obviously, continue Fill in 1. Fill in 1 until the second B of Y. Because when it comes to BDCAB, they have another common subsequence, AB.
x
""
B
D
C
A
B
A
#""
0
0
0
0
0
0
0
##A
0
0
0
0
1
1
1
##B
0
1
1
1
1
2
#C
0
D
0
A
0
B
0
At this point, we can summarize some rules. Then we will verify our ideas through calculations and add new rules or limiting conditions to improve them.
Y Send all the characters, X is still 2 characters, after careful observation, still fill in 2.
Look at the five lines, send more X If the set of subsequences of C and ABC is larger than the set of subsequences of AB, then it and the set of B subsequences of Y are larger. Even if they are not larger, they cannot be smaller than the original ones. Obviously the newly added C cannot become a combat power and is not a common character between the two, so the value should be equal to the subsequence set of AB.
##×
""
B
D
C
A
B
A
##""
0
0
00
0
0
0
##A
0
0
0
0
1
1
1
##B
0
1
1
1
1
2
2
#C
0
1
##D
0
A
0
B
0
And we can be sure that if the characters to be compared between the two strings are different, then the grid to be filled in is related to the left or upper side, and the larger side will be used.
If the compared characters are the same, don’t worry, it happens that the C of X needs to be compared with the C of Y, that is, the subsequence set of ABC {"",A,B,C,AB,BC, ABC} is compared with the subsequence set {"",B,D,C,BD,DC,BDC} of BDC, and the common substrings obtained are "",B,D. At this time, the conclusion is still the same as before. When the characters are equal, its corresponding grid value is equal to the value of the left, right, and upper left corners, and the left, upper, and upper left sides are always equal. These mysteries require more rigorous mathematical knowledge to demonstrate.
Suppose there are two arrays, A and B. A[i] is the i-th element of A, and A(i) is the prefix consisting of the first element to the i-th element of A. m(i, j) is the longest common subsequence length of A(i) and B(j).
Due to the recursive nature of the algorithm itself, in fact, we only need to prove that for a certain i and j:
The first formula is easy to prove, that is, when A[i] = B[j]. You can use counter-proof, assuming m(i, j) > m(i-1, j-1) + 1 (m(i, j) cannot be less than m(i-1, j-1) + 1 for a very good reason Obviously), then we can deduce the contradictory result that m(i-1, j-1) is not the longest.
The second one is a bit tricky. When A[i] != B[j], it is still a disproof, assuming m(i, j) > max( m(i-1, j), m(i, j-1) ).
By disproving the hypothesis, it can be obtained that m(i, j) > m(i-1, j). This can be deduced that A[i] must be in the LCS sequence corresponding to m(i, j) (contradictory evidence is available). And since A[i] != B[j], B[j] must not be in the LCS sequence corresponding to m(i, j). So it can be deduced that m(i, j) = m(i, j-1). This leads to results that contradict the hypothesis anyway.
Get certified.
We now use the following equation to continue filling in the table.
Program implementation
//by 司徒正美
function LCS(str1, str2){
var rows = str1.split("")
rows.unshift("")
var cols = str2.split("")
cols.unshift("")
var m = rows.length
var n = cols.length
var dp = []
for(var i = 0; i < m; i++){
dp[i] = []
for(var j = 0; j < n; j++){
if(i === 0 || j === 0){
dp[i][j] = 0
continue
}
if(rows[i] === cols[j]){
dp[i][j] = dp[i-1][j-1] + 1 //对角＋1
}else{
dp[i][j] = Math.max( dp[i-1][j], dp[i][j-1]) //对左边，上边取最大
}
}
console.log(dp[i].join(""))//调试
}
return dp[i-1][j-1]
}
LCS can be further simplified, just by moving the position, eliminating the need to generate a new array
//by司徒正美
function LCS(str1, str2){
var m = str1.length
var n = str2.length
var dp = [new Array(n+1).fill(0)] //第一行全是0
for(var i = 1; i <= m; i++){ //一共有m+1行
dp[i] = [0] //第一列全是0
for(var j = 1; j <= n; j++){//一共有n+1列
if(str1[i-1] === str2[j-1]){
//注意这里，str1的第一个字符是在第二列中，因此要减1，str2同理
dp[i][j] = dp[i-1][j-1] + 1 //对角＋1
} else {
dp[i][j] = Math.max( dp[i-1][j], dp[i][j-1])
}
}
}
return dp[m][n];
}
Print an LCS
#We will give the printing function and first look at how to print one. We start looking from the lower right corner and end at the top line. Therefore the target string is constructed in reverse order. In order to avoid using troublesome intermediate quantities such as stringBuffer, we can implement it recursively. Each time the program is executed, only one string is returned, otherwise an empty string is returned, using printLCS(x,y,...) + str[ i] are added to get the string we require.
We write another method to verify whether the string we get is a real LCS string. As a person who is already working, I cannot write code like a student in school, putting it online without doing unit testing and letting others step on it.
