On Greedy Algorithms for Binary de Bruijn Sequences

TL;DR

This paper introduces a versatile greedy algorithm, GPO, for generating binary de Bruijn sequences, unifying many existing algorithms under a common framework and highlighting their underlying connections.

Contribution

The paper presents the GPO algorithm, a general framework that encompasses and extends previous greedy algorithms for binary de Bruijn sequences.

Findings

Most known greedy algorithms are special cases of GPO

GPO can generate a wide range of binary de Bruijn sequences

Identification of specific feedback functions and initial states enhances sequence generation

Abstract

We propose a general greedy algorithm for binary de Bruijn sequences, called Generalized Prefer-Opposite (GPO) Algorithm, and its modifications. By identifying specific feedback functions and initial states, we demonstrate that most previously-known greedy algorithms that generate binary de Bruijn sequences are particular cases of our new algorithm.

Tables

Table 1. Table 1 : All 16 16 16 de Bruijn Sequences in ℱ 1 ​ ( 6 ; h , 4 ) subscript ℱ 1 6 ℎ 4 \mathcal{F}_{1}(6;h,4) with h ​ ( x 0 , x 1 , x 2 , x 3 ) ℎ subscript 𝑥 0 subscript 𝑥 1 subscript 𝑥 2 subscript 𝑥 3 h(x_{0},x_{1},x_{2},x_{3}) in Equation  2 .

$𝐛$	The Resulting de Bruijn Sequence
$000011$	$\left(0000001010010111100001110110011111101010110111001000100110001101\right)$
$000110$	$\left(0000001010010111100001100111011000100011111101010110111001001101\right)$
$001101$	$\left(0000001010010111100001101100111011100100010011000111111010101101\right)$
$011010$	$\left(0000001010010111100001101010110011101101110010001001100011111101\right)$
$110100$	$\left(0000001010100111011001100011111101011011100100101111000011010001\right)$
$101001$	$\left(0000001011110000110100111011001100010001111110101011011100100101\right)$
$010010$	$\left(0000001011110000110100100010011101100110001111110101011011100101\right)$
$100101$	$\left(0000001011001110110111001001100011111101011110000110100101010001\right)$
$001011$	$\left(0000001010000110100101100111011011100100010011000111111010101111\right)$
$010111$	$\left(0000001010000110100101110110011100100010011000111111010101101111\right)$
$101111$	$\left(0000001010000110100101111110110011101010110111001000100110001111\right)$
$011110$	$\left(0000001010000110100101111011001110101011011100100010011000111111\right)$
$111100$	$\left(0000001010000110100101111001110110010001001100011111101010110111\right)$
$111000$	$\left(0000001010000110100101111000100011101100111111010101101110010011\right)$
$110000$	$\left(0000001000101010000111011001111110101101110010011000110100101111\right)$
$100001$	$\left(0000001110110011111101011011100100110001101001011110000100010101\right)$

Table 2. Table 2: More Families of Feedback Functions with 𝐛 = 𝟎 n 𝐛 superscript 0 𝑛 {\mathbf{b}}={\mathbf{0}}^{n} for GPO.

No.	$f$	Condition(s)
$1$	$x_{n-1}+x_t\cdot x_{n-1}$	for $n\ge 3$
$2$	$x_{n-1}+\prod _{t=1}^{n-1}{x}_t$	$n\ge 3$
$3$	$x_t\cdot x_{t+1}+x_{t+1}\cdot x_{n-1}$	for $n\ge 4$
$4$	$x_k\cdot x_{n-1}+x_{\ell }\cdot x_{n-1}$	for $n\ge 4$

Table 3. Table 3 : The family ℱ 4 ​ ( 6 ) subscript ℱ 4 6 \mathcal{F}_{4}(6) obtained from Prefer-No .

$t$	The Resulting de Bruijn Sequence
$1$	$\left(0000001010100101101011101000100110110010000110001110011110111111\right)$
$2$	$\left(0000001101101011001100011101110010111101001111110000101010001001\right)$
$3$	$\left(0000001110111001111010111111000110110100110010110000101010001001\right)$
$4$	$\left(0000001111011111100111010111000110110100110010110000101010001001\right)$
$5$	$\left(0000001111110111100111010111000110110100110010110000101010001001\right)$

Table 4. Table 4 : Some de Bruijn Sequences in ℱ 5 ​ ( 6 ) subscript ℱ 5 6 \mathcal{F}_{5}(6) : one from each primitive polynomial g ​ ( x ) 𝑔 𝑥 g(x) of degree 1 ≤ m ≤ 5 1 𝑚 5 1\leq m\leq 5 .

$m$	$g\left(x\right)$	The Resulting de Bruijn Sequence
$1$	$11$	$\left(0000001111110101011010010100010111011001001101111001110001100001\right)$
$2$	$111$	$\left(0000001101010110111111001001010011110100010000101110001110110011\right)$
$3$	$1011$	$\left(0000001010111001001011010001101100010000111111010100110011101111\right)$
	$1101$	$\left(0000001111011101001001110001011000010001100101011111100110110101\right)$
$4$	$10011$	$\left(0000001011011100111111010010011010101111000100001110110010100011\right)$
	$11001$	$\left(0000001100010010111111001111010101100100011100001010011011101101\right)$
$5$	$100101$	$\left(0000001110010001001011001111110001101110101000010101111011010011\right)$
	$101001$	$\left(0000001100101101111010100010101110110001111110011010010000100111\right)$
	$101111$	$\left(0000001101100111101001010111000101101010001110111111001001100001\right)$
	$110111$	$\left(0000001011101101010011110001110011011111101000100101011000011001\right)$
	$111011$	$\left(0000001001100011010100100010111111011001110000111100101011011101\right)$
	$111101$	$\left(0000001000110010011111101110001010110100001110101001011110011011\right)$

Table 5. Table 5: GPO List for n = 3 𝑛 3 n=3

No.	$f$ for $𝐒=\left(00010111\right)$	$𝐛\in$	$f$ for $𝐒=\left(00011101\right)$	$𝐛\in$
$1$	$x_1\cdot x_2+x_1+x_2+1$	$\left\{010\right\}$	$x_1\cdot x_2+x_1+x_2+1$	$\{100,001\}$
$2$	$x_1+1$	$\{100,001\}$	$x_2\cdot x_1+x_1+1$	$\left\{111\right\}$
$3$	$x_2+1$	$\left\{101\right\}$	$x_1+1$	$\{011,110\}$
$4$	$x_2\cdot x_1+1$	$\{011,110\}$	$x_2+1$	$\left\{010\right\}$
$5$	$x_2\cdot x_1+x_2$	$\left\{000\right\}$	$x_2\cdot x_1+1$	$\left\{101\right\}$
$6$	$1$	$\left\{111\right\}$	$0$	$\left\{000\right\}$

Table 6. Table 6: Order of appearance in a run of the GPO algorithm.

