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# A collection of constructors of common words

AUTHORS:

- Franco Saliola (2008-12-17): merged into sage
- Sebastien Labbe (2008-12-17): merged into sage
- Arnaud Bergeron (2008-12-17): merged into sage
- Amy Glen (2008-12-17): merged into sage
- Sebastien Labbe (2009-12-19): Added S-adic words (trac ticket #7543)

USE:

To see a list of all word constructors, type `words.` and then press the tab key. The documentation for each constructor includes information about each word, which provides a useful reference.

REFERENCES:

[AC03] | B. Adamczewski, J. Cassaigne, On the transcendence of real numbers with a regular expansion, J. Number Theory 103 (2003) 27–37.

[BmBGL07] | A. Blondin-Masse, S. Brlek, A. Glen, and S. Labbe. On the critical exponent of generalized Thue-Morse words. Discrete Math. Theor. Comput. Sci. 9 (1):293–304, 2007.

[BmBGL09] | (1, 2) A. Blondin-Masse, S. Brlek, A. Garon, and S. Labbe. Christoffel and Fibonacci Tiles, DGCI 2009, Montreal, to appear in LNCS.

[Loth02] | (1, 2, 3) M. Lothaire, Algebraic Combinatorics On Words, vol. 90 of Encyclopedia of Mathematics and its Applications, Cambridge University Press, U.K., 2002.

[Fogg] | Pytheas Fogg, https://www.lirmm.fr/arith/wiki/PytheasFogg/S-adiques.

EXAMPLES:

    sage: t = words.ThueMorseWord(); t
    word: 0110100110010110100101100110100110010110...

class `sage.combinat.words.word_generators.``LowerChristoffelWord`(p, q, alphabet=(0, 1), algorithm='cf')

  Bases: `sage.combinat.words.word.FiniteWord_list`

  Returns the lower Christoffel word of slope \(p/q\), where \(p\) and \(q\) are relatively prime non-negative integers, over the given two-letter alphabet.

  The Christoffel word of slope `p/q` is obtained from the Cayley graph of \(\ZZ/(p+q)\ZZ\) with generator \(q\) as follows. If \(u \rightarrow v\) is an edge in the Cayley graph, then \(v = u + p \mod{p+q}\). Label the edge \(u \rightarrow v\) by `alphabet[1]` if \(u < v\) and `alphabet[0]` otherwise. The Christoffel word is the word obtained by reading the edge labels along the cycle beginning from 0.

  EXAMPLES:

      sage: words.LowerChristoffelWord(4,7)
      word: 00100100101

      sage: words.LowerChristoffelWord(4,7,alphabet='ab')
      word: aabaabaabab

  TESTS:

      sage: words.LowerChristoffelWord(1,0)
      word: 1
      sage: words.LowerChristoffelWord(0,1,'xy')
      word: x
      sage: words.LowerChristoffelWord(1,1)
      word: 01

  `markoff_number`()

    Returns the Markoff number associated to the Christoffel word self.

    The Markoff number of a Christoffel word \(w\) is \(trace(M(w))/3\), where \(M(w)\) is the \(2\times 2\) matrix obtained by applying the morphism: 0 -> matrix(2,[2,1,1,1]) 1 -> matrix(2,[5,2,2,1])

    EXAMPLES:

        sage: w0 = words.LowerChristoffelWord(4,7)
        sage: w1, w2 = w0.standard_factorization()
        sage: (m0,m1,m2) = (w.markoff_number() for w in (w0,w1,w2))
        sage: (m0,m1,m2)
        (294685, 13, 7561)
        sage: m0**2 + m1**2 + m2**2 == 3*m0*m1*m2
        True

  `standard_factorization`()

    Returns the standard factorization of the Christoffel word `self`.

    The standard factorization of a Christoffel word \(w\) is the unique factorization of \(w\) into two Christoffel words.

    EXAMPLES:

        sage: w = words.LowerChristoffelWord(5,9)
        sage: w
        word: 00100100100101
        sage: w1, w2 = w.standard_factorization()
        sage: w1
        word: 001
        sage: w2
        word: 00100100101

        sage: w = words.LowerChristoffelWord(51,37)
        sage: w1, w2 = w.standard_factorization()
        sage: w1
        word: 0101011010101101011
        sage: w2
        word: 0101011010101101011010101101010110101101...
        sage: w1 * w2 == w
        True

class `sage.combinat.words.word_generators.``WordGenerator`

  Bases: `object`

  Constructor of several famous words.

  EXAMPLES:

      sage: words.ThueMorseWord()
      word: 0110100110010110100101100110100110010110...

      sage: words.FibonacciWord()
      word: 0100101001001010010100100101001001010010...

      sage: words.ChristoffelWord(5, 8)
      word: 0010010100101

      sage: words.RandomWord(10, 4)    # not tested random
      word: 1311131221

      sage: words.CodingOfRotationWord(alpha=0.618, beta=0.618)
      word: 1010110101101101011010110110101101101011...

      sage: tm = WordMorphism('a->ab,b->ba')
      sage: fib = WordMorphism('a->ab,b->a')
      sage: tmword = words.ThueMorseWord([0, 1])
      sage: from itertools import repeat
      sage: words.s_adic(tmword, repeat('a'), {0:tm, 1:fib})
      word: abbaababbaabbaabbaababbaababbaabbaababba...

