$\Shift(\sa{001011101}, \sa{0111}) = \sa{010010111}$ and $\Shift(\sa{001011101}, \sa{1010}) = \sa{010111010}$.
With $n=2$, $\sa{0010} \oplus \sa{1101} = \sa{0010}\,\sa{1110}$, since $\Shift(\sa{1101}, \sa{10}) = \sa{1110}$.
$\Psi(\sa{0010111011}) = \sa{0011010010}$.
For the de Bruijn word $\sa{0011}$, we have $\Psi(\sa{0011})=\sa{0010}$ and $\overline{\Psi(\sa{0011})}=\sa{1101}$. Then $\sa{0010} \oplus \sa{1101} = \sa{0010}\,\sa{1110}$, which is a de Bruijn word of length $8$.
Problem 118: Recursive Generation of de Bruijn Words |
Following Problem 117, the present problem provides another method to generate a de Bruijn word on the alphabet $\Bi = \{\sa{0},\sa{1}\}$. The method is recursive and its description requires a few preliminary definitions.
Let us first define $\Shift$. When a word $u$ occurs circularly exactly once in a word $x$, $|u|\lt |x|$, $\Shift(x,u)$ denotes the conjugate of $x$ admitting $u$ as a suffix.
Let us then define the operation $\oplus$ between two words $x,y\in \Bi^N$, $N={2^n}$, for which the suffix $u$ of length $n$ of $x$ has a unique circular factor occurrence in $y$: $x \oplus y$ denotes $x\cdot \Shift(y,u)$.
Eventually, for a binary word $w$, let $\Psi(w)$ be the binary word $v$ of length $|w|$ defined, for $0 \leq i \leq |w|-1$, by $$v[i] = (w[0]+w[1]+ \cdots +w[i]) \bmod 2.$$ Also denote by $\overline{x}$ the complement of $x$, that is, its bitwise negation.