${𝐒}_5$ with $𝐛={𝐮}_0=11000$
No.	State	No.	State	No.	State	No.	State
$1$	${𝐮}_0$	$9$	$10100\in T_1$	$17$	$11011\in T_5$	$25$	${𝐮}_2$
$2$	$\widehat{{𝐮}_1}\in T_2$	$10$	$01001\in T_3$	$18$	$\widehat{{𝐮}_5}\in T_0$	$26$	${𝐮}_3$
$3$	${\mathrm{𝟎}}^5\in T_2$	$11$	$10011\in T_6$	$19$	$01100\in T_0$	$27$	$\widehat{{𝐮}_4}\in T_4$
$4$	${\mathrm{𝟎}}^{4}1\in T_2$	$12$	$00111\in T_6$	$20$	$\widehat{{𝐮}_0}\in T_3$	$28$	$10101\in T_4$
$5$	$\widehat{{𝐮}_2}\in T_6$	$13$	$\widehat{{𝐮}_6}\in T_7$	$21$	$10010\in T_3$	$29$	${𝐮}_4$
$6$	$00110\in T_0$	$14$	${\mathrm{𝟏}}^5\in T_7$	$22$	$\widehat{{𝐮}_3}\in T_1$	$30$	${𝐮}_5$
$7$	$01101\in T_5$	$15$	${\mathrm{𝟏}}^{4}0\in T_7$	$23$	$01000\in T_1$	$31$	${𝐮}_6$
$8$	$11010\in T_4$	$16$	$\widehat{{𝐮}_7}\in T_5$	$24$	${𝐮}_1$	$32$	${𝐮}_7$

