Longest common subsequence
You are encouraged to solve this task according to the task description, using any language you may know.
Introduction
Define a subsequence to be any string obtained by deleting zero or more symbols from an input string.
The Longest Common Subsequence or LCS is a subsequence of maximum length common to two (or more) strings.
Let A = A[0]… A[m-1] and B = B[0]… B[n-1], m <= n be strings drawn from an alphabet Σ of size s, containing every distinct symbol in A + B.
An ordered pair (i, j) will be called a match if A[i] == B[j], where 0 <= i < m and 0 <= j < n.
Define the strict Cartesian product-order (<) over matches, such that (i1, j1) < (i2, j2) iff i1 < j1 and i2 < j2. Defining (>) similarly, we can write m2 < m1 as m1 > m2.
If i1 <= j1 and i2 >= j2 (or i1 >= i2 and i2 <= j2) then neither m1 < m2 nor m1 > m2 are possible; and m1, m2 are incomparable. Defining (#) to denote this case, we write m1 # m2. Note: This includes the possibility that m1 == m2, i.e., that i1 == i2 and j1 == j2.
We write m1 <> m2 to mean that either m1 < m2 or m1 > m2 holds. Thus, the (<>) operator is the inverse of (#). In this case, m1 != m2, i.e., each of tuples differ in some component.
Given a product-order over the set of matches M, a chain C is any subset of M where m1 <> m2 for every pair of distinct elements m1 and m2 of C.
Finding an LCS can then be restated as the problem of finding a chain of maximum cardinality p over the set of matches M.
The set of matches M can be visualized as an m*n bit matrix, where each bit M[i, j] is True iff there is a match at the corresponding positions of strings A and B.
Then any chain C can be visualized as a strictly increasing curve through those match bits which are set to True.
Background
Where the number of symbols appearing in matches is small relative to the length of the input strings, reuse of the symbols increases; and the number of matches will tend towards quadratic, O(m*n) growth. This occurs, for example, in the Bioinformatics application of nucleotide and protein sequencing.
The "divide and conquer" approach by 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 may be reduced to the length of the LCS and the number of matches may tend towards linear, O(m+n) growth.
In this case, a binary search optimization due to Hunt and Szymanski can be applied to the basic Dynamic Programming approach which results in expected performance of O(n log m). In the worst case, performance may degrade 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).
Note: Claus Rick has described a linear-space algorithm with a time bound of O(n*s + min(p*m, p*(n-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]
- "Simple and fast linear space computation of longest common subsequences" by Claus Rick, received 17 March 2000, Information Processing Letters, Elsevier Science [Volume 75, pp. 275–281]
Examples
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.
- Metrics
- Counting
- Word frequency
- Letter frequency
- Jewels and stones
- I before E except after C
- Bioinformatics/base count
- Count occurrences of a substring
- Count how many vowels and consonants occur in a string
- Remove/replace
- XXXX redacted
- Conjugate a Latin verb
- Remove vowels from a string
- String interpolation (included)
- Strip block comments
- Strip comments from a string
- Strip a set of characters from a string
- Strip whitespace from a string -- top and tail
- Strip control codes and extended characters from a string
- Anagrams/Derangements/shuffling
- Word wheel
- ABC problem
- Sattolo cycle
- Knuth shuffle
- Ordered words
- Superpermutation minimisation
- Textonyms (using a phone text pad)
- Anagrams
- Anagrams/Deranged anagrams
- Permutations/Derangements
- Find/Search/Determine
- ABC words
- Odd words
- Word ladder
- Semordnilap
- Word search
- Wordiff (game)
- String matching
- Tea cup rim text
- Alternade words
- Changeable words
- State name puzzle
- String comparison
- Unique characters
- Unique characters in each string
- Extract file extension
- Levenshtein distance
- Palindrome detection
- Common list elements
- Longest common suffix
- Longest common prefix
- Compare a list of strings
- Longest common substring
- Find common directory path
- Words from neighbour ones
- Change e letters to i in words
- Non-continuous subsequences
- Longest common subsequence
- Longest palindromic substrings
- Longest increasing subsequence
- Words containing "the" substring
- Sum of the digits of n is substring of n
- Determine if a string is numeric
- Determine if a string is collapsible
- Determine if a string is squeezable
- Determine if a string has all unique characters
- Determine if a string has all the same characters
- Longest substrings without repeating characters
- Find words which contains all the vowels
- Find words which contains most consonants
- Find words which contains more than 3 vowels
- Find words which first and last three letters are equals
- Find words which odd letters are consonants and even letters are vowels or vice_versa
- Formatting
- Substring
- Rep-string
- Word wrap
- String case
- Align columns
- Literals/String
- Repeat a string
- Brace expansion
- Brace expansion using ranges
- Reverse a string