  Note

  To see a list of all word constructors, type `words.` and then hit the TAB key. The documentation for each constructor includes information about each word, which provides a useful reference.

  TESTS:

      sage: from sage.combinat.words.word_generators import WordGenerator
      sage: words2 = WordGenerator()
      sage: type(loads(dumps(words2)))
      <class 'sage.combinat.words.word_generators.WordGenerator'>

  `CharacteristicSturmianWord`(slope, alphabet=(0, 1), bits=None)

    Returns the characteristic Sturmian word (also called standard Sturmian word) of given slope.

    Over a binary alphabet \(\{a,b\}\), the characteristic Sturmian word \(c_\alpha\) of irrational slope \(\alpha\) is the infinite word satisfying \(s_{\alpha,0} = ac_\alpha\) and \(s'_{\alpha,0} = bc_\alpha\), where \(s_{\alpha,0}\) and \(s'_{\alpha,0}\) are respectively the lower and upper mechanical words with slope \(\alpha\) and intercept \(0\). Equivalently, for irrationnal \(\alpha\), \(c_\alpha = s_{\alpha,\alpha} = s'_{\alpha,\alpha}\).

    Let \(\alpha = [0, d_1 + 1, d_2, d_3, \ldots]\) be the continued fraction expansion of \(\alpha\). It has been shown that the characteristic Sturmian word of slope \(\alpha\) is also the limit of the sequence: \(s_0 = b, s_1 = a, \ldots, s_{n+1} = s_n^{d_n} s_{n-1}\) for \(n > 0\).

    See Section 2.1 of [Loth02] for more details.

    INPUT:

    - `slope` - the slope of the word. It can be one of the following :
      - real number in \(]0, 1[\)
      - iterable over the continued fraction expansion of a real number in \(]0, 1[\)
    - `alphabet` - any container of length two that is suitable to build an instance of OrderedAlphabet (list, tuple, str, ...)
    - `bits` - integer (optional and considered only if `slope` is a real number) the number of bits to consider when computing the continued fraction.

    OUTPUT:

    word

    ALGORITHM:

    Let \([0, d_1 + 1, d_2, d_3, \ldots]\) be the continued fraction expansion of \(\alpha\). Then, the characteristic Sturmian word of slope \(\alpha\) is the limit of the sequence: \(s_0 = b\), \(s_1 = a\) and \(s_{n+1} = s_n^{d_n} s_{n-1}\) for \(n > 0\).

    EXAMPLES:

    From real slope:

        sage: words.CharacteristicSturmianWord(1/golden_ratio^2)
        word: 0100101001001010010100100101001001010010...
        sage: words.CharacteristicSturmianWord(4/5)
        word: 11110
        sage: words.CharacteristicSturmianWord(5/14)
        word: 01001001001001
        sage: words.CharacteristicSturmianWord(pi-3)
        word: 0000001000000100000010000001000000100000...

    From an iterator of the continued fraction expansion of a real:

        sage: def cf():
        ...     yield 0
        ...     yield 2
        ...     while True: yield 1
        sage: F = words.CharacteristicSturmianWord(cf()); F
        word: 0100101001001010010100100101001001010010...
        sage: Fib = words.FibonacciWord(); Fib
        word: 0100101001001010010100100101001001010010...
        sage: F[:10000] == Fib[:10000]
        True

    The alphabet may be specified:

        sage: words.CharacteristicSturmianWord(cf(), 'rs')
        word: rsrrsrsrrsrrsrsrrsrsrrsrrsrsrrsrrsrsrrsr...

    The characteristic sturmian word of slope \((\sqrt{3}-1)/2\):

        sage: words.CharacteristicSturmianWord((sqrt(3)-1)/2)
        word: 0100100101001001001010010010010100100101...

    The same word defined from the continued fraction expansion of \((\sqrt{3}-1)/2\):

        sage: from itertools import cycle, chain
        sage: it = chain([0], cycle([2, 1]))
        sage: words.CharacteristicSturmianWord(it)
        word: 0100100101001001001010010010010100100101...

    The first terms of the standard sequence of the characteristic sturmian word of slope \((\sqrt{3}-1)/2\):

        sage: words.CharacteristicSturmianWord([0,2])
        word: 01
        sage: words.CharacteristicSturmianWord([0,2,1])
        word: 010
        sage: words.CharacteristicSturmianWord([0,2,1,2])
        word: 01001001
        sage: words.CharacteristicSturmianWord([0,2,1,2,1])
        word: 01001001010
        sage: words.CharacteristicSturmianWord([0,2,1,2,1,2])
        word: 010010010100100100101001001001
        sage: words.CharacteristicSturmianWord([0,2,1,2,1,2,1])
        word: 0100100101001001001010010010010100100101...