//by 司徒正美，打印一个LCS
function printLCS(dp, str1, str2, i, j){
if (i == 0 || j == 0){
return "";
}
if( str1[i-1] == str2[j-1] ){
return printLCS(dp, str1, str2, i-1, j-1) + str1[i-1];
}else{
if (dp[i][j-1] > dp[i-1][j]){
return printLCS(dp, str1, str2, i, j-1);
}else{
return printLCS(dp, str1, str2, i-1, j);
}
}
}
//by司徒正美， 将目标字符串转换成正则，验证是否为之前两个字符串的LCS
function validateLCS(el, str1, str2){
var re = new RegExp( el.split("").join(".*") )
console.log(el, re.test(str1),re.test(str2))
return re.test(str1) && re.test(str2)
}
Use:
function LCS(str1, str2){
var m = str1.length
var n = str2.length
//....略，自行补充
var s = printLCS(dp, str1, str2, m, n)
validateLCS(s, str1, str2)
return dp[m][n]
}
var c1 = LCS( "ABCBDAB","BDCABA");
console.log(c1) //4 BCBA、BCAB、BDAB
var c2 = LCS("13456778" , "357486782" );
console.log(c2) //5 34678
var c3 = LCS("ACCGGTCGAGTGCGCGGAAGCCGGCCGAA" ,"GTCGTTCGGAATGCCGTTGCTCTGTAAA" );
console.log(c3) //20 GTCGTCGGAAGCCGGCCGAA
Print all LCS
The idea is similar to the above , let us note that there is a Math.max value in the LCS method, which actually integrates three situations, so three strings can be forked. Our method will return an es6 collection object for automatic removal. Then each time the new set is used to merge the strings of the old set.
//by 司徒正美 打印所有LCS
function printAllLCS(dp, str1, str2, i, j){
if (i == 0 || j == 0){
return new Set([""])
}else if(str1[i-1] == str2[j-1]){
var newSet = new Set()
printAllLCS(dp, str1, str2, i-1, j-1).forEach(function(el){
newSet.add(el + str1[i-1])
})
return newSet
}else{
var set = new Set()
if (dp[i][j-1] >= dp[i-1][j]){
printAllLCS(dp, str1, str2, i, j-1).forEach(function(el){
set.add(el)
})
}
if (dp[i-1][j] >= dp[i][j-1]){//必须用>=，不能简单一个else搞定
printAllLCS(dp, str1, str2, i-1, j).forEach(function(el){
set.add(el)
})
}
return set
}
}
Using:
function LCS(str1, str2){
var m = str1.length
var n = str2.length
//....略，自行补充
var s = printAllLCS(dp, str1, str2, m, n)
console.log(s)
s.forEach(function(el){
validateLCS(el,str1, str2)
console.log("输出LCS",el)
})
return dp[m][n]
}
var c1 = LCS( "ABCBDAB","BDCABA");
console.log(c1) //4 BCBA、BCAB、BDAB
var c2 = LCS("13456778" , "357486782" );
console.log(c2) //5 34678
var c3 = LCS("ACCGGTCGAGTGCGCGGAAGCCGGCCGAA" ,"GTCGTTCGGAATGCCGTTGCTCTGTAAA" );
console.log(c3) //20 GTCGTCGGAAGCCGGCCGAA
Space optimization
Using rolling array:
function LCS(str1, str2){
var m = str1.length
var n = str2.length
var dp = [new Array(n+1).fill(0)],now = 1，row //第一行全是0
for(var i = 1; i <= m; i++){ //一共有2行
row = dp[now] = [0] //第一列全是0
for(var j = 1; j <= n; j++){//一共有n+1列
if(str1[i-1] === str2[j-1]){
//注意这里，str1的第一个字符是在第二列中，因此要减1，str2同理
dp[now][j] = dp[i-now][j-1] + 1 //对角＋1
} else {
dp[now][j] = Math.max( dp[i-now][j], dp[now][j-1])
}
}
now = 1- now; //1-1=>0;1-0=>1; 1-1=>0 ...
}
return row ? row[n]: 0
}
Dangerous recursive solution
A subsequence of str1 corresponds to a subsequence of the subscript sequence {1, 2, …, m} sequence, therefore, str1 has a total of ${2^m}$ different subsequences (the same is true for str2, such as ${2^n}$), so the complexity reaches an astonishing exponential time (${2^m * 2^ n}$).
//警告，字符串的长度一大就会爆栈
function LCS(str1, str2, a, b) {
if(a === void 0){
a = str1.length - 1
}
if(b === void 0){
b = str2.length - 1
}
if(a == -1 || b == -1){
return 0
}
if(str1[a] == str2[b]) {
return LCS(str1, str2, a-1, b-1)+1;
}
if(str1[a] != str2[b]) {
var x = LCS(str1, str2, a, b-1)
var y = LCS(str1, str2, a-1, b)
return x >= y ? x : y
}
}
I believe you have mastered the method after reading the case in this article. For more exciting information, please pay attention to other related articles on the php Chinese website!
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