- Phrase reversals
- Comma quibbling
- Special characters
- String concatenation
- Substring/Top and tail
- Commatizing numbers
- Reverse words in a string
- Suffixation of decimal numbers
- Long literals, with continuations
- Numerical and alphabetical suffixes
- Abbreviations, easy
- Abbreviations, simple
- Abbreviations, automatic
- Song lyrics/poems/Mad Libs/phrases
- Mad Libs
- Magic 8-ball
- 99 Bottles of Beer
- The Name Game (a song)
- The Old lady swallowed a fly
- The Twelve Days of Christmas
- Tokenize
- Text between
- Tokenize a string
- Word break problem
- Tokenize a string with escaping
- Split a character string based on change of character
- Sequences
11l
<lang 11l>F lcs(a, b)
V lengths = [[0] * (b.len+1)] * (a.len+1)
L(x) a
V i = L.index
L(y) b
V j = L.index
I x == y
lengths[i + 1][j + 1] = lengths[i][j] + 1
E
lengths[i + 1][j + 1] = max(lengths[i + 1][j], lengths[i][j + 1])
V result = ‘’
V j = b.len
L(i) 1..a.len
I lengths[i][j] != lengths[i - 1][j]
result ‘’= a[i - 1]
R result
print(lcs(‘1234’, ‘1224533324’)) print(lcs(‘thisisatest’, ‘testing123testing’))</lang>
- Output:
1234 tisitst
Ada
Using recursion: <lang ada>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;</lang>
- Output:
tsitest
Non-recursive solution: <lang ada>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;</lang>
- Output:
tsitest
ALGOL 68
<lang algol68>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))
)</lang>
- Output:
tsitest
APL
<lang 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
}</lang>
AutoHotkey
using dynamic programming
ahk forum: discussion <lang AutoHotkey>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
}</lang>
BASIC
<lang qbasic>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</lang>
BASIC256
<lang BASIC256>function LCS(a, b) if length(a) = 0 or length(b) = 0 then return "" if right(a, 1) = right(b, 1) then LCS = LCS(left(a, length(a) - 1), left(b, length(b) - 1)) + right(a, 1) else x = LCS(a, left(b, length(b) - 1)) y = LCS(left(a, length(a) - 1), b) if length(x) > length(y) then return x else return y end if end function
print LCS("1234", "1224533324") print LCS("thisisatest", "testing123testing") end</lang>
BBC BASIC
This makes heavy use of BBC BASIC's shortcut LEFT$(a$) and RIGHT$(a$) functions. <lang bbcbasic> 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$</lang>
Output:
1234 tsitest
Bracmat
<lang bracmat> ( LCS
= A a ta B b tb prefix
. !arg:(?prefix.@(?A:%?a ?ta).@(?B:%?b ?tb))
& ( !a:!b&LCS$(!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);</lang>
- Output:
max 7 lcs t s i t e s t
C
<lang c>#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;
} </lang> Testing <lang c>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;
}</lang>
C#
With recursion
<lang csharp>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;
}
}
}
}</lang>
C++
Hunt and Szymanski algorithm <lang cpp>#include <stdint.h>
- include <string>
- include <memory> // for shared_ptr<>
- include <iostream>
- include <deque>
- include <map>
- include <algorithm> // for lower_bound()
- include <iterator> // for next() and prev()
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;
// The following implements the Hunt and Szymanski algorithm:
uint32_t 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.rbegin(); it2 != dq2.rend(); it2++) {
// Each of the index1, index2 pairs considered here correspond to a match
auto index2 = *it2;
//
// Note: The reverse iterator it2 visits index2 values in descending order,
// allowing thresholds to be updated in-place. std::lower_bound() is used
// to perform a binary search.
//
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 = next(it2) != dq2.rend() &&
(limit == threshold.begin() || *prev(limit) < *next(it2));
if (skip) continue;
if (limit == threshold.end()) {
// insert case
threshold.push_back(index2);
// Refresh limit iterator:
limit = prev(threshold.end());
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) {
// Return the LCS as a linked list of matched index pairs:
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_back(index++);
for (const auto& it : s1) {
auto& dq2 = indexes[it];
matches.push_back(&dq2);
}
}
string Select(shared_ptr<Pair> pairs, uint32_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);
}
};</lang> Example: <lang cpp> LCS lcs;
auto s = lcs.Correspondence(s1, s2); cout << s << endl;</lang>
Clojure
Based on algorithm from Wikipedia. <lang Clojure>(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)))))))</lang>
CoffeeScript
<lang coffeescript> 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)</lang>
Common Lisp
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. <lang lisp>(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)))))</lang>
For example,
<lang lisp>(longest-common-subsequence "123456" "1a2b3c")</lang>
produces the two values
<lang lisp>"123" 3</lang>
An alternative adopted from Clojure
Here is another version with its own memoization macro:
<lang lisp>(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)))))))</lang>
When we test it, we get:
<lang lisp>(coerce (lcs (coerce "thisisatest" 'list) (coerce "testing123testing" 'list)) 'string))))
"tsitest"</lang>
D
Both versions don't work correctly with Unicode text.
