# Factorize string into Lyndon words

Factorize string into Lyndon words
You are encouraged to solve this task according to the task description, using any language you may know.

Given a finite alphabet, we can lexicographically order all strings in the alphabet. If two strings have different lengths, then pad the shorter string on the right with the smallest letter. For example, we have 01 > 001, but 01 = 010. As we see, this order is a total preorder, but not a total order, since it identifies different strings.

A Lyndon word is a non-empty string that is strictly lower in lexicographic order than all its circular rotations. In particular, if a string is equal to a circular rotation, then it is not a Lyndon word. For example, since 0101 = 0101 (rotation by 2), it is not a Lyndon word.

The first Lyndon words on the binary alphabet {0, 1} are:

0, 1, 01, 001, 011, 0001, 0011, 0111, 00001, 00011, 00101, 00111, 01011, 01111, ...

Some basic properties:

• The only Lyndon word that ends with 0 is 0.
• Proof. If s0 is a Lyndon word, and s is not empty, then s0 < 0s. If s contains 1 somewhere, then s0 > 0s. Therefore s has only 0. But then s0 = 0s, contradiction.
• The lexicographic order is a total order on the Lyndon words.
• Proof. For, the only way for two different strings s, s' to have the same lexicographic ordering is for one of them to pad to the other. We can assume that s00...0 = s'. If that is so, then s00...0 is a Lyndon word that ends with 0, so it is just 0, and so s is a Lyndon word that is also empty, contradiction.

The Chen–Fox–Lyndon theorem states that any string is uniquely factorizable into a sequence of Lyndon words non-decreasing in lexicographic order. Duval's algorithm computes this in O(length of input) time and and O(1) space.[1] See [2] for a description of Duval's algorithm.

## ALGOL 68

Translation of: Python – with multiline output as with a number of other samples
```BEGIN # Factorize a string into Lyndon words - translated of the Python sample #

PROC chen fox lyndon factorization = ( STRING s )[]STRING:
BEGIN
INT n  = UPB s;
INT i := LWB s;
FLEX[ 1 : 100 ]STRING factorization;
INT f max := LWB factorization;
WHILE i <= n DO
INT j := i + 1, k := i;
WHILE IF j <= n THEN s[ k ] <= s[ j ] ELSE FALSE FI DO
IF s[ k ] < s[ j ] THEN
k  := i
ELSE
k +:= 1
FI;
j +:= 1
OD;
WHILE i <= k DO
f max +:= 1;
IF f max > UPB factorization THEN
[ 1 : UPB factorization + 20 ]STRING new factorization;
new factorization[ 1 : UPB factorization ] := factorization;
factorization := new factorization
FI;
factorization[ f max ] := s[ i : i + j - k - 1 ];
i +:= j - k
OD
OD;
STRING check := "";
FOR i TO f max DO check +:= factorization[ i ] OD;
IF check /= s THEN print( ( "Incorrect factorization", newline ) ); stop FI;
factorization[ 1 : f max ]
END # chen fox lyndon factorization # ;

# Example use with Thue-Morse sequence                                     #
STRING m := "0";
TO 7 DO
STRING m0 = m;
FOR i FROM LWB m TO UPB m DO
IF m[ i ]  = "0" THEN m[ i ] := "1" ELIF m[ i ]  = "1" THEN m[ i ] := "0" FI
OD;
m0 +=: m
OD;

[]STRING m factors = chen fox lyndon factorization( m );
FOR i FROM LWB m factors TO UPB m factors DO
print( ( m factors[ i ], newline ) )
OD

END```
Output:
```011
01
0011
00101101
0010110011010011
00101100110100101101001100101101
001011001101001011010011001011001101001100101101
001011001101
001
```

## C++

Copied from [2], under Creative Commons Attribution Share Alike 4.0 International License and converted to a runnable C++ program.

```#include <iostream>
#include <vector>
#include <string>
#include <algorithm>

using namespace std;

vector<string> duval(string const& s) {
int n = s.size();
int i = 0;
vector<string> factorization;
while (i < n) {
int j = i + 1, k = i;
while (j < n && s[k] <= s[j]) {
if (s[k] < s[j])
k = i;
else
k++;
j++;
}
while (i <= k) {
factorization.push_back(s.substr(i, j - k));
i += j - k;
}
}
return factorization;
}

