Longest common subsequence

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Task
Longest common subsequence
You are encouraged to solve this task according to the task description, using any language you may know.

The longest common subsequence (or LCS) of groups A and B is the longest group of elements from A and B that are common between the two groups and in the same order in each group. For example, the sequences "1234" and "1224533324" have an LCS of "1234":

1234
1224533324

For a string example, consider the sequences "thisisatest" and "testing123testing". An LCS would be "tsitest":

thisisatest
testing123testing

In this puzzle, your code only needs to deal with strings. Write a function which returns an LCS of two strings (case-sensitive). You don't need to show multiple LCS's.

For more information on this problem please see Wikipedia.

Ada[edit]

Using recursion:

with Ada.Text_IO;  use Ada.Text_IO;
 
procedure Test_LCS is
function LCS (A, B : String) return String is
begin
if A'Length = 0 or else B'Length = 0 then
return "";
elsif A (A'Last) = B (B'Last) then
return LCS (A (A'First..A'Last - 1), B (B'First..B'Last - 1)) & A (A'Last);
else
declare
X : String renames LCS (A, B (B'First..B'Last - 1));
Y : String renames LCS (A (A'First..A'Last - 1), B);
begin
if X'Length > Y'Length then
return X;
else
return Y;
end if;
end;
end if;
end LCS;
begin
Put_Line (LCS ("thisisatest", "testing123testing"));
end Test_LCS;
Output:
tsitest

Non-recursive solution:

with Ada.Text_IO;  use Ada.Text_IO;
 
procedure Test_LCS is
function LCS (A, B : String) return String is
L : array (A'First..A'Last + 1, B'First..B'Last + 1) of Natural;
begin
for I in L'Range (1) loop
L (I, B'First) := 0;
end loop;
for J in L'Range (2) loop
L (A'First, J) := 0;
end loop;
for I in A'Range loop
for J in B'Range loop
if A (I) = B (J) then
L (I + 1, J + 1) := L (I, J) + 1;
else
L (I + 1, J + 1) := Natural'Max (L (I + 1, J), L (I, J + 1));
end if;
end loop;
end loop;
declare
I : Integer := L'Last (1);
J : Integer := L'Last (2);
R : String (1..Integer'Max (A'Length, B'Length));
K : Integer := R'Last;
begin
while I > L'First (1) and then J > L'First (2) loop
if L (I, J) = L (I - 1, J) then
I := I - 1;
elsif L (I, J) = L (I, J - 1) then
J := J - 1;
else
I := I - 1;
J := J - 1;
R (K) := A (I);
K := K - 1;
end if;
end loop;
return R (K + 1..R'Last);
end;
end LCS;
begin
Put_Line (LCS ("thisisatest", "testing123testing"));
end Test_LCS;
Output:
tsitest

ALGOL 68[edit]

Translation of: Ada
Works with: ALGOL 68 version Standard - no extensions to language used
Works with: ALGOL 68G version Any - tested with release mk15-0.8b.fc9.i386
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8.8d.fc9.i386
main:(
PROC lcs = (STRING a, b)STRING:
BEGIN
IF UPB a = 0 OR UPB b = 0 THEN
""
ELIF a [UPB a] = b [UPB b] THEN
lcs (a [:UPB a - 1], b [:UPB b - 1]) + a [UPB a]
ELSE
STRING x = lcs (a, b [:UPB b - 1]);
STRING y = lcs (a [:UPB a - 1], b);
IF UPB x > UPB y THEN x ELSE y FI
FI
END # lcs #;
print((lcs ("thisisatest", "testing123testing"), new line))
)
Output:
tsitest

APL[edit]

Works with: Dyalog APL
lcs←{
⎕IO←0
betterof←{⊃(</+/¨⍺ ⍵)⌽⍺ ⍵} ⍝ better of 2 selections
cmbn←{↑,⊃∘.,/(⊂⊂⍬),⍵} ⍝ combine lists
rr←{∧/↑>/1 ¯1↓[1]¨⊂⍵} ⍝ rising rows
hmrr←{∨/(rr ⍵)∧∧/⍵=⌈\⍵} ⍝ has monotonically rising rows
rnbc←{{⍵/⍳⍴⍵}¨↓[0]×⍵} ⍝ row numbers by column
valid←hmrr∘cmbn∘rnbc ⍝ any valid solutions?
a w←(</⊃∘⍴¨⍺ ⍵)⌽⍺ ⍵ ⍝ longest first
matches←a∘.=w
aps←{⍵[;⍒+⌿⍵]}∘{(⍵/2)⊤⍳2*⍵} ⍝ all possible subsequences
swps←{⍵/⍨∧⌿~(~∨⌿⍺)⌿⍵} ⍝ subsequences with possible solns
sstt←matches swps aps⊃⍴w ⍝ subsequences to try
w/⍨{
⍺←0⍴⍨⊃⍴⍵ ⍝ initial selection
(+/⍺)≥+/⍵[;0]:⍺ ⍝ no scope to improve
this←⍺ betterof{⍵×valid ⍵/matches}⍵[;0] ⍝ try to improve
1=1⊃⍴⍵:this ⍝ nothing left to try
this ∇ 1↓[1]⍵ ⍝ keep looking
}sstt
}

AutoHotkey[edit]

Translation of: Java
using dynamic programming

ahk forum: discussion

lcs(a,b) { ; Longest Common Subsequence of strings, using Dynamic Programming
Loop % StrLen(a)+2 { ; Initialize
i := A_Index-1
Loop % StrLen(b)+2
j := A_Index-1, len%i%_%j% := 0
}
Loop Parse, a ; scan a
{
i := A_Index, i1 := i+1, x := A_LoopField
Loop Parse, b ; scan b
{
j := A_Index, j1 := j+1, y := A_LoopField
len%i1%_%j1% := x=y ? len%i%_%j% + 1
 : (u:=len%i1%_%j%) > (v:=len%i%_%j1%) ? u : v
}
}
x := StrLen(a)+1, y := StrLen(b)+1
While x*y { ; construct solution from lengths
x1 := x-1, y1 := y-1
If (len%x%_%y% = len%x1%_%y%)
x := x1
Else If (len%x%_%y% = len%x%_%y1%)
y := y1
Else
x := x1, y := y1, t := SubStr(a,x,1) t
}
Return t
}


BASIC[edit]

Works with: QuickBasic version 4.5
Translation of: Java
FUNCTION lcs$ (a$, b$)
IF LEN(a$) = 0 OR LEN(b$) = 0 THEN
lcs$ = ""
ELSEIF RIGHT$(a$, 1) = RIGHT$(b$, 1) THEN
lcs$ = lcs$(LEFT$(a$, LEN(a$) - 1), LEFT$(b$, LEN(b$) - 1)) + RIGHT$(a$, 1)
ELSE
x$ = lcs$(a$, LEFT$(b$, LEN(b$) - 1))
y$ = lcs$(LEFT$(a$, LEN(a$) - 1), b$)
IF LEN(x$) > LEN(y$) THEN
lcs$ = x$
ELSE
lcs$ = y$
END IF
END IF
END FUNCTION





BBC BASIC[edit]

This makes heavy use of BBC BASIC's shortcut LEFT$(a$) and RIGHT$(a$) functions.

      PRINT FNlcs("1234", "1224533324")
PRINT FNlcs("thisisatest", "testing123testing")
END
 
DEF FNlcs(a$, b$)
IF a$="" OR b$="" THEN = ""
IF RIGHT$(a$) = RIGHT$(b$) THEN = FNlcs(LEFT$(a$), LEFT$(b$)) + RIGHT$(a$)
LOCAL x$, y$
x$ = FNlcs(a$, LEFT$(b$))
y$ = FNlcs(LEFT$(a$), b$)
IF LEN(y$) > LEN(x$) SWAP x$,y$
= x$

Output:

1234
tsitest

Bracmat[edit]

  ( LCS
= A a ta B b tb prefix
.  !arg:(?prefix.@(?A:%?a ?ta).@(?B:%?b ?tb))
& ( !a:!b&LCS$(!prefix !a.!ta.!tb)
| LCS$(!prefix.!A.!tb)&LCS$(!prefix.!ta.!B)
)
| !prefix:? ([>!max:[?max):?lcs
|
)
& 0:?max
& :?lcs
& LCS$(.thisisatest.testing123testing)
& out$(max !max lcs !lcs);
Output:
max 7 lcs t s i t e s t

C[edit]

#include <stdio.h>
#include <stdlib.h>
 
#define MAX(a, b) (a > b ? a : b)
 
int lcs (char *a, int n, char *b, int m, char **s) {
int i, j, k, t;
int *z = calloc((n + 1) * (m + 1), sizeof (int));
int **c = calloc((n + 1), sizeof (int *));
for (i = 0; i <= n; i++) {
c[i] = &z[i * (m + 1)];
}
for (i = 1; i <= n; i++) {
for (j = 1; j <= m; j++) {
if (a[i - 1] == b[j - 1]) {
c[i][j] = c[i - 1][j - 1] + 1;
}
else {
c[i][j] = MAX(c[i - 1][j], c[i][j - 1]);
}
}
}
t = c[n][m];
*s = malloc(t);
for (i = n, j = m, k = t - 1; k >= 0;) {
if (a[i - 1] == b[j - 1])
(*s)[k] = a[i - 1], i--, j--, k--;
else if (c[i][j - 1] > c[i - 1][j])
j--;
else
i--;
}
free(c);
free(z);
return t;
}
 

Testing

int main () {
char a[] = "thisisatest";
char b[] = "testing123testing";
int n = sizeof a - 1;
int m = sizeof b - 1;
char *s = NULL;
int t = lcs(a, n, b, m, &s);
printf("%.*s\n", t, s); // tsitest
return 0;
}

C++[edit]

The Longest Common Subsequence (LCS) Problem

Defining a subsequence to be a string obtained by deleting zero or more symbols from an input string, the LCS Problem is to find a subsequence of maximum length that is common to two input strings.

Background

Where the number of symbols appearing in matches is small relative to the length of the input strings, reuse of the symbols necessarily increases; and the number of matches will tend towards quadratic, O(m*n) growth.

This occurs, for example, in Bioinformatics applications of nucleotide and protein sequencing.

Here the "divide and conquer" approach of Hirschberg limits the space required to O(m+n). However, this approach requires O(m*n) time even in the best case.

This quadratic time dependency may become prohibitive, given very long input strings. Thus, heuristics are often favored over optimal Dynamic Programming solutions.

In the application of comparing file revisions, records from the input files form a large symbol space; and the number of symbols used for matches may approach the length of the LCS.

Assuming a uniform distribution of symbols, the number of matches may tend only towards linear, O(m+n) growth.

A binary search optimization due to Hunt and Szymanski can be applied in this case, which results in expected performance of O(n log m), given m <= n. In the worst case, performance degrades to O(m*n log m) time if the number of matches, and the space required to represent them, should grow to O(m*n).

More recent improvements by Rick and by Goeman and Clausen reduce the time bound to O(n*s + min(p*m, p*(n-p))) where the alphabet is of size s and the LCS is of length p.

References

"A linear space algorithm for computing maximal common subsequences"
by Daniel S. Hirschberg, published June 1975
Communications of the ACM [Volume 18, Number 6, pp. 341–343]

"An Algorithm for Differential File Comparison"
by James W. Hunt and M. Douglas McIlroy, June 1976
Computing Science Technical Report, Bell Laboratories 41

"A Fast Algorithm for Computing Longest Common Subsequences"
by James W. Hunt and Thomas G. Szymanski, published May 1977
Communications of the ACM [Volume 20, Number 5, pp. 350-353]

"A new flexible algorithm for the longest common subsequence problem"
by Claus Rick, published 1995, Proceedings, 6th Annual Symposium on
Combinatorial Pattern Matching [Lecture Notes in Computer Science,
Springer Verlag, Volume 937, pp. 340-351]

"A New Practical Linear Space Algorithm for the Longest Common
Subsequence Problem" by Heiko Goeman and Michael Clausen,
published 2002, Kybernetika [volume 38, Issue 1, pp. 45-66]

Hunt and Szymanski algorithm

#include <stdint.h>
#include <string>
#include <memory> // for shared_ptr<>
#include <iostream>
#include <deque>
#include <map>
#include <algorithm> // for lower_bound()
 
using namespace std;
 
class LCS {
protected:
// This linked list class is used to trace the LCS candidates
class Pair {
public:
uint32_t index1;
uint32_t index2;
shared_ptr<Pair> next;
 
Pair(uint32_t index1, uint32_t index2, shared_ptr<Pair> next = nullptr)
: index1(index1), index2(index2), next(next) {
}
 
static shared_ptr<Pair> Reverse(const shared_ptr<Pair> pairs) {
shared_ptr<Pair> head = nullptr;
for (auto next = pairs; next != nullptr; next = next->next)
head = make_shared<Pair>(next->index1, next->index2, head);
return head;
}
};
 
typedef deque<shared_ptr<Pair>> PAIRS;
typedef deque<uint32_t> THRESHOLD;
typedef deque<uint32_t> INDEXES;
typedef map<char, INDEXES> CHAR2INDEXES;
typedef deque<INDEXES*> MATCHES;
 
// return the LCS as a linked list of matched index pairs
uint64_t LCS::Pairs(MATCHES& matches, shared_ptr<Pair> *pairs) {
auto trace = pairs != nullptr;
PAIRS traces;
THRESHOLD threshold;
 
//
//[Assert]After each index1 iteration threshold[index3] is the least index2
// such that the LCS of s1[0:index1] and s2[0:index2] has length index3 + 1
//
uint32_t index1 = 0;
for (const auto& it1 : matches) {
if (!it1->empty()) {
auto dq2 = *it1;
auto limit = threshold.end();
for (auto it2 = dq2.begin(); it2 != dq2.end(); it2++) {
// Each of the index1, index2 pairs considered here correspond to a match
auto index2 = *it2;
 
//
// Note: The index2 values are monotonically decreasing, which allows the
// thresholds to be updated in place. Montonicity allows a binary search,
// implemented here by std::lower_bound()
//
limit = lower_bound(threshold.begin(), limit, index2);
auto index3 = distance(threshold.begin(), limit);
 
//
// Look ahead to the next index2 value to optimize space used in the Hunt
// and Szymanski algorithm. If the next index2 is also an improvement on
// the value currently held in threshold[index3], a new Pair will only be
// superseded on the next index2 iteration.
//
// Depending on match redundancy, the number of Pair constructions may be
// divided by factors ranging from 2 up to 10 or more.
//
auto skip = it2 + 1 != dq2.end() &&
(limit == threshold.begin() || *(limit - 1) < *(it2 + 1));
 
if (skip) continue;
 
if (limit == threshold.end()) {
// insert case
threshold.push_back(index2);
if (trace) {
auto prefix = index3 > 0 ? traces[index3 - 1] : nullptr;
auto last = make_shared<Pair>(index1, index2, prefix);
traces.push_back(last);
}
}
else if (index2 < *limit) {
// replacement case
*limit = index2;
if (trace) {
auto prefix = index3 > 0 ? traces[index3 - 1] : nullptr;
auto last = make_shared<Pair>(index1, index2, prefix);
traces[index3] = last;
}
}
} // next index2
}
 
index1++;
} // next index1
 
if (trace) {
auto last = traces.size() > 0 ? traces.back() : nullptr;
// Reverse longest back-trace
*pairs = Pair::Reverse(last);
}
 
auto length = threshold.size();
return length;
}
 
//
// Match() avoids incurring m*n comparisons by using the associative
// memory implemented by CHAR2INDEXES to achieve O(m+n) performance,
// where m and n are the input lengths.
//
// The lookup time can be assumed constant in the case of characters.
// The symbol space is larger in the case of records; but the lookup
// time will be O(log(m+n)), at most.
//
void Match(CHAR2INDEXES& indexes, MATCHES& matches,
const string& s1, const string& s2) {
uint32_t index = 0;
for (const auto& it : s2)
indexes[it].push_front(index++);
 
for (const auto& it : s1) {
auto& dq2 = indexes[it];
matches.push_back(&dq2);
}
}
 
string Select(shared_ptr<Pair> pairs, uint64_t length,
bool right, const string& s1, const string& s2) {
string buffer;
buffer.reserve(length);
for (auto next = pairs; next != nullptr; next = next->next) {
auto c = right ? s2[next->index2] : s1[next->index1];
buffer.push_back(c);
}
return buffer;
}
 
public:
string Correspondence(const string& s1, const string& s2) {
CHAR2INDEXES indexes;
MATCHES matches; // holds references into indexes
Match(indexes, matches, s1, s2);
shared_ptr<Pair> pairs; // obtain the LCS as index pairs
auto length = Pairs(matches, &pairs);
return Select(pairs, length, false, s1, s2);
}
};

Example:

    LCS lcs;
auto s = lcs.Correspondence(s1, s2);
cout << s << endl;

C#[edit]

With recursion[edit]

using System;
 
namespace LCS
{
class Program
{
static void Main(string[] args)
{
string word1 = "thisisatest";
string word2 = "testing123testing";
 
Console.WriteLine(lcsBack(word1, word2));
Console.ReadKey();
}
 
public static string lcsBack(string a, string b)
{
string aSub = a.Substring(0, (a.Length - 1 < 0) ? 0 : a.Length - 1);
string bSub = b.Substring(0, (b.Length - 1 < 0) ? 0 : b.Length - 1);
 
if (a.Length == 0 || b.Length == 0)
return "";
else if (a[a.Length - 1] == b[b.Length - 1])
return lcsBack(aSub, bSub) + a[a.Length - 1];
else
{
string x = lcsBack(a, bSub);
string y = lcsBack(aSub, b);
return (x.Length > y.Length) ? x : y;
}
}
}
}

Clojure[edit]

Based on algorithm from Wikipedia.

(defn longest [xs ys] (if (> (count xs) (count ys)) xs ys))
 
 
(def lcs
(memoize
(fn [[x & xs] [y & ys]]
(cond
(or (= x nil) (= y nil) ) nil
(= x y) (cons x (lcs xs ys))
 :else (longest (lcs (cons x xs) ys) (lcs xs (cons y ys)))))))

CoffeeScript[edit]

 
lcs = (s1, s2) ->
len1 = s1.length
len2 = s2.length
 
# Create a virtual matrix that is (len1 + 1) by (len2 + 1),
# where m[i][j] is the longest common string using only
# the first i chars of s1 and first j chars of s2. The
# matrix is virtual, because we only keep the last two rows
# in memory.
prior_row = ('' for i in [0..len2])
 
for i in [0...len1]
row = ['']
for j in [0...len2]
if s1[i] == s2[j]
row.push prior_row[j] + s1[i]
else
subs1 = row[j]
subs2 = prior_row[j+1]
if subs1.length > subs2.length
row.push subs1
else
row.push subs2
prior_row = row
 
row[len2]
 
s1 = "thisisatest"
s2 = "testing123testing"
console.log lcs(s1, s2)

Common Lisp[edit]

Here's a memoizing/dynamic-programming solution that uses an n × m array where n and m are the lengths of the input arrays. The first return value is a sequence (of the same type as array1) which is the longest common subsequence. The second return value is the length of the longest common subsequence.

(defun longest-common-subsequence (array1 array2)
(let* ((l1 (length array1))
(l2 (length array2))
(results (make-array (list l1 l2) :initial-element nil)))
(declare (dynamic-extent results))
(labels ((lcs (start1 start2)
;; if either sequence is empty, return (() 0)
(if (or (eql start1 l1) (eql start2 l2)) (list '() 0)
;; otherwise, return any memoized value
(let ((result (aref results start1 start2)))
(if (not (null result)) result
;; otherwise, compute and store a value
(setf (aref results start1 start2)
(if (eql (aref array1 start1) (aref array2 start2))
;; if they start with the same element,
;; move forward in both sequences
(destructuring-bind (seq len)
(lcs (1+ start1) (1+ start2))
(list (cons (aref array1 start1) seq) (1+ len)))
;; otherwise, move ahead in each separately,
;; and return the better result.
(let ((a (lcs (1+ start1) start2))
(b (lcs start1 (1+ start2))))
(if (> (second a) (second b))
a
b)))))))))
(destructuring-bind (seq len) (lcs 0 0)
(values (coerce seq (type-of array1)) len)))))

For example,

(longest-common-subsequence "123456" "1a2b3c")

produces the two values

"123"
3

An alternative adopted from Clojure[edit]

Here is another version with its own memoization macro:

(defmacro mem-defun (name args body)
(let ((hash-name (gensym)))
`(let ((,hash-name (make-hash-table :test 'equal)))
(defun ,name ,args
(or (gethash (list ,@args) ,hash-name)
(setf (gethash (list ,@args) ,hash-name)
,body))))))
 
(mem-defun lcs (xs ys)
(labels ((longer (a b) (if (> (length a) (length b)) a b)))
(cond ((or (null xs) (null ys)) nil)
((equal (car xs) (car ys)) (cons (car xs) (lcs (cdr xs) (cdr ys))))
(t (longer (lcs (cdr xs) ys)
(lcs xs (cdr ys)))))))

When we test it, we get:

(coerce (lcs (coerce "thisisatest" 'list) (coerce "testing123testing" 'list)) 'string))))
 
"tsitest"

D[edit]

Both versions don't work correctly with Unicode text.