    TESTS:

        sage: words.CharacteristicSturmianWord([1,1,1],'xyz')
        Traceback (most recent call last):
        ...
        TypeError: alphabet does not contain two distinct elements

        sage: words.CharacteristicSturmianWord(5/4)
        Traceback (most recent call last):
        ...
        ValueError: The argument slope (=5/4) must be in ]0,1[.

        sage: words.CharacteristicSturmianWord(1/golden_ratio^2, bits=30)
        word: 0100101001001010010100100101001001010010...
        sage: _.length()
        6765

        sage: a = words.LowerMechanicalWord(1/pi)[1:]
        sage: b = words.UpperMechanicalWord(1/pi)[1:]
        sage: c = words.CharacteristicSturmianWord(1/pi)
        sage: n = 500; a[:n] == b[:n] == c[:n]
        True

        sage: alpha = random()
        sage: c = words.CharacteristicSturmianWord(alpha)
        sage: l = words.LowerMechanicalWord(alpha)[1:]
        sage: u = words.UpperMechanicalWord(alpha)[1:]
        sage: i = 10000; j = i + 500; c[i:j] == l[i:j] == u[i:j]
        True

        sage: a, b = 207, 232
        sage: u = words.ChristoffelWord(a, b)
        sage: v = words.CharacteristicSturmianWord(a/(a+b))
        sage: u[1:-1] == v[:-2]
        True

  `ChristoffelWord`

    alias of `LowerChristoffelWord`

  `CodingOfRotationWord`(alpha, beta, x=0, alphabet=(0, 1))

    Returns the infinite word obtained from the coding of rotation of parameters \((\alpha,\beta, x)\) over the given two-letter alphabet.

    The coding of rotation corresponding to the parameters \((\alpha,\beta, x)\) is the symbolic sequence \(u = (u_n)_{n\geq 0}\) defined over the binary alphabet \(\{0, 1\}\) by \(u_n = 1\) if \(x+n\alpha\in[0, \beta[\) and \(u_n = 0\) otherwise. See [AC03].

    EXAMPLES:

        sage: alpha = 0.45
        sage: beta = 0.48
        sage: words.CodingOfRotationWord(0.45, 0.48)
        word: 1101010101001010101011010101010010101010...

        sage: words.CodingOfRotationWord(0.45, 0.48, alphabet='xy')
        word: yyxyxyxyxyxxyxyxyxyxyyxyxyxyxyxxyxyxyxyx...

    TESTS:

        sage: words.CodingOfRotationWord(0.51,0.43,alphabet=[1,0,2])
        Traceback (most recent call last):
        ...
        TypeError: alphabet does not contain two distinct elements

  `FibonacciWord`(alphabet=(0, 1), construction_method='recursive')

    Returns the Fibonacci word on the given two-letter alphabet.

    INPUT:

    - `alphabet` – any container of length two that is suitable to build an instance of OrderedAlphabet (list, tuple, str, ...)
    - `construction_method` – can be any of the following: “recursive”, “fixed point”, “function” (see below for definitions).

    Recursive construction: the Fibonacci word is the limit of the following sequence of words: \(S_0 = 0\), \(S_1 = 01\), \(S_n = S_{n-1} S_{n-2}\) for \(n \geq 2\).

    Fixed point construction: the Fibonacci word is the fixed point of the morphism: \(0 \mapsto 01\) and \(1 \mapsto 0\). Hence, it can be constructed by the following read-write process:

    1. beginning at the first letter of \(01\),
    2. if the next letter is \(0\), append \(01\) to the word;
    3. if the next letter is \(1\), append \(1\) to the word;
    4. move to the next letter of the word.

    Function: Over the alphabet \(\{1, 2\}\), the n-th letter of the Fibonacci word is \(\lfloor (n+2) \varphi \rfloor - \lfloor (n+1) \varphi \rfloor\) where \(\varphi=(1+\sqrt{5})/2\) is the golden ratio.

    EXAMPLES:

        sage: w = words.FibonacciWord(construction_method="recursive"); w
        word: 0100101001001010010100100101001001010010...

        sage: v = words.FibonacciWord(construction_method="recursive", alphabet='ab'); v
        word: abaababaabaababaababaabaababaabaababaaba...

        sage: u = words.FibonacciWord(construction_method="fixed point"); u
        word: 0100101001001010010100100101001001010010...

        sage: words.FibonacciWord(construction_method="fixed point", alphabet=[4, 1])
        word: 4144141441441414414144144141441441414414...

        sage: words.FibonacciWord([0,1], 'function')
        word: 0100101001001010010100100101001001010010...
        sage: words.FibonacciWord('ab', 'function')
        word: abaababaabaababaababaabaababaabaababaaba...

    TESTS:

        sage: from math import floor, sqrt
        sage: golden_ratio = (1 + sqrt(5))/2.0
        sage: a = golden_ratio / (1  + 2*golden_ratio)
        sage: wn = lambda n : int(floor(a*(n+2)) - floor(a*(n+1)))
        sage: f = Words([0,1])(wn); f
        word: 0100101001001010010100100101001001010010...
        sage: f[:10000] == w[:10000]
        True
        sage: f[:10000] == u[:10000] #long time
        True
        sage: words.FibonacciWord("abc")
        Traceback (most recent call last):
        ...
        TypeError: alphabet does not contain two distinct elements

  `FixedPointOfMorphism`(morphism, first_letter)

    Returns the fixed point of the morphism beginning with `first_letter`.