int main() {
// Thue-Morse example
string m = "0";
for (int i = 0; i < 7; ++i) {
string m0 = m;
replace(m.begin(), m.end(), '0', 'a');
replace(m.begin(), m.end(), '1', '0');
replace(m.begin(), m.end(), 'a', '1');
m = m0 + m;
}
for (string s : duval(m)) cout << s << endl;
return 0;
}
```
Output:
```011
01
0011
00101101
0010110011010011
00101100110100101101001100101101
001011001101001011010011001011001101001100101101
001011001101
001
```

## EasyLang

Translation of: Julia
```proc lyndonfact s\$ . .
n = len s\$
s\$[] = strchars s\$
i = 1
while i <= n
j = i + 1
k = i
while j <= n and strcode s\$[k] <= strcode s\$[j]
if strcode s\$[k] < strcode s\$[j]
k = i
else
k += 1
.
j += 1
.
while i <= k
print substr s\$ i (j - k)
i += j - k
.
.
.
m\$ = "0"
for i to 7
for c\$ in strchars m\$
if c\$ = "0"
m\$ &= "1"
else
m\$ &= "0"
.
.
.
lyndonfact m\$```

## Java

```import java.util.ArrayList;
import java.util.List;

public class FactorizeStringIntoLyndonWords {

public static void main(String[] args) {
// Create a 128 character Thue-Morse word
String thueMorse = "0";
for ( int i = 0; i < 7; i++ ) {
String thueMorseCopy = thueMorse;
thueMorse = thueMorse.replace("0", "a");
thueMorse = thueMorse.replace("1", "0");
thueMorse = thueMorse.replace("a", "1");
thueMorse = thueMorseCopy + thueMorse;
}

System.out.println("The Thue-Morse word to be factorised:");
System.out.println(thueMorse);

System.out.println();
System.out.println("The factors are:");
for ( String factor : duval(thueMorse) ) {
System.out.println(factor);
}
}

// Duval's algorithm
private static List<String> duval(String text) {
List<String> factorisation = new ArrayList<String>();
int i = 0;

while ( i < text.length() ) {
int j = i + 1;
int k = i;

while ( j < text.length() && text.charAt(k) <= text.charAt(j) ) {
if ( text.charAt(k) < text.charAt(j) ) {
k = i;
} else {
k += 1;
}

j += 1;
}

while ( i <= k ) {
factorisation.addLast(text.substring(i, i + j - k));
i += j - k;
}
}

return factorisation;
}

}
```
```The Thue-Morse word to be factorised:
01101001100101101001011001101001100101100110100101101001100101101001011001101001011010011001011001101001100101101001011001101001

The factors are:
011
01
0011
00101101
0010110011010011
00101100110100101101001100101101
001011001101001011010011001011001101001100101101
001011001101
```

## jq

Works with: jq

Works with gojq, the Go implementation of jq

```def duval:
. as \$s
| def a(\$i): \$s[\$i:\$i+1];
length as \$n
| { i: 0, factorization: []}
| until (.i >= \$n;
.j = .i + 1
| .k = .i
| until (.j >= \$n or a(.k) > a(.j);
if a(.k) < a(.j) then .k = .i else .k += 1 end
| .j += 1 )
| until (.i > .k;
.factorization += [\$s[.i: .i+.j-.k]]
| .i += .j - .k ) )
| .factorization ;

# Thue-Morse example
def m:
reduce range(0; 7) as \$i ( "0";
. as \$m0
| gsub("0"; "a")
| gsub("1"; "0")
| gsub("a"; "1")
| \$m0 + .) ;

m | duval | join("\n")```
Output:

As for Wren.

## Julia

Translation of: Phix
Translation of: Python
```function chenfoxlyndonfactorization(s)
n = length(s)
i = 1
factorization = String[]
while i <= n
j = i + 1
k = i
while j <= n && s[k] <= s[j]
if s[k] < s[j]
k = i
else
k += 1
end
j += 1
end
while i <= k
push!(factorization, s[i:i+j-k-1])
i += j - k
end
end
@assert s == prod(factorization)
return factorization
end

let m = "0"
for i in 1:7
m *= replace(m, '0' => '1', '1' => '0')
end
println(chenfoxlyndonfactorization(m))
end
```
Output:
```["011", "01", "0011", "00101101", "0010110011010011", "00101100110100101101001100101101", "001011001101001011010011001011001101001100101101", "001011001101", "001"]
```

## MATLAB

Translation of: Python
```clear all;close all;clc;
m = '0';
for i = 1:7
m0 = m;
m = strrep(m, '0', 'a');
m = strrep(m, '1', '0');
m = strrep(m, 'a', '1');
m = strcat(m0, m);
end

factorization = chenFoxLyndonFactorization(m);
for index=1:length(factorization)
disp(factorization(index));
end

function factorization = chenFoxLyndonFactorization(s)
n = length(s);
i = 1;
factorization = {};
while i <= n
j = i + 1;
k = i;
while j <= n && s(k) <= s(j)
if s(k) < s(j)
k = i;
else
k = k + 1;
end
j = j + 1;
end
while i <= k
factorization{end+1} = s(i:i + j - k - 1);
i = i + j - k;
end
end
assert(strcmp(join(factorization, ''), s));
end
```
Output:
```    {'011'}

