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Factorize string into Lyndon words: Difference between revisions
Factorize string into Lyndon words (view source)
Revision as of 21:59, 30 January 2024
, 3 months agoC program
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imported>CosmiaNebula (C program) |
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{{task}}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 [[wp:Lyndon_word|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.
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* '''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.<ref>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</ref> See <ref name=":0">https://cp-algorithms.com/string/lyndon_factorization.html</ref> for a description of Duval's algorithm.
== [[C]] ==
Copied from <ref name=":0" />, under Creative Commons Attribution Share Alike 4.0 International License.<syntaxhighlight lang="c">
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;
}
</syntaxhighlight>
== [[Python]] ==
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