Recursive version
<lang d>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;
}</lang>
- Output:
tsitest
Faster dynamic programming version
The output is the same. <lang d>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;
}</lang>
Hirschberg algorithm version
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
<lang d>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;
}</lang>
Dart
<lang dart>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')}");
} </lang>
- 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
<lang egison> (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)) </lang> Output: <lang egison> > (lcs "thisisatest" "testing123testing")) "tsitest" </lang>
Elixir
Simple recursion
This solution is Brute force. It is slow
<lang elixir>defmodule LCS do
def lcs(a, b) do lcs(to_charlist(a), to_charlist(b), []) |> to_string end defp lcs([h|at], [h|bt], res), do: lcs(at, bt, [h|res]) defp lcs([_|at]=a, [_|bt]=b, res) do Enum.max_by([lcs(a, bt, res), lcs(at, b, res)], &length/1) end defp lcs(_, _, res), do: res |> Enum.reverse
end
IO.puts LCS.lcs("thisisatest", "testing123testing") IO.puts LCS.lcs('1234','1224533324')</lang>
Dynamic Programming
<lang elixir>defmodule LCS do
def lcs_length(s,t), do: lcs_length(s,t,Map.new) |> elem(0)
defp lcs_length([],t,cache), do: {0,Map.put(cache,{[],t},0)}
defp lcs_length(s,[],cache), do: {0,Map.put(cache,{s,[]},0)}
defp lcs_length([h|st]=s,[h|tt]=t,cache) do
{l,c} = lcs_length(st,tt,cache)
{l+1,Map.put(c,{s,t},l+1)}
end
defp lcs_length([_sh|st]=s,[_th|tt]=t,cache) do
if Map.has_key?(cache,{s,t}) do
{Map.get(cache,{s,t}),cache}
else
{l1,c1} = lcs_length(s,tt,cache)
{l2,c2} = lcs_length(st,t,c1)
l = max(l1,l2)
{l,Map.put(c2,{s,t},l)}
end
end
def lcs(s,t) do
{s,t} = {to_charlist(s),to_charlist(t)}
{_,c} = lcs_length(s,t,Map.new)
lcs(s,t,c,[]) |> to_string
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 Map.get(cache,{s,tt}) > Map.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")</lang>
- Output:
tsitest 1234
Referring to LCS here.
Erlang
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. <lang erlang> 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.
</lang> Output: <lang erlang> 77> lcs:lcs("thisisatest","testing123testing"). "tsitest" 78> lcs:lcs("1234","1224533324"). "1234" </lang>
We can also use the process dictionary to memoize the recursive implementation:
<lang erlang> 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.
</lang>
Similar to the above, but without using the process dictionary: <lang erlang> -module(lcs).
%% API exports -export([
lcs/2
]).
%%==================================================================== %% API functions %%====================================================================
lcs(A, B) ->
{LCS, _Cache} = get_lcs(A, B, [], #{}),
lists:reverse(LCS).
%%==================================================================== %% Internal functions %%=====================================================
get_lcs(A, B, Acc, Cache) ->
case maps:find({A, B, Acc}, Cache) of
{ok, LCS} -> {LCS, Cache};
error ->
{NewLCS, NewCache} = compute_lcs(A, B, Acc, Cache),
{NewLCS, NewCache#{ {A, B, Acc} => NewLCS }}
end.
compute_lcs(A, B, Acc, Cache) when length(A) == 0 orelse length(B) == 0 ->
{Acc, Cache};
compute_lcs([Token |ATail], [Token |BTail], Acc, Cache) ->
get_lcs(ATail, BTail, [Token |Acc], Cache);
compute_lcs([_AToken |ATail]=A, [_BToken |BTail]=B, Acc, Cache) ->
{LCSA, CacheA} = get_lcs(A, BTail, Acc, Cache),
{LCSB, CacheB} = get_lcs(ATail, B, Acc, CacheA),
LCS = case length(LCSA) > length(LCSB) of
true -> LCSA;
false -> LCSB
end,
{LCS, CacheB}.
</lang>
Output: <lang erlang> 48> lcs:lcs("thisisatest", "testing123testing"). "tsitest" </lang>
F#
Copied and slightly adapted from OCaml (direct recursion) <lang fsharp>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
</lang>
Factor
<lang factor>USE: lcs "thisisatest" "testing123testing" lcs print</lang>
- Output:
tsitest
Fortran
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, ==, =)
<lang fortran>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</lang>
FreeBASIC
<lang freebasic>Function LCS(a As String, b As String) As String
Dim As String x, y
If Len(a) = 0 Or Len(b) = 0 Then
Return ""
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 Return x Else Return y
End If
End Function
Print LCS("1234", "1224533324") Print LCS("thisisatest", "testing123testing") Sleep</lang>
Go
Recursion
Brute force <lang go>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
}</lang>
Dynamic Programming
<lang go>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) }</lang>
Groovy
Recursive solution: <lang groovy>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"));</lang>
- Output:
1234 tsitest
Haskell
The Wikipedia solution translates directly into Haskell, with the only difference that equal characters are added in front:
<lang haskell>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))</lang>
A Memoized version of the naive algorithm.
<lang haskell>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</lang>
Memoization (aka dynamic programming) of that uses zip to make both the index and the character available:
<lang haskell>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))</lang>
All 3 solutions work of course not only with strings, but also with any other list. Example: <lang haskell>*Main> lcs "thisisatest" "testing123testing" "tsitest"</lang> The dynamic programming version without using arrays: <lang haskell>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]</lang>
Simple and slow solution:
<lang haskell>import Data.Ord import Data.List
-- longest common lcs xs ys = maximumBy (comparing length) $ intersect (subsequences xs) (subsequences ys)
main = print $ lcs "thisisatest" "testing123testing"</lang>
- Output:
"tsitest"
Icon and Unicon
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.
