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Identifier

St000875: Binary words ⟶ ℤ

Values

0 => 0

1 => 0

00 => 0

01 => 0

10 => 1

11 => 0

000 => 0

001 => 0

010 => 1

011 => 0

100 => 1

101 => 1

110 => 1

111 => 0

0000 => 0

0001 => 0

0010 => 1

0011 => 0

0100 => 1

0101 => 1

0110 => 1

0111 => 0

1000 => 1

1001 => 1

1010 => 2

1011 => 1

1100 => 2

1101 => 1

1110 => 1

1111 => 0

00000 => 0

00001 => 0

00010 => 1

00011 => 0

00100 => 1

00101 => 1

00110 => 1

00111 => 0

01000 => 1

01001 => 1

01010 => 2

01011 => 1

01100 => 2

01101 => 1

01110 => 1

01111 => 0

10000 => 1

10001 => 1

10010 => 1

10011 => 1

10100 => 2

10101 => 2

10110 => 1

10111 => 1

11000 => 2

11001 => 2

11010 => 2

11011 => 1

11100 => 2

11101 => 1

11110 => 1

11111 => 0

000000 => 0

000001 => 0

000010 => 1

000011 => 0

000100 => 1

000101 => 1

000110 => 1

000111 => 0

001000 => 1

001001 => 1

001010 => 2

001011 => 1

001100 => 2

001101 => 1

001110 => 1

001111 => 0

010000 => 1

010001 => 1

010010 => 1

010011 => 1

010100 => 2

010101 => 2

010110 => 1

010111 => 1

011000 => 2

011001 => 2

011010 => 2

011011 => 1

011100 => 2

011101 => 1

011110 => 1

011111 => 0

100000 => 1

100001 => 1

100010 => 1

100011 => 1

100100 => 1

100101 => 1

100110 => 1

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Description

The semilength of the longest Dyck word in the Catalan factorisation of a binary word.
Every binary word can be written in a unique way as $(\mathcal D 0)^\ell \mathcal D (1 \mathcal D)^m$, where $\mathcal D$ is the set of Dyck words. This is the Catalan factorisation, see [1, sec.9.1.2].
This statistic records the semilength of the longest Dyck word in this factorisation.

References

[1] Lothaire, M. Applied combinatorics on words MathSciNet:2165687

Code

def statistic(w):
    c = catalan_factorization(w)
    return max(len(f)/2 for f in c[0] + [c[1]] + c[2])

def first_passage_decomposition(self):
    r"""
    Decompose this word in the form (D 0)* P, where D is the family
    of complete Dyck words and P is the family of Dyck prefixes.

    The result is returned as a pair.

    EXAMPLES::

        sage: first_passage_decomposition([0,1,1,0,0,1])
        ([[]], [1, 1, 0, 0, 1])

        sage: first_passage_decomposition([0,1,1,0,0,0,1])
        ([[], [1, 1, 0, 0]), [1])
    """
    down = []
    i = 0  # start of current piece
    h = 0  # current height
    s = -1 # next minimal height
    for j, e in enumerate(self):
        if e == 1:
            h += 1
        else:
            h -= 1
        if h == s:
            down.append(self[i:j])
            i = j+1
            s -= 1
    return (down, self[i:])

def catalan_factorization(self):
    r"""
    Decompose this word in the form (D 0)* D (1 D)*, where D is the
    family of complete Dyck words.

    The result is returned as a triple.

    EXAMPLES::

        sage: catalan_factorization([])
        ([], [], [])
        sage: catalan_factorization([1])
        ([], [], [[]])
        sage: catalan_factorization([1,0,1])
        ([], [1, 0], [[]])

        sage: catalan_factorization([1,0,0,1,0,1,1,1,0,1,0,0])
        ([[1, 0]], [1, 0], [[1, 1, 0, 1, 0, 0]])

        sage: catalan_factorization([0,1,1,0,0,1,1,0,1])
        ([[]], [1, 1, 0, 0], [[1, 0], []])

    """
    down, rest = first_passage_decomposition(self)
    up, middle = first_passage_decomposition([1-e for e in rest[::-1]])
    return down, [1-e for e in middle[::-1]], [[1-e for e in p[::-1]] for p in up[::-1]]

Created

Jul 03, 2017 at 08:46 by Martin Rubey

Updated

Jan 20, 2018 at 20:12 by Martin Rubey