Recursive version[edit]

import std.stdio, std.array;
 
T[] lcs(T)(in T[] a, in T[] b) pure nothrow @safe {
if (a.empty || b.empty) return null;
if (a[0] == b[0])
return a[0] ~ lcs(a[1 .. $], b[1 .. $]);
const longest = (T[] x, T[] y) => x.length > y.length ? x : y;
return longest(lcs(a, b[1 .. $]), lcs(a[1 .. $], b));
}
 
void main() {
lcs("thisisatest", "testing123testing").writeln;
}
Output:
tsitest

Faster dynamic programming version[edit]

The output is the same.

import std.stdio, std.algorithm, std.traits;
 
T[] lcs(T)(in T[] a, in T[] b) pure /*nothrow*/ {
auto L = new uint[][](a.length + 1, b.length + 1);
 
foreach (immutable i; 0 .. a.length)
foreach (immutable j; 0 .. b.length)
L[i + 1][j + 1] = (a[i] == b[j]) ? (1 + L[i][j]) :
max(L[i + 1][j], L[i][j + 1]);
 
Unqual!T[] result;
for (auto i = a.length, j = b.length; i > 0 && j > 0; ) {
if (a[i - 1] == b[j - 1]) {
result ~= a[i - 1];
i--;
j--;
} else
if (L[i][j - 1] < L[i - 1][j])
i--;
else
j--;
}
 
result.reverse(); // Not nothrow.
return result;
}
 
void main() {
lcs("thisisatest", "testing123testing").writeln;
}

Hirschberg algorithm version[edit]

See: http://en.wikipedia.org/wiki/Hirschberg_algorithm

This is currently a little slower than the classic dynamic programming version, but it uses a linear amount of memory, so it's usable for much larger inputs. To speed up this code on dmd remove the memory allocations from lensLCS, and do not use the retro range (replace it with foreach_reverse). The output is the same.

Adapted from Python code: http://wordaligned.org/articles/longest-common-subsequence

import std.stdio, std.algorithm, std.range, std.array, std.string, std.typecons;
 
uint[] lensLCS(R)(R xs, R ys) pure nothrow @safe {
auto prev = new typeof(return)(1 + ys.length);
auto curr = new typeof(return)(1 + ys.length);
 
foreach (immutable x; xs) {
swap(curr, prev);
size_t i = 0;
foreach (immutable y; ys) {
curr[i + 1] = (x == y) ? prev[i] + 1 : max(curr[i], prev[i + 1]);
i++;
}
}
 
return curr;
}
 
void calculateLCS(T)(in T[] xs, in T[] ys, bool[] xs_in_lcs,
in size_t idx=0) pure nothrow @safe {
immutable nx = xs.length;
immutable ny = ys.length;
 
if (nx == 0)
return;
 
if (nx == 1) {
if (ys.canFind(xs[0]))
xs_in_lcs[idx] = true;
} else {
immutable mid = nx / 2;
const xb = xs[0.. mid];
const xe = xs[mid .. $];
immutable ll_b = lensLCS(xb, ys);
 
const ll_e = lensLCS(xe.retro, ys.retro); // retro is slow with dmd.
 
//immutable k = iota(ny + 1)
// .reduce!(max!(j => ll_b[j] + ll_e[ny - j]));
immutable k = iota(ny + 1)
.minPos!((i, j) => tuple(ll_b[i] + ll_e[ny - i]) >
tuple(ll_b[j] + ll_e[ny - j]))[0];
 
calculateLCS(xb, ys[0 .. k], xs_in_lcs, idx);
calculateLCS(xe, ys[k .. $], xs_in_lcs, idx + mid);
}
}
 
const(T)[] lcs(T)(in T[] xs, in T[] ys) pure /*nothrow*/ @safe {
auto xs_in_lcs = new bool[xs.length];
calculateLCS(xs, ys, xs_in_lcs);
return zip(xs, xs_in_lcs).filter!q{ a[1] }.map!q{ a[0] }.array; // Not nothrow.
}
 
string lcsString(in string s1, in string s2) pure /*nothrow*/ @safe {
return lcs(s1.representation, s2.representation).assumeUTF;
}
 
void main() {
lcsString("thisisatest", "testing123testing").writeln;
}

Dart[edit]

import 'dart:math';
 
String lcsRecursion(String a, String b) {
int aLen = a.length;
int bLen = b.length;
 
if (aLen == 0 || bLen == 0) {
return "";
} else if (a[aLen-1] == b[bLen-1]) {
return lcsRecursion(a.substring(0,aLen-1),b.substring(0,bLen-1)) + a[aLen-1];
} else {
var x = lcsRecursion(a, b.substring(0,bLen-1));
var y = lcsRecursion(a.substring(0,aLen-1), b);
return (x.length > y.length) ? x : y;
}
}
 
String lcsDynamic(String a, String b) {
var lengths = new List<List<int>>.generate(a.length + 1,
(_) => new List.filled(b.length+1, 0), growable: false);
 
// row 0 and column 0 are initialized to 0 already
for (int i = 0; i < a.length; i++) {
for (int j = 0; j < b.length; j++) {
if (a[i] == b[j]) {
lengths[i+1][j+1] = lengths[i][j] + 1;
} else {
lengths[i+1][j+1] = max(lengths[i+1][j], lengths[i][j+1]);
}
}
}
 
// read the substring out from the matrix
StringBuffer reversedLcsBuffer = new StringBuffer();
for (int x = a.length, y = b.length; x != 0 && y != 0;) {
if (lengths[x][y] == lengths[x-1][y]) {
x--;
} else if (lengths[x][y] == lengths[x][y-1]) {
y--;
} else {
assert(a[x-1] == b[y-1]);
reversedLcsBuffer.write(a[x-1]);
x--;
y--;
}
}
 
// reverse String
var reversedLCS = reversedLcsBuffer.toString();
var lcsBuffer = new StringBuffer();
for(var i = reversedLCS.length - 1; i>=0; i--) {
lcsBuffer.write(reversedLCS[i]);
}
return lcsBuffer.toString();
}
 
void main() {
print("lcsDynamic('1234', '1224533324') = ${lcsDynamic('1234', '1224533324')}");
print("lcsDynamic('thisisatest', 'testing123testing') = ${lcsDynamic('thisisatest', 'testing123testing')}");
print("lcsDynamic('', 'x') = ${lcsDynamic('', 'x')}");
print("lcsDynamic('x', 'x') = ${lcsDynamic('x', 'x')}");
print('');
print("lcsRecursion('1234', '1224533324') = ${lcsRecursion('1234', '1224533324')}");
print("lcsRecursion('thisisatest', 'testing123testing') = ${lcsRecursion('thisisatest', 'testing123testing')}");
print("lcsRecursion('', 'x') = ${lcsRecursion('', 'x')}");
print("lcsRecursion('x', 'x') = ${lcsRecursion('x', 'x')}");
}
 
Output:
lcsDynamic('1234', '1224533324') = 1234
lcsDynamic('thisisatest', 'testing123testing') = tsitest
lcsDynamic('', 'x') = 
lcsDynamic('x', 'x') = x

lcsRecursion('1234', '1224533324') = 1234
lcsRecursion('thisisatest', 'testing123testing') = tsitest
lcsRecursion('', 'x') = 
lcsRecursion('x', 'x') = x

Egison[edit]

 
(define $common-seqs
(lambda [$xs $ys]
(match-all [xs ys] [(list char) (list char)]
[[(loop $i [1 $n] <join _ <cons $c_i ...>> _)
(loop $i [1 ,n] <join _ <cons ,c_i ...>> _)]
(map (lambda [$i] c_i) (between 1 n))])))
 
(define $lcs (compose common-seqs rac))
 

Output:

 
> (lcs "thisisatest" "testing123testing"))
"tsitest"
 

Elixir[edit]

Translation of: Erlang
defmodule LCS do
def lcs_length(s,t) do
{l,_c} = lcs_length(s,t,Map.new)
l
end
 
defp lcs_length([],t,cache), do: {0,Dict.put(cache,{[],t},0)}
defp lcs_length(s,[],cache), do: {0,Dict.put(cache,{s,[]},0)}
defp lcs_length([h|st]=s,[h|tt]=t,cache) do
{l,c} = lcs_length(st,tt,cache)
{l+1,Dict.put(c,{s,t},l+1)}
end
defp lcs_length([_sh|st]=s,[_th|tt]=t,cache) do
if Dict.has_key?(cache,{s,t}) do
{Dict.get(cache,{s,t}),cache}
else
{l1,c1} = lcs_length(s,tt,cache)
{l2,c2} = lcs_length(st,t,c1)
l = Enum.max([l1,l2])
{l,Dict.put(c2,{s,t},l)}
end
end
 
def lcs(s,t) do
{_,c} = lcs_length(s,t,Map.new)
lcs(s,t,c,[])
end
 
defp lcs([],_,_,acc), do: Enum.reverse(acc)
defp lcs(_,[],_,acc), do: Enum.reverse(acc)
defp lcs([h|st],[h|tt],cache,acc), do: lcs(st,tt,cache,[h|acc])
defp lcs([_sh|st]=s,[_th|tt]=t,cache,acc) do
if Dict.get(cache,{s,tt}) > Dict.get(cache,{st,t}) do
lcs(s,tt,cache,acc)
else
lcs(st,t,cache,acc)
end
end
end
 
IO.puts LCS.lcs('thisisatest','testing123testing')
IO.puts LCS.lcs('1234','1224533324')
Output:
tsitest
1234

Erlang[edit]

This implementation also includes the ability to calculate the length of the longest common subsequence. In calculating that length, we generate a cache which can be traversed to generate the longest common subsequence.

 
module(lcs).
-compile(export_all).
 
lcs_length(S,T) ->
{L,_C} = lcs_length(S,T,dict:new()),
L.
 
lcs_length([]=S,T,Cache) ->
{0,dict:store({S,T},0,Cache)};
lcs_length(S,[]=T,Cache) ->
{0,dict:store({S,T},0,Cache)};
lcs_length([H|ST]=S,[H|TT]=T,Cache) ->
{L,C} = lcs_length(ST,TT,Cache),
{L+1,dict:store({S,T},L+1,C)};
lcs_length([_SH|ST]=S,[_TH|TT]=T,Cache) ->
case dict:is_key({S,T},Cache) of
true -> {dict:fetch({S,T},Cache),Cache};
false ->
{L1,C1} = lcs_length(S,TT,Cache),
{L2,C2} = lcs_length(ST,T,C1),
L = lists:max([L1,L2]),
{L,dict:store({S,T},L,C2)}
end.
 
lcs(S,T) ->
{_,C} = lcs_length(S,T,dict:new()),
lcs(S,T,C,[]).
 
lcs([],_,_,Acc) ->
lists:reverse(Acc);
lcs(_,[],_,Acc) ->
lists:reverse(Acc);
lcs([H|ST],[H|TT],Cache,Acc) ->
lcs(ST,TT,Cache,[H|Acc]);
lcs([_SH|ST]=S,[_TH|TT]=T,Cache,Acc) ->
case dict:fetch({S,TT},Cache) > dict:fetch({ST,T},Cache) of
true ->
lcs(S,TT,Cache, Acc);
false ->
lcs(ST,T,Cache,Acc)
end.
 