    A fixed point of a morphism \(\varphi\) is a word \(w\) such that \(\varphi(w) = w\).

    INPUT:

    - `morphism` – endomorphism prolongable on `first_letter`. It must be something that WordMorphism’s constructor understands (dict, str, ...).
    - `first_letter` – the first letter of the fixed point

    OUTPUT:

    The fixed point of the morphism beginning with `first_letter`

    EXAMPLES:

        sage: mu = {0:[0,1], 1:[1,0]}
        sage: tm = words.FixedPointOfMorphism(mu,0); tm
        word: 0110100110010110100101100110100110010110...
        sage: TM = words.ThueMorseWord()
        sage: tm[:1000] == TM[:1000]
        True

        sage: mu = {0:[0,1], 1:[0]}
        sage: f = words.FixedPointOfMorphism(mu,0); f
        word: 0100101001001010010100100101001001010010...
        sage: F = words.FibonacciWord(); F
        word: 0100101001001010010100100101001001010010...
        sage: f[:1000] == F[:1000]
        True

        sage: fp = words.FixedPointOfMorphism('a->abc,b->,c->','a'); fp
        word: abc

  `KolakoskiWord`(alphabet=(1, 2))

    Returns the Kolakoski word over the given alphabet and starting with the first letter of the alphabet.

    Let \(A = \{a,b\}\) be an alphabet, where \(a\) and \(b\) are two distinct positive integers. The Kolakoski word \(K_{a,b}\) over \(A\) and starting with \(a\) is the unique infinite word \(w\) such that \(w = \Delta(w)\), where \(\Delta(w)\) is the word encoding the runs of \(w\) (see `delta()` method on words for more details).

    Note that \(K_{a,b} \neq K_{b,a}\). On the other hand, the words \(K_{a,b}\) and \(K_{b,a}\) are the unique two words over \(A\) that are fixed by \(\Delta\).

    INPUT:

    - `alphabet` - (default: (1,2)) an iterable of two positive integers

    OUTPUT:

    infinite word

    EXAMPLES:

    The usual Kolakoski word:

        sage: w = words.KolakoskiWord()
        sage: w
        word: 1221121221221121122121121221121121221221...
        sage: w.delta()
        word: 1221121221221121122121121221121121221221...

    The other Kolakoski word on the same alphabet:

        sage: w = words.KolakoskiWord(alphabet = (2,1))
        sage: w
        word: 2211212212211211221211212211211212212211...
        sage: w.delta()
        word: 2211212212211211221211212211211212212211...

    It is naturally generalized to any two integers alphabet:

        sage: w = words.KolakoskiWord(alphabet = (2,5))
        sage: w
        word: 2255222225555522552255225555522222555552...
        sage: w.delta()
        word: 2255222225555522552255225555522222555552...

    TESTS:

        sage: for i in range(1,10):
        ...       for j in range(1,10):
        ...           if i != j:
        ...               w = words.KolakoskiWord(alphabet=(i,j))
        ...               assert w[:50] == w.delta()[:50]

        sage: words.KolakoskiWord((0, 2))
        Traceback (most recent call last):
        ...
        ValueError: The alphabet (=(0, 2)) must consist of two distinct positive integers

    REFERENCES:

    [Kolakoski66] | William Kolakoski, proposal 5304, American Mathematical Monthly 72 (1965), 674; for a partial solution, see “Self Generating Runs,” by Necdet Üçoluk, Amer. Math. Mon. 73 (1966), 681-2.

  `LowerChristoffelWord`

    alias of `LowerChristoffelWord`

  `LowerMechanicalWord`(alpha, rho=0, alphabet=None)

    Returns the lower mechanical word with slope \(\alpha\) and intercept \(\rho\)

    The lower mechanical word \(s_{\alpha,\rho}\) with slope \(\alpha\) and intercept \(\rho\) is defined by \(s_{\alpha,\rho}(n) = \lfloor\alpha(n+1) + \rho\rfloor - \lfloor\alpha n + \rho\rfloor\). [Loth02]

    INPUT:

    - `alpha` – real number such that \(0 \leq\alpha\leq 1\)
    - `rho` – real number (optional, default: 0)
    - `alphabet` – iterable of two elements or `None` (optional, default: `None`)

    OUTPUT:

    infinite word

    EXAMPLES:

        sage: words.LowerMechanicalWord(1/golden_ratio^2)
        word: 0010010100100101001010010010100100101001...
        sage: words.LowerMechanicalWord(1/5)
        word: 0000100001000010000100001000010000100001...
        sage: words.LowerMechanicalWord(1/pi)
        word: 0001001001001001001001000100100100100100...

    TESTS:

        sage: m = words.LowerMechanicalWord(1/golden_ratio^2)[1:]
        sage: s = words.CharacteristicSturmianWord(1/golden_ratio^2)
        sage: m[:500] == s[:500]
        True

    Check that this returns a word in an alphabet (trac ticket #10054):

        sage: words.UpperMechanicalWord(1/golden_ratio^2).parent()
        Words over {0, 1}

  `MinimalSmoothPrefix`(n)

    This function finds and returns the minimal smooth prefix of length `n`.

    See [BMP07] for a definition.

    INPUT:

    - `n` – the desired length of the prefix

    OUTPUT:

    word – the prefix

    Note

    Be patient, this function can take a really long time if asked for a large prefix.

    EXAMPLES:

        sage: words.MinimalSmoothPrefix(10)
        word: 1212212112

    REFERENCES:

    [BMP07] | S. Brlek, G. Melançon, G. Paquin, Properties of the extremal infinite smooth words, Discrete Math. Theor. Comput. Sci. 9 (2007) 33–49.

  `PalindromicDefectWord`(k=1, alphabet='ab')

    Returns the finite word \(w = a b^k a b^{k-1} a a b^{k-1} a b^{k} a\).

    As described by Brlek, Hamel, Nivat and Reuteunaer in [BHNR04], this finite word \(w\) is such that the infinite periodic word \(w^{\omega}\) have palindromic defect `k`.

    INPUT:

    - `k` – positive integer (optional, default: 1)
    - `alphabet` – iterable (optional, default: `'ab'`) of size two

    OUTPUT:

    finite word

    EXAMPLES:

        sage: words.PalindromicDefectWord(10)
        word: abbbbbbbbbbabbbbbbbbbaabbbbbbbbbabbbbbbb...

        sage: w = words.PalindromicDefectWord(3)
        sage: w
        word: abbbabbaabbabbba
        sage: w.defect()
        0
        sage: (w^2).defect()
        3
        sage: (w^3).defect()
        3

    On other alphabets:

        sage: words.PalindromicDefectWord(3, alphabet='cd')
        word: cdddcddccddcdddc
        sage: words.PalindromicDefectWord(3, alphabet=['c', 3])
        word: c333c33cc33c333c

    TESTS:

        sage: k = 25
        sage: (words.PalindromicDefectWord(k)^2).defect()
        25

    If k is negative or zero, then we get the same word:

        sage: words.PalindromicDefectWord(0)
        word: aaaaaa
        sage: words.PalindromicDefectWord(-3)
        word: aaaaaa

  `RandomWord`(n, m=2, alphabet=None)

    Returns a random word of length \(n\) over the given \(m\)-letter alphabet.

    INPUT:

    - `n` - integer, the length of the word
    - `m` - integer (default 2), the size of the output alphabet
    - `alphabet` - (default is \(\{0,1,...,m-1\}\)) any container of length m that is suitable to build an instance of OrderedAlphabet (list, tuple, str, ...)

    EXAMPLES:

        sage: words.RandomWord(10)         # random results
        word: 0110100101
        sage: words.RandomWord(10, 4)      # random results
        word: 0322313320
        sage: words.RandomWord(100, 7)     # random results
        word: 2630644023642516442650025611300034413310...
        sage: words.RandomWord(100, 7, range(-3,4))  # random results
        word: 1,3,-1,-1,3,2,2,0,1,-2,1,-1,-3,-2,2,0,3,0,-3,0,3,0,-2,-2,2,0,1,-3,2,-2,-2,2,0,2,1,-2,-3,-2,-1,0,...
        sage: words.RandomWord(100, 5, "abcde") # random results
        word: acebeaaccdbedbbbdeadeebbdeeebeaaacbadaac...
        sage: words.RandomWord(17, 5, "abcde")     # random results
        word: dcacbbecbddebaadd

    TESTS:

        sage: words.RandomWord(2,3,"abcd")
        Traceback (most recent call last):
        ...
        TypeError: alphabet does not contain 3 distinct elements

  `StandardEpisturmianWord`(directive_word)

    Returns the standard episturmian word (or epistandard word) directed by directive_word. Over a 2-letter alphabet, this function gives characteristic Sturmian words.

    An infinite word \(w\) over a finite alphabet \(A\) is said to be standard episturmian (or epistandard) iff there exists an infinite word \(x_1x_2x_3\cdots\) over \(A\) (called the directive word of \(w\)) such that \(w\) is the limit as \(n\) goes to infinity of \(Pal(x_1\cdots x_n)\), where \(Pal\) is the iterated palindromic closure function.

    Note that an infinite word is episturmian if it has the same set of factors as some epistandard word.

    See for instance [DJP01], [JP02], and [GJ07].

    INPUT:

    - `directive_word` - an infinite word or a period of a periodic infinite word

    EXAMPLES:

        sage: Fibonacci = words.StandardEpisturmianWord(Words('ab')('ab')); Fibonacci
        word: abaababaabaababaababaabaababaabaababaaba...
        sage: Tribonacci = words.StandardEpisturmianWord(Words('abc')('abc')); Tribonacci
        word: abacabaabacababacabaabacabacabaabacababa...
        sage: S = words.StandardEpisturmianWord(Words('abcd')('aabcabada')); S
        word: aabaacaabaaabaacaabaabaacaabaaabaacaabaa...
        sage: S = words.StandardEpisturmianWord(Fibonacci); S
        word: abaabaababaabaabaababaabaababaabaabaabab...
        sage: S[:25]
        word: abaabaababaabaabaababaaba
        sage: S = words.StandardEpisturmianWord(Tribonacci); S
        word: abaabacabaabaabacabaababaabacabaabaabaca...
        sage: words.StandardEpisturmianWord(123)
        Traceback (most recent call last):
        ...
        TypeError: directive_word is not a word, so it cannot be used to build an episturmian word
        sage: words.StandardEpisturmianWord(Words('ab'))
        Traceback (most recent call last):
        ...
        TypeError: directive_word is not a word, so it cannot be used to build an episturmian word

    REFERENCES:

    [JP02] | J. Justin, G. Pirillo, Episturmian words and episturmian morphisms, Theoret. Comput. Sci. 276 (2002) 281–313.

    [GJ07] | A. Glen, J. Justin, Episturmian words: a survey, Preprint, 2007, arXiv:0801.1655.

  `ThueMorseWord`(alphabet=(0, 1), base=2)

    Returns the (Generalized) Thue-Morse word over the given alphabet.

    There are several ways to define the Thue-Morse word \(t\). We use the following definition: \(t[n]\) is the sum modulo \(m\) of the digits in the given base expansion of \(n\).

    See [BmBGL07], [Brlek89], and [MH38].

    INPUT:

    - `alphabet` - (default: (0, 1) ) any container that is suitable to build an instance of OrderedAlphabet (list, tuple, str, ...)
    - `base` - an integer (default : 2) greater or equal to 2

    EXAMPLES:

    Thue-Morse word:

        sage: t = words.ThueMorseWord(); t
        word: 0110100110010110100101100110100110010110...

    Thue-Morse word on other alphabets:

        sage: t = words.ThueMorseWord('ab'); t
        word: abbabaabbaababbabaababbaabbabaabbaababba...

        sage: t = words.ThueMorseWord(['L1', 'L2'])
        sage: t[:8]
        word: L1,L2,L2,L1,L2,L1,L1,L2

    Generalized Thue Morse word:

        sage: words.ThueMorseWord(alphabet=(0,1,2), base=2)
        word: 0112122012202001122020012001011212202001...
        sage: t = words.ThueMorseWord(alphabet=(0,1,2), base=5); t
        word: 0120112012201200120112012120122012001201...
        sage: t[100:130].critical_exponent()
        10/3

    TESTS:

        sage: words.ThueMorseWord(alphabet='ab', base=1)
        Traceback (most recent call last):
        ...
        ValueError: base (=1) and len(alphabet) (=2) must be at least 2

    REFERENCES:

    [Brlek89] | Brlek, S. 1989. «Enumeration of the factors in the Thue-Morse word», Discrete Appl. Math., vol. 24, p. 83–96.

    [MH38] | Morse, M., et G. A. Hedlund. 1938. «Symbolic dynamics», American Journal of Mathematics, vol. 60, p. 815–866.

  `UpperChristoffelWord`(p, q, alphabet=(0, 1))

    Returns the upper Christoffel word of slope \(p/q\), where \(p\) and \(q\) are relatively prime non-negative integers, over the given alphabet.

    The upper Christoffel word of slope `p/q` is equal to the reversal of the lower Christoffel word of slope \(p/q\). Equivalently, if \(xuy\) is the lower Christoffel word of slope \(p/q\), where \(x\) and \(y\) are letters, then \(yux\) is the upper Christoffel word of slope \(p/q\) (because \(u\) is a palindrome).

    INPUT:

    - `alphabet` - any container of length two that is suitable to build an instance of OrderedAlphabet (list, tuple, str, ...)

    EXAMPLES:

        sage: words.UpperChristoffelWord(1,0)
        word: 1

        sage: words.UpperChristoffelWord(0,1)
        word: 0

        sage: words.UpperChristoffelWord(1,1)
        word: 10

        sage: words.UpperChristoffelWord(4,7)
        word: 10100100100

    TESTS::

        sage: words.UpperChristoffelWord(51,43,"abc")
        Traceback (most recent call last):
        ...
        ValueError: alphabet must contain exactly two distinct elements

  `UpperMechanicalWord`(alpha, rho=0, alphabet=None)

    Returns the upper mechanical word with slope \(\alpha\) and intercept \(\rho\)

    The upper mechanical word \(s'_{\alpha,\rho}\) with slope \(\alpha\) and intercept \(\rho\) is defined by \(s'_{\alpha,\rho}(n) = \lceil\alpha(n+1) + \rho\rceil - \lceil\alpha n + \rho\rceil\). [Loth02]

    INPUT:

    - `alpha` – real number such that \(0 \leq\alpha\leq 1\)
    - `rho` – real number (optional, default: 0)
    - `alphabet` – iterable of two elements or `None` (optional, default: `None`)

    OUTPUT:

    infinite word

    EXAMPLES:

        sage: words.UpperMechanicalWord(1/golden_ratio^2)
        word: 1010010100100101001010010010100100101001...
        sage: words.UpperMechanicalWord(1/5)
        word: 1000010000100001000010000100001000010000...
        sage: words.UpperMechanicalWord(1/pi)
        word: 1001001001001001001001000100100100100100...

