Algorithm Implementation/Strings/Longest common subsequence
C#
[edit | edit source]Finding the LCS
[edit | edit source]The following C# program calculates the longest common subsequence (note the singular) of 2 strings. For example, for the strings "computer" and "houseboat" this algorithm returns a value of 3, specifically the string "out".
Computing the length table of the LCS
[edit | edit source] static int[,] LCS(string s1, string s2)
{
int[,] c = new int[s1.Length + 1, s2.Length + 1];
for (int i = 1; i <= s1.Length; i++)
for (int j = 1; j <= s2.Length; j++)
{
if (s1[i - 1] == s2[j - 1])
c[i, j] = c[i - 1, j - 1] + 1;
else
c[i, j] = c[i - 1, j] > c[i, j - 1]? c[i - 1, j] : c[i, j - 1];
}
return c;
}
Reading out an LCS
[edit | edit source] static string BackTrack(int[,] c, string s1, string s2, int i, int j)
{
if (i == 0 || j == 0)
return "";
else if (s1[i - 1] == s2[j - 1])
return BackTrack(c, s1, s2, i - 1, j - 1) + s1[i - 1];
else if (c[i, j - 1] > c[i - 1, j])
return BackTrack(c, s1, s2, i, j - 1);
else
return BackTrack(c, s1, s2, i - 1, j);
}
Print the diff
[edit | edit source] static string PrintDiff(int[,] c, string s1, string s2, int i, int j)
{
var a = "";
if (i > 0 && j > 0 && s1[i - 1] == s2[j - 1])
{
a = PrintDiff(c, s1, s2, i - 1, j - 1);
return a + "" + s1[i - 1];
}
else if (j > 0 && (i == 0 || (c[i, j - 1] > c[i - 1, j])))
{
a = PrintDiff(c, s1, s2, i, j - 1);
return a + "+" + s2[j - 1];
}
else if (i > 0 && (j == 0 || (c[i, j - 1] <= c[i - 1, j])))
{
a = PrintDiff(c, s1, s2, i - 1, j);
return a + "-" + s1[i - 1];
}
return a;
}
Example usage
[edit | edit source] static void Main(string[] args)
{
string s1 = "computer";
string s2 = "houseboat";
var c = LCS(s1, s2);
Console.WriteLine(BackTrack(c, s1, s2, s1.Length, s2.Length)); //out
Console.WriteLine(PrintDiff(c, s1, s2, s1.Length, s2.Length)); //+h-co-m-pu+s+e+b+o+at-e-r
Console.ReadLine();
}
Common Lisp
[edit | edit source]This version only works for lists but can be generalized for all sequences.
(defun lcs-list (list-1 list-2 &key (test #'eql))
"Find the longest common subsequence of LIST-1 and LIST-2 using TEST."
(cond
((null list-1) nil)
((null list-2) nil)
((funcall test (first list-1) (first list-2))
(cons (first list-1) (lcs-list (rest list-1) (rest list-2) :test test)))
(t (let ((lcs-1 (lcs-list list-1 (rest list-2) :test test))
(lcs-2 (lcs-list (rest list-1) list-2 :test test)))
(if (> (length lcs-1) (length lcs-2))
lcs-1
lcs-2)))))
(defun diff (list1 list2 &key (test #'eql))
"Find the differences between LIST1 and LIST2 using TEST."