Output:

 
77> lcs:lcs("thisisatest","testing123testing").
"tsitest"
78> lcs:lcs("1234","1224533324").
"1234"
 

We can also use the process dictionary to memoize the recursive implementation:

 
lcs(Xs0, Ys0) ->
CacheKey = {lcs_cache, Xs0, Ys0},
case get(CacheKey)
of undefined ->
Result =
case {Xs0, Ys0}
of {[], _} -> []
; {_, []} -> []
; {[Same | Xs], [Same | Ys]} ->
[Same | lcs(Xs, Ys)]
; {[_ | XsRest]=XsAll, [_ | YsRest]=YsAll} ->
A = lcs(XsRest, YsAll),
B = lcs(XsAll , YsRest),
case length(A) > length(B)
of true -> A
; false -> B
end
end,
undefined = put(CacheKey, Result),
Result
; Result ->
Result
end.
 

Fortran[edit]

Works with: Fortran version 95

Using the iso_varying_string module which can be found here (or equivalent module conforming to the ISO/IEC 1539-2:2000 API or to a subset according to the need of this code: char, len, //, extract, ==, =)

program lcstest
use iso_varying_string
implicit none
 
type(varying_string) :: s1, s2
 
s1 = "thisisatest"
s2 = "testing123testing"
print *, char(lcs(s1, s2))
 
s1 = "1234"
s2 = "1224533324"
print *, char(lcs(s1, s2))
 
contains
 
recursive function lcs(a, b) result(l)
type(varying_string) :: l
type(varying_string), intent(in) :: a, b
 
type(varying_string) :: x, y
 
l = ""
if ( (len(a) == 0) .or. (len(b) == 0) ) return
if ( extract(a, len(a), len(a)) == extract(b, len(b), len(b)) ) then
l = lcs(extract(a, 1, len(a)-1), extract(b, 1, len(b)-1)) // extract(a, len(a), len(a))
else
x = lcs(a, extract(b, 1, len(b)-1))
y = lcs(extract(a, 1, len(a)-1), b)
if ( len(x) > len(y) ) then
l = x
else
l = y
end if
end if
end function lcs
 
end program lcstest

F#[edit]

Copied and slightly adapted from OCaml (direct recursion)

open System
 
let longest xs ys = if List.length xs > List.length ys then xs else ys
 
let rec lcs a b =
match a, b with
| [], _
| _, [] -> []
| x::xs, y::ys ->
if x = y then
x :: lcs xs ys
else
longest (lcs a ys) (lcs xs b)
 
[<EntryPoint>]
let main argv =
let split (str:string) = List.init str.Length (fun i -> str.[i])
printfn "%A" (String.Join("",
(lcs (split "thisisatest") (split "testing123testing"))))
0
 

Go[edit]

Translation of: Java

Recursion[edit]

Brute force

func lcs(a, b string) string {
aLen := len(a)
bLen := len(b)
if aLen == 0 || bLen == 0 {
return ""
} else if a[aLen-1] == b[bLen-1] {
return lcs(a[:aLen-1], b[:bLen-1]) + string(a[aLen-1])
}
x := lcs(a, b[:bLen-1])
y := lcs(a[:aLen-1], b)
if len(x) > len(y) {
return x
}
return y
}

Dynamic Programming[edit]

func lcs(a, b string) string {
arunes := []rune(a)
brunes := []rune(b)
aLen := len(arunes)
bLen := len(brunes)
lengths := make([][]int, aLen+1)
for i := 0; i <= aLen; i++ {
lengths[i] = make([]int, bLen+1)
}
// row 0 and column 0 are initialized to 0 already
 
for i := 0; i < aLen; i++ {
for j := 0; j < bLen; j++ {
if arunes[i] == brunes[j] {
lengths[i+1][j+1] = lengths[i][j] + 1
} else if lengths[i+1][j] > lengths[i][j+1] {
lengths[i+1][j+1] = lengths[i+1][j]
} else {
lengths[i+1][j+1] = lengths[i][j+1]
}
}
}
 
// read the substring out from the matrix
s := make([]rune, 0, lengths[aLen][bLen])
for x, y := aLen, bLen; x != 0 && y != 0; {
if lengths[x][y] == lengths[x-1][y] {
x--
} else if lengths[x][y] == lengths[x][y-1] {
y--
} else {
s = append(s, arunes[x-1])
x--
y--
}
}
// reverse string
for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
s[i], s[j] = s[j], s[i]
}
return string(s)
}

Groovy[edit]

Recursive solution:

def lcs(xstr, ystr) {
if (xstr == "" || ystr == "") {
return "";
}
 
def x = xstr[0];
def y = ystr[0];
 
def xs = xstr.size() > 1 ? xstr[1..-1] : "";
def ys = ystr.size() > 1 ? ystr[1..-1] : "";
 
if (x == y) {
return (x + lcs(xs, ys));
}
 
def lcs1 = lcs(xstr, ys);
def lcs2 = lcs(xs, ystr);
 
lcs1.size() > lcs2.size() ? lcs1 : lcs2;
}
 
println(lcs("1234", "1224533324"));
println(lcs("thisisatest", "testing123testing"));
Output:
1234
tsitest

Haskell[edit]

The Wikipedia solution translates directly into Haskell, with the only difference that equal characters are added in front:

longest xs ys = if length xs > length ys then xs else ys
 
lcs [] _ = []
lcs _ [] = []
lcs (x:xs) (y:ys)
| x == y = x : lcs xs ys
| otherwise = longest (lcs (x:xs) ys) (lcs xs (y:ys))

A Memoized version of the naive algorithm.

import qualified Data.MemoCombinators as M
 
lcs = memoize lcsm
where
lcsm [] _ = []
lcsm _ [] = []
lcsm (x:xs) (y:ys)
| x == y = x : lcs xs ys
| otherwise = maxl (lcs (x:xs) ys) (lcs xs (y:ys))
 
maxl x y = if length x > length y then x else y
memoize = M.memo2 mString mString
mString = M.list M.char -- Chars, but you can specify any type you need for the memo

Memoization (aka dynamic programming) of that uses zip to make both the index and the character available:

import Data.Array
 
lcs xs ys = a!(0,0) where
n = length xs
m = length ys
a = array ((0,0),(n,m)) $ l1 ++ l2 ++ l3
l1 = [((i,m),[]) | i <- [0..n]]
l2 = [((n,j),[]) | j <- [0..m]]
l3 = [((i,j), f x y i j) | (x,i) <- zip xs [0..], (y,j) <- zip ys [0..]]
f x y i j
| x == y = x : a!(i+1,j+1)
| otherwise = longest (a!(i,j+1)) (a!(i+1,j))

All 3 solutions work of course not only with strings, but also with any other list. Example:

*Main> lcs "thisisatest" "testing123testing"
"tsitest"

The dynamic programming version without using arrays:

import Data.List
 
longest xs ys = if length xs > length ys then xs else ys
 
lcs xs ys = head $ foldr(\xs -> map head. scanr1 f. zipWith (\x y -> [x,y]) xs) e m where
m = map (\x -> flip (++) [[]] $ map (\y -> [x | x==y]) ys) xs
e = replicate (length ys) []
f [a,b] [c,d]
| null a = longest b c: [b]
| otherwise = (a++d):[b]


Simple and slow solution:

import Data.Ord
import Data.List
 
-- longest common
lcs xs ys = maximumBy (comparing length) $ intersect (subsequences xs) (subsequences ys)
 
main = print $ lcs "thisisatest" "testing123testing"
Output:
"tsitest"

Icon and Unicon[edit]

This solution is a modified variant of the recursive solution. The modifications include (a) deleting all characters not common to both strings and (b) stripping off common prefixes and suffixes in a single step.

Uses deletec from strings
procedure main()
LCSTEST("thisisatest","testing123testing")
LCSTEST("","x")
LCSTEST("x","x")
LCSTEST("beginning-middle-ending","beginning-diddle-dum-ending")
end
 
link strings
 
procedure LCSTEST(a,b) #: helper to show inputs and results
write("lcs( ",image(a),", ",image(b)," ) = ",image(res := lcs(a,b)))
return res
end
 
procedure lcs(a,b) #: return longest common sub-sequence of characters (modified recursive method)
local i,x,y
local c,nc
 
if *(a|b) = 0 then return "" # done if either string is empty
if a == b then return a # done if equal
 
if *(a ++ b -- (c := a ** b)) > 0 then { # find all characters not in common
a := deletec(a,nc := ~c) # .. remove
b := deletec(b,nc) # .. remove
} # only unequal strings and shared characters beyond
 
i := 0 ; while a[i+1] == b[i+1] do i +:=1 # find common prefix ...
if *(x := a[1+:i]) > 0 then # if any
return x || lcs(a[i+1:0],b[i+1:0]) # ... remove and process remainder
 
i := 0 ; while a[-(i+1)] == b[-(i+1)] do i +:=1 # find common suffix ...
if *(y := a[0-:i]) > 0 then # if any
return lcs(a[1:-i],b[1:-i]) || y # ... remove and process remainder
 
return if *(x := lcs(a,b[1:-1])) > *(y := lcs(a[1:-1],b)) then x else y # divide, discard, and keep longest
end
Output:
lcs( "thisisatest", "testing123testing" ) = "tsitest"
lcs( "", "x" ) = ""
lcs( "x", "x" ) = "x"
lcs( "beginning-middle-ending", "beginning-diddle-dum-ending" ) = "beginning-iddle-ending"

J[edit]

lcs=: dyad define
|.x{~ 0{"1 cullOne^:_ (\: +/"1)(\:{."1) 4$.$. x =/ y
)
 
cullOne=: ({~[: <@<@< [: (i. 0:)1,[: *./[: |: 2>/\]) :: ]

Here's another approach:

mergeSq=: ;@}:  ~.@, {.@;@{. ,&.> 3 {:: 4&{.
common=: 2 2 <@mergeSq@,;.3^:_ [: (<@#&.> i.@$) =/
lcs=: [ {~ 0 {"1 ,&$ #: 0 ({:: (#~ [: (= >./) #@>)) 0 ({:: ,) common

Example use (works with either definition of lcs):

   'thisisatest' lcs 'testing123testing'
tsitest

Dynamic programming version

longest=: ]`[@.(>&#)
upd=:{:@[,~ ({.@[ ,&.> {:@])`({:@[ longest&.> {.@])@.(0 = #&>@{.@[)
lcs=: 0{:: [: ([: {.&> [: upd&.>/\.<"1@:,.)/ a:,.~a:,~=/{"1 a:,.<"0@[

Output:

   '1234' lcs '1224533324'
1234
 
'thisisatest' lcs 'testing123testing'
tsitest

Recursion

lcs=:;(($:}.) longest }.@[ $: ])`({.@[,$:&}.)@.(=&{.)`((i.0)"_)@.(+.&(0=#))&((e.#[)&>/) ;~

Java[edit]

Recursion[edit]

This is not a particularly fast algorithm, but it gets the job done eventually. The speed is a result of many recursive function calls.

public static String lcs(String a, String b){
int aLen = a.length();
int bLen = b.length();
if(aLen == 0 || bLen == 0){
return "";
}else if(a.charAt(aLen-1) == b.charAt(bLen-1)){
return lcs(a.substring(0,aLen-1),b.substring(0,bLen-1))
+ a.charAt(aLen-1);
}else{
String x = lcs(a, b.substring(0,bLen-1));
String y = lcs(a.substring(0,aLen-1), b);
return (x.length() > y.length()) ? x : y;
}
}