{'01'}

{'0011'}

{'00101101'}

{'0010110011010011'}

{'00101100110100101101001100101101'}

{'001011001101001011010011001011001101001100101101'}

{'001011001101'}

{'001'}
```

## Phix

Translation of: Python
```with javascript_semantics
function chen_fox_lyndon_factorization(string s)
integer n = length(s), i = 1
sequence factorization = {}
while i <= n do
integer j = i + 1, k = i
while j <= n and s[k] <= s[j] do
if s[k] < s[j] then
k = i
else
k += 1
end if
j += 1
end while
while i <= k do
factorization &= {s[i..i+j-k-1]}
i += j - k
end while
end while
assert(join(factorization,"") == s)
return factorization
end function
-- Example use with Thue-Morse sequence
string m = "0"
for i=1 to 7 do
m &= sq_sub('0'+'1',m)
end for
?chen_fox_lyndon_factorization(m)
-- alt:
printf(1,"\n%s\n",join(chen_fox_lyndon_factorization(m),"\n"))
```
Output:
```{"011","01","0011","00101101","0010110011010011","00101100110100101101001100101101","001011001101001011010011001011001101001100101101","001011001101","001"}

011
01
0011
00101101
0010110011010011
00101100110100101101001100101101
001011001101001011010011001011001101001100101101
001011001101
001
```

## Python

Duval's algorithm:

```def chen_fox_lyndon_factorization(s):
n = len(s)
i = 0
factorization = []
while i < n:
j, k = i + 1, i
while j < n and s[k] <= s[j]:
if s[k] < s[j]:
k = i
else:
k += 1
j += 1
while i <= k:
factorization.append(s[i:i + j - k])
i += j - k
assert ''.join(factorization) == s
return factorization
```

Example use with Thue-Morse sequence

```m='0'
for i in range(0,7):
m0=m
m=m.replace('0','a')
m=m.replace('1','0')
m=m.replace('a','1')
m=m0+m

print(chen_fox_lyndon_factorization(m))
```

Output:

```['011', '01', '0011', '00101101', '0010110011010011', '00101100110100101101001100101101', '001011001101001011010011001011001101001100101101', '001011001101', '001']
```

## Raku

Translation of: Julia
```# 20240213 Raku programming solution

sub chenfoxlyndonfactorization(Str \$s) {
my (\$n, \$i, @factorization) = \$s.chars, 1;
while \$i <= \$n {
my (\$j, \$k) = \$i X+ (1, 0);
while \$j <= \$n && substr(\$s, \$k-1, 1) <= substr(\$s, \$j-1, 1) {
substr(\$s, \$k-1, 1) < substr(\$s, \$j++ -1, 1) ?? ( \$k = \$i ) !! \$k++;
}
while \$i <= \$k {
@factorization.push: substr(\$s, \$i-1, \$j-\$k);
\$i += \$j - \$k
}
}
die unless \$s eq [~] @factorization;
return @factorization
}

my \$m = "0";
for ^7 { \$m ~= \$m.trans('0' => '1', '1' => '0') }
say chenfoxlyndonfactorization(\$m);
```
Output:

Same as Julia example.

You may Attempt This Online!

## RPL

Based on Duval's algorithm. `THUEM` is defined at Thue-Morse.

Works with: HP version 48G
```« DUP SIZE { } → s n res
« 1
WHILE DUP n ≤ REPEAT
DUP 1 + OVER
WHILE s OVER DUP SUB s 4 PICK DUP SUB DUP2 ≤ 5 PICK n ≤ AND REPEAT
IF < THEN DROP OVER ELSE 1 + END
SWAP 1 + SWAP
END DROP2
SWAP OVER - SWAP ROT
WHILE DUP2 ≥ REPEAT
'res' s 3 PICK DUP 7 PICK + 1 - SUB STO+
3 PICK +
END ROT ROT DROP2
END
DROP res
» » 'LYNDON' STO     @ ( "string" → { "f1".."fn" } )

« 2 7 ^ THUEM « →STR + » STREAM
LYNDON
```
Output:
```1: { "011" "01" "0011" "00101101" "0010110011010011" "00101100110100101101001100101101" "001011001101001011010011001011001101001100101101" "001011001101" "001" }
```

Using local variables

The version below favors local variables to the stack, which makes it a little bit slower - but easier to read.