Table 7. Table 7: GPO List for n = 4 𝑛 4 n=4 ; Part One of Two

No.	$f$ for Sequence $\left(0000111101100101\right)$	$𝐛\in$	$f$ for Sequence $\left(0000101111010011\right)$	$𝐛\in$
$1$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_1+x_2\cdot x_3+x_2+x_3+1$	$\{1000,0001\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_1+x_2\cdot x_3+x_2+x_3+1$	$\left\{0010\right\}$
$2$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_1+x_2+1$	$\left\{1100\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_1+x_2+x_3+1$	$\left\{0100\right\}$
$3$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1+x_2\cdot x_3+x_2+1$	$\{0111,1110\}$	$x_1\cdot x_2+x_1\cdot x_3+x_1+x_2\cdot x_3+x_3+1$	$\left\{1010\right\}$
$4$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_1+x_2\cdot x_3+x_2+1$	$\left\{1011\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1+x_2\cdot x_3+x_3+1$	$\left\{1101\right\}$
$5$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_2\cdot x_3+x_2+x_3+1$	$\left\{0010\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_3+x_2+1$	$\left\{0011\right\}$
$6$	$x_1\cdot x_2+x_1+x_2\cdot x_3+x_2+1$	$\left\{1111\right\}$	$x_1\cdot x_2\cdot x_3+x_1+x_2\cdot x_3+x_3+1$	$\left\{1111\right\}$
$7$	$x_1\cdot x_2+x_1+x_2+1$	$\left\{0011\right\}$	$x_1\cdot x_2+x_1+x_2+1$	$\{1000,0001\}$
$8$	$x_1\cdot x_3+x_1+x_2+1$	$\left\{0110\right\}$	$x_1\cdot x_3+x_1+x_3+1$	$\left\{0101\right\}$
$9$	$x_1+x_2\cdot x_3+x_2+1$	$\left\{1101\right\}$	$x_1+x_2\cdot x_3+x_3+1$	$\{0111,1110\}$
$10$	$x_2\cdot x_3+x_2+x_3+1$	$\left\{0100\right\}$	$x_1\cdot x_2\cdot x_3+x_1+x_3+1$	$\left\{1011\right\}$
$11$	$x_1\cdot x_2+x_1\cdot x_3+x_3+1$	$\left\{0101\right\}$	$x_1\cdot x_2+x_2+x_3+1$	$\left\{1001\right\}$
$12$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_3+1$	$\left\{1010\right\}$	$x_1\cdot x_2+x_1\cdot x_3+x_2+1$	$\left\{0110\right\}$
$13$	$x_2+1$	$\left\{0011\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_2+1$	$\left\{1100\right\}$
$14$	$0$	$\left\{0000\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_2\cdot x_3+x_3$	$\left\{0000\right\}$
	$f$ for Sequence $\left(0000101001111011\right)$		$f$ for Sequence $\left(0000111101011001\right)$
$1$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_1+x_2\cdot x_3+x_2+x_3+1$	$\left\{0100\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_1+x_2\cdot x_3+x_2+1$	$\left\{1111\right\}$
$2$	$x_1\cdot x_2+x_1+x_2\cdot x_3+x_2+x_3+1$	$\left\{0010\right\}$	$x_1\cdot x_2+x_1+x_2\cdot x_3+x_2+x_3+1$	$\{1000,0001\}$
$3$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_1+x_2+1$	$\{1000,0001\}$	$x_1\cdot x_2+x_1\cdot x_3+x_1+x_2\cdot x_3+x_2+1$	$\{0111,1110\}$
$4$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_3+x_2+1$	$\left\{1101\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_1+x_2+1$	$\left\{0011\right\}$
$5$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_2\cdot x_3+x_2+1$	$\left\{1111\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_1+x_2\cdot x_3+x_2+1$	$\left\{1101\right\}$
$6$	$x_1\cdot x_3+x_1+x_3+1$	$\left\{1010\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_1+x_2\cdot x_3+1$	$\left\{0110\right\}$
$7$	$x_1\cdot x_2\cdot x_3+x_1+x_3+1$	$\left\{0101\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_2\cdot x_3+x_2+x_3+1$	$\left\{0100\right\}$
$8$	$x_2\cdot x_3+x_2+x_3+1$	$\left\{1001\right\}$	$x_1+x_2\cdot x_3+x_2+1$	$\left\{1010\right\}$
$9$	$x_1\cdot x_2+x_2\cdot x_3+x_2+1$	$\left\{1011\right\}$	$x_1\cdot x_2+x_1\cdot x_3+x_1+1$	$\left\{1011\right\}$
$10$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_2+1$	$\left\{0110\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1+1$	$\left\{0101\right\}$
$11$	$x_1\cdot x_3+x_2\cdot x_3+x_2+1$	$\{0111,1110\}$	$x_1\cdot x_3+x_1+x_2\cdot x_3+1$	$\left\{1100\right\}$
$12$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_2+1$	$\left\{0011\right\}$	$x_1\cdot x_2+x_1\cdot x_3+x_3+1$	$\left\{0010\right\}$
$13$	$x_2+1$	$\left\{1100\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_2\cdot x_3+1$	$\left\{1001\right\}$
$14$	$x_2\cdot x_3+x_3$	$\left\{0000\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3$	$\left\{0000\right\}$
	$f$ for Sequence $\left(0000110111100101\right)$		$f$ for Sequence $\left(0000101101001111\right)$
$1$	$x_1\cdot x_2+x_1\cdot x_3+x_1+x_2+x_3+1$	$\{1000,0001\}$	$x_1\cdot x_2+x_1\cdot x_3+x_1+x_2+x_3+1$	$\left\{0010\right\}$
$2$	$x_1\cdot x_2+x_1\cdot x_3+x_1+x_2\cdot x_3+x_2+1$	$\left\{1100\right\}$	$x_1\cdot x_3+x_1+x_2\cdot x_3+x_2+x_3+1$	$\left\{0100\right\}$
$3$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1+x_2\cdot x_3+x_2+1$	$\left\{0011\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1+x_2\cdot x_3+x_2+1$	$\{1000,0001\}$
$4$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_1+x_2\cdot x_3+x_2+1$	$\{0111,1110\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_1+x_3+1$	$\left\{1010\right\}$
$5$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_3+x_3+1$	$\left\{0101\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_1+x_2\cdot x_3+x_3+1$	$\left\{0101\right\}$
$6$	$x_1\cdot x_3+x_1+x_2\cdot x_3+x_2+1$	$\left\{1111\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_2\cdot x_3+x_2+x_3+1$	$\left\{1001\right\}$
$7$	$x_1\cdot x_2+x_1+x_2+1$	$\left\{0110\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_3+x_2+1$	$\{0111,1110\}$
$8$	$x_1\cdot x_3+x_1+x_2+1$	$\left\{1011\right\}$	$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_3+x_2+1$	$\left\{1111\right\}$
$9$	$x_1\cdot x_2\cdot x_3+x_1+x_2+1$	$\left\{1101\right\}$	$x_1\cdot x_2+x_1+x_3+1$	$\left\{1101\right\}$
$10$	$x_1\cdot x_3+x_2+x_3+1$	$\left\{0010\right\}$	$x_1+x_2\cdot x_3+x_3+1$	$\left\{1011\right\}$
$11$	$x_1\cdot x_2\cdot x_3+x_2+x_3+1$	$\left\{0100\right\}$	$x_1\cdot x_2\cdot x_3+x_1+x_3+1$	$\left\{0110\right\}$
$12$	$x_1\cdot x_2\cdot x_3+x_2\cdot x_3+x_2+1$	$\left\{1001\right\}$	$x_1\cdot x_2+x_1\cdot x_3+x_2+1$	$\left\{0011\right\}$
$13$	$x_1\cdot x_2+x_2\cdot x_3+x_3+1$	$\left\{1010\right\}$	$x_1\cdot x_3+x_2\cdot x_3+x_2+1$	$\left\{1100\right\}$
$14$	$x_1\cdot x_2\cdot x_3+x_2\cdot x_3$	$\left\{0000\right\}$	$x_1\cdot x_3+x_3$	$\left\{0000\right\}$
	$f$ for Sequence $\left(0000101001101111\right)$		$f$ for Sequence $\left(0000100111101011\right)$
$1$	$x_1\cdot x_2+x_1\cdot x_3+x_1+x_2+x_3+1$	$\left\{0100\right\}$	$x_1\cdot x_2+x_1+x_2\cdot x_3+x_2+x_3+1$	$\left\{0100\right\}$
$2$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1+x_2+x_3+1$	$\left\{0010\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_2\cdot x_3+x_2+x_3+1$	$\left\{1001\right\}$
$3$	$x_1\cdot x_2+x_1\cdot x_3+x_1+x_2\cdot x_3+x_2+1$	$\{1000,0001\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_3+x_2+1$	$\left\{1010\right\}$
$4$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_1+x_2\cdot x_3+x_3+1$	$\left\{1010\right\}$	$x_1\cdot x_2\cdot x_3+x_1+x_3+1$	$\left\{0010\right\}$
$5$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_2\cdot x_3+x_2+1$	$\left\{1111\right\}$	$x_1\cdot x_3+x_1+x_2\cdot x_3+1$	$\{1000,0001\}$
$6$	$x_1+x_2\cdot x_3+x_3+1$	$\left\{0101\right\}$	$x_1\cdot x_2+x_2\cdot x_3+x_2+1$	$\left\{1101\right\}$
$7$	$x_1\cdot x_2\cdot x_3+x_2+x_3+1$	$\left\{1001\right\}$	$x_1\cdot x_2\cdot x_3+x_2\cdot x_3+x_2+1$	$\{0111,1110\}$
$8$	$x_1\cdot x_2+x_1\cdot x_3+x_2+1$	$\left\{1101\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_2\cdot x_3+1$	$\left\{1100\right\}$
$9$	$x_1\cdot x_2+x_2\cdot x_3+x_2+1$	$\{0111,1110\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_2+x_3$	$\left\{0000\right\}$
$10$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_2+1$	$\left\{1011\right\}$	$x_2\cdot x_3+x_2+1$	$\left\{1111\right\}$
$11$	$x_1\cdot x_3+x_2\cdot x_3+x_2+1$	$\left\{0011\right\}$	$x_2+1$	$\left\{0011\right\}$
$12$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_2+1$	$\left\{0110\right\}$	$x_1\cdot x_3+1$	$\left\{0101\right\}$
$13$	$x_1\cdot x_2\cdot x_3+x_2\cdot x_3+x_2+1$	$\left\{1100\right\}$	$x_2\cdot x_3+1$	$\left\{0110\right\}$
$14$	$x_1\cdot x_2\cdot x_3+x_3$	$\left\{0000\right\}$	$x_1\cdot x_2\cdot x_3+1$	$\left\{1011\right\}$