Dynamic Programming[edit]

public static String lcs(String a, String b) {
int[][] lengths = new int[a.length()+1][b.length()+1];
 
// row 0 and column 0 are initialized to 0 already
 
for (int i = 0; i < a.length(); i++)
for (int j = 0; j < b.length(); j++)
if (a.charAt(i) == b.charAt(j))
lengths[i+1][j+1] = lengths[i][j] + 1;
else
lengths[i+1][j+1] =
Math.max(lengths[i+1][j], lengths[i][j+1]);
 
// read the substring out from the matrix
StringBuffer sb = new StringBuffer();
for (int x = a.length(), y = b.length();
x != 0 && y != 0; ) {
if (lengths[x][y] == lengths[x-1][y])
x--;
else if (lengths[x][y] == lengths[x][y-1])
y--;
else {
assert a.charAt(x-1) == b.charAt(y-1);
sb.append(a.charAt(x-1));
x--;
y--;
}
}
 
return sb.reverse().toString();
}

JavaScript[edit]

Recursion[edit]

Translation of: Java

This is more or less a translation of the recursive Java version above.

function lcs(a, b) {
var aSub = a.substr(0, a.length - 1);
var bSub = b.substr(0, b.length - 1);
 
if (a.length === 0 || b.length === 0) {
return '';
} else if (a.charAt(a.length - 1) === b.charAt(b.length - 1)) {
return lcs(aSub, bSub) + a.charAt(a.length - 1);
} else {
var x = lcs(a, bSub);
var y = lcs(aSub, b);
return (x.length > y.length) ? x : y;
}
}

ES6 recursive implementation

 
var longest = (xs, ys) => (xs.length > ys.length) ? xs : ys;
 
var lcs = (xx, yy) => {
if (!xx.length || !yy.length) { return ''; }
 
var x = xx[0],
y = yy[0];
xs = xx.slice(1);
ys = yy.slice(1);
 
return (x === y) ? lcs(xs, ys) :
longest(lcs(xx, ys), lcs(xs, yy));
};

Dynamic Programming[edit]

This version runs in O(mn) time and consumes O(mn) space. Factoring out loop edge cases could get a small constant time improvement, and it's fairly trivial to edit the final loop to produce a full diff in addition to the lcs.

function lcs(x,y){
var s,i,j,m,n,
lcs=[],row=[],c=[],
left,diag,latch;
//make sure shorter string is the column string
if(m<n){s=x;x=y;y=s;}
m = x.length;
n = y.length;
//build the c-table
for(j=0;j<n;row[j++]=0);
for(i=0;i<m;i++){
c[i] = row = row.slice();
for(diag=0,j=0;j<n;j++,diag=latch){
latch=row[j];
if(x[i] == y[j]){row[j] = diag+1;}
else{
left = row[j-1]||0;
if(left>row[j]){row[j] = left;}
}
}
}
i--,j--;
//row[j] now contains the length of the lcs
//recover the lcs from the table
while(i>-1&&j>-1){
switch(c[i][j]){
default: j--;
lcs.unshift(x[i]);
case (i&&c[i-1][j]): i--;
continue;
case (j&&c[i][j-1]): j--;
}
}
return lcs.join('');
}

BUG note: In line 6, m and n are not yet initialized, and so x and y are never swapped. Swapping is useless here, and becomes wrong when extending the algorithm to produce a diff.

The final loop can be modified to concatenate maximal common substrings rather than individual characters:

	var t=i;
while(i>-1&&j>-1){
switch(c[i][j]){
default:i--,j--;
continue;
case (i&&c[i-1][j]):
if(t!==i){lcs.unshift(x.substring(i+1,t+1));}
t=--i;
continue;
case (j&&c[i][j-1]): j--;
if(t!==i){lcs.unshift(x.substring(i+1,t+1));}
t=i;
}
}
if(t!==i){lcs.unshift(x.substring(i+1,t+1));}

Greedy Algorithm[edit]

This is an heuristic algorithm that won't always return the correct answer, but is significantly faster and less memory intensive than the dynamic programming version, in exchange for giving up the ability to re-use the table to find alternate solutions and greater complexity in generating diffs. Note that this implementation uses a binary buffer for additional efficiency gains, but it's simple to transform to use string or array concatenation;

function lcs_greedy(x,y){
var p1, i, idx,
symbols = {},
r = 0,
p = 0,
l = 0,
m = x.length,
n = y.length,
s = new Buffer((m < n) ? n : m);
 
p1 = popsym(0);
 
for (i = 0; i < m; i++) {
p = (r === p) ? p1 : popsym(i);
p1 = popsym(i + 1);
if (p > p1) {
i += 1;
idx = p1;
} else {
idx = p;
}
 
if (idx === n) {
p = popsym(i);
} else {
r = idx;
s[l] = x.charCodeAt(i);
l += 1;
}
}
return s.toString('utf8', 0, l);
 
function popsym(index) {
var s = x[index],
pos = symbols[s] + 1;
 
pos = y.indexOf(s, ((pos > r) ? pos : r));
if (pos === -1) { pos = n; }
symbols[s] = pos;
return pos;
}
}
Note that it won't return the correct answer for all inputs. For example:
lcs_greedy('bcaaaade', 'deaaaabc'); // 'bc' instead of 'aaaa'

jq[edit]

Naive recursive version:

def lcs(xstr; ystr):
if (xstr == "" or ystr == "") then ""
else
xstr[0:1] as $x
| xstr[1:] as $xs
| ystr[1:] as $ys
| if ($x == ystr[0:1]) then ($x + lcs($xs; $ys))
else
lcs(xstr; $ys) as $one
| lcs($xs; ystr) as $two
| if ($one|length) > ($two|length) then $one else $two end
end
end;

Example:

lcs("1234"; "1224533324"),
lcs("thisisatest"; "testing123testing")
Output:
 
# jq -n -f lcs-recursive.jq
"1234"
"tsitest"

Liberty BASIC[edit]

 
'variation of BASIC example
w$="aebdef"
z$="cacbc"
print lcs$(w$,z$)
 
'output:
'ab
 
wait
 
FUNCTION lcs$(a$, b$)
IF LEN(a$) = 0 OR LEN(b$) = 0 THEN
lcs$ = ""
exit function
end if
 
IF RIGHT$(a$, 1) = RIGHT$(b$, 1) THEN
lcs$ = lcs$(LEFT$(a$, LEN(a$) - 1), LEFT$(b$, LEN(b$) - 1)) + RIGHT$(a$, 1)
exit function
ELSE
x$ = lcs$(a$, LEFT$(b$, LEN(b$) - 1))
y$ = lcs$(LEFT$(a$, LEN(a$) - 1), b$)
IF LEN(x$) > LEN(y$) THEN
lcs$ = x$
exit function
ELSE
lcs$ = y$
exit function
END IF
END IF
END FUNCTION
 

[edit]

This implementation works on both words and lists.

to longest :s :t
output ifelse greater? count :s count :t [:s] [:t]
end
to lcs :s :t
if empty? :s [output :s]
if empty? :t [output :t]
if equal? first :s first :t [output combine first :s lcs bf :s bf :t]
output longest lcs :s bf :t lcs bf :s :t
end

Lua[edit]

function LCS( a, b )    
if #a == 0 or #b == 0 then
return ""
elseif string.sub( a, -1, -1 ) == string.sub( b, -1, -1 ) then
return LCS( string.sub( a, 1, -2 ), string.sub( b, 1, -2 ) ) .. string.sub( a, -1, -1 )
else
local a_sub = LCS( a, string.sub( b, 1, -2 ) )
local b_sub = LCS( string.sub( a, 1, -2 ), b )
 
if #a_sub > #b_sub then
return a_sub
else
return b_sub
end
end
end
 
print( LCS( "thisisatest", "testing123testing" ) )

M4[edit]

define(`set2d',`define(`$1[$2][$3]',`$4')')
define(`get2d',`defn($1[$2][$3])')
define(`tryboth',
`pushdef(`x',lcs(`$1',substr(`$2',1),`$1 $2'))`'pushdef(`y',
lcs(substr(`$1',1),`$2',`$1 $2'))`'ifelse(eval(len(x)>len(y)),1,
`x',`y')`'popdef(`x')`'popdef(`y')')
define(`checkfirst',
`ifelse(substr(`$1',0,1),substr(`$2',0,1),
`substr(`$1',0,1)`'lcs(substr(`$1',1),substr(`$2',1))',
`tryboth(`$1',`$2')')')
define(`lcs',
`ifelse(get2d(`c',`$1',`$2'),`',
`pushdef(`a',ifelse(
`$1',`',`',
`$2',`',`',
`checkfirst(`$1',`$2')'))`'a`'set2d(`c',`$1',`$2',a)`'popdef(`a')',
`get2d(`c',`$1',`$2')')')
 
lcs(`1234',`1224533324')
 
lcs(`thisisatest',`testing123testing')

Note: the caching (set2d/get2d) obscures the code even more than usual, but is necessary in order to get the second test to run in a reasonable amount of time.

Maple[edit]

 
> StringTools:-LongestCommonSubSequence( "thisisatest", "testing123testing" );
"tsitest"
 

Mathematica[edit]

A built-in function can do this for us:

a = "thisisatest";
b = "testing123testing";
LongestCommonSequence[a, b]

gives:

tsitest

Note that Mathematica also has a built-in function called LongestCommonSubsequence[a,b]:

finds the longest contiguous subsequence of elements common to the strings or lists a and b.

which would give "test" as the result for LongestCommonSubsequence[a, b].

The description for LongestCommonSequence[a,b] is:

finds the longest sequence of contiguous or disjoint elements common to the strings or lists a and b.

I added this note because the name of this article suggests LongestCommonSubsequence does the job, however LongestCommonSubsequence performs the puzzle-description.

