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.
C
Copied from [2], under Creative Commons Attribution Share Alike 4.0 International License.
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;
}
MATLAB
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
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)
- Output:
{"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']
Rust
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
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
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
- ↑ 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
- ↑ 2.0 2.1 https://cp-algorithms.com/string/lyndon_factorization.html