(let ((lcs (lcs-list list1 list2 :test test))
result)
(dolist (c lcs)
(let* ((sync-list1 (position c list1 :test test))
(sync-list2 (position c list2 :test test))
(removed (subseq list1 0 sync-list1))
(added (subseq list2 0 sync-list2)))
(setf list1 (subseq list1 (1+ sync-list1)))
(setf list2 (subseq list2 (1+ sync-list2)))
(when removed
(push (cons :removed removed) result))
(when added
(push (cons :added added) result))
(push c result)))
(when list1
(push (cons :removed list1) result))
(when list2
(push (cons :added list2) result))
(nreverse result)))
Python
[edit | edit source]Computing the length table of the LCS
[edit | edit source]def LCS(X, Y):
m = len(X)
n = len(Y)
# An (m+1) times (n+1) matrix
C = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(1, m+1):
for j in range(1, n+1):
if X[i-1] == Y[j-1]:
C[i][j] = C[i-1][j-1] + 1
else:
C[i][j] = max(C[i][j-1], C[i-1][j])
return C
Reading out an LCS
[edit | edit source]def backTrack(C, X, Y, i, j):
if i == 0 or j == 0:
return ""
elif X[i-1] == Y[j-1]:
return backTrack(C, X, Y, i-1, j-1) + X[i-1]
else:
if C[i][j-1] > C[i-1][j]:
return backTrack(C, X, Y, i, j-1)
else:
return backTrack(C, X, Y, i-1, j)
Reading out all LCSs
[edit | edit source]def backTrackAll(C, X, Y, i, j):
if i == 0 or j == 0:
return set([""])
elif X[i-1] == Y[j-1]:
return set([Z + X[i-1] for Z in backTrackAll(C, X, Y, i-1, j-1)])
else:
R = set()
if C[i][j-1] >= C[i-1][j]:
R.update(backTrackAll(C, X, Y, i, j-1))
if C[i-1][j] >= C[i][j-1]:
R.update(backTrackAll(C, X, Y, i-1, j))
return R
Usage example
[edit | edit source]X = "AATCC"
Y = "ACACG"
m = len(X)
n = len(Y)
C = LCS(X, Y)
print "Some LCS: '%s'" % backTrack(C, X, Y, m, n)
print "All LCSs: %s" % backTrackAll(C, X, Y, m, n)
It prints the following:
Some LCS: 'AAC' All LCSs: set(['ACC', 'AAC'])
Print the diff
[edit | edit source]def printDiff(C, X, Y, i, j):
if i > 0 and j > 0 and X[i-1] == Y[j-1]:
printDiff(C, X, Y, i-1, j-1)
print " " + X[i-1]
else:
if j > 0 and (i == 0 or C[i][j-1] >= C[i-1][j]):
printDiff(C, X, Y, i, j-1)
print "+ " + Y[j-1]
elif i > 0 and (j == 0 or C[i][j-1] < C[i-1][j]):
printDiff(C, X, Y, i-1, j)
print "- " + X[i-1]
Usage example
[edit | edit source]X = [
"This part of the document has stayed",
"the same from version to version.",
"",
"This paragraph contains text that is",
"outdated - it will be deprecated '''and'''",
"deleted '''in''' the near future.",
"",
"It is important to spell check this",
"dokument. On the other hand, a misspelled",
"word isn't the end of the world.",
]
Y = [
"This is an important notice! It should",
"therefore be located at the beginning of",
"this document!",
"",
"This part of the document has stayed",
"the same from version to version.",
"",
"It is important to spell check this",
"document. On the other hand, a misspelled",
"word isn't the end of the world. This",
"paragraph contains important new",
"additions to this document.",
]
C = LCS(X, Y)
printDiff(C, X, Y, len(X), len(Y))
It prints the following:
+ This is an important notice! It should + therefore be located at the beginning of + this document! + This part of the document has stayed the same from version to version. - - This paragraph contains text that is - outdated - it will be deprecated and - deleted in the near future. It is important to spell check this - dokument. On the other hand, a misspelled - word isn't the end of the world. + document. On the other hand, a misspelled + word isn't the end of the world. This + paragraph contains important new + additions to this document.
VB.NET
[edit | edit source]The following VB.NET program calculates the longest common subsequence (note the singular) of 2 strings. For example, for the strings "computer" and "houseboat" this algorithm returns a value of 3, specifically the string "out".