Nim[edit]

Recursion[edit]

Translation of: Python
proc lcs(x, y): string =
if x == "" or y == "":
return ""
 
if x[0] == y[0]:
return x[0] & lcs(x[1..x.high], y[1..y.high])
 
let a = lcs(x, y[1..y.high])
let b = lcs(x[1..x.high], y)
result = if a.len > b.len: a else: b
 
echo lcs("1234", "1224533324")
echo lcs("thisisatest", "testing123testing")

Dynamic Programming[edit]

Translation of: Python
proc lcs(a, b): string =
var ls = newSeq[seq[int]] a.len+1
for i in 0 .. a.len:
ls[i].newSeq b.len+1
 
for i, x in a:
for j, y in b:
if x == y:
ls[i+1][j+1] = ls[i][j] + 1
else:
ls[i+1][j+1] = max(ls[i+1][j], ls[i][j+1])
 
result = ""
var x = a.len
var y = b.len
while x > 0 and y > 0:
if ls[x][y] == ls[x-1][y]:
dec x
elif ls[x][y] == ls[x][y-1]:
dec y
else:
assert a[x-1] == b[y-1]
result = a[x-1] & result
dec x
dec y
 
echo lcs("1234", "1224533324")
echo lcs("thisisatest", "testing123testing")

OCaml[edit]

Recursion[edit]

from Haskell

let longest xs ys = if List.length xs > List.length ys then xs else ys
 
let rec lcs a b = match a, b with
[], _
| _, [] -> []
| x::xs, y::ys ->
if x = y then
x :: lcs xs ys
else
longest (lcs a ys) (lcs xs b)

Memoized recursion[edit]

 
let lcs xs ys =
let cache = Hashtbl.create 16 in
let rec lcs xs ys =
try Hashtbl.find cache (xs, ys) with
| Not_found ->
let result =
match xs, ys with
| [], _ -> []
| _, [] -> []
| x :: xs, y :: ys when x = y ->
x :: lcs xs ys
| _ :: xs_rest, _ :: ys_rest ->
let a = lcs xs_rest ys in
let b = lcs xs ys_rest in
if (List.length a) > (List.length b) then a else b
in
Hashtbl.add cache (xs, ys) result;
result
in
lcs xs ys

Dynamic programming[edit]

let lcs xs' ys' =
let xs = Array.of_list xs'
and ys = Array.of_list ys' in
let n = Array.length xs
and m = Array.length ys in
let a = Array.make_matrix (n+1) (m+1) [] in
for i = n-1 downto 0 do
for j = m-1 downto 0 do
a.(i).(j) <- if xs.(i) = ys.(j) then
xs.(i) :: a.(i+1).(j+1)
else
longest a.(i).(j+1) a.(i+1).(j)
done
done;
a.(0).(0)

Because both solutions only work with lists, here are some functions to convert to and from strings:

let list_of_string str =
let result = ref [] in
String.iter (fun x -> result := x :: !result)
str;
List.rev !result
 
let string_of_list lst =
let result = String.create (List.length lst) in
ignore (List.fold_left (fun i x -> result.[i] <- x; i+1) 0 lst);
result

Both solutions work. Example:

# string_of_list (lcs (list_of_string "thisisatest")
                      (list_of_string "testing123testing"));;
- : string = "tsitest"

Oz[edit]

Translation of: Haskell

Recursive solution:

declare
fun {LCS Xs Ys}
case [Xs Ys]
of [nil _] then nil
[] [_ nil] then nil
[] [X|Xr Y|Yr] andthen X==Y then X|{LCS Xr Yr}
[] [_|Xr _|Yr] then {Longest {LCS Xs Yr} {LCS Xr Ys}}
end
end
 
fun {Longest Xs Ys}
if {Length Xs} > {Length Ys} then Xs else Ys end
end
in
{System.showInfo {LCS "thisisatest" "testing123testing"}}

Pascal[edit]

Translation of: Fortran
Program LongestCommonSubsequence(output);
 
function lcs(a, b: string): string;
var
x, y: string;
lenga, lengb: integer;
begin
lenga := length(a);
lengb := length(b);
lcs := '';
if (lenga > 0) and (lengb > 0) then
if a[lenga] = b[lengb] then
lcs := lcs(copy(a, 1, lenga-1), copy(b, 1, lengb-1)) + a[lenga]
else
begin
x := lcs(a, copy(b, 1, lengb-1));
y := lcs(copy(a, 1, lenga-1), b);
if length(x) > length(y) then
lcs := x
else
lcs := y;
end;
end;
 
var
s1, s2: string;
begin
s1 := 'thisisatest';
s2 := 'testing123testing';
writeln (lcs(s1, s2));
s1 := '1234';
s2 := '1224533324';
writeln (lcs(s1, s2));
end.
Output:
:> ./LongestCommonSequence
tsitest
1234

Perl[edit]

sub lcs {
my ($a, $b) = @_;
if (!length($a) || !length($b)) {
return "";
}
if (substr($a, 0, 1) eq substr($b, 0, 1)) {
return substr($a, 0, 1) . lcs(substr($a, 1), substr($b, 1));
}
my $c = lcs(substr($a, 1), $b) || "";
my $d = lcs($a, substr($b, 1)) || "";
return length($c) > length($d) ? $c : $d;
}
 
print lcs("thisisatest", "testing123testing") . "\n";

Perl 6[edit]

Recursion[edit]

Works with: rakudo version 2015-09-16

This solution is similar to the Haskell one. It is slow.

say lcs("thisisatest", "testing123testing");sub lcs(Str $xstr, Str $ystr) {
return "" unless $xstr && $ystr;
 
my ($x, $xs, $y, $ys) = $xstr.substr(0, 1), $xstr.substr(1), $ystr.substr(0, 1), $ystr.substr(1);
return $x eq $y
?? $x ~ lcs($xs, $ys)
!! max(:by{ $^a.chars }, lcs($xstr, $ys), lcs($xs, $ystr) );
}
 
say lcs("thisisatest", "testing123testing");

Dynamic Programming[edit]

Translation of: Java
 
sub lcs(Str $xstr, Str $ystr) {
my ($xlen, $ylen) = ($xstr, $ystr)>>.chars;
my @lengths = map {[(0) xx ($ylen+1)]}, 0..$xlen;
 
for $xstr.comb.kv -> $i, $x {
for $ystr.comb.kv -> $j, $y {
@lengths[$i+1][$j+1] = $x eq $y ?? @lengths[$i][$j]+1 !! (@lengths[$i+1][$j], @lengths[$i][$j+1]).max;
}
}
 
my @x = $xstr.comb;
my ($x, $y) = ($xlen, $ylen);
my $result = "";
while $x != 0 && $y != 0 {
if @lengths[$x][$y] == @lengths[$x-1][$y] {
$x--;
}
elsif @lengths[$x][$y] == @lengths[$x][$y-1] {
$y--;
}
else {
$result = @x[$x-1] ~ $result;
$x--;
$y--;
}
}
 
return $result;
}
 
say lcs("thisisatest", "testing123testing");

Bit Vector[edit]

Bit parallel dynamic programming with nearly linear complexity O(n). It is fast.

sub lcs(Str $xstr, Str $ystr) {
my ($a,$b) = ([$xstr.comb],[$ystr.comb]);
 
my $positions;
for $a.kv -> $i,$x { $positions{$x} +|= 1 +< $i };
 
my $S = +^0;
my $Vs = [];
my ($y,$u);
for (0..+$b-1) -> $j {
$y = $positions{$b[$j]} // 0;
$u = $S +& $y;
$S = ($S + $u) +| ($S - $u);
$Vs[$j] = $S;
}
 
my ($i,$j) = (+$a-1, +$b-1);
my $result = "";
while ($i >= 0 && $j >= 0) {
if ($Vs[$j] +& (1 +< $i)) { $i-- }
else {
unless ($j && +^$Vs[$j-1] +& (1 +< $i)) {
$result = $a[$i] ~ $result;
$i--;
}
$j--;
}
}
return $result;
}
 
say lcs("thisisatest", "testing123testing");

PicoLisp[edit]

(de commonSequences (A B)
(when A
(conc
(when (member (car A) B)
(mapcar '((L) (cons (car A) L))
(cons NIL (commonSequences (cdr A) (cdr @))) ) )
(commonSequences (cdr A) B) ) ) )
 
(maxi length
(commonSequences
(chop "thisisatest")
(chop "testing123testing") ) )
Output:
-> ("t" "s" "i" "t" "e" "s" "t")

PowerShell[edit]

Returns a sequence (array) of a type:

 
function Get-Lcs ($ReferenceObject, $DifferenceObject)
{
$longestCommonSubsequence = @()
$x = $ReferenceObject.Length
$y = $DifferenceObject.Length
 
$lengths = New-Object -TypeName 'System.Object[,]' -ArgumentList ($x + 1), ($y + 1)
 
for($i = 0; $i -lt $x; $i++)
{
for ($j = 0; $j -lt $y; $j++)
{
if ($ReferenceObject[$i] -ceq $DifferenceObject[$j])
{
$lengths[($i+1),($j+1)] = $lengths[$i,$j] + 1
}
else
{
$lengths[($i+1),($j+1)] = [Math]::Max(($lengths[($i+1),$j]),($lengths[$i,($j+1)]))
}
}
}
 
while (($x -ne 0) -and ($y -ne 0))
{
if ( $lengths[$x,$y] -eq $lengths[($x-1),$y])
{
--$x
}
elseif ($lengths[$x,$y] -eq $lengths[$x,($y-1)])
{
--$y
}
else
{
if ($ReferenceObject[($x-1)] -ceq $DifferenceObject[($y-1)])
{
$longestCommonSubsequence = ,($ReferenceObject[($x-1)]) + $longestCommonSubsequence
}
 
--$x
--$y
}
}
 
$longestCommonSubsequence
}
 

Returns the character array as a string:

 
(Get-Lcs -ReferenceObject "thisisatest" -DifferenceObject "testing123testing") -join ""
 
Output:
tsitest

Returns an array of integers:

 
Get-Lcs -ReferenceObject @(1,2,3,4) -DifferenceObject @(1,2,2,4,5,3,3,3,2,4)
 
Output:
1
2
3
4

Given two lists of objects, return the LCS of the ID property:

 
$list1
 
ID X Y
-- - -
1 101 201
2 102 202
3 103 203
4 104 204
5 105 205
6 106 206
7 107 207
8 108 208
9 109 209
 
$list2
 
ID X Y
-- - -
1 101 201
3 103 203
5 105 205
7 107 207
9 109 209
 
Get-Lcs -ReferenceObject $list1.ID -DifferenceObject $list2.ID
 
Output:
1
3
5
7
9

Prolog[edit]

Recursive Version[edit]

First version:

test :-
time(lcs("thisisatest", "testing123testing", Lcs)),
writef('%s',[Lcs]).
 
 
lcs([ H|L1],[ H|L2],[H|Lcs]) :- !,
lcs(L1,L2,Lcs).
 
lcs([H1|L1],[H2|L2],Lcs):-
lcs( L1 ,[H2|L2],Lcs1),
lcs([H1|L1], L2 ,Lcs2),
longest(Lcs1,Lcs2,Lcs),!.
 
lcs(_,_,[]).
 
 
longest(L1,L2,Longest) :-
length(L1,Length1),
length(L2,Length2),
((Length1 > Length2) -> Longest = L1; Longest = L2).

Second version, with memoization:

%declare that we will add lcs_db facts during runtime
:- dynamic lcs_db/3.
 
test :-
retractall(lcs_db(_,_,_)), %clear the database of known results
time(lcs("thisisatest", "testing123testing", Lcs)),
writef('%s',[Lcs]).
 
