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 output string obtained by deleting zero or more symbols from an input string.
The Longest Common Subsequence (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 referred to as a match if A[i] = B[j], where 0 ≤ i < m and 0 ≤ j < n.
The set of matches M defines a relation over matches: M[i, j] ⇔ (i, j) ∈ M.
Define a non-strict product-order (≤) over ordered pairs, such that (i1, j1) ≤ (i2, j2) ⇔ i1 ≤ i2 and j1 ≤ j2. We define (≥) similarly.
We say ordered pairs p1 and p2 are comparable if either p1 ≤ p2 or p1 ≥ p2 holds. If i1 < i2 and j2 < j1 (or i2 < i1 and j1 < j2) then neither p1 ≤ p2 nor p1 ≥ p2 are possible, and we say p1 and p2 are incomparable.
Define the strict product-order (<) over ordered pairs, such that (i1, j1) < (i2, j2) ⇔ i1 < i2 and j1 < j2. We define (>) similarly.
A chain C is a subset of M consisting of at least one element m; and where either m1 < m2 or m1 > m2 for every pair of distinct elements m1 and m2. An antichain D is any subset of M in which every pair of distinct elements m1 and m2 are incomparable.
A chain can be visualized as a strictly increasing curve that passes through matches (i, j) in the m*n coordinate space of M[i, j].
Every Common Sequence of length q corresponds to a chain of cardinality q, over the set of matches M. Thus, finding an LCS can be restated as the problem of finding a chain of maximum cardinality p.
According to [Dilworth 1950], this cardinality p equals the minimum number of disjoint antichains into which M can be decomposed. Note that such a decomposition into the minimal number p of disjoint antichains may not be unique.
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 O(m*n) quadratic growth. This occurs, for example, in the Bioinformatics application of nucleotide and protein sequencing.
The divide-and-conquer approach of [Hirschberg 1975] limits the space required to O(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 approaches the length of the LCS. In this case the number of matches reduces to linear, O(n) growth.
A binary search optimization due to [Hunt and Szymanski 1977] can be applied to the basic Dynamic Programming approach, resulting in an expected performance of O(n log m). Performance can degrade to O(m*n log m) time in the worst case, as the number of matches grows to O(m*n).
Note
[Rick 2000] describes a linear-space algorithm with a time bound of O(n*s + p*min(m, n - p)).
Legend
A, B are input strings of lengths m, n respectively p is the length of the LCS M is the set of matches (i, j) such that A[i] = B[j] r is the magnitude of M s is the magnitude of the alphabet Σ of distinct symbols in A + B
References
[Dilworth 1950] "A decomposition theorem for partially ordered sets" by Robert P. Dilworth, published January 1950, Annals of Mathematics [Volume 51, Number 1, pp. 161-166]
[Goeman and Clausen 2002] "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]
[Hirschberg 1975] "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]
[Hunt and McIlroy 1976] "An Algorithm for Differential File Comparison" by James W. Hunt and M. Douglas McIlroy, June 1976 Computing Science Technical Report, Bell Laboratories 41
[Hunt and Szymanski 1977] "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]
[Rick 2000] "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.
Other tasks related to string operations:
- 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 contain the most consonants
- Find words which contains more than 3 vowels
- Find words whose first and last three letters are equal
- Find words with alternating vowels and consonants
- 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
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’))- Output:
1234 tisitst
Ada
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
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
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
}
Arturo
lcs: function [a,b][
ls: new array.of: @[inc size a, inc size b] 0
loop.with:'i a 'x [
loop.with:'j b 'y [
ls\[i+1]\[j+1]: (x=y)? -> ls\[i]\[j] + 1
-> max @[ls\[i+1]\[j], ls\[i]\[j+1]]
]
]
[result, x, y]: @[new "", size a, size b]
while [and? [x > 0][y > 0]][
if? ls\[x]\[y] = ls\[x-1]\[y] -> x: x-1
else [
if? ls\[x]\[y] = ls\[x]\[y-1] -> y: y-1
else [
result: a\[x-1] ++ result
x: x-1
y: y-1
]
]
]
return result
]
print lcs "1234" "1224533324"
print lcs "thisisatest" "testing123testing"
- Output:
1234 tsitest
AutoHotkey
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
QuickBASIC
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
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
BBC BASIC
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
BQN
It's easier and faster to get only the length of the longest common subsequence, using LcsLen ← ¯1 ⊑ 0¨∘⊢ {𝕩⌈⌈`𝕨+»𝕩}˝ =⌜⟜⌽. This function can be expanded by changing ⌈ to ⊣⍟(>○≠) (choosing from two arguments one that has the greatest length), and replacing the empty length 0 with the empty string "" in the right places.