Public Function LongestCommonSubsequence(ByVal s1 As String, ByVal s2 As String) As Integer
'Bulletproofing - 1 or both inputs contains nothing
If s1.Length.Equals(0) Or s2.Length.Equals(0) Then
Return "0"
End If
'*** Actual Algorithm From Here ***
Dim num(s1.Length - 1, s2.Length - 1) As Long '2D Array
Dim letter1 As Char = Nothing
Dim letter2 As Char = Nothing
For i As Integer = 0 To s1.Length - 1
For j As Integer = 0 To s2.Length - 1
letter1 = s1.Chars(i)
letter2 = s2.Chars(j)
If letter1.Equals(letter2) Then
If i.Equals(0) Or j.Equals(0) Then 'The first elements respectively
num(i, j) = 1
Else
num(i, j) = 1 + num(i - 1, j - 1)
End If
Else
If i.Equals(0) And j.Equals(0) Then
num(i, j) = 0
ElseIf i.Equals(0) And Not j.Equals(0) Then 'First ith element
num(i, j) = Math.Max(0, num(i, j - 1))
ElseIf j.Equals(0) And Not i.Equals(0) Then 'First jth element
num(i, j) = Math.Max(num(i - 1, j), 0)
ElseIf i <> 0 And j <> 0 Then
num(i, j) = Math.Max(num(i - 1, j), num(i, j - 1))
End If
End If
Next j
Next i
Return num(s1.Length - 1, s2.Length - 1)
End Function
Usage: LongestCommonSubsequence("computer", "houseboat")
Java
[edit | edit source]public static <E> List<E> LongestCommonSubsequence(E[] s1, E[] s2)
{
int[][] num = new int[s1.length+1][s2.length+1]; //2D array, initialized to 0
//Actual algorithm
for (int i = 1; i <= s1.length; i++)
for (int j = 1; j <= s2.length; j++)
if (s1[i-1].equals(s2[j-1]))
num[i][j] = 1 + num[i-1][j-1];
else
num[i][j] = Math.max(num[i-1][j], num[i][j-1]);
System.out.println("length of LCS = " + num[s1.length][s2.length]);
int s1position = s1.length, s2position = s2.length;
List<E> result = new LinkedList<E>();
while (s1position != 0 && s2position != 0)
{
if (s1[s1position - 1].equals(s2[s2position - 1]))
{
result.add(s1[s1position - 1]);
s1position--;
s2position--;
}
else if (num[s1position][s2position - 1] >= num[s1position][s2position])
{
s2position--;
}
else
{
s1position--;
}
}
Collections.reverse(result);
return result;
}
The Java implementation uses two classes. The first, an abstract class that implements the algorithm. The second, an concrete class implementing this for a string. Obviously, you could use this so that instead of comparing characters in a string, it could compare lines in a file, blocks of code, nodes in an XML document, or whatever you choose.
import static java.lang.Math.*;
import java.util.ArrayList;
import java.util.List;
import static java.lang.Math.*;
/**
* A class to compute the longest common subsequence in two strings.
* Algorithms from Wikipedia:
* http://en.wikipedia.org/wiki/Longest_common_subsequence_problem
*
* @author jhess
*
*/
public abstract class LongestCommonSubsequence<VALUE> {
private int[][] c;
private ArrayList<DiffEntry<VALUE>> diff;
private ArrayList<VALUE> backtrack;
/**
* A constructor for classes inheriting this one, allowing them to
* do some initialization before setting the values of X and Y. Once
* the initialization is complete, the inheriting class must call
* initValues(VALUE[] x, VALUE[] y)
*/
protected LongestCommonSubsequence() {
}
protected abstract int lengthOfY() ;
protected abstract int lengthOfX() ;
protected abstract VALUE valueOfX(int index) ;
protected abstract VALUE valueOfY(int index) ;
protected boolean equals(VALUE x1, VALUE y1) {
return (null == x1 && null == y1) || x1.equals(y1);
}
private boolean isXYEqual(int i, int j) {
return equals(valueOfXInternal(i),valueOfYInternal(j));
}
private VALUE valueOfXInternal(int i) {
return valueOfX(i-1);
}
private VALUE valueOfYInternal(int j) {
return valueOfY(j-1);
}
public void calculateLcs() {
if(c != null) {
return;
}
c = new int[lengthOfX()+1][];
for(int i = 0; i < c.length; i++) {
c[i] = new int[lengthOfY()+1];
}
for(int i = 1; i < c.length; i++) {
for(int j = 1; j < c[i].length; j++) {