```« DUP SIZE 1 → s n ii
« { }
WHILE ii n ≤ REPEAT
ii DUP 1 + → k j
« WHILE s k DUP SUB s j DUP SUB DUP2 ≤ ii n ≤ AND REPEAT
< ii k 1 + IFTE 'k' STO
'j' 1 STO+
»
'j' k STO-
WHILE ii k ≤ REPEAT
s ii DUP j + 1 - SUB +
'ii' j STO+
END
»
END
» » 'LYNDON' STO     @ ( "string" → { "f1".."fn" } )
```

## Rust

Translation of: Python
```fn chen_fox_lyndon_factorization(s: &str) -> Vec<String> {
let n = s.len();
let mut i = 0;
let mut factorization = Vec::new();
while i < n {
let (mut j, mut k) = (i + 1, i);
while j < n && s.as_bytes()[k] <= s.as_bytes()[j] {
if s.as_bytes()[k] < s.as_bytes()[j] {
k = i;
} else {
k += 1;
}
j += 1;
}
while i <= k {
factorization.push(s[i..i + j - k].to_string());
i += j - k;
}
}
factorization
}

fn main() {
let mut m = String::from("0");
for _ in 0..7 {
let m0 = m.clone();
m = m.replace("0", "a");
m = m.replace("1", "0");
m = m.replace("a", "1");
m = m0 + &m;
}

let factorization = chen_fox_lyndon_factorization(&m);
println!("{:?}", factorization);
}
```
Output:
```["011", "01", "0011", "00101101", "0010110011010011", "00101100110100101101001100101101", "001011001101001011010011001011001101001100101101", "001011001101", "001"]
```

## Scala

Translation of: Python
```object ChenFoxLyndonFactorization extends App {
def chenFoxLyndonFactorization(s: String): List[String] = {
val n = s.length
var i = 0
var factorization = List[String]()
while (i < n) {
var j = i + 1
var k = i
while (j < n && s.charAt(k) <= s.charAt(j)) {
if (s.charAt(k) < s.charAt(j)) {
k = i
} else {
k += 1
}
j += 1
}
while (i <= k) {
factorization = factorization :+ s.substring(i, i + j - k)
i += j - k
}
}
assert(s == factorization.mkString)
factorization
}

var m = "0"
for (i <- 0 until 7) {
val m0 = m
m = m.replace('0', 'a').replace('1', '0').replace('a', '1')
m = m0 + m
}

println(chenFoxLyndonFactorization(m))
}
```
Output:
```List(011, 01, 0011, 00101101, 0010110011010011, 00101100110100101101001100101101, 001011001101001011010011001011001101001100101101, 001011001101, 001)

```

## SETL

Translation of: C++
```program lyndon_factorization;
tm := "01101001100101101001011001101001100101100110"
+ "10010110100110010110100101100110100101101001"
+ "1001011001101001100101101001011001101001";

loop for part in duval(tm) do
print(part);
end loop;

proc duval(s);
i := 1;
fact := [];
loop while i <= #s do
j := i + 1;
k := i;
loop while j <= #s and s(k) <= s(j) do
k := if s(k) < s(j) then i else k+1 end;
j +:= 1;
end loop;
loop while i <= k do
fact with:= s(i..i+j-k-1);
i +:= j-k;
end loop;
end loop;
return fact;
end proc;
end program;```
Output:
```011
01
0011
00101101
0010110011010011
00101100110100101101001100101101
001011001101001011010011001011001101001100101101
001011001101
001```

## Wren

Translation of: Python
Library: Wren-str
```import "./str" for Str

var duval = Fn.new { |s|
var n = s.count
var i = 0
var factorization = []
while (i < n) {
var j = i + 1
var k = i
while (j < n && Str.le(s[k], s[j])) {
if (Str.lt(s[k], s[j])) k = i else k = k + 1
j = j + 1
}
while (i <= k) {
i = i + j - k
}
}
return factorization
}

// Thue-Morse example
var m = "0"
for (i in 0..6) {
var m0 = m
m = m.replace("0", "a")
m = m.replace("1", "0")
m = m.replace("a", "1")
m = m0 + m
}
System.print(duval.call(m).join("\n"))>
```
Output:
```011
01
0011
00101101
0010110011010011
00101100110100101101001100101101
001011001101001011010011001011001101001100101101
001011001101
001
```
1. Duval, Jean-Pierre (1983), "Factorizing words over an ordered alphabet", Journal of Algorithms, 4 (4): 363–381, doi:10.1016/0196-6774(83)90017-2
2. https://cp-algorithms.com/string/lyndon_factorization.html