LCS ← ¯1 ⊑ "" <⊸∾ ""¨∘⊢ ⊣⍟(>○≠){𝕩𝔽¨𝔽`𝕨∾¨""<⊸»𝕩}˝ (=⌜⥊¨⊣)⟜⌽
Output:
"1234" LCS "1224533324"
"1234"
"thisisatest" LCS "testing123testing"
"tsitest"
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);- Output:
max 7 lcs t s i t e s t
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;
}
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#
With recursion
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;
}
}
}
}
C++
Hunt and Szymanski algorithm
#include <stdint.h>
#include <string>
#include <memory> // for shared_ptr<>
#include <iostream>
#include <deque>
#include <unordered_map> //[C++11]
#include <algorithm> // for lower_bound()
#include <iterator> // for next() and prev()
using namespace std;
class LCS {
protected:
// Instances of the Pair linked list class are used to recover the LCS:
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> INDEXES;
typedef unordered_map<char, INDEXES> CHAR_TO_INDEXES_MAP;
typedef deque<INDEXES*> MATCHES;
static uint32_t FindLCS(
MATCHES& indexesOf2MatchedByIndex1, shared_ptr<Pair>* pairs) {
auto traceLCS = pairs != nullptr;
PAIRS chains;
INDEXES prefixEnd;
//
//[Assert]After each index1 iteration prefixEnd[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 : indexesOf2MatchedByIndex1) {
auto dq2 = *it1;
auto limit = prefixEnd.end();
for (auto it2 = dq2.rbegin(); it2 != dq2.rend(); it2++) {
// Each index1, index2 pair corresponds to a match
auto index2 = *it2;
//
// Note: The reverse iterator it2 visits index2 values in descending order,
// allowing in-place update of prefixEnd[]. std::lower_bound() is used to
// perform a binary search.
//
limit = lower_bound(prefixEnd.begin(), limit, index2);
//
// Look ahead to the next index2 value to optimize Pairs used by the Hunt
// and Szymanski algorithm. If the next index2 is also an improvement on
// the value currently held in prefixEnd[index3], a new Pair will only be
// superseded on the next index2 iteration.
//
// Verify that a next index2 value exists; and that this value is greater
// than the final index2 value of the LCS prefix at prev(limit):
//
auto preferNextIndex2 = next(it2) != dq2.rend() &&
(limit == prefixEnd.begin() || *prev(limit) < *next(it2));
//
// Depending on match redundancy, this optimization may reduce the number
// of Pair allocations by factors ranging from 2 up to 10 or more.
//
if (preferNextIndex2) continue;
auto index3 = distance(prefixEnd.begin(), limit);
if (limit == prefixEnd.end()) {
// Insert Case
prefixEnd.push_back(index2);
// Refresh limit iterator:
limit = prev(prefixEnd.end());
if (traceLCS) {
chains.push_back(pushPair(chains, index3, index1, index2));
}
}
else if (index2 < *limit) {
// Update Case
// Update limit value:
*limit = index2;
if (traceLCS) {
chains[index3] = pushPair(chains, index3, index1, index2);
}
}
} // next index2
index1++;
} // next index1
if (traceLCS) {
// Return the LCS as a linked list of matched index pairs:
auto last = chains.empty() ? nullptr : chains.back();
// Reverse longest chain
*pairs = Pair::Reverse(last);
}
auto length = prefixEnd.size();
return length;
}
private:
static shared_ptr<Pair> pushPair(
PAIRS& chains, const ptrdiff_t& index3, uint32_t& index1, uint32_t& index2) {
auto prefix = index3 > 0 ? chains[index3 - 1] : nullptr;
return make_shared<Pair>(index1, index2, prefix);
}
protected:
//
// Match() avoids m*n comparisons by using CHAR_TO_INDEXES_MAP 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.