if(isXYEqual(i, j)) {
c[i][j] = c[i-1][j-1] + 1;
} else {
c[i][j] = max(c[i][j-1], c[i-1][j]);
}
}
}
}
public int getLcsLength() {
calculateLcs();
return c[lengthOfX()][lengthOfY()];
}
public int getMinEditDistance() {
calculateLcs();
return lengthOfX()+lengthOfY()-2*abs(getLcsLength());
}
public List<VALUE> backtrack() {
calculateLcs();
if(this.backtrack == null) {
this.backtrack = new ArrayList<VALUE>();
backtrack(lengthOfX(),lengthOfY());
}
return this.backtrack;
}
public void backtrack(int i, int j) {
calculateLcs();
if (i == 0 || j == 0) {
return;
}
else if (isXYEqual(i, j)) {
backtrack(i-1,j-1);
backtrack.add(valueOfXInternal(i));
}
else {
if(c[i][j-1] > c[i-1][j]) {
backtrack(i,j-1);
} else {
backtrack(i-1,j);
}
}
}
public List<DiffEntry<VALUE>> diff() {
calculateLcs();
if(this.diff == null) {
this.diff = new ArrayList<DiffEntry<VALUE>>();
diff(lengthOfX(),lengthOfY());
}
return this.diff;
}
private void diff(int i, int j) {
calculateLcs();
while(!(i==0 && j==0)) {
if (i > 0 && j > 0 && isXYEqual(i, j)) {
this.diff.add(new DiffEntry<VALUE>(DiffType.NONE,
valueOfXInternal(i)));
i--;j--;
} else {
if (j > 0 && (i == 0 || c[i][j - 1] >= c[i - 1][j])) {
this.diff.add(new DiffEntry<VALUE>(DiffType.ADD,
valueOfYInternal(j)));
j--;
} else if (i > 0 && (j == 0 || c[i][j - 1] < c[i - 1][j])) {
this.diff.add(new DiffEntry<VALUE>(DiffType.REMOVE,
valueOfXInternal(i)));
i--;
}
}
}
Collections.reverse(this.diff);
}
@Override
public String toString() {
calculateLcs();
StringBuffer buf = new StringBuffer();
buf.append(" ");
for(int j = 1; j <= lengthOfY(); j++) {
buf.append(valueOfYInternal(j));
}
buf.append("\n");
buf.append(" ");
for(int j = 0; j < c[0].length; j++) {
buf.append(Integer.toString(c[0][j]));
}
buf.append("\n");
for(int i = 1; i < c.length; i++) {
buf.append(valueOfXInternal(i));
for(int j = 0; j < c[i].length; j++) {
buf.append(Integer.toString(c[i][j]));
}
buf.append("\n");
}
return buf.toString();
}
public static enum DiffType {
ADD("+","add"),REMOVE("-","remove"),NONE(" ","none");
private String val;
private String name;
DiffType(String val, String name) {
this.val = val;
this.name = name;
}
@Override
public String toString() {
return val;
}
public String getName() {
return name;
}
public String getVal() {
return val;
}
}
public static class DiffEntry<VALUE> {
private DiffType type;
private VALUE value;
public DiffEntry(DiffType type, VALUE value) {
super();
this.type = type;
this.value = value;
}
public DiffType getType() {
return type;
}
public void setType(DiffType type) {
this.type = type;
}
public VALUE getValue() {
return value;
}
public void setValue(VALUE value) {
this.value = value;
}
@Override
public String toString() {
return type.toString()+value.toString();
}
}
}
import java.util.List;
public class LcsString extends LongestCommonSubsequence<Character> {
private String x;
private String y;
public LcsString(String from, String to) {
this.x = from;
this.y = to;
}
protected int lengthOfY() {
return y.length();
}
protected int lengthOfX() {
return x.length();
}
protected Character valueOfX(int index) {
return x.charAt(index);
}
protected Character valueOfY(int index) {
return y.charAt(index);
}
public String getHtmlDiff() {
DiffType type = null;
List<DiffEntry<Character>> diffs = diff();
StringBuffer buf = new StringBuffer();
for(DiffEntry<Character> entry : diffs) {
if(type != entry.getType()) {
if(type != null) {
buf.append("</span>");
}
buf.append("<span class=\""+entry.getType().getName()+"\">");
type = entry.getType();
}
buf.append(escapeHtml(entry.getValue()));
}
buf.append("</span>");
return buf.toString();
}
private String escapeHtml(Character ch) {
switch(ch) {
case '<' : return "<";
case '>' : return ">";
case '"' : return "\\"";
default : return ch.toString();
}
}
// EXAMPLE. Here's how you use it.
