# Lyndon words¶

sage.combinat.words.lyndon_word.LyndonWord(data, check=True)

Construction of a Lyndon word.

INPUT:

• data – list

• check – bool (optional, default: True) if True, check that the input data represents a Lyndon word.

OUTPUT:

A Lyndon word.

EXAMPLES:

sage: LyndonWord([1,2,2])
word: 122
sage: LyndonWord([1,2,3])
word: 123
sage: LyndonWord([2,1,2,3])
Traceback (most recent call last):
...
ValueError: not a Lyndon word

If check is False, then no verification is done:

sage: LyndonWord([2,1,2,3], check=False)
word: 2123
sage.combinat.words.lyndon_word.LyndonWords(e=None, k=None)

Return the combinatorial class of Lyndon words.

A Lyndon word $$w$$ is a word that is lexicographically less than all of its rotations. Equivalently, whenever $$w$$ is split into two non-empty substrings, $$w$$ is lexicographically less than the right substring.

INPUT:

• no input at all

or

• e – integer, size of alphabet

• k – integer, length of the words

or

• e – a composition

OUTPUT:

A combinatorial class of Lyndon words.

EXAMPLES:

sage: LyndonWords()
Lyndon words

If e is an integer, then e specifies the length of the alphabet; k must also be specified in this case:

sage: LW = LyndonWords(3, 4); LW
Lyndon words from an alphabet of size 3 of length 4
sage: LW.first()
word: 1112
sage: LW.last()
word: 2333
sage: LW.random_element() # random
word: 1232
sage: LW.cardinality()
18

If e is a (weak) composition, then it returns the class of Lyndon words that have evaluation e:

sage: LyndonWords([2, 0, 1]).list()
[word: 113]
sage: LyndonWords([2, 0, 1, 0, 1]).list()
[word: 1135, word: 1153, word: 1315]
sage: LyndonWords([2, 1, 1]).list()
[word: 1123, word: 1132, word: 1213]
class sage.combinat.words.lyndon_word.LyndonWords_class(alphabet=None)

The set of all Lyndon words.

class sage.combinat.words.lyndon_word.LyndonWords_evaluation(e)

The set of Lyndon words on a fixed multiset of letters.

EXAMPLES:

sage: L = LyndonWords([1,2,1])
sage: L
Lyndon words with evaluation [1, 2, 1]
sage: L.list()
[word: 1223, word: 1232, word: 1322]
cardinality()

Return the number of Lyndon words with the evaluation e.

EXAMPLES:

sage: LyndonWords([]).cardinality()
0
sage: LyndonWords([2,2]).cardinality()
1
sage: LyndonWords([2,3,2]).cardinality()
30

Check to make sure that the count matches up with the number of Lyndon words generated:

sage: comps = [[],[2,2],[3,2,7],[4,2]] + Compositions(4).list()
sage: lws = [LyndonWords(comp) for comp in comps]
sage: all(lw.cardinality() == len(lw.list()) for lw in lws)
True
class sage.combinat.words.lyndon_word.LyndonWords_nk(n, k)

Lyndon words of fixed length $$k$$ over the alphabet $$\{1, 2, \ldots, n\}$$.

INPUT:

• n – the size of the alphabet

• k – the length of the words

EXAMPLES:

sage: L = LyndonWords(3, 4)
sage: L.list()
[word: 1112,
word: 1113,
word: 1122,
word: 1123,
...
word: 1333,
word: 2223,
word: 2233,
word: 2333]
cardinality()
sage.combinat.words.lyndon_word.StandardBracketedLyndonWords(n, k)

Return the combinatorial class of standard bracketed Lyndon words from [1, …, n] of length k.

These are in one to one correspondence with the Lyndon words and form a basis for the subspace of degree k of the free Lie algebra of rank n.

EXAMPLES:

sage: SBLW33 = StandardBracketedLyndonWords(3,3); SBLW33
Standard bracketed Lyndon words from an alphabet of size 3 of length 3
sage: SBLW33.first()
[1, [1, 2]]
sage: SBLW33.last()
[[2, 3], 3]
sage: SBLW33.cardinality()
8
sage: SBLW33.random_element() in SBLW33
True
class sage.combinat.words.lyndon_word.StandardBracketedLyndonWords_nk(n, k)
cardinality()

EXAMPLES:

sage: StandardBracketedLyndonWords(3, 3).cardinality()
8
sage: StandardBracketedLyndonWords(3, 4).cardinality()
18
sage.combinat.words.lyndon_word.standard_bracketing(lw)

Return the standard bracketing of a Lyndon word lw.

EXAMPLES:

sage: import sage.combinat.words.lyndon_word as lyndon_word
sage: [lyndon_word.standard_bracketing(u) for u in LyndonWords(3,3)]
[[1, [1, 2]],
[1, [1, 3]],
[[1, 2], 2],
[1, [2, 3]],
[[1, 3], 2],
[[1, 3], 3],
[2, [2, 3]],
[[2, 3], 3]]
sage.combinat.words.lyndon_word.standard_unbracketing(sblw)

Return flattened sblw if it is a standard bracketing of a Lyndon word, otherwise raise an error.

EXAMPLES:

sage: from sage.combinat.words.lyndon_word import standard_unbracketing
sage: standard_unbracketing([1, [2, 3]])
word: 123
sage: standard_unbracketing([[1, 2], 3])
Traceback (most recent call last):
...
ValueError: not a standard bracketing of a Lyndon word