//
static void Match(
CHAR_TO_INDEXES_MAP& indexesOf2MatchedByChar, MATCHES& indexesOf2MatchedByIndex1,
const string& s1, const string& s2) {
uint32_t index = 0;
for (const auto& it : s2)
indexesOf2MatchedByChar[it].push_back(index++);
for (const auto& it : s1) {
auto& dq2 = indexesOf2MatchedByChar[it];
indexesOf2MatchedByIndex1.push_back(&dq2);
}
}
static 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:
static string Correspondence(const string& s1, const string& s2) {
CHAR_TO_INDEXES_MAP indexesOf2MatchedByChar;
MATCHES indexesOf2MatchedByIndex1; // holds references into indexesOf2MatchedByChar
Match(indexesOf2MatchedByChar, indexesOf2MatchedByIndex1, s1, s2);
shared_ptr<Pair> pairs; // obtain the LCS as index pairs
auto length = FindLCS(indexesOf2MatchedByIndex1, &pairs);
return Select(pairs, length, false, s1, s2);
}
};
Example:
auto s = LCS::Correspondence(s1, s2);
cout << s << endl;
More fully featured examples are available at Samples/C++/LCS.
Clojure
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
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
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
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
Both versions don't work correctly with Unicode text.
Recursive version
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
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
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
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
EasyLang
func$ right a$ n .
return substr a$ (len a$ - n + 1) n
.
func$ left a$ n .
if n < 0
n = len a$ + n
.
return substr a$ 1 n
.
func$ lcs a$ b$ .
if len a$ = 0 or len b$ = 0
return ""
.
if right a$ 1 = right b$ 1
return lcs left a$ -1 left b$ -1 & right a$ 1
.
x$ = lcs a$ left b$ -1
y$ = lcs left a$ -1 b$
if len x$ > len y$
return x$
else
return y$
.
.
print lcs "1234" "1224533324"
print lcs "thisisatest" "testing123testing"
- Output:
1234 tsitest
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))Output:
> (lcs "thisisatest" "testing123testing"))
"tsitest"Elixir
Simple recursion
This solution is Brute force. It is slow
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')
Dynamic Programming
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")
- 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.
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.
Similar to the above, but without using the process dictionary:
-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}.
Output:
48> lcs:lcs("thisisatest", "testing123testing").
"tsitest"
F#
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
Factor
USE: lcs
"thisisatest" "testing123testing" lcs print
- 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, ==, =)
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
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
Go
Recursion
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
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
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
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
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.
- Output:
lcs( "thisisatest", "testing123testing" ) = "tsitest" lcs( "", "x" ) = "" lcs( "x", "x" ) = "x" lcs( "beginning-middle-ending", "beginning-diddle-dum-ending" ) = "beginning-iddle-ending"
J
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
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.
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
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
Recursion
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
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));
};
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.
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
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
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"
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")
- 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
// 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))
}
- Output:
tsitest
Liberty BASIC
'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 FUNCTIONLogo
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
endLua
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
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
> StringTools:-LongestCommonSubSequence( "thisisatest", "testing123testing" );
"tsitest"Mathematica /Wolfram Language
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 LongestCommonSequence performs the puzzle-description.
Nim
Recursion
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")
This recursive version is not efficient but the execution time can be greatly improved by using memoization.
Dynamic Programming
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")
OCaml
Recursion
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
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
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
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
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
PascalABC.NET
function Lengths(x,y: string): array[,] of integer;
begin
var (m,n) := (x.Length,y.Length);
var C := new integer[m+1, n+1]; // filled with zeroes
for var i:=1 to m do
for var j:=1 to n do
if x[i] = y[j] then
C[i,j] := C[i-1,j-1] + 1
else C[i,j] := max(C[i,j-1], C[i-1,j]);
Result := C;
end;
function lcshelper(x,y: string; i,j: integer): string;
begin
var C := Lengths(x,y);
if (i = 0) or (j = 0) then
Result := ''
else if X[i] = Y[j] then
Result := lcshelper(X, Y, i-1, j-1) + X[i]
else if C[i,j-1] > C[i-1,j] then
Result := lcshelper(X, Y, i, j-1)
else Result := lcshelper(X, Y, i-1, j)
end;
function lcs(x,y: string) := lcshelper(x,y,x.Length,y.Length);
begin
Println(lcs('1234','1224533324'));
Println(lcs('thisisatest','testing123testing'));
end.