public static void main(String[] args) {
LcsString seq = new LcsString("<p>the quick brown fox</p>","<p>the <b>Fast</b> brown dog</p>");
System.out.println("LCS: "+seq.getLcsLength());
System.out.println("Edit Dist: "+seq.getMinEditDistance());
System.out.println("Backtrack: "+seq.backtrack());
System.out.println("HTML Diff: "+seq.getHtmlDiff());
}
}
C++
[edit | edit source]#include <algorithm>
#include <string>
#include <vector>
#include <stdio.h>
#include <string.h>
// See http://www-igm.univ-mlv.fr/~lecroq/seqcomp/node4.html.
class LCS
{
class LCSTable
{
size_t m_;
size_t n_;
size_t* data_;
public:
LCSTable(size_t m, size_t n)
: m_(m)
, n_(n)
{
data_ = new size_t[(m_ + 1) * (n_ + 1)];
}
~LCSTable()
{
delete [] data_;
}
void setAt(size_t i, size_t j, size_t value)
{
data_[i + j * (m_ + 1)] = value;
}
size_t getAt(size_t i, size_t j) const
{
return data_[i + j * (m_ + 1)];
}
template<typename T> void
build(const T* X, const T* Y)
{
for (size_t i=0; i<=m_; ++i)
setAt(i, 0, 0);
for (size_t j=0; j<=n_; ++j)
setAt(0, j, 0);
for (size_t i = 0; i < m_; ++i)
{
for (size_t j = 0; j < n_; ++j)
{
if (X[i] == Y[j])
setAt(i+1, j+1, getAt(i, j)+1);
else
setAt(i+1, j+1, std::max(getAt(i+1, j), getAt(i, j+1)));
}
}
}
};
template<typename T> static void
backtrackOne(const LCSTable& table,
const T* X, const T* Y, size_t i, size_t j,
std::vector<T>& result)
{
result.clear();
if (i == 0 || j == 0)
return;
if (X[i - 1] == Y[j - 1])
{
backtrackOne(table, X, Y, i - 1, j - 1, result);
result.push_back(X[i - 1]);
return;
}
if (table.getAt(i, j - 1) > table.getAt(i -1, j))
backtrackOne(table, X, Y, i, j - 1, result);
else
backtrackOne(table, X, Y, i - 1, j, result);
}
public:
template<typename T> static void
findOne(T* X, size_t m, T* Y, size_t n,
std::vector<T>& result)
{
LCSTable table(m, n);
table.build(X, Y);
backtrackOne(table, X, Y, m, n, result);
}
};
Usage:
char X[] = "XMJYAUZ";
char Y[] = "MZJAWXU";
std::vector<char> result;
LCS::findOne(X, strlen(X), Y, strlen(Y), result);
std::string resultString(&result.front(), result.size());
Ruby
[edit | edit source]def subsequence(s1, s2)
return 0 if s1.empty? || s2.empty?