Table 8. Table 8: GPO List for n = 4 𝑛 4 n=4 ; Part Two of Two

No.	$f$ for Sequence $\left(0000110101111001\right)$	$𝐛\in$	$f$ for Sequence $\left(0000100110101111\right)$	$𝐛\in$
$1$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1+x_2+x_3+1$	$\{1000,0001\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1+x_2+x_3+1$	$\left\{0100\right\}$
$2$	$x_1\cdot x_2+x_1\cdot x_3+x_1+x_2\cdot x_3+x_2+1$	$\left\{0011\right\}$	$x_1+x_2\cdot x_3+x_3+1$	$\left\{0010\right\}$
$3$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_1+x_2+1$	$\left\{0110\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_1+1$	$\{1000,0001\}$
$4$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_1+x_2\cdot x_3+1$	$\left\{1011\right\}$	$x_1\cdot x_3+x_2+x_3+1$	$\left\{1001\right\}$
$5$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_3+x_3+1$	$\left\{0010\right\}$	$x_1\cdot x_2+x_1\cdot x_3+x_2+1$	$\left\{1010\right\}$
$6$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_1+1$	$\left\{1111\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_2+1$	$\left\{1101\right\}$
$7$	$x_1\cdot x_3+x_1+x_2+1$	$\left\{1101\right\}$	$x_1\cdot x_2\cdot x_3+x_2\cdot x_3+x_2+1$	$\left\{0011\right\}$
$8$	$x_1\cdot x_2\cdot x_3+x_1+x_2+1$	$\left\{1010\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_2\cdot x_3+1$	$\{0101,0000\}$
$9$	$x_1\cdot x_2+x_1\cdot x_3+x_1+1$	$\{0111,1110\}$	$x_2+1$	$\{0110,1100\}$
$10$	$x_1\cdot x_2+x_1+x_2\cdot x_3+1$	$\left\{0101\right\}$	$x_2\cdot x_3+1$	$\left\{1011\right\}$
$11$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_1+1$	$\left\{1100\right\}$	$x_1\cdot x_2\cdot x_3+1$	$\{0111,1110\}$
$12$	$x_1\cdot x_3+x_2+x_3+1$	$\left\{0100\right\}$	$1$	$\left\{1111\right\}$
$13$	$x_1\cdot x_2+1$	$\left\{1001\right\}$
$14$	$x_1\cdot x_3+x_2\cdot x_3$	$\left\{0000\right\}$
	$f$ for Sequence $\left(0000111100101101\right)$		$f$ for Sequence $\left(0000101111001101\right)$
$1$	$x_1\cdot x_3+x_1+x_2\cdot x_3+x_2+x_3+1$	$\{1000,0001\}$	$x_1\cdot x_3+x_1+x_2\cdot x_3+x_2+x_3+1$	$\left\{0010\right\}$
$2$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1+x_2\cdot x_3+x_2+1$	$\left\{1100\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_1+x_3+1$	$\left\{0101\right\}$
$3$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_2\cdot x_3+x_2+x_3+1$	$\left\{0100\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1+x_2\cdot x_3+x_3+1$	$\{0111,1110\}$
$4$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_3+x_3+1$	$\left\{1011\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_3+x_3+1$	$\left\{1001\right\}$
$5$	$x_1\cdot x_2\cdot x_3+x_1+x_2\cdot x_3+x_2+1$	$\left\{1111\right\}$	$x_1\cdot x_2+x_1+x_2\cdot x_3+x_3+1$	$\left\{111\right\}$
$6$	$x_1+x_2\cdot x_3+x_2+1$	$\{0111,1110\}$	$x_1\cdot x_2\cdot x_3+x_1+x_2+1$	$\{1000,0001\}$
$7$	$x_1\cdot x_2\cdot x_3+x_1+x_2+1$	$\left\{0011\right\}$	$x_1\cdot x_2+x_1+x_3+1$	$\left\{1011\right\}$
$8$	$x_1\cdot x_2\cdot x_3+x_2+x_3+1$	$\left\{0010\right\}$	$x_1+x_2\cdot x_3+x_3+1$	$\left\{1100\right\}$
$9$	$x_1\cdot x_3+x_2\cdot x_3+x_2+1$	$\left\{1001\right\}$	$x_1\cdot x_2+x_1\cdot x_3+x_2+1$	$\left\{0100\right\}$
$10$	$x_1\cdot x_2+x_1\cdot x_3+x_3+1$	$\left\{0110\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_2\cdot x_3+1$	$\left\{0110\right\}$
$11$	$x_1\cdot x_2+x_2\cdot x_3+x_3+1$	$\left\{0101\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_2\cdot x_3+1$	$\left\{1010\right\}$
$12$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_3+1$	$\left\{1101\right\}$	$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_3+x_3$	$\left\{0000\right\}$
$13$	$x_3+1$	$\left\{1010\right\}$	$x_1\cdot x_2+1$	$\left\{0011\right\}$
$14$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2$	$\left\{0000\right\}$	$x_2\cdot x_3+1$	$\left\{1101\right\}$
	$f$ for Sequence $\left(0000110010111101\right)$		$f$ for Sequence $\left(0000101100111101\right)$
$1$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_1+x_2+x_3+1$	$\{1000,0001\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_1+x_2+x_3+1$	$\left\{0010\right\}$
$2$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_3+x_3+1$	$\{0111,1110\}$	$x_1\cdot x_2+x_1\cdot x_3+x_1+x_2\cdot x_3+x_3+1$	$\left\{0101\right\}$
$3$	$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_3+x_3+1$	$\left\{1111\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1+x_2\cdot x_3+x_3+1$	$\left\{1011\right\}$
$4$	$x_1\cdot x_2+x_1+x_2+1$	$\left\{1100\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_3+x_2+1$	$\left\{0100\right\}$
$5$	$x_1+x_2\cdot x_3+x_2+1$	$\left\{0011\right\}$	$x_1+x_2\cdot x_3+x_2+1$	$\{1000,0001\}$
$6$	$x_1\cdot x_2\cdot x_3+x_1+x_2+1$	$\left\{0110\right\}$	$x_1\cdot x_2+x_1+x_3+1$	$\left\{0110\right\}$
$7$	$x_1\cdot x_2+x_2+x_3+1$	$\left\{0100\right\}$	$x_1\cdot x_2\cdot x_3+x_1+x_3+1$	$\left\{1100\right\}$
$8$	$x_2\cdot x_3+x_2+x_3+1$	$\left\{0010\right\}$	$x_1\cdot x_2+x_1\cdot x_3+x_3+1$	$\left\{1001\right\}$
$9$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_2+1$	$\left\{1001\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_2\cdot x_3+1$	$\left\{0011\right\}$
$10$	$x_1\cdot x_2+x_1\cdot x_3+x_3+1$	$\left\{1011\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_3$	$\left\{0000\right\}$
$11$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_3+1$	$\left\{0101\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+1$	$\left\{1111\right\}$
$12$	$x_1\cdot x_3+x_2\cdot x_3+x_3+1$	$\left\{1101\right\}$	$x_1\cdot x_2+1$	$\{0111,1110\}$
$13$	$x_1\cdot x_2\cdot x_3+x_2\cdot x_3+x_3+1$	$\left\{1010\right\}$	$x_1\cdot x_3+1$	$\left\{1010\right\}$
$14$	$x_1\cdot x_2+x_2\cdot x_3$	$\left\{0000\right\}$	$x_1\cdot x_2\cdot x_3+1$	$\left\{1101\right\}$
	$f$ for Sequence $\left(0000110100101111\right)$		$f$ for Sequence $\left(0000111101001011\right)$
$1$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_1+x_2\cdot x_3+1$	$\left\{1101\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_3+x_2+1$	$\left\{1001\right\}$
$2$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_2\cdot x_3+x_2+x_3+1$	$\left\{0010\right\}$	$x_1+x_2\cdot x_3+x_2+1$	$\left\{0100\right\}$
$3$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_3+x_3+1$	$\left\{1100\right\}$	$x_1\cdot x_2\cdot x_3+x_1+x_3+1$	$\{1000,0001\}$
$4$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_2\cdot x_3+x_3+1$	$\left\{1111\right\}$	$x_1\cdot x_2+x_1\cdot x_3+x_1+1$	$\left\{1101\right\}$
$5$	$x_1\cdot x_2\cdot x_3+x_1+x_2+1$	$\left\{0100\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1+1$	$\left\{1010\right\}$
$6$	$x_1+x_2\cdot x_3+x_3+1$	$\{1000,0001\}$	$x_1\cdot x_3+x_1+x_2\cdot x_3+1$	$\left\{0011\right\}$
$7$	$x_1\cdot x_2+x_1+x_2\cdot x_3+1$	$\left\{1010\right\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_1+1$	$\{0111,1110\}$
$8$	$x_1\cdot x_3+x_1+x_2\cdot x_3+1$	$\left\{0110\right\}$	$x_1\cdot x_2+x_2+x_3+1$	$\left\{0010\right\}$
$9$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_1+1$	$\left\{0011\right\}$	$x_1\cdot x_2+x_1\cdot x_3+x_3+1$	$\left\{1100\right\}$
$10$	$x_1\cdot x_2+x_1\cdot x_3+x_2+1$	$\left\{1001\right\}$	$x_1\cdot x_3+x_2\cdot x_3+x_3+1$	$\left\{1011\right\}$
$11$	$x_1\cdot x_3+x_2\cdot x_3+x_3+1$	$\{0111,1110\}$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_3+1$	$\left\{0110\right\}$
$12$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_3+x_3+1$	$\left\{1011\right\}$	$x_1\cdot x_2\cdot x_3+x_2\cdot x_3+x_3+1$	$\left\{0101\right\}$
$13$	$x_1\cdot x_2\cdot x_3+x_1\cdot x_2+x_1\cdot x_3+x_2$	$\left\{0000\right\}$	$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_3+x_2$	$\left\{0000\right\}$
$14$	$x_3+1$	$\left\{0101\right\}$	$x_1\cdot x_3+x_1+1$	$\left\{1111\right\}$