<lang Icon>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</lang>
- Output:
lcs( "thisisatest", "testing123testing" ) = "tsitest" lcs( "", "x" ) = "" lcs( "x", "x" ) = "x" lcs( "beginning-middle-ending", "beginning-diddle-dum-ending" ) = "beginning-iddle-ending"
J
<lang j>lcs=: dyad define
|.x{~ 0{"1 cullOne^:_ (\: +/"1)(\:{."1) 4$.$. x =/ y
)
cullOne=: ({~[: <@<@< [: (i. 0:)1,[: *./[: |: 2>/\]) :: ]</lang>
Here's another approach:
<lang J>mergeSq=: ;@}: ~.@, {.@;@{. ,&.> 3 {:: 4&{. common=: 2 2 <@mergeSq@,;.3^:_ [: (<@#&.> i.@$) =/ lcs=: [ {~ 0 {"1 ,&$ #: 0 ({:: (#~ [: (= >./) #@>)) 0 ({:: ,) common</lang>
Example use (works with either definition of lcs):
<lang J> 'thisisatest' lcs 'testing123testing' tsitest</lang>
Dynamic programming version <lang j>longest=: ]`[@.(>&#) upd=:{:@[,~ ({.@[ ,&.> {:@])`({:@[ longest&.> {.@])@.(0 = #&>@{.@[) lcs=: 0{:: [: ([: {.&> [: upd&.>/\.<"1@:,.)/ a:,.~a:,~=/{"1 a:,.<"0@[</lang> Output: <lang j> '1234' lcs '1224533324' 1234
'thisisatest' lcs 'testing123testing'
tsitest</lang>
Recursion <lang j>lcs=:;(($:}.) longest }.@[ $: ])`({.@[,$:&}.)@.(=&{.)`((i.0)"_)@.(+.&(0=#))&((e.#[)&>/) ;~</lang>
Java
Recursion
This is not a particularly fast algorithm, but it gets the job done eventually. The speed is a result of many recursive function calls. <lang java>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;
}
}</lang>
Dynamic Programming
<lang java>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();
}</lang>
JavaScript
Recursion
This is more or less a translation of the recursive Java version above. <lang javascript>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;
}
}</lang>
ES6 recursive implementation
<lang javascript> const longest = (xs, ys) => (xs.length > ys.length) ? xs : ys;
const lcs = (xx, yy) => {
if (!xx.length || !yy.length) { return ; }
const [x, ...xs] = xx;
const [y, ...ys] = yy;
return (x === y) ? (x + lcs(xs, ys)) : longest(lcs(xx, ys), lcs(xs, yy));
};</lang>
Dynamic Programming
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. <lang javascript>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(); }</lang>
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: <lang javascript> 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));}</lang>
Greedy Algorithm
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; <lang javascript>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;
}
}</lang>
Note that it won't return the correct answer for all inputs. For example: <lang javascript>lcs_greedy('bcaaaade', 'deaaaabc'); // 'bc' instead of 'aaaa'</lang>
jq
Naive recursive version: <lang jq>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;</lang>
Example: <lang jq>lcs("1234"; "1224533324"), lcs("thisisatest"; "testing123testing")</lang> Output:<lang sh>
- jq -n -f lcs-recursive.jq
"1234" "tsitest"</lang>
Julia
<lang julia>longest(a::String, b::String) = length(a) ≥ length(b) ? a : b
""" julia> lcsrecursive("thisisatest", "testing123testing") "tsitest" """
- Recursive
function lcsrecursive(xstr::String, ystr::String)
if length(xstr) == 0 || length(ystr) == 0
return ""
end
x, xs, y, ys = xstr[1], xstr[2:end], ystr[1], ystr[2:end]
if x == y
return string(x, lcsrecursive(xs, ys))
else
return longest(lcsrecursive(xstr, ys), lcsrecursive(xs, ystr))
end
end
- Dynamic
function lcsdynamic(a::String, b::String)
lengths = zeros(Int, length(a) + 1, length(b) + 1)
# row 0 and column 0 are initialized to 0 already
for (i, x) in enumerate(a), (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])
end
end
# read the substring out from the matrix
result = ""
x, y = length(a) + 1, length(b) + 1
while x > 1 && y > 1
if lengths[x, y] == lengths[x-1, y]
x -= 1
elseif lengths[x, y] == lengths[x, y-1]
y -= 1
else
@assert a[x-1] == b[y-1]
result = string(a[x-1], result)
x -= 1
y -= 1
end
end
return result
end
@show lcsrecursive("thisisatest", "testing123testing")
@time lcsrecursive("thisisatest", "testing123testing")
@show lcsdynamic("thisisatest", "testing123testing")
@time lcsdynamic("thisisatest", "testing123testing")</lang>
- Output:
lcsrecursive("thisisatest", "testing123testing") = "tsitest"
0.038153 seconds (537.77 k allocations: 16.415 MiB)
lcsdynamic("thisisatest", "testing123testing") = "tsitest"
0.000004 seconds (12 allocations: 2.141 KiB)
Kotlin
<lang scala>// version 1.1.2
fun lcs(x: String, y: String): String {
if (x.length == 0 || y.length == 0) return "" val x1 = x.dropLast(1) val y1 = y.dropLast(1) if (x.last() == y.last()) return lcs(x1, y1) + x.last() val x2 = lcs(x, y1) val y2 = lcs(x1, y) return if (x2.length > y2.length) x2 else y2
}
fun main(args: Array<String>) {
val x = "thisisatest" val y = "testing123testing" println(lcs(x, y))
}</lang>
- Output:
tsitest
Liberty BASIC
<lang lb> '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 </lang>
Logo
This implementation works on both words and lists. <lang logo>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</lang>
Lua
<lang lua>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" ) )</lang>
M4
<lang M4>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')</lang> 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
<lang Maple> > StringTools:-LongestCommonSubSequence( "thisisatest", "testing123testing" );
"tsitest"
</lang>
Mathematica/Wolfram Language
A built-in function can do this for us: <lang Mathematica>a = "thisisatest"; b = "testing123testing"; LongestCommonSequence[a, b]</lang> gives: <lang Mathematica>tsitest</lang> 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 LongestCommonSequence performs the puzzle-description.