num=Array.new(s1.size){Array.new(s2.size)}
s1.scan(/./).each_with_index{|letter1,i|
s2.scan(/./).each_with_index{|letter2,j|
if s1[i]==s2[j]
if i==0||j==0
num[i][j] = 1
else
num[i][j] = 1 + num[i - 1][ j - 1]
end
else
if i==0 && j==0
num[i][j] = 0
elsif i==0 && j!=0 #First ith element
num[i][j] = [0, num[i][j - 1]].max
elsif j==0 && i!=0 #First jth element
num[i][j] = [0, num[i - 1][j]].max
elsif i != 0 && j!= 0
num[i][j] = [num[i - 1][j], num[i][j - 1]].max
end
end
}
}
num[s1.length - 1][s2.length - 1]
end
puts subsequence("houseboat","computer")
JavaScript
[edit | edit source]function LCS(a, b) {
var m = a.length, n = b.length,
C = [], i, j;
for (i = 0; i <= m; i++) C.push([0]);
for (j = 0; j < n; j++) C[0].push(0);
for (i = 0; i < m; i++)
for (j = 0; j < n; j++)
C[i+1][j+1] = a[i] === b[j] ? C[i][j]+1 : Math.max(C[i+1][j], C[i][j+1]);
return (function bt(i, j) {
if (i*j === 0) { return ""; }
if (a[i-1] === b[j-1]) { return bt(i-1, j-1) + a[i-1]; }
return (C[i][j-1] > C[i-1][j]) ? bt(i, j-1) : bt(i-1, j);
}(m, n));
}
Phix
[edit | edit source]function lcs(sequence a, b)
sequence res = ""
if length(a) and length(b) then
if a[$]=b[$] then
res = lcs(a[1..-2],b[1..-2])&a[$]
else
sequence l = lcs(a,b[1..-2]),
r = lcs(a[1..-2],b)
res = iff(length(l)>length(r)?l:r)
end if
end if
return res
end function
PHP
[edit | edit source]<?php
/**
* Class LCS - Using this class to calculate the LCS(Longest common subsequence)
* You can pass both the strings in String format in the constructor when you are initializing the class
* All the subsequent method calls are in the constructor and you can access the results using the $remains and $differences
* public variables
*/
class LCS{
// Define the class variables
public $remains;
public $differences;
public function __construct($str1, $str2)
{
// The algorithm requires us to have m+1 by n+1 matrix starting with leading 0
// for the initial matrix. Here we convert string to array and add the 0 as the first element
$this->remains = "";
$this->differences = "";
$arr1 = array_merge([0], str_split($str1));
$arr2 = array_merge([0], str_split($str2));
$matrix = $this->generateMatrix($arr1, $arr2);
$this->calculateDifference($matrix, $arr1, $arr2, count($arr1)-1, count($arr2)-1);
return;
}
/**
* @function calculateDifference - To use recursively and build the ending result of
* the backtrace using the LCS matrix
*
* @input $matrix - 2D array of the values generated with the generateMatrix function
* @input $first - Array of the characters for the first string with required leading 0th element
* @input $second - Array of the characters for the second(result) string with required leading 0th element
* @input $i - Position of the current element for the $first array
* @input $j - Position of the current element for the $second array
*
* @return null - Building the data into a public class object that we access later
*/
private function calculateDifference($matrix, $first, $second, $i, $j){
if($i>0 && $j>0 && ($first[$i] == $second[$j])){
// Match found go diagonally
$this->calculateDifference($matrix, $first, $second, $i-1, $j-1);
$this->remains .= $first[$i];
$this->differences .= $first[$i];
} else if($j > 0 && ($i == 0 || ($matrix[$i][$j-1] >= $matrix[$i-1][$j]))){
// column end reached or row above value greater than the value in the left column
$this->calculateDifference($matrix, $first, $second, $i, $j-1);
$this->differences .= "+".$second[$j];
} else if($i > 0 && ($j == 0 || $matrix[$i][$j-1] < $matrix[$i-1][$j])){
// column end reached or row above value lesser than the value in the left column
$this->calculateDifference($matrix, $first, $second, $i-1, $j);
$this->differences .= "-".$first[$i];
} else{
// both zeroes reached the end
return;
}
}
/**
* @function generateMatrix - Creates a traceback matrix for the LCS
*
* @input $first - Array of the characters for the first string with required leading 0th element
* @input $second - Array of the characters for the second(result) string with required leading 0th element
*
* @return $matrix - 2D array of numeric values for tree matching
*/
private function generateMatrix($first, $second)
{
// Set the 0th row and column to have numeric value of 0
for($i=0; $i<=count($first); ++$i){
$matrix[$i][0] = 0;
}
for($j=0; $j<=count($second); ++$j){
$matrix[0][$j] = 0;
}
// Populate the matrix
for($i=1; $i<count($first); ++$i){
for($j=1; $j<count($second); ++$j){
// if current i-th and j-th element match increment the value in the matrix from the previous
if($first[$i] == $second[$j])
$matrix[$i][$j] = $matrix[$i-1][$j-1]+1;
// get the better adjacent calculated value so far from the matrix and use that value
else
$matrix[$i][$j] = max($matrix[$i][$j-1], $matrix[$i-1][$j]);
}
}
return($matrix);
}
}
Usage example
[edit | edit source]Include your class file and instantiate a new object of the class by passing both strings to the constructor Access the results by using $obj->remains and $obj->differences
$obj= new LCS($str1, $str2);
echo $obj->remains;
echo $obj->differences;
APL
[edit | edit source]Computing the length of the LCS
[edit | edit source] C←LCSLength rarg;X;Y;i;j;shape;lastalike;words;B;jb
//⍝ Computes the length of the Longest common sequence
X Y←rarg
//⍝ Optimize by working with numbers instead of strings
words←X∪Y
X←words⍳X
Y←words⍳Y
//⍝ Optimize and reduce sets by removing similar trailing objects.