Equations

$$c_1,c_2,\dots ,c_{n-1},\overline{f(c_0,c_1,\dots ,c_{n-1})}$$

$$f(x_0,x_1,\dots ,x_{n-1}):=h(x_{n-m},x_{n-m+1},\dots ,x_{n-1})$$

$$h(x_0,x_1,x_2,x_3):=1+x_0+x_2\cdot x_3+x_1\cdot x_2\cdot x_3.$$

$$|{\mathcal{F}}_1(n;h,m)|=\{\begin{array}{ll}{2}^m-2& \text{mbox}ifn=m+1,\\ 2^m& \text{mbox}ifn>m+1.\end{array}$$

$$f(x_0,x_1,\dots ,x_{n-1}):=1+\prod _{i=t}^{n-1}{x}_i$$

$$\mathbf{b}=$$

$$\to$$

$$f(x_0,x_1,\dots ,x_{n-1})=1+x_{n-3}+x_{n-1}+x_{n-3}\cdot x_{n-2}+x_{n-2}\cdot x_{n-1}+x_{n-3}\cdot x_{n-2}\cdot x_{n-1}$$

$$\mathbf{b}=$$

$$\to$$

$$f(x_0,x_1,\dots ,x_{n-1}):=\prod _{j=t}^{n-1}{x}_j=x_t\cdot x_{t+1}\cdot \dots \cdot x_{n-1}.$$

$$\mathbf{b}$$

$$\to$$

$$f(x_0,x_1,\dots ,x_{n-1}):=x_{n-m}+a_1{x}_{n-m+1}+\dots +a_{m-1}{x}_{n-1}.$$

$$|{\mathcal{F}}_5(n)|=\frac{\phi \left(2^{n-1}-1\right)}{n-1}(2^{n-1}-2)+\sum _{m=1}^{n-2}\frac{\phi \left(2^m-1\right)}{m}(2^m-1).$$

$$\tilde{f}(x_0,x_1,\dots ,x_{n-1}):=f(x_0,x_1,\dots ,x_{n-1})+1=1+x_{n-m}+a_1{x}_{n-m+1}+\dots +a_{m-1}{x}_{n-1}$$

$$f(x_0,x_1,\dots ,x_{n-1}):=x_{n-3}+x_{n-2}+x_{n-3}\cdot x_{n-2}+x_{n-3}\cdot x_{n-2}\cdot x_{n-1}$$

$$\mathbf{b}=$$

$$\to$$

$$g(\mathbf{v})=1\text{mbox}for\mathbf{v}\in \{111,011,001,000\}\text{mbox}andg(\mathbf{v})=0\text{mbox}for\mathbf{v}\in \{110,101,100,010\}.$$

$${\mathbf{u}}_0,\widehat{{\mathbf{u}}_1},\widehat{{\mathbf{u}}_3},\widehat{{\mathbf{u}}_6},\widehat{{\mathbf{u}}_0},\widehat{{\mathbf{u}}_5},\widehat{{\mathbf{u}}_4},{\mathbf{u}}_1,\widehat{{\mathbf{u}}_2},{\mathbf{u}}_2,{\mathbf{u}}_3,{\mathbf{u}}_4,{\mathbf{u}}_5,{\mathbf{u}}_6,\widehat{{\mathbf{u}}_7},{\mathbf{u}}_7.$$

$${\mathbf{S}}_5:=(11000001101001111101100100010101).$$

$$V_1$$

$$V_2$$

$$V_3$$

$${\mathbf{u}}_i,{\mathbf{u}}_{i+1},\dots ,{\mathbf{u}}_{2^m-1},{\mathbf{u}}_0,\dots ,{\mathbf{u}}_{i-1}.$$

$${\mathbf{u}}_{\ell }:=u_0,u_1,\dots ,u_{m-1},\dots ,u_{n-1}⟷{\mathbf{m}}_{\ell }:=u_0,u_1,\dots ,u_{m-1}.$$

$$v_{n-m+1},\dots ,v_{n-1},h(v_{n-m},v_{n-m+1},\dots ,v_{n-1})+1.$$

$$T_{\widehat{{\mathbf{m}}_{i+2}}},T_{\widehat{{\mathbf{m}}_{i+3}}},\dots ,T_{\widehat{{\mathbf{m}}_{i+1+n-m}}}.$$

$$\bigcup_{\ell\in\left\llbracket{i+1,j}\right\rrbracket}\{T_{\widehat{{\mathbf{m}}_{\ell+1}}},T_{\widehat{{\mathbf{m}}_{\ell+2}}},\ldots,T_{\widehat{{\mathbf{m}}_{\ell+n-m}}}\}.$$

$$j+n-m-(i+2)+1=j-i+(n-m-1)$$

$$u_0\ne u_1=u_2=\dots =u_{n-1}\ne u_n.$$

$${\mathbf{m}}_i=\widehat{{\mathbf{m}}_{i+1}}\text{mbox}and{\mathbf{m}}_{i+1}=\widehat{{\mathbf{m}}_{i+2}}⟹{\mathbf{m}}_i={\mathbf{m}}_{i+2},$$