Nim
Recursion
<lang nim>proc lcs(x, y: string): 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")</lang>
This recursive version is not efficient but the execution time can be greatly improved by using memoization.
Dynamic Programming
<lang nim>proc lcs(a, b: string): 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")</lang>
OCaml
Recursion
from Haskell <lang ocaml>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)</lang>
Memoized recursion
<lang ocaml> 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</lang>
Dynamic programming
<lang ocaml>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)</lang>
Because both solutions only work with lists, here are some functions to convert to and from strings: <lang ocaml>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</lang>
Both solutions work. Example:
# string_of_list (lcs (list_of_string "thisisatest")
(list_of_string "testing123testing"));;
- : string = "tsitest"
Oz
Recursive solution: <lang oz>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"}}</lang>
Pascal
<lang pascal>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.</lang>
- Output:
:> ./LongestCommonSequence tsitest 1234
Perl
<lang perl>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 : $d;
}
print lcs("thisisatest", "testing123testing") . "\n";</lang>
Alternate letting regex do all the work
<lang perl>#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Longest_common_subsequence use warnings;
print "lcs is ", lcs('thisisatest', 'testing123testing'), "\n";
sub lcs
{
my ($c, $d) = @_;
for my $len ( reverse 1 .. length($c & $d) )
{
"$c\n$d" =~ join '.*', ('(.)') x $len, "\n", map "\\$_", 1 .. $len and
return join , @{^CAPTURE};
}
return ;
}</lang>
- Output:
lcs is tsitest
Phix
If you want this to work with (utf8) Unicode text, just chuck the inputs through utf8_to_utf32() first (and the output through utf32_to_utf8()). <lang Phix>function lcs(sequence a, b) sequence res = ""
if length(a) and length(b) then
if a[$]=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
constant tests = {{"1234","1224533324"},
{"thisisatest","testing123testing"}}
for i=1 to length(tests) do
string {a,b} = tests[i]
?lcs(a,b)
end for</lang>
- Output:
"1234" "tsitest"
Alternate version
same output <lang Phix>function LCSLength(sequence X, sequence Y) sequence C = repeat(repeat(0,length(Y)+1),length(X)+1)
for i=1 to length(X) do
for j=1 to length(Y) do
if X[i]=Y[j] then
C[i+1][j+1] := C[i][j]+1
else
C[i+1][j+1] := max(C[i+1][j], C[i][j+1])
end if
end for
end for
return C
end function
function backtrack(sequence C, sequence X, sequence Y, integer i, integer j)
if i=0 or j=0 then
return ""
elsif X[i]=Y[j] then
return backtrack(C, X, Y, i-1, j-1) & X[i]
else
if C[i+1][j]>C[i][j+1] then
return backtrack(C, X, Y, i, j-1)
else
return backtrack(C, X, Y, i-1, j)
end if
end if
end function
function lcs(sequence a, sequence b)
return backtrack(LCSLength(a,b),a,b,length(a),length(b))
end function
constant tests = {{"1234","1224533324"},
{"thisisatest","testing123testing"}}
for i=1 to length(tests) do
string {a,b} = tests[i]
?lcs(a,b)
end for</lang>
PicoLisp
<lang PicoLisp>(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") ) )</lang>
- Output:
-> ("t" "s" "i" "t" "e" "s" "t")
PowerShell
Returns a sequence (array) of a type: <lang PowerShell> 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
} </lang> Returns the character array as a string: <lang PowerShell> (Get-Lcs -ReferenceObject "thisisatest" -DifferenceObject "testing123testing") -join "" </lang>
- Output:
tsitest
Returns an array of integers: <lang PowerShell> Get-Lcs -ReferenceObject @(1,2,3,4) -DifferenceObject @(1,2,2,4,5,3,3,3,2,4) </lang>
- Output:
1 2 3 4
Given two lists of objects, return the LCS of the ID property: <lang PowerShell> $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 $list2.ID </lang>
- Output:
1 3 5 7 9
Prolog
Recursive Version
First version: <lang Prolog>test :-
time(lcs("thisisatest", "testing123testing", Lcs)),
writef('%s',[Lcs]).