 
% check if the result is known
lcs(L1,L2,Lcs) :-
lcs_db(L1,L2,Lcs),!.
 
lcs([ H|L1],[ H|L2],[H|Lcs]) :- !,
lcs(L1,L2,Lcs).
 
lcs([H1|L1],[H2|L2],Lcs) :-
lcs( L1 ,[H2|L2],Lcs1),
lcs([H1|L1], L2 ,Lcs2),
longest(Lcs1,Lcs2,Lcs),!,
assert(lcs_db([H1|L1],[H2|L2],Lcs)).
 
lcs(_,_,[]).
 
 
longest(L1,L2,Longest) :-
length(L1,Length1),
length(L2,Length2),
((Length1 > Length2) -> Longest = L1; Longest = L2).
Demonstrating:

Example for "beginning-middle-ending" and "beginning-diddle-dum-ending"
First version :

?- time(lcs("beginning-middle-ending","beginning-diddle-dum-ending", Lcs)),writef('%s', [Lcs]).
% 10,875,184 inferences, 1.840 CPU in 1.996 seconds (92% CPU, 5910426 Lips)
beginning-iddle-ending

Second version which is much faster :

?- time(lcs("beginning-middle-ending","beginning-diddle-dum-ending", Lcs)),writef('%s', [Lcs]).
% 2,376 inferences, 0.010 CPU in 0.003 seconds (300% CPU, 237600 Lips)
beginning-iddle-ending

PureBasic[edit]

Translation of: Basic
Procedure.s lcs(a$, b$)
Protected x$ , lcs$
If Len(a$) = 0 Or Len(b$) = 0
lcs$ = ""
ElseIf Right(a$, 1) = Right(b$, 1)
lcs$ = lcs(Left(a$, Len(a$) - 1), Left(b$, Len(b$) - 1)) + Right(a$, 1)
Else
x$ = lcs(a$, Left(b$, Len(b$) - 1))
y$ = lcs(Left(a$, Len(a$) - 1), b$)
If Len(x$) > Len(y$)
lcs$ = x$
Else
lcs$ = y$
EndIf
EndIf
ProcedureReturn lcs$
EndProcedure
OpenConsole()
PrintN( lcs("thisisatest", "testing123testing"))
PrintN("Press any key to exit"): Repeat: Until Inkey() <> ""

Python[edit]

The simplest way is to use LCS within mlpy package

Recursion[edit]

This solution is similar to the Haskell one. It is slow.

def lcs(xstr, ystr):
"""
>>> lcs('thisisatest', 'testing123testing')
'tsitest'
"""

if not xstr or not ystr:
return ""
x, xs, y, ys = xstr[0], xstr[1:], ystr[0], ystr[1:]
if x == y:
return x + lcs(xs, ys)
else:
return max(lcs(xstr, ys), lcs(xs, ystr), key=len)

Test it:

if __name__=="__main__":
import doctest; doctest.testmod()

Dynamic Programming[edit]

Translation of: Java
def lcs(a, b):
lengths = [[0 for j in range(len(b)+1)] for i in range(len(a)+1)]
# row 0 and column 0 are initialized to 0 already
for i, x in enumerate(a):
for j, y in enumerate(b):
if x == y:
lengths[i+1][j+1] = lengths[i][j] + 1
else:
lengths[i+1][j+1] = max(lengths[i+1][j], lengths[i][j+1])
# read the substring out from the matrix
result = ""
x, y = len(a), len(b)
while x != 0 and y != 0:
if lengths[x][y] == lengths[x-1][y]:
x -= 1
elif lengths[x][y] == lengths[x][y-1]:
y -= 1
else:
assert a[x-1] == b[y-1]
result = a[x-1] + result
x -= 1
y -= 1
return result

Racket[edit]

#lang racket
(define (longest xs ys)
(if (> (length xs) (length ys))
xs ys))
 
(define memo (make-hash))
(define (lookup xs ys)
(hash-ref memo (cons xs ys) #f))
(define (store xs ys r)
(hash-set! memo (cons xs ys) r)
r)
 
(define (lcs/list sx sy)
(or (lookup sx sy)
(store sx sy
(match* (sx sy)
[((cons x xs) (cons y ys))
(if (equal? x y)
(cons x (lcs/list xs ys))
(longest (lcs/list sx ys) (lcs/list xs sy)))]
[(_ _) '()]))))
 
(define (lcs sx sy)
(list->string (lcs/list (string->list sx) (string->list sy))))
 
(lcs "thisisatest" "testing123testing")
Output:
"tsitest">

REXX[edit]

/*REXX program to test the  LCS (Longest Common Subsequence) subroutine.*/
parse arg aaa bbb . /*get two arguments (strings). */
say 'string A = 'aaa /*echo string A to screen. */
say 'string B = 'bbb /*echo string B to screen. */
say ' LCS = 'lcs(aaa,bbb) /*tell Longest Common Sequence. */
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────LCS subroutine──────────────────────*/
lcs: procedure; parse arg a,b,z /*Longest Common Subsequence. */
/*reduce recursions, removes the */
/*chars in A ¬ in B, & vice-versa*/
if z=='' then return lcs( lcs(a,b,0), lcs(b,a,0), 9)
j=length(a)
if z==0 then do /*special invocation: shrink Z. */
do j=1 for j; _=substr(a,j,1)
if pos(_,b)\==0 then z=z||_
end /*j*/
return substr(z,2)
end
k=length(b)
if j==0 | k==0 then return '' /*Either string null? Bupkis. */
_=substr(a,j,1)
if _==substr(b,k,1) then return lcs(substr(a,1,j-1),substr(b,1,k-1),9)_
x=lcs(a,substr(b,1,k-1),9)
y=lcs(substr(a,1,j-1),b,9)
if length(x)>length(y) then return x
return y
Output with input “ 1234 1224533324 ”:
string A=1234
string B=1224533324
     LCS=1234
Output with input “ thisisatest testing123testing ”:
string A=thisisatest
string B=testing123testing
     LCS=tsitest

Ring[edit]

 
see longest("1267834", "1224533324") + nl
 
func longest a, b
if a = "" or b = "" return "" ok
if right(a, 1) = right(b, 1)
lcs = longest(left(a, len(a) - 1), left(b, len(b) - 1)) + right(a, 1)
return lcs
else
x1 = longest(a, left(b, len(b) - 1))
x2 = longest(left(a, len(a) - 1), b)
if len(x1) > len(x2)
lcs = x1
return lcs
else
lcs = x2
return lcs ok ok
 

Output:

1234

Ruby[edit]

Recursion[edit]

This solution is similar to the Haskell one. It is slow (The time complexity is exponential.)

Works with: Ruby version 1.9
=begin
irb(main):001:0> lcs('thisisatest', 'testing123testing')
=> "tsitest"
=end

def lcs(xstr, ystr)
return "" if xstr.empty? || ystr.empty?
 
x, xs, y, ys = xstr[0..0], xstr[1..-1], ystr[0..0], ystr[1..-1]
if x == y
x + lcs(xs, ys)
else
[lcs(xstr, ys), lcs(xs, ystr)].max_by {|x| x.size}
end
end

Dynamic programming[edit]

Works with: Ruby version 1.9

Walker class for the LCS matrix:

class LCS
SELF, LEFT, UP, DIAG = [0,0], [0,-1], [-1,0], [-1,-1]
 
def initialize(a, b)
@m = Array.new(a.length) { Array.new(b.length) }
a.each_char.with_index do |x, i|
b.each_char.with_index do |y, j|
match(x, y, i, j)
end
end
end
 
def match(c, d, i, j)
@i, @j = i, j
@m[i][j] = compute_entry(c, d)
end
 
def lookup(x, y) [@i+x, @j+y] end
def valid?([email protected], [email protected]) i >= 0 && j >= 0 end
 
def peek(x, y)
i, j = lookup(x, y)
valid?(i, j) ? @m[i][j] : 0
end
 
def compute_entry(c, d)
c == d ? peek(*DIAG) + 1 : [peek(*LEFT), peek(*UP)].max
end
 
def backtrack
@i, @j = @m.length-1, @m[0].length-1
y = []
y << @i+1 if backstep? while valid?
y.reverse
end
 
def backtrack2
@i, @j = @m.length-1, @m[0].length-1
y = []
y << @j+1 if backstep? while valid?
[backtrack, y.reverse]
end
 
def backstep?
backstep = compute_backstep
@i, @j = lookup(*backstep)
backstep == DIAG
end
 
def compute_backstep
case peek(*SELF)
when peek(*LEFT) then LEFT
when peek(*UP) then UP
else DIAG
end
end
end
 
def lcs(a, b)
walker = LCS.new(a, b)
walker.backtrack.map{|i| a[i]}.join
end
 
if $0 == __FILE__
puts lcs('thisisatest', 'testing123testing')
puts lcs("rosettacode", "raisethysword")
end
Output:
tsitest
rsetod

Referring to LCS here and here.

Run BASIC[edit]

a$	= "aebdaef"
b$ = "cacbac"
print lcs$(a$,b$)
end
 
FUNCTION lcs$(a$, b$)
IF a$ = "" OR b$ = "" THEN
lcs$ = ""
goto [ext]
end if
 
IF RIGHT$(a$, 1) = RIGHT$(b$, 1) THEN
lcs$ = lcs$(LEFT$(a$, LEN(a$) - 1), LEFT$(b$, LEN(b$) - 1)) + RIGHT$(a$, 1)
goto [ext]
ELSE
x1$ = lcs$(a$, LEFT$(b$, LEN(b$) - 1))
x2$ = lcs$(LEFT$(a$, LEN(a$) - 1), b$)
IF LEN(x1$) > LEN(x2$) THEN
lcs$ = x1$
goto [ext]
ELSE
lcs$ = x2$
goto [ext]
END IF
END IF
[ext]
END FUNCTION
aba

Scala[edit]

This example is in need of improvement.
Translation of: Java
Works with: Scala 2.9.1
object LCS extends App {
 
// recursive version:
def lcsr(a: String, b: String): String = {
if (a.size==0 || b.size==0) ""
else if (a==b) a
else
if(a(a.size-1)==b(b.size-1)) lcsr(a.substring(0,a.size-1),b.substring(0,b.size-1))+a(a.size-1)
else {
val x = lcsr(a,b.substring(0,b.size-1))
val y = lcsr(a.substring(0,a.size-1),b)
if (x.size > y.size) x else y
}
}
 
// dynamic programming version:
def lcsd(a: String, b: String): String = {
if (a.size==0 || b.size==0) ""
else if (a==b) a
else {
val lengths = Array.ofDim[Int](a.size+1,b.size+1)
for (i <- 0 until a.size)
for (j <- 0 until b.size)
if (a(i) == b(j))
lengths(i+1)(j+1) = lengths(i)(j) + 1
else
lengths(i+1)(j+1) = scala.math.max(lengths(i+1)(j),lengths(i)(j+1))
 
// read the substring out from the matrix
val sb = new StringBuilder()
var x = a.size
var y = b.size
do {
if (lengths(x)(y) == lengths(x-1)(y))
x -= 1
else if (lengths(x)(y) == lengths(x)(y-1))
y -= 1
else {
assert(a(x-1) == b(y-1))
sb += a(x-1)
x -= 1
y -= 1
}
} while (x!=0 && y!=0)
sb.toString.reverse
}
}
 
val elapsed: (=> Unit) => Long = f => {val s = System.currentTimeMillis; f; (System.currentTimeMillis - s)/1000}
 
val pairs = List(("thisiaatest","testing123testing")
,("","x")
,("x","x")
,("beginning-middle-ending", "beginning-diddle-dum-ending"))
 
var s = ""
println("recursive version:")
pairs foreach {p =>
println{val t = elapsed(s = lcsr(p._1,p._2))
"lcsr(\""+p._1+"\",\""+p._2+"\") = \""+s+"\" ("+t+" sec)"}
}
 
println("\n"+"dynamic programming version:")
pairs foreach {p =>
println{val t = elapsed(s = lcsd(p._1,p._2))
"lcsd(\""+p._1+"\",\""+p._2+"\") = \""+s+"\" ("+t+" sec)"}
}
}
Output:
recursive version:
lcsr("thisiaatest","testing123testing") = "tsitest"   (0 sec)
lcsr("","x") = ""   (0 sec)
lcsr("x","x") = "x"   (0 sec)
lcsr("beginning-middle-ending","beginning-diddle-dum-ending") = "beginning-iddle-ending"   (29 sec)

dynamic programming version:
lcsd("thisiaatest","testing123testing") = "tsitest"   (0 sec)
lcsd("","x") = ""   (0 sec)
lcsd("x","x") = "x"   (0 sec)
lcsd("beginning-middle-ending","beginning-diddle-dum-ending") = "beginning-iddle-ending"   (0 sec)

Scheme[edit]

Port from Clojure.

