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Identifier

St001372: Binary words ⟶ ℤ

Values

0 => 0

1 => 1

00 => 0

01 => 1

10 => 1

11 => 2

000 => 0

001 => 1

010 => 1

011 => 2

100 => 1

101 => 2

110 => 2

111 => 3

0000 => 0

0001 => 1

0010 => 1

0011 => 2

0100 => 1

0101 => 1

0110 => 2

0111 => 3

1000 => 1

1001 => 2

1010 => 1

1011 => 3

1100 => 2

1101 => 3

1110 => 3

1111 => 4

00000 => 0

00001 => 1

00010 => 1

00011 => 2

00100 => 1

00101 => 1

00110 => 2

00111 => 3

01000 => 1

01001 => 1

01010 => 2

01011 => 2

01100 => 2

01101 => 2

01110 => 3

01111 => 4

10000 => 1

10001 => 2

10010 => 1

10011 => 3

10100 => 1

10101 => 2

10110 => 2

10111 => 4

11000 => 2

11001 => 3

11010 => 2

11011 => 4

11100 => 3

11101 => 4

11110 => 4

11111 => 5

000000 => 0

000001 => 1

000010 => 1

000011 => 2

000100 => 1

000101 => 1

000110 => 2

000111 => 3

001000 => 1

001001 => 1

001010 => 2

001011 => 2

001100 => 2

001101 => 2

001110 => 3

001111 => 4

010000 => 1

010001 => 1

010010 => 2

010011 => 2

010100 => 2

010101 => 1

010110 => 3

010111 => 3

011000 => 2

011001 => 2

011010 => 3

011011 => 2

011100 => 3

011101 => 3

011110 => 4

011111 => 5

100000 => 1

100001 => 2

100010 => 1

100011 => 3

100100 => 1

100101 => 2

100110 => 2

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Description

The length of a longest cyclic run of ones of a binary word.
Consider the binary word as a cyclic arrangement of ones and zeros. Then this statistic is the length of the longest continuous sequence of ones in this arrangement.

Code

def statistic(w):
    if len(w) == 0:
        return 0

    if w[0] == 1:
        c = Composition(w.delta())[::2]
    else:
        c = Composition(w.delta())[1::2]

    if len(c) > 1 and w[0] == w[-1]:
        c[-1] += c[0]
        c = c[1:]

    return max(c + [0])

Created

Mar 19, 2019 at 22:50 by Martin Rubey

Updated

Jun 22, 2019 at 13:50 by Martin Rubey