AUTHORS:
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...
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
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
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
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'>
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:
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)
doctest:...: DeprecationWarning: the argument 'bits' is deprecated
See http://trac.sagemath.org/14567 for details.
word: 0100101001001010010100100101001001010010...
sage: _.length()
+Infinity
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
alias of LowerChristoffelWord
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
Returns the Fibonacci word on the given two-letter alphabet.
INPUT:
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:
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
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:
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
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:
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. |
alias of LowerChristoffelWord
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:
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}
This function finds and returns the minimal smooth prefix of length n.
See [BMP07] for a definition.
INPUT:
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. |
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:
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
Returns a random word of length \(n\) over the given \(m\)-letter alphabet.
INPUT:
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
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:
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. |
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:
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. |
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:
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
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:
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}
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...
Returns the \(n\)-th Fibonacci Tile [BmBGL09].
EXAMPLES:
sage: for i in range(3): words.fibonacci_tile(i)
Path: 3210
Path: 323030101212
Path: 3230301030323212323032321210121232121010...
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:
Given a set of substitutions \(S\), we say that the representation is \(S\)-adic standard if the subtitutions are chosen in \(S\).
INPUT:
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: