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Identifier

Mp00268: Binary words —zeros to flag zeros⟶ Binary words

Images

0 => 0

1 => 1

00 => 10

01 => 00

10 => 01

11 => 11

000 => 010

001 => 110

010 => 100

011 => 000

100 => 101

101 => 001

110 => 011

111 => 111

0000 => 1010

0001 => 0010

0010 => 0110

0011 => 1110

0100 => 0100

0101 => 1100

0110 => 1000

0111 => 0000

1000 => 0101

1001 => 1101

1010 => 1001

1011 => 0001

1100 => 1011

1101 => 0011

1110 => 0111

1111 => 1111

00000 => 01010

00001 => 11010

00010 => 10010

00011 => 00010

00100 => 10110

00101 => 00110

00110 => 01110

00111 => 11110

01000 => 10100

01001 => 00100

01010 => 01100

01011 => 11100

01100 => 01000

01101 => 11000

01110 => 10000

01111 => 00000

10000 => 10101

10001 => 00101

10010 => 01101

10011 => 11101

10100 => 01001

10101 => 11001

10110 => 10001

10111 => 00001

11000 => 01011

11001 => 11011

11010 => 10011

11011 => 00011

11100 => 10111

11101 => 00111

11110 => 01111

11111 => 11111

000000 => 101010

000001 => 001010

000010 => 011010

000011 => 111010

000100 => 010010

000101 => 110010

000110 => 100010

000111 => 000010

001000 => 010110

001001 => 110110

001010 => 100110

001011 => 000110

001100 => 101110

001101 => 001110

001110 => 011110

001111 => 111110

010000 => 010100

010001 => 110100

010010 => 100100

010011 => 000100

010100 => 101100

010101 => 001100

010110 => 011100

010111 => 111100

011000 => 101000

011001 => 001000

011010 => 011000

011011 => 111000

011100 => 010000

011101 => 110000

011110 => 100000

011111 => 000000

100000 => 010101

100001 => 110101

100010 => 100101

100011 => 000101

100100 => 101101

100101 => 001101

>>> Load all 2584 entries. <<<

click to show experimental identities (only identities of compositions of up to three maps are shown)

Description

Return a binary word of the same length, such that the number of occurrences of

$10$ in the word obtained by prepending the reverse of the complement equals the number of

$0$ s in the original word.
For example, the image of the word

$w=1\dots1$ is

$1\dots1$ , because

$w$ has no zeros, and

$1\dots1$ is the only word such that prepending the reverse of its complement has no occurrence of the factor

$10$ .
On the other hand,

$0\dots0$ must be mapped to

$10\dots10$ if the length is even, and

$010\dots10$ if it is odd.

Properties

bijective, graded

Sage code

def mapping(w):
    n = len(w)
    zeros = [-1] + [i for i in range(n) if w[i] == 0]
    zeros = [zeros[i] - zeros[i-1] - 1 for i in range(1, len(zeros))]
    result = []
    letter = 1
    for b in zeros:
        result += [letter]*b
        if letter == 0:
            result += [0, 1]
        letter = 1 - letter
    result += [letter]*(n-len(result))
    return Words([0,1])(reversed(result))

Weight

Created

Dec 01, 2022 at 22:36 by Martin Rubey

Updated

Dec 01, 2022 at 22:36 by Martin Rubey