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BW Transform of Thue-Morse Words

kgoto
March 07, 2022

BW Transform of Thue-Morse Words

Chapter94 in the book 125 problems in Text Algorithms.

kgoto

March 07, 2022
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  1. BW Transform of Thue-Morse Words
    Chapter 94 in 125 problems book.
    Speaker: @kgoto
    Book Reading Seminar #18, March 7, 2022
    1
    1

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  2. We call BW-string of
    BW Transform
    The Burrows-Wheeler transform of length- string is the word composed
    of the last letters of the sorted conjugates of .
    BW(x) n x
    x
    BW(x) x
    2
    2

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  3. Thue-Morse Words
    Thue-Morse morphism from
    -th Thue-Morse words
    for
    μ {a, b}∗
    μ(a) = ab
    μ(b) = ba
    n τ

    n
    τ

    =
    0
    a
    τ

    =
    1
    ab
    τ

    =
    2
    abba
    τ

    =
    3
    abbabaab
    τ

    =
    n
    μ(τ

    )
    n−1
    n > 0
    Today's topic

    The recursive structure of BW-string of Thue-Morse words
    ,

    where ,
    BW (τ

    ) =
    n+1
    b

    a
    n−1 BW (τ

    )
    n
    n−1
    =
    a b =
    b a 3
    3

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  4. Observation
    : matrix whose rows are sorted rotations of
    is split to three parts
    : top rows
    : middle rows, which corresponds to
    : bottom rows
    S

    n+1 2 ×
    n+1 2n+1 τ

    n+1
    S

    n+1
    T

    n+1 2n−1
    M

    n+1 2n μ(S

    )
    n
    B

    n+1 2n−1
    4
    4

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  5. Observation
    and occur in , respectively.
    in produces in , then such occurs in
    is composed of the rotation by splitting the middle of such
    If , then
    If , then
    Since thre is no occurrence of , then does not contain , and the bottom
    of starts with
    Top of starts with
    a b 2n−1 τ

    n
    b τ

    n ba τ

    n+1 ba 2n−1 τ

    n+1
    T

    n+1 ba
    s <
    1 s

    2 μ(s

    ) <
    1 μ(s

    )
    2
    s = bt aμ(t)b < μ(t)ba < baμ(t)
    bbb T

    n+1 ababa
    T

    n+1 abaa
    M

    n+1 abab
    5
    5

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  6. Observation
    Symmetically, is composed of the rotation by splitting the middle
    of produced by in
    starts with and ends with
    starts with and ends with
    The last characters of are flipped characters of ones of
    B

    n+1
    ab a τ

    n
    T

    n+1 a b
    B

    n+1 b a
    M

    n+1 S

    n+1
    Conclusion
    ,

    where ,
    BW (τ

    ) =
    n+1
    b

    a
    n−1 BW (τ

    )
    n
    n−1
    =
    a b =
    b a
    6
    6

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  7. Reference
    Brlek, S., Frosini, A., Mancini, I., Pergola, E. and Rinaldi, S., 2019, July.
    Burrows-Wheeler transform of words defined by morphisms. In
    International Workshop on Combinatorial Algorithms (pp. 393-404).
    Springer, Cham.

    https://link.springer.com/chapter/10.1007/978-3-030-25005-8_32
    7
    7

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