- Output:
1234 tsitest
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 : $d;
}
print lcs("thisisatest", "testing123testing") . "\n";
Alternate letting regex do all the work
use strict;
use warnings;
use feature 'bitwise';
print "lcs is ", lcs('thisisatest', 'testing123testing'), "\n";
sub lcs
{
my ($c, $d) = @_;
for my $len ( reverse 1 .. length($c &. $d) )
{
"$c\n$d" =~ join '.*', ('(.)') x $len, "\n", map "\\$_", 1 .. $len and
return join '', @{^CAPTURE};
}
return '';
}
- Output:
lcs is tastiest
Phix
If you want this to work with (utf8) Unicode text, just chuck the inputs through utf8_to_utf32() first (and the output through utf32_to_utf8()).
with javascript_semantics function lcs(sequence a, b) sequence res = "" if length(a) and length(b) then if a[$]=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
- Output:
"1234" "tsitest"
Alternate version
same output
with javascript_semantics 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
Picat
Wikipedia algorithm
With some added trickery for a 1-based language.
lcs_wiki(X,Y) = V =>
[C, _Len] = lcs_length(X,Y),
V = backTrace(C,X,Y,X.length+1,Y.length+1).
lcs_length(X, Y) = V=>
M = X.length,
N = Y.length,
C = [[0 : J in 1..N+1] : I in 1..N+1],
foreach(I in 2..M+1,J in 2..N+1)
if X[I-1] == Y[J-1] then
C[I,J] := C[I-1,J-1] + 1
else
C[I,J] := max([C[I,J-1], C[I-1,J]])
end
end,
V = [C, C[M+1,N+1]].
backTrace(C, X, Y, I, J) = V =>
if I == 1; J == 1 then
V = ""
elseif X[I-1] == Y[J-1] then
V = backTrace(C, X, Y, I-1, J-1) ++ [X[I-1]]
else
if C[I,J-1] > C[I-1,J] then
V = backTrace(C, X, Y, I, J-1)
else
V = backTrace(C, X, Y, I-1, J)
end
end.Rule-based
table
lcs_rule(A, B) = "", (A == ""; B == "") => true.
lcs_rule(A, B) = [A[1]] ++ lcs_rule(butfirst(A), butfirst(B)), A[1] == B[1] => true.
lcs_rule(A, B) = longest(lcs_rule(butfirst(A), B), lcs_rule(A, butfirst(B))) => true.
% Return the longest string of A and B
longest(A, B) = cond(A.length > B.length, A, B).
% butfirst (everything except first element)
butfirst(A) = [A[I] : I in 2..A.length].Test
go =>
Tests = [["thisisatest","testing123testing"],
["XMJYAUZ", "MZJAWXU"],
["1234", "1224533324"],
["beginning-middle-ending","beginning-diddle-dum-ending"]
],
Funs = [lcs_wiki,lcs_rule],
foreach(Fun in Funs)
println(fun=Fun),
foreach(Test in Tests)
printf("%w : %w\n", Test, apply(Fun,Test[1],Test[2]))
end,
nl
end,
nl.- Output:
fun = lcs_wiki [thisisatest,testing123testing] : tsitest [XMJYAUZ,MZJAWXU] : MJAU [1234,1224533324] : 1234 [beginning-middle-ending,beginning-diddle-dum-ending] : beginning-iddle-ending fun = lcs_rule [thisisatest,testing123testing] : tsitest [XMJYAUZ,MZJAWXU] : MJAU [1234,1224533324] : 1234 [beginning-middle-ending,beginning-diddle-dum-ending] : beginning-iddle-ending
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") ) )- Output:
-> ("t" "s" "i" "t" "e" "s" "t")
PowerShell
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 $list2.ID
- Output:
1 3 5 7 9
Prolog
Recursive Version
First version:
test :-
time(lcs("thisisatest", "testing123testing", Lcs)),
writef('%s',[Lcs]).
lcs([ H|L1],[ H|L2],[H|Lcs]) :- !,
lcs(L1,L2,Lcs).
lcs([H1|L1],[H2|L2],Lcs):-
lcs( L1 ,[H2|L2],Lcs1),
lcs([H1|L1], L2 ,Lcs2),
longest(Lcs1,Lcs2,Lcs),!.
lcs(_,_,[]).
longest(L1,L2,Longest) :-
length(L1,Length1),
length(L2,Length2),
((Length1 > Length2) -> Longest = L1; Longest = L2).
Second version, with memoization:
%declare that we will add lcs_db facts during runtime
:- dynamic lcs_db/3.
test :-
retractall(lcs_db(_,_,_)), %clear the database of known results
time(lcs("thisisatest", "testing123testing", Lcs)),
writef('%s',[Lcs]).