lcs([ H|L1],[ H|L2],[H|Lcs]) :- !,
lcs(L1,L2,Lcs).
lcs([H1|L1],[H2|L2],Lcs):-
lcs( L1 ,[H2|L2],Lcs1), lcs([H1|L1], L2 ,Lcs2), longest(Lcs1,Lcs2,Lcs),!.
lcs(_,_,[]).
longest(L1,L2,Longest) :-
length(L1,Length1), length(L2,Length2), ((Length1 > Length2) -> Longest = L1; Longest = L2).</lang>
Second version, with memoization: <lang Prolog>%declare that we will add lcs_db facts during runtime
- - dynamic lcs_db/3.
test :-
retractall(lcs_db(_,_,_)), %clear the database of known results
time(lcs("thisisatest", "testing123testing", Lcs)),
writef('%s',[Lcs]).
% check if the result is known
lcs(L1,L2,Lcs) :-
lcs_db(L1,L2,Lcs),!.
lcs([ H|L1],[ H|L2],[H|Lcs]) :- !,
lcs(L1,L2,Lcs).
lcs([H1|L1],[H2|L2],Lcs) :-
lcs( L1 ,[H2|L2],Lcs1), lcs([H1|L1], L2 ,Lcs2), longest(Lcs1,Lcs2,Lcs),!, assert(lcs_db([H1|L1],[H2|L2],Lcs)).
lcs(_,_,[]).
longest(L1,L2,Longest) :-
length(L1,Length1), length(L2,Length2), ((Length1 > Length2) -> Longest = L1; Longest = L2).</lang>
- Demonstrating:
Example for "beginning-middle-ending" and "beginning-diddle-dum-ending"
First version :
<lang Prolog>?- time(lcs("beginning-middle-ending","beginning-diddle-dum-ending", Lcs)),writef('%s', [Lcs]).
% 10,875,184 inferences, 1.840 CPU in 1.996 seconds (92% CPU, 5910426 Lips)
beginning-iddle-ending</lang>
Second version which is much faster :
<lang Prolog>?- time(lcs("beginning-middle-ending","beginning-diddle-dum-ending", Lcs)),writef('%s', [Lcs]).
% 2,376 inferences, 0.010 CPU in 0.003 seconds (300% CPU, 237600 Lips)
beginning-iddle-ending</lang>
PureBasic
<lang PureBasic>Procedure.s lcs(a$, 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() <> ""</lang>
Python
The simplest way is to use LCS within mlpy package
Recursion
This solution is similar to the Haskell one. It is slow. <lang python>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 str(lcs(xs, ys)) + x
else:
return max(lcs(xstr, ys), lcs(xs, ystr), key=len)</lang>
Test it: <lang python>if __name__=="__main__":
import doctest; doctest.testmod()</lang>
Dynamic Programming
<lang python>def lcs(a, b):
# generate matrix of length of longest common subsequence for substrings of both words
lengths = [[0] * (len(b)+1) for _ in range(len(a)+1)]
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 a substring from the matrix
result =
j = len(b)
for i in range(1, len(a)+1):
if lengths[i][j] != lengths[i-1][j]:
result += a[i-1]
return result</lang>
Racket
<lang racket>#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")</lang>
- Output:
"tsitest">
Raku
(formerly Perl 6)
Recursion
This solution is similar to the Haskell one. It is slow. <lang perl6>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");</lang>
Dynamic Programming
<lang perl6> 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");</lang>
Bit Vector
Bit parallel dynamic programming with nearly linear complexity O(n). It is fast. <lang perl6>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");</lang>
ReasonML
<lang ocaml> let longest = (xs, ys) =>
if (List.length(xs) > List.length(ys)) {
xs;
} else {
ys;
};
let rec lcs = (a, b) =>
switch (a, b) {
| ([], _)
| (_, []) => []
| ([x, ...xs], [y, ...ys]) =>
if (x == y) {
[x, ...lcs(xs, ys)];
} else {
longest(lcs(a, ys), lcs(xs, b));
}
};
</lang>
REXX
<lang rexx>/*REXX program tests the LCS (Longest Common Subsequence) subroutine. */ parse arg aaa bbb . /*obtain optional arguments from the CL*/ say 'string A =' aaa /*echo the string A to the screen. */ say 'string B =' bbb /* " " " B " " " */ say ' LCS =' LCS(aaa, bbb) /*tell the Longest Common Sequence. */ exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ LCS: procedure; parse arg a,b,z /*Longest Common Subsequence. */
/*reduce recursions, removes the */
/*chars in A ¬ in B, and vice─versa.*/
if z== then return LCS( LCS(a,b,0), LCS(b,a,0), 9) /*Is Z null? Do recurse. */
j= length(a)
if z==0 then do /*a 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 /*Is 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</lang>
- output when using the input of: 1234 1224533324
string A = 1234
string B = 1224533324
LCS = 1234
- output when using the input of: thisisatest testing123testing
string A = thisisatest
string B = testing123testing
LCS = tsitest
Ring
<lang ring> 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
</lang> Output:
1234
Ruby
Recursion
This solution is similar to the Haskell one. It is slow (The time complexity is exponential.)