Full text

1 Introduction

There are $2^{2^{n-1}-n}$ binary de Bruijn sequence of order $n$ [1]. Each has period $N=2^{n}$ in which every binary $n$-tuple occurs exactly once.

Relatively fast generation of a few de Bruijn sequences has a long history in the literature. Any primitive polynomial of degree $n$ generates a maximal length sequence (also known as $m$-sequence) that can be easily modified to a de Bruijn sequence [2]. There are at least three different algorithms to generate a Ford sequence, which is the lexicograpically smallest de Bruijn sequence. They were treated, respectively, by Fredricksen and Maiorana in [3], by Fredricksen in [4], and by Ralston in [5]. There are frameworks for construction via successor rules. More recent, necklace-based, successor rules and their co-necklace variants can be found in several works, e.g., [6], [7], [8], and [9].

Many, often less efficient, methods that produce much larger numbers of de Bruijn sequences are known. The majority of them work by joining smaller cycles into de Bruijn sequences. Prominent examples include methods to join cycles generated by pure cycling register or pure summing registers given by Fredricksen in [10], Etzion and Lempel in [11], and Huang in [12]. Examples of recent works on more general component cycles are the work of Li et al. [13] and Chang et al. [14].

Here we focus on greedy algorithms to generate de Bruijn sequences. There have also been many known ones, e.g., Prefer-One and Prefer-Same in [4] and Prefer-Opposite in [15]. They have since been generalized using the notion of preference functions in [16]. A paper of Wang et al. [17], which was presented at SETA 2018, discusses a greedy algorithm based on the feedback function $x_{n-2}+x_{n-1}$ for $n\geq 3$.

In general one can come up with numerous feedback functions to generate de Bruijn sequences using greedy algorithms. It had unfortunately been rather challenging to confirm which ones of these functions actually work. We show how to circumvent this by specifying feedback functions that come with a certificate of correctness.

We state the Generalized Prefer-Opposite (GPO) Algorithm and prove sufficient conditions on the feedback functions to ensure that the algorithm indeed generates de Bruijn sequences. This leads us to numerous classes of special feedback functions that can be used to generate de Bruijn sequences via the algorithm. As a corollary, we show that Prefer-One, Prefer-Zero, as well as others based on preference functions are special cases of the GPO algorithm.

To include even more classes of feedback functions, we put forward suitable modifications of the GPO Algorithm. Several new families of de Bruijn sequences can then be generated in a greedy manner.

After this introduction comes preliminary notions and results in Section 2. We introduce the GPO Algorithm in Section 3. Sufficient conditions for the algorithm to produce de Bruijn sequences are then proved. The three subsections describe three respective families of de Bruijn sequences, with some analysis on their properties. Section 4 shows how to modify the GPO Algorithm when the sufficient condition is not met. The modification results in numerous instances of successful construction of de Bruijn sequences. Three more families of such sequences are then showcased. We end with a conclusion that, for any de Bruijn sequence ${\mathbf{S}}$ of order $n>2$ and any $n$-string initial state ${\mathbf{b}}$, one can always find a feedback function $f$ that the GPO Algorithm can take as an input to produce ${\mathbf{S}}$.

2 Preliminaries

Let $0<k<\ell$ be integers. We denote $\{0,1,2,\ldots,\ell\}$ by \left\llbracket{\ell}\right\rrbracket and $\{k,k+1,\ldots,\ell\}$ by \left\llbracket{k,\ell}\right\rrbracket. An $n$-stage shift register is a clock-regulated circuit with $n$ consecutive storage units. Each of the units holds a bit and, as the clock pulses, the bit is shifted to the next stage in line. The output is a new bit $s_{n}$ based on the $n$ bits ${\mathbf{s}}_{0}=s_{0},\ldots,s_{n-1}$ called the initial state. The corresponding feedback function $f(x_{0},\ldots,x_{n-1})$ is the Boolean function that, on input ${\mathbf{s}}_{0}$, outputs $s_{n}$.

With a function $f(x_{0},x_{1},\ldots,x_{n-1})$, a feedback shift register (FSR) outputs a sequence ${\mathbf{s}}=s_{0},s_{1},\ldots,s_{n},\ldots$ satisfying $s_{n+\ell}=f(s_{\ell},s_{\ell+1},\ldots,s_{\ell+n-1})$ for $\ell=0,1,2,\ldots$. Let $N$ be the smallest positive integer satisfying $s_{i+N}=s_{i}$ for all $i\geq 0$. Then ${\mathbf{s}}$ is $N$-periodic or with period $N$ and one writes ${\mathbf{s}}=(s_{0},s_{1},s_{2},\ldots,s_{N-1})$. We call ${\mathbf{s}}_{i}=s_{i},s_{i+1},\ldots,s_{i+n-1}$ the $i$-th state of ${\mathbf{s}}$ and states ${\mathbf{s}}_{i-1}$ and ${\mathbf{s}}_{i+1}$ the predecessor and successor of ${\mathbf{s}}_{i}$, respectively.

An $n$-string $c_{0},c_{1},\ldots,c_{n-1}$ is often written concisely as $c_{0}c_{1}\ldots c_{n-1}$, especially in algorithms and tables. The complement $\overline{c}$ of $c\in{\mathbb{F}}_{2}$ is $1+c$. The complement of a string is the corresponding string produced by taking the complement of each element. The string of zeroes of length $\ell$ is denoted by ${\mathbf{0}}^{\ell}$. Analogously, ${\mathbf{1}}^{\ell}$ denotes the string of ones of length $\ell$. Given a state ${\mathbf{c}}=c_{0},c_{1},\ldots,c_{n-1}$, we say that $\overline{{\mathbf{c}}}:=\overline{c_{0}},c_{1},\ldots,c_{n-1}$ and $\widehat{{\mathbf{c}}}:=c_{0},c_{1},\ldots,\overline{c_{n-1}}$ are, respectively, the conjugate state and the companion state of ${\mathbf{c}}$.