    TESTS:

        sage: m = words.UpperMechanicalWord(1/golden_ratio^2)[1:]
        sage: s = words.CharacteristicSturmianWord(1/golden_ratio^2)
        sage: m[:500] == s[:500]
        True

    Check that this returns a word in an alphabet (trac ticket #10054):

        sage: words.UpperMechanicalWord(1/golden_ratio^2).parent()
        Words over {0, 1}

  `dual_fibonacci_tile`(n)

    Returns the \(n\)-th dual Fibonacci Tile [BmBGL09].

    EXAMPLES:

        sage: for i in range(4): words.dual_fibonacci_tile(i)
        Path: 3210
        Path: 32123032301030121012
        Path: 3212303230103230321232101232123032123210...
        Path: 3212303230103230321232101232123032123210...

  `fibonacci_tile`(n)

    Returns the \(n\)-th Fibonacci Tile [BmBGL09].

    EXAMPLES:

        sage: for i in range(3): words.fibonacci_tile(i)
        Path: 3210
        Path: 323030101212
        Path: 3230301030323212323032321210121232121010...

  `s_adic`(sequence, letters, morphisms=None)

    Returns the \(s\)-adic infinite word obtained from a sequence of morphisms applied on a letter.

    DEFINITION (from [Fogg]):

    Let \(w\) be a infinite word over an alphabet \(A = A_0\). A standard representation of \(w\) is obtained from a sequence of substitutions \(\sigma_k : A_{k+1} \to A_k\) and a sequence of letters \(a_k \in A_k\) such that:

    \[\lim_{k\to\infty} \sigma_0 \circ \sigma_1 \circ \cdots \sigma_k(a_k).\]

    Given a set of substitutions \(S\), we say that the representation is \(S\)-adic standard if the subtitutions are chosen in \(S\).

    INPUT:

    - `sequence` - An iterable sequence of indices or of morphisms. It may be finite or infinite. If `sequence` is infinite, the image of the \((i+1)\)-th letter under the \((i+1)\)-th morphism must start with the \(i\)-th letter.
    - `letters` - A letter or a sequence of letters.
    - `morphisms` - dict, list, callable or `None` (optional, default `None`) an object that maps indices to morphisms. If `None`, then `sequence` must consist of morphisms.

    OUTPUT:

    A word.

    EXAMPLES:

    Let’s define three morphisms and compute the first nested succesive prefixes of the \(s\)-adic word:

        sage: m1 = WordMorphism('e->gh,f->hg')
        sage: m2 = WordMorphism('c->ef,d->e')
        sage: m3 = WordMorphism('a->cd,b->dc')
        sage: words.s_adic([m1],'e')
        word: gh
        sage: words.s_adic([m1,m2],'ec')
        word: ghhg
        sage: words.s_adic([m1,m2,m3],'eca')
        word: ghhggh

    When the given sequence of morphism is finite, one may simply give the last letter, i.e. `'a'`, instead of giving all of them, i.e. `'eca'`:

        sage: words.s_adic([m1,m2,m3],'a')
        word: ghhggh
        sage: words.s_adic([m1,m2,m3],'b')
        word: ghghhg

    If the letters don’t satisfy the hypothesis of the algorithm (nested prefixes), an error is raised:

        sage: words.s_adic([m1,m2,m3],'ecb')
        Traceback (most recent call last):
        ...
        ValueError: The hypothesis of the algorithm used is not satisfied: the image of the 3-th letter (=b) under the 3-th morphism (=a->cd, b->dc) should start with the 2-th letter (=c).

    Let’s define the Thue-Morse morphism and the Fibonacci morphism which will be used below to illustrate more examples and let’s import the `repeat` tool from the `itertools`:

        sage: tm = WordMorphism('a->ab,b->ba')
        sage: fib = WordMorphism('a->ab,b->a')
        sage: from itertools import repeat

    Two trivial examples of infinite \(s\)-adic words:

        sage: words.s_adic(repeat(tm),repeat('a'))
        word: abbabaabbaababbabaababbaabbabaabbaababba...

        sage: words.s_adic(repeat(fib),repeat('a'))
        word: abaababaabaababaababaabaababaabaababaaba...

    A less trivial infinite \(s\)-adic word:

        sage: t = words.ThueMorseWord([tm,fib])
        sage: words.s_adic(t, repeat('a'))
        word: abbaababbaabbaabbaababbaababbaabbaababba...

    The same thing using a sequence of indices:

        sage: tmword = words.ThueMorseWord([0,1])
        sage: words.s_adic(tmword, repeat('a'), [tm,fib])
        word: abbaababbaabbaabbaababbaababbaabbaababba...

    The correspondance of the indices may be given as a dict:

        sage: words.s_adic(tmword, repeat('a'), {0:tm,1:fib})
        word: abbaababbaabbaabbaababbaababbaabbaababba...

    because dict are more versatile for indices:

        sage: tmwordTF = words.ThueMorseWord('TF')
        sage: words.s_adic(tmwordTF, repeat('a'), {'T':tm,'F':fib})
        word: abbaababbaabbaabbaababbaababbaabbaababba...

    or by a callable:

        sage: f = lambda n: tm if n == 0 else fib
        sage: words.s_adic(words.ThueMorseWord(), repeat('a'), f)
        word: abbaababbaabbaabbaababbaababbaabbaababba...

    Random infinite \(s\)-adic words:

        sage: from sage.misc.prandom import randint
        sage: def it():
        ...     while True: yield randint(0,1)
        sage: words.s_adic(it(), repeat('a'), [tm,fib])
        word: abbaabababbaababbaabbaababbaabababbaabba...
        sage: words.s_adic(it(), repeat('a'), [tm,fib])
        word: abbaababbaabbaababbaababbaabbaababbaabba...
        sage: words.s_adic(it(), repeat('a'), [tm,fib])
        word: abaaababaabaabaaababaabaaababaaababaabaa...

    An example where the sequences cycle on two morphisms and two letters:

        sage: G = WordMorphism('a->cd,b->dc')
        sage: H = WordMorphism('c->ab,d->ba')
        sage: from itertools import cycle
        sage: words.s_adic([G,H],'ac')
        word: cddc
        sage: words.s_adic(cycle([G,H]),cycle('ac'))
        word: cddcdccddccdcddcdccdcddccddcdccddccdcddc...

    The morphism \(\sigma: a\mapsto ba, b\mapsto b\) can’t satisfy the hypothesis of the nested prefixes, but one may compute arbitrarily long finite words having the limit \(\sigma^\omega(a)\):

        sage: sigma = WordMorphism('a->ba,b->b')
        sage: words.s_adic(repeat(sigma),repeat('a'))
        Traceback (most recent call last):
        ...
        ValueError: The hypothesis of the algorithm used is not satisfied: the image of the 2-th letter (=a) under the 2-th morphism (=a->ba, b->b) should start with the 1-th letter (=a).
        sage: words.s_adic([sigma],'a')
        word: ba
        sage: words.s_adic([sigma,sigma],'a')
        word: bba
        sage: words.s_adic([sigma]*3,'a')
        word: bbba
        sage: words.s_adic([sigma]*4,'a')
        word: bbbba
        sage: words.s_adic([sigma]*5,'a')
        word: bbbbba
        sage: words.s_adic([sigma]*6,'a')
        word: bbbbbba
        sage: words.s_adic([sigma]*7,'a')
        word: bbbbbbba

    The following examples illustrates an \(S\)-adic word defined over an infinite set \(S\) of morphisms \(x_h\):

        sage: x = lambda h:WordMorphism({1:[2],2:[3]+[1]*(h+1),3:[3]+[1]*h})
        sage: for h in [0,1,2,3]: print h, x(h)
        0 1->2, 2->31, 3->3
        1 1->2, 2->311, 3->31
        2 1->2, 2->3111, 3->311
        3 1->2, 2->31111, 3->3111
        sage: w = Word(lambda n : valuation(n+1, 2) ); w
        word: 0102010301020104010201030102010501020103...
        sage: s = words.s_adic(w, repeat(3), x); s
        word: 3232232232322322322323223223232232232232...
        sage: prefixe = s[:10000]
        sage: map(prefixe.number_of_factors, range(15))
        [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
        sage: [_[i+1] - _[i] for i in range(len(_)-1)]
        [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]

    TESTS:

        sage: tm = WordMorphism('a->ab,b->ba')
        sage: fib = WordMorphism('a->ab,b->a')
        sage: w = words.s_adic([fib,tm,tm,fib,tm,fib]*3,'a')
        sage: w
        word: abaaabaababaabaaababaaababaaabaababaabaa...
        sage: w.length()
        32400
        sage: w.parent()
        Words over {'a', 'b'}
        sage: type(w)
        <class 'sage.combinat.words.word.FiniteWord_iter_with_caching'>

        sage: words.s_adic([fib,tm,tm,fib,tm,fib],'aaaaaaa')
        word: abaaabaababaabaaababaaababaaabaababa

        sage: words.s_adic([0,1,0,1,0,1,0,1],'a',[tm,fib])
        word: abbaabababbaabbaababbaababbaabababbaabba...

        sage: words.s_adic([fib,fib],'bb')
        Traceback (most recent call last):
        ...
        ValueError: The hypothesis of the algorithm used is not satisfied: the image of the 2-th letter (=b) under the 2-th morphism (=a->ab, b->a) should start with the 1-th letter (=b).

    Test on different letters:

        sage: tm = WordMorphism({0:[0,1], 1:[1,0]})
        sage: fib = WordMorphism({0:[0,1], 1:[0]})
        sage: f = lambda n: tm if n == 0 else fib
        sage: words.s_adic(words.ThueMorseWord(), repeat(0), f)
        word: 0110010110011001100101100101100110010110...

    Testing the message error for the third argument:

        sage: words.s_adic(words.ThueMorseWord(), repeat(0), 5)
        Traceback (most recent call last):
        ...
        TypeError: morphisms (=5) must be None, callable or provide a __getitem__ method.

    AUTHORS:

    - Sebastien Labbe (2009-12-18): initial version

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