<lang ruby>=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</lang>
Dynamic programming
Walker class for the LCS matrix:
<lang ruby>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?(i=@i, j=@j) 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 $0 == __FILE__
puts lcs('thisisatest', 'testing123testing')
puts lcs("rosettacode", "raisethysword")
end</lang>
- Output:
tsitest rsetod
Referring to LCS here and here.
Run BASIC
<lang runbasic>a$ = "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</lang>
aba
Rust
Dynamic programming version: <lang rust> use std::cmp;
fn lcs(string1: String, string2: String) -> (usize, String){
let total_rows = string1.len() + 1; let total_columns = string2.len() + 1; // rust doesn't allow accessing string by index let string1_chars = string1.as_bytes(); let string2_chars = string2.as_bytes();
let mut table = vec![vec![0; total_columns]; total_rows];
for row in 1..total_rows{
for col in 1..total_columns {
if string1_chars[row - 1] == string2_chars[col - 1]{
table[row][col] = table[row - 1][col - 1] + 1;
} else {
table[row][col] = cmp::max(table[row][col-1], table[row-1][col]);
}
}
}
let mut common_seq = Vec::new(); let mut x = total_rows - 1; let mut y = total_columns - 1;
while x != 0 && y != 0 {
// Check element above is equal
if table[x][y] == table[x - 1][y] {
x = x - 1;
}
// check element to the left is equal
else if table[x][y] == table[x][y - 1] {
y = y - 1;
}
else {
// check the two element at the respective x,y position is same
assert_eq!(string1_chars[x-1], string2_chars[y-1]);
let char = string1_chars[x - 1];
common_seq.push(char);
x = x - 1;
y = y - 1;
}
}
common_seq.reverse();
(table[total_rows - 1][total_columns - 1], String::from_utf8(common_seq).unwrap())
}
fn main() {
let res = lcs("abcdaf".to_string(), "acbcf".to_string());
assert_eq!((4 as usize, "abcf".to_string()), res);
let res = lcs("thisisatest".to_string(), "testing123testing".to_string());
assert_eq!((7 as usize, "tsitest".to_string()), res);
// LCS for input Sequences “AGGTAB” and “GXTXAYB” is “GTAB” of length 4.
let res = lcs("AGGTAB".to_string(), "GXTXAYB".to_string());
assert_eq!((4 as usize, "GTAB".to_string()), res);
}</lang>
Scala
Using lazily evaluated lists: <lang scala> def lcsLazy[T](u: IndexedSeq[T], v: IndexedSeq[T]): IndexedSeq[T] = {
def su = subsets(u).to(LazyList)
def sv = subsets(v).to(LazyList)
su.intersect(sv).headOption match{
case Some(sub) => sub
case None => IndexedSeq[T]()
}
}
def subsets[T](u: IndexedSeq[T]): Iterator[IndexedSeq[T]] = {
u.indices.reverseIterator.flatMap{n => u.indices.combinations(n + 1).map(_.map(u))}
}</lang>
Using recursion: <lang scala> def lcsRec[T]: (IndexedSeq[T], IndexedSeq[T]) => IndexedSeq[T] = {
case (a +: as, b +: bs) if a == b => a +: lcsRec(as, bs)
case (as, bs) if as.isEmpty || bs.isEmpty => IndexedSeq[T]()
case (a +: as, b +: bs) =>
val (s1, s2) = (lcsRec(a +: as, bs), lcsRec(as, b +: bs))
if(s1.length > s2.length) s1 else s2
}</lang>
- Output:
scala> lcsLazy("thisisatest", "testing123testing").mkString
res0: String = tsitest
scala> lcsRec("thisisatest", "testing123testing").mkString
res1: String = tsitest
Recursive version: <lang scala> def lcs[T]: (List[T], List[T]) => List[T] = {
case (_, Nil) => Nil
case (Nil, _) => Nil
case (x :: xs, y :: ys) if x == y => x :: lcs(xs, ys)
case (x :: xs, y :: ys) => {
(lcs(x :: xs, ys), lcs(xs, y :: ys)) match {
case (xs, ys) if xs.length > ys.length => xs
case (xs, ys) => ys
}
}
}</lang>
The dynamic programming version:
<lang scala> case class Memoized[A1, A2, B](f: (A1, A2) => B) extends ((A1, A2) => B) {
val cache = scala.collection.mutable.Map.empty[(A1, A2), B] def apply(x: A1, y: A2) = cache.getOrElseUpdate((x, y), f(x, y)) }
lazy val lcsM: Memoized[List[Char], List[Char], List[Char]] = Memoized {
case (_, Nil) => Nil
case (Nil, _) => Nil
case (x :: xs, y :: ys) if x == y => x :: lcsM(xs, ys)
case (x :: xs, y :: ys) => {
(lcsM(x :: xs, ys), lcsM(xs, y :: ys)) match {
case (xs, ys) if xs.length > ys.length => xs
case (xs, ys) => ys
}
}
}</lang>
- Output:
scala> lcsM("thisiaatest".toList, "testing123testing".toList).mkString
res0: String = tsitest
Scheme
Port from Clojure.