A function $f(x_{0},x_{1},\ldots,x_{n-1})=x_{0}+g(x_{1},\ldots,x_{n-1})$, where $g$ is a Boolean function, is said to be non-singular. An FSR with non-singular feedback function will generate periodic sequences [2, p. 116]. Otherwise $f$ is said to be singular and some states would have two distinct preimages under $f$.

The state graph of the FSR with feedback function $f(x_{0},x_{1},\ldots,x_{n-1})$ is a directed graph ${\mathcal{G}}_{f}$ whose vertices are all of the $n$-stage states. There is a directed edge from a state ${\mathbf{u}}:=u_{0},u_{1},\ldots,u_{n-1}$ to a state ${\mathbf{v}}:=u_{1},u_{2},\ldots,f(u_{0},u_{1},\ldots,u_{n-1})$. We call ${\mathbf{u}}$ a child of ${\mathbf{v}}$ and ${\mathbf{v}}$ the parent of ${\mathbf{u}}$. We allow ${\mathbf{u}}={\mathbf{v}}$, in which case ${\mathcal{G}}_{f}$ contains a loop. A leaf in ${\mathcal{G}}_{f}$ is a vertex with no child, i.e., a leaf has outdegree $1$ and indegree [math]. We say that a vertex ${\mathbf{w}}$ is a descendant of ${\mathbf{u}}$ is there is a directed path that starts at ${\mathbf{w}}$ and ends at ${\mathbf{u}}$, which in turn is called an ancestor of ${\mathbf{w}}$. A rooted tree $T_{f,{\mathbf{b}}}$ in ${\mathcal{G}}_{f}$ is the largest tree in ${\mathcal{G}}_{f}$ in which one vertex ${\mathbf{b}}$ has been designated the root with the edge that emanates out of ${\mathbf{b}}$ removed from ${\mathcal{G}}_{f}$. In this work the orientation is towards the root ${\mathbf{b}}$, i.e., $T_{f,{\mathbf{b}}}$ is an in-tree or an arborescence converging to a root as defined in [18, Chapter 6].

3 Generalized Prefer-Opposite Algorithm

This section describes the GPO algorithm and proves sufficient conditions on the feedback function to guarantee that the algorithm generates de Bruijn sequences.

For a given feedback function $f(x_{0},x_{1},\ldots,x_{n-1})$ and initial state ${\mathbf{b}}=b_{0},b_{1},\ldots,b_{n-1}$, the GPO algorithm is presented here as Algorithm 1. Notice that it does not always produce de Bruijn sequences. The following Theorem 1 provides sufficient conditions on the feedback function and the initial state for the GPO algorithm to generate de Bruijn sequence(s) of order $n$.

Theorem 1**.**

The GPO Algorithm, on input a function $f(x_{0},x_{1},\ldots,x_{n-1})$ and an initial state ${\mathbf{b}}=b_{0},b_{1},\ldots,b_{n-1}$, generates a binary de Bruijn sequence of order $n$ if the state graph ${\mathcal{G}}_{f}$ of the FSR satisfies the following two conditions.

All of the states, except for the leaves, have exactly two children. 2. 2.

There is a unique directed path from any state ${\mathbf{v}}$ to ${\mathbf{b}}$.

Before proving the theorem, we establish an important lemma that will be frequently invoked.

Lemma 2**.**

Let the state graph ${\mathcal{G}}_{f}$ of the FSR with feedback function $f$ satisfy the first condition in the statement of Theorem 1. Let the state ${\mathbf{s}}$ be a vertex with two children. Let the GPO Algorithm starts with an initial state ${\mathbf{b}}\neq{\mathbf{s}}$, with ${\mathbf{b}}$ not a leaf. By the time the algorithm visits ${\mathbf{s}}$ it must have visited both children of ${\mathbf{s}}$.

Proof.

Suppose that ${\mathbf{b}}$ is a leaf. To visit ${\mathbf{b}}$ for the second time, one of its possible predecessors, say ${\mathbf{u}}=u_{0},u_{1},\ldots,u_{n-1}$, must have been visited before. At this point, ${\mathbf{b}}$ must be of the form $u_{1},\ldots,u_{n-1},\overline{f(u_{0},u_{1},\ldots,u_{n-1})}$. By the rule of the algorithm, since ${\mathbf{b}}$ had appeared at the very beginning, the successor of ${\mathbf{u}}$ must then be $u_{1},\ldots,u_{n-1},f(u_{0},u_{1},\ldots,u_{n-1})$. This rules ${\mathbf{b}}$ out from being visited again. The algorithm will not terminate since Line 11 in Algorithm 1 is always satisfied and the iteration continues. Thus, to ensure that the output is a periodic sequence, ${\mathbf{b}}$ must not be a leaf.

Consider the state ${\mathbf{s}}$ and let ${\mathbf{v}}$ and ${\mathbf{w}}$ be the two children of ${\mathbf{s}}$ in ${\mathcal{G}}_{f}$. Since ${\mathbf{s}}$ is not a leaf, one of the children, say ${\mathbf{w}}$, must have been its predecessor in the sequence ${\mathbf{S}}$ produced by the GPO Algorithm thus far. Suppose, for a contradiction, that the algorithm visits ${\mathbf{s}}$ without having visited ${\mathbf{v}}$. Consider the other possible successor of both ${\mathbf{v}}$ and ${\mathbf{w}}$, which is the companion state $\widehat{{\mathbf{s}}}$ of ${\mathbf{s}}$. Since ${\mathbf{v}}$ has not been visited, then $\widehat{{\mathbf{s}}}$ must have not been visited as well. The fact that ${\mathbf{s}}$ is the actual successor of ${\mathbf{w}}$ contradicts the rule of the algorithm since the successor of ${\mathbf{w}}$ ought to have been $\widehat{{\mathbf{s}}}$. ∎

Proof of Theorem 1 1**.**

We first show that it is impossible for Algorithm 1 to visit any state twice before it reaches the initial state ${\mathbf{b}}$ the second time. For a contradiction, suppose that ${\mathbf{u}}\neq{\mathbf{b}}$ is the first state to be visited twice. The assignment rule precludes visiting the state of the form

[TABLE]

twice. Hence, ${\mathbf{u}}$ must be of the form $c_{1},\ldots,c_{n-1},f(c_{0},c_{1},\ldots,c_{n-1})$, implying that ${\mathbf{u}}$ has as a child the state $c_{0},c_{1},\ldots,c_{n-1}$ in ${\mathcal{G}}_{f}$ and ${\mathbf{u}}$ is not a leaf. By Lemma 2, when ${\mathbf{u}}$ is first visited by the algorithm, its two children must have been visited. The second time ${\mathbf{u}}$ is visited, one of its two children must have also been visited twice. This contradicts the assumption that ${\mathbf{u}}$ is the first vertex to have been visited twice.