 
;; using srfi-69
(define (memoize proc)
(let ((results (make-hash-table)))
(lambda args
(or (hash-table-ref results args (lambda () #f))
(let ((r (apply proc args)))
(hash-table-set! results args r)
r)))))
 
(define (longest xs ys)
(if (> (length xs)
(length ys))
xs ys))
 
(define lcs
(memoize
(lambda (seqx seqy)
(if (pair? seqx)
(let ((x (car seqx))
(xs (cdr seqx)))
(if (pair? seqy)
(let ((y (car seqy))
(ys (cdr seqy)))
(if (equal? x y)
(cons x (lcs xs ys))
(longest (lcs seqx ys)
(lcs xs seqy))))
'()))
'()))))
 

Testing:

 
 
(test-group
"lcs"
(test '() (lcs '(a b c) '(A B C)))
(test '(a) (lcs '(a a a) '(A A a)))
(test '() (lcs '() '(a b c)))
(test '() (lcs '(a b c) '()))
(test '(a c) (lcs '(a b c) '(a B c)))
(test '(b) (lcs '(a b c) '(A b C)))
 
(test '( b d e f g h j)
(lcs '(a b d e f g h i j)
'(A b c d e f F a g h j))))
 

Seed7[edit]

$ include "seed7_05.s7i";
 
const func string: lcs (in string: a, in string: b) is func
result
var string: lcs is "";
local
var string: x is "";
var string: y is "";
begin
if a <> "" and b <> "" then
if a[length(a)] = b[length(b)] then
lcs := lcs(a[.. pred(length(a))], b[.. pred(length(b))]) & str(a[length(a)]);
else
x := lcs(a, b[.. pred(length(b))]);
y := lcs(a[.. pred(length(a))], b);
if length(x) > length(y) then
lcs := x;
else
lcs := y;
end if;
end if;
end if;
end func;
 
const proc: main is func
begin
writeln(lcs("thisisatest", "testing123testing"));
writeln(lcs("1234", "1224533324"));
end func;

Output:

tsitest
1234

SETL[edit]

Recursive; Also works on tuples (vectors)

    op .longest(a, b);
return if #a > #b then a else b end;
end .longest;
 
procedure lcs(a, b);
if exists empty in {a, b} | #empty = 0 then
return empty;
elseif a(1) = b(1) then
return a(1) + lcs(a(2..), b(2..));
else
return lcs(a(2..), b) .longest lcs(a, b(2..));
end;
end lcs;

Sidef[edit]

func lcs(xstr, ystr) is cached {
 
xstr.is_empty && return xstr;
ystr.is_empty && return ystr;
 
var(x, xs, y, ys) = (xstr.ft(0,0), xstr.ft(1),
ystr.ft(0,0), ystr.ft(1));
 
if (x == y) {
x + lcs(xs, ys)
} else {
[lcs(xstr, ys), lcs(xs, ystr)].max_by { .len };
}
}
 
say lcs("thisisatest", "testing123testing");
Output:
tsitest

Slate[edit]

We define this on the Sequence type since there is nothing string-specific about the concept.

Recursion[edit]

Translation of: Java
s1@(Sequence traits) longestCommonSubsequenceWith: s2@(Sequence traits)
[
s1 isEmpty \/ s2 isEmpty ifTrue: [^ {}].
s1 last = s2 last
ifTrue: [(s1 allButLast longestCommonSubsequenceWith: s2 allButLast) copyWith: s1 last]
ifFalse: [| x y |
x: (s1 longestCommonSubsequenceWith: s2 allButLast).
y: (s1 allButLast longestCommonSubsequenceWith: s2).
x length > y length ifTrue: [x] ifFalse: [y]]
].

Dynamic Programming[edit]

Translation of: Ruby
s1@(Sequence traits) longestCommonSubsequenceWith: s2@(Sequence traits)
[| lengths |
lengths: (ArrayMD newWithDimensions: {s1 length `cache. s2 length `cache} defaultElement: 0).
s1 doWithIndex: [| :elem1 :index1 |
s2 doWithIndex: [| :elem2 :index2 |
elem1 = elem2
ifTrue: [lengths at: {index1 + 1. index2 + 1} put: (lengths at: {index1. index2}) + 1]
ifFalse: [lengths at: {index1 + 1. index2 + 1} put:
((lengths at: {index1 + 1. index2}) max: (lengths at: {index1. index2 + 1}))]]].
([| :result index1 index2 |
index1: s1 length.
index2: s2 length.
[index1 isPositive /\ index2 isPositive]
whileTrue:
[(lengths at: {index1. index2}) = (lengths at: {index1 - 1. index2})
ifTrue: [index1: index1 - 1]
ifFalse: [(lengths at: {index1. index2}) = (lengths at: {index1. index2 - 1})]
ifTrue: [index2: index2 - 1]
ifFalse: ["assert: (s1 at: index1 - 1) = (s2 at: index2 - 1)."
result nextPut: (s1 at: index1 - 1).
index1: index1 - 1.
index2: index2 - 1]]
] writingAs: s1) reverse
].

Swift[edit]

Translation of: Java

Swift 2.2

Recursion[edit]

func rlcs(s1:String, _ s2:String) -> String {
let x = s1.characters.count
let y = s2.characters.count
 
if x == 0 || y == 0 {
return ""
} else if s1[s1.startIndex.advancedBy(x - 1)] == s2[s2.startIndex.advancedBy(y - 1)] {
return rlcs(s1[s1.startIndex..<s1.startIndex.advancedBy(x - 1)],
s2[s2.startIndex..<s2.startIndex.advancedBy(y - 1)]) + String(s1[s1.startIndex.advancedBy(x - 1)])
} else {
let xstr = rlcs(s1, s2[s2.startIndex..<s2.startIndex.advancedBy(y - 1)])
let ystr = rlcs(s1[s1.startIndex..<s1.startIndex.advancedBy(x - 1)], s2)
 
return xstr.characters.count > ystr.characters.count ? xstr : ystr
}
}

Dynamic Programming[edit]

func lcs(s1:String, _ s2:String) -> String {
var x = s1.characters.count
var y = s2.characters.count
var lens = Array(count: x + 1, repeatedValue:
Array(count: y + 1, repeatedValue: 0))
var returnStr = ""
 
for i in 0..<x {
for j in 0..<y {
if s1[s1.startIndex.advancedBy(i)] == s2[s2.startIndex.advancedBy(j)] {
lens[i + 1][j + 1] = lens[i][j] + 1
} else {
lens[i + 1][j + 1] = max(lens[i + 1][j], lens[i][j + 1])
}
}
}
 
while x != 0 && y != 0 {
if lens[x][y] == lens[x - 1][y] {
--x
} else if lens[x][y] == lens[x][y - 1] {
--y
} else {
returnStr += String(s1[s1.startIndex.advancedBy(x - 1)])
--x
--y
}
}
 
return String(returnStr.characters.reverse())
}

Tcl[edit]

Recursive[edit]

Translation of: Java
proc r_lcs {a b} {
if {$a eq "" || $b eq ""} {return ""}
set a_ [string range $a 1 end]
set b_ [string range $b 1 end]
if {[set c [string index $a 0]] eq [string index $b 0]} {
return "$c[r_lcs $a_ $b_]"
} else {
set x [r_lcs $a $b_]
set y [r_lcs $a_ $b]
return [expr {[string length $x] > [string length $y] ? $x :$y}]
}
}

Dynamic[edit]

Translation of: Java
Works with: Tcl version 8.5
package require Tcl 8.5
namespace import ::tcl::mathop::+
namespace import ::tcl::mathop::-
namespace import ::tcl::mathfunc::max
 
proc d_lcs {a b} {
set la [string length $a]
set lb [string length $b]
set lengths [lrepeat [+ $la 1] [lrepeat [+ $lb 1] 0]]
 
for {set i 0} {$i < $la} {incr i} {
for {set j 0} {$j < $lb} {incr j} {
if {[string index $a $i] eq [string index $b $j]} {
lset lengths [+ $i 1] [+ $j 1] [+ [lindex $lengths $i $j] 1]
} else {
lset lengths [+ $i 1] [+ $j 1] [max [lindex $lengths [+ $i 1] $j] [lindex $lengths $i [+ $j 1]]]
}
}
}
 
set result ""
set x $la
set y $lb
while {$x >0 && $x > 0} {
if {[lindex $lengths $x $y] == [lindex $lengths [- $x 1] $y]} {
incr x -1
} elseif {[lindex $lengths $x $y] == [lindex $lengths $x [- $y 1]]} {
incr y -1
} else {
if {[set c [string index $a [- $x 1]]] ne [string index $b [- $y 1]]} {
error "assertion failed: a.charAt(x-1) == b.charAt(y-1)"
}
append result $c
incr x -1
incr y -1
}
}
return [string reverse $result]
}

Performance Comparison[edit]

% time {d_lcs thisisatest testing123testing} 10
637.5 microseconds per iteration
% time {r_lcs thisisatest testing123testing} 10
1275566.8 microseconds per iteration

Ursala[edit]

This uses the same recursive algorithm as in the Haskell example, and works on lists of any type.

#import std
 
lcs = ~&alrB^& ~&E?abh/~&alh2fabt2RC @faltPrXlrtPXXPW leql?/~&r ~&l

test program:

#cast %s
 
example = lcs('thisisatest','testing123testing')
Output:
'tsitest'

zkl[edit]

This is quite vile in terms of [time] efficiency, another algorithm should be used for real work.

Translation of: D
fcn lcs(a,b){
if(not a or not b) return("");
if (a[0]==b[0]) return(a[0] + self.fcn(a[1,*],b[1,*]));
return(fcn(x,y){if(x.len()>y.len())x else y}(lcs(a,b[1,*]),lcs(a[1,*],b)))
}

The last line looks strange but it is just return(lambda longest(lcs.lcs))

Output:
zkl: lcs("thisisatest", "testing123testing")
tsitest