<lang scheme>
- 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))))
'()))
'()))))
</lang>
Testing: <lang scheme>
(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))))
</lang>
Seed7
<lang seed7>$ 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;</lang>
Output:
tsitest 1234
SequenceL
It is interesting to note that x and y are computed in parallel, dividing work across threads repeatedly down through the recursion.
<lang sequencel>import <Utilities/Sequence.sl>;
lcsBack(a(1), b(1)) :=
let
aSub := allButLast(a);
bSub := allButLast(b);
x := lcsBack(a, bSub);
y := lcsBack(aSub, b);
in
[] when size(a) = 0 or size(b) = 0
else
lcsBack(aSub, bSub) ++ [last(a)] when last(a) = last(b)
else
x when size(x) > size(y)
else
y;
main(args(2)) :=
lcsBack(args[1], args[2]) when size(args) >=2
else
lcsBack("thisisatest", "testing123testing");</lang>
- Output:
"tsitest"
SETL
Recursive; Also works on tuples (vectors) <lang setl> 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;</lang>
Sidef
<lang ruby>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");</lang>
- Output:
tsitest
Slate
We define this on the Sequence type since there is nothing string-specific about the concept.
Recursion
<lang slate>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]]
].</lang>
Dynamic Programming
<lang slate>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
].</lang>
Swift
Swift 5.1
Recursion
<lang Swift>rlcs(_ s1: String, _ s2: String) -> String {
if s1.count == 0 || s2.count == 0 {
return ""
} else if s1[s1.index(s1.endIndex, offsetBy: -1)] == s2[s2.index(s2.endIndex, offsetBy: -1)] {
return rlcs(String(s1[s1.startIndex..<s1.index(s1.endIndex, offsetBy: -1)]),
String(s2[s2.startIndex..<s2.index(s2.endIndex, offsetBy: -1)])) + String(s1[s1.index(s1.endIndex, offsetBy: -1)])
} else {
let str1 = rlcs(s1, String(s2[s2.startIndex..<s2.index(s2.endIndex, offsetBy: -1)]))
let str2 = rlcs(String(s1[s1.startIndex..<s1.index(s1.endIndex, offsetBy: -1)]), s2)
return str1.count > str2.count ? str1 : str2 }
}</lang>
Dynamic Programming
<lang Swift>func lcs(_ s1: String, _ s2: String) -> String {
var lens = Array(
repeating:Array(repeating: 0, count: s2.count + 1),
count: s1.count + 1
)
for i in 0..<s1.count {
for j in 0..<s2.count {
if s1[s1.index(s1.startIndex, offsetBy: i)] == s2[s2.index(s2.startIndex, offsetBy: 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])
}
}
}
var returnStr = ""
var x = s1.count
var y = s2.count
while x != 0 && y != 0 {
if lens[x][y] == lens[x - 1][y] {
x -= 1
} else if lens[x][y] == lens[x][y - 1] {
y -= 1
} else {
returnStr += String(s1[s1.index(s1.startIndex, offsetBy: x - 1)])
x -= 1
y -= 1
}
}
return String(returnStr.reversed())
}</lang>
Tcl
Recursive
<lang tcl>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}]
}
}</lang>
Dynamic
<lang tcl>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 && $y > 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]
}</lang>
Performance Comparison
<lang tcl>% time {d_lcs thisisatest testing123testing} 10 637.5 microseconds per iteration % time {r_lcs thisisatest testing123testing} 10 1275566.8 microseconds per iteration</lang>
Ursala
This uses the same recursive algorithm as in the Haskell example, and works on lists of any type. <lang Ursala>#import std
lcs = ~&alrB^& ~&E?abh/~&alh2fabt2RC @faltPrXlrtPXXPW leql?/~&r ~&l</lang> test program: <lang Ursala>#cast %s
example = lcs('thisisatest','testing123testing')</lang>
- Output:
'tsitest'
Wren
<lang ecmascript>var lcs // recursive lcs = Fn.new { |x, y|
if (x.count == 0 || y.count == 0) return "" var x1 = x[0...-1] var y1 = y[0...-1] if (x[-1] == y[-1]) return lcs.call(x1, y1) + x[-1] var x2 = lcs.call(x, y1) var y2 = lcs.call(x1, y) return (x2.count > y2.count) ? x2 : y2
}
var x = "thisisatest" var y = "testing123testing" System.print(lcs.call(x, y))</lang>
- Output:
tsitest
zkl
This is quite vile in terms of [time] efficiency, another algorithm should be used for real work.
<lang zkl>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)))
}</lang> The last line looks strange but it is just return(lambda longest(lcs.lcs))
- Output:
zkl: lcs("thisisatest", "testing123testing")
tsitest
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