shape←-(⍴X)⌊(⍴Y)
lastalike←-+/∧\⌽(shape↑X)=(shape↑Y)
X←lastalike↓X
Y←lastalike↓Y
//⍝ C is a numeric matrix where the height is the shape of X
//⍝ and the width is the shape of Y
C←(1+⊃,/⍴¨X Y)⍴0
//⍝ Fill C with LCSlengths
j←1+⍳⍴Y
jb←j-1
:For i :In 1+⍳⍴X
B←X[i-1]=Y
C[i;B/j]←1+C[i-1;B/jb]
C[i;(~B)/j]←⌈\(~B)/C[i;jb]⌈C[i-1;j]
:EndFor
Reading out all LCSs
[edit | edit source] R←MakeDiff rarg;do;⎕IO;C;X;Y;i;j;t1;t2
//⍝ Get the diff between two vectors given the LCS in rarg1
//⍝ \ret [;1] 1=Added item, 0=Deleted item
//⍝ [;2] index in rarg2
//⍝ [;3] index in rarg3
//⍝ [;4] The item
⎕IO←0
C X Y←rarg
//⍝ Number of elements in each vector. This is the starting
//⍝ point for the calculation below
i j←⊃,/⍴¨,¨X Y
//⍝ Add an empty item before each item... Needed
X Y←' ',¨X Y
//⍝ Test 1
t1←{C i j←⍵ ⋄ j>0:{i=0:1 ⋄ C[i;j-1]≥C[i-1;j]:1 ⋄ 0}0 ⋄ 0}
//⍝ Test 2
t2←{C i j←⍵ ⋄ i>0:{j=0:1 ⋄ C[i;j-1]<C[i-1;j]:1 ⋄ 0}0 ⋄ 0}
//⍝ Walk the path and trace additions or removals of items recursivly
do←{
C X Y i j sofar←⍵
(i>0)∧(j>0)∧(X[i]≡Y[j]):∇ C X Y(i-1)(j-1)sofar
t1 C i j:∇ C X Y(i)(j-1)(sofar⍪1 i j(j⊃Y))
t2 C i j:∇ C X Y(i-1)(j)(sofar⍪0 i j(i⊃X))
sofar
}
//⍝ Shoot
R←do C X Y i j(0 4⍴0)
Usage example
[edit | edit source]new←'MZJAWXU'
old←'XMJYAUZ'
C←LCSLength old new
D←MakeDiff C old new
D←⊖D
Variable D is:
0 1 0 X
1 2 2 Z
0 4 3 Y
1 5 5 W
1 5 6 X
0 7 7 Z
J
[edit | edit source] lcs=: [: >./@, [: (* * 1 + >./\@:(_1&|.)@:|:^:2)^:_ [: ,&0@:,.&0 =/
'XMJYAUZ' lcs 'MZJAWXU'
4
See [1] for details.
Scala
[edit | edit source]import scalaz.Memo
def longestCommonSubsequence(s1: String, s2: String): String = {
def longest(a: String, b: String) = if (a.length > b.length) a else b
lazy val lcs: ((Int, Int)) => String = Memo.mutableHashMapMemo[(Int, Int), String] {
case (0, _) => ""
case (_, 0) => ""
case (i, j) if s1(i - 1) == s2(j - 1) => lcs((i - 1, j - 1)) + s1(i - 1)
case (i, j) => longest(lcs((i, j - 1)), lcs((i - 1, j)))
}
lcs((s1.length, s2.length))
}