% check if the result is known
lcs(L1,L2,Lcs) :-
lcs_db(L1,L2,Lcs),!.
lcs([ H|L1],[ H|L2],[H|Lcs]) :- !,
lcs(L1,L2,Lcs).
lcs([H1|L1],[H2|L2],Lcs) :-
lcs( L1 ,[H2|L2],Lcs1),
lcs([H1|L1], L2 ,Lcs2),
longest(Lcs1,Lcs2,Lcs),!,
assert(lcs_db([H1|L1],[H2|L2],Lcs)).
lcs(_,_,[]).
longest(L1,L2,Longest) :-
length(L1,Length1),
length(L2,Length2),
((Length1 > Length2) -> Longest = L1; Longest = L2).
- Demonstrating:
Example for "beginning-middle-ending" and "beginning-diddle-dum-ending"
First version :
?- time(lcs("beginning-middle-ending","beginning-diddle-dum-ending", Lcs)),writef('%s', [Lcs]).
% 10,875,184 inferences, 1.840 CPU in 1.996 seconds (92% CPU, 5910426 Lips)
beginning-iddle-ending
Second version which is much faster :
?- time(lcs("beginning-middle-ending","beginning-diddle-dum-ending", Lcs)),writef('%s', [Lcs]).
% 2,376 inferences, 0.010 CPU in 0.003 seconds (300% CPU, 237600 Lips)
beginning-iddle-ending
PureBasic
Procedure.s lcs(a$, 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
The simplest way is to use LCS within mlpy package
Recursion
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 str(lcs(xs, ys)) + x
else:
return max(lcs(xstr, ys), lcs(xs, ystr), key=len)
Test it:
if __name__=="__main__":
import doctest; doctest.testmod()
Dynamic Programming
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
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")
- Output:
"tsitest">
Raku
(formerly Perl 6)
Recursion
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
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
Bit parallel dynamic programming with nearly linear complexity O(n). It is fast.
sub lcs(Str $xstr, Str $ystr) {
my (@a, @b) := ($xstr, $ystr)».comb;
my (%positions, @Vs, $lcs);
for @a.kv -> $i, $x { %positions{$x} +|= 1 +< ($i % @a) }
my $S = +^ 0;
for (0 ..^ @b) -> $j {
my $u = $S +& (%positions{@b[$j]} // 0);
@Vs[$j] = $S = ($S + $u) +| ($S - $u)
}
my ($i, $j) = @a-1, @b-1;
while ($i ≥ 0 and $j ≥ 0) {
unless (@Vs[$j] +& (1 +< $i)) {
$lcs [R~]= @a[$i] unless $j and ^@Vs[$j-1] +& (1 +< $i);
$j--
}
$i--
}
$lcs
}
say lcs("thisisatest", "testing123testing");
ReasonML
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));
}
};
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
- 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
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 okOutput:
1234
Ruby
Recursion
This solution is similar to the Haskell one. It is slow (The time complexity is exponential.)
=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
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?(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 $0 == __FILE__
puts lcs('thisisatest', 'testing123testing')
puts lcs("rosettacode", "raisethysword")
end
- Output:
tsitest rsetod
Referring to LCS here and here.
Run BASIC
a$ = "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 FUNCTIONaba
Rust
Dynamic programming version:
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);
}
Scala
Using lazily evaluated lists:
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))}
}
Using recursion:
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
}- Output:
scala> lcsLazy("thisisatest", "testing123testing").mkString
res0: String = tsitest
scala> lcsRec("thisisatest", "testing123testing").mkString
res1: String = tsitest
Recursive version:
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
}
}
}The dynamic programming version:
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
}
}
}- Output:
scala> lcsM("thisiaatest".toList, "testing123testing".toList).mkString
res0: String = tsitest
Scheme
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
$ 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
SequenceL
It is interesting to note that x and y are computed in parallel, dividing work across threads repeatedly down through the recursion.
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");- Output:
"tsitest"
SETL
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
func lcs(xstr, ystr) is cached {
xstr.is_empty && return xstr
ystr.is_empty && return ystr
var(x, xs, y, ys) = (xstr.first(1), xstr.slice(1),
ystr.first(1), ystr.slice(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
We define this on the Sequence type since there is nothing string-specific about the concept.
Recursion
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
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
Swift 5.1
Recursion
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
}
}Dynamic Programming
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())
}Tcl
Recursive
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
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]
}Performance Comparison
% time {d_lcs thisisatest testing123testing} 10
637.5 microseconds per iteration
% time {r_lcs thisisatest testing123testing} 10
1275566.8 microseconds per iterationUrsala
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 ~<est program:
#cast %s
example = lcs('thisisatest','testing123testing')- Output:
'tsitest'
Wren
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))- Output:
tsitest
zkl
This is quite vile in terms of [time] efficiency, another algorithm should be used for real work.
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
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