The algorithm continues until it visits ${\mathbf{b}}$ again. Any other state, say ${\mathbf{v}}\neq{\mathbf{u}}$, has two possible successors in ${\mathbf{S}}$. At least one of the two must have not been visited yet since their two possible predecessors in ${\mathbf{S}}$ include ${\mathbf{v}}$.

Now it suffices to show that Algorithm 1 visits all states. Since there is unique directed path to ${\mathbf{b}}$ from any other state, all of the other states are descendants of ${\mathbf{b}}$. By Lemma 2, by the time the algorithm revisits ${\mathbf{b}}$, it must have visited ${\mathbf{b}}$’s two children, even if one of the two is ${\mathbf{b}}$ itself. The same lemma implies that the respective child(ren) of these two children of ${\mathbf{b}}$ must have been visited beforehand. By repeated applications of the lemma we confirm that all of the descendants of ${\mathbf{b}}$ must have been covered in the running of Algorithm 1.

The initial state in Prefer-One is ${\mathbf{0}}^{n}$. The next bit of the sequence is $1$ if the newly formed $n$-stage state has not previously appeared in the sequence. If otherwise, then the next bit is [math]. For $n=4$, for example, the algorithm outputs ${\mathbf{S}}:=(0000~{}1111~{}0110~{}0101)$. Prefer-Zero, introduced by Martin in [19], works in a similar manner, interchanging [math] and $1$ in the rule of Prefer-One. Knuth nicknamed Prefer-Zero the granddaddy construction of de Bruijn sequences [20]. For $n=4$, the granddaddy outputs $(1111~{}0000~{}1001~{}1010)$, which is the complement of ${\mathbf{S}}$.

Corollary 3**.**

Let $n\geq 2$. Prefer-One is the special case with $f(x_{0},x_{1}\ldots,x_{n-1})=0$ and ${\mathbf{b}}={\mathbf{0}}^{n}$. Prefer-Zero is the special case with $f(x_{0},x_{1}\ldots,x_{n-1})=1$ and ${\mathbf{b}}={\mathbf{1}}^{n}$.

To generate de Bruijn sequences by Algorithm 1 it suffices to find families of feedback functions and initial states that satisfy the conditions in Theorem 1. There are numerous such combinations with only three families of them explicitly discussed in the present work.

3.1 Family One

Our next result gives a family of de Bruijn sequences that can be greedily constructed from de Bruijn sequences of lower orders. It provides an alternative proof to [16, Theorem 2.4] in the binary case.

Theorem 4**.**

Let $h(x_{0},x_{1},\ldots,x_{m-1})$ be a feedback function whose FSR generates a de Bruijn sequence ${\mathbf{S}}_{m}$ of order $m$. We fix a positive integer $n>m\geq 2$. The GPO Algorithm generates the family $\mathcal{F}_{1}(n;h,m)$ of de Bruijn sequences of order $n$ on input any string of length $n$ in ${\mathbf{S}}_{m}$ as the initial state ${\mathbf{b}}$ and

[TABLE]

Proof.

It suffices to prove that ${\mathcal{G}}_{f}$ with $f$ in Equation (1) satisfies the two conditions in Theorem 1. Since the coefficient of $x_{0}$ in $f$ is [math], any non-leaf state has two children. It is immediate to confirm that ${\mathcal{G}}_{f}$ contains a directed cycle whose vertices are the $n$-stage states of ${\mathbf{S}}_{m}$. All other states are vertices in the trees whose respective roots are the states of the above cycle. Since there is a unique directed path from any state to some state ${\mathbf{u}}$ in the cycle and there is a unique directed path from ${\mathbf{u}}$ to ${\mathbf{b}}$, the second condition is satisfied. ∎

Example 1**.**

${\mathbf{S}}_{4}:=(0000~{}1101~{}0010~{}1111)$* is de Bruijn with*

[TABLE]

Table 1 provides the $16$ distinct de Bruijn sequences in the $\mathcal{F}_{1}(6;h,4)$ family. $\square$

When $n>m+1$, distinct initial states yield distinct de Bruijn sequences of order $n$, yielding $2^{m}$ distinct de Bruijn sequences for a valid $f$. But if $n=m+1$, then there are collisions in the output. Two pairs of initial states, namely $({\mathbf{1}}^{m}0,0{\mathbf{1}}^{m})$ and $({\mathbf{0}}^{m}1,1{\mathbf{0}}^{m})$, generate the same sequence. We formalize the observation in Theorem 5 and supply its proof in the appendix.

Theorem 5**.**

The number of distinct de Bruijn sequences in $\mathcal{F}_{1}(n;h,m)$ with $n>m\geq 2$ is

[TABLE]

3.2 Family Two

Here is another family of de Bruijn sequences that can be greedily constructed.

Theorem 6**.**

Given $n$ and an integer $0<t<n$, let the feedback function be

[TABLE]

and the initial state ${\mathbf{b}}$ be any $n$-stage state of the sequence $(0,{\mathbf{1}}^{n-t})$. Then the GPO Algorithm generates the $\mathcal{F}_{2}(n)$ family of de Bruijn sequences of order $n$.

Proof.

First, since the coefficient of $x_{0}$ in $f$ as given in Equation (4) is [math], any non-leaf vertex has two children in ${\mathcal{G}}_{f}$. Second, we confirm that ${\mathcal{G}}_{f}$ contains a cycle whose vertices are all of the $n$-stage states of $(0,{\mathbf{1}}^{n-t})$. All other states are vertices in the trees whose respective roots are the states of this cycle. Checking that the two conditions in Theorem 1 are satisfied is in the end rather straightforward. ∎

Remark 1**.**

We make two notes on the $\mathcal{F}_{2}(n)$ family. First, the produced sequence is the same as the output of Prefer-Same when $t=n-1$. Our description here simplifies the one given in [4] on the said algorithm. Second, we exclude $t=0$ since the resulting ${\mathcal{G}}_{f}$ does not satisfy the two conditions in Theorem 1, although, with the initial state ${\mathbf{1}}^{n}$, the GPO Algorithm generates the same de Bruijn sequence as the one produced by Prefer-Zero.

Example 2**.**

Figure 1 presents ${\mathcal{G}}_{f}$ for $f=x_{1}\cdot x_{2}\cdot x_{3}+1$. Running Algorithm 1 on input $f$ and ${\mathbf{b}}=1110$ yields the following sequence of states

[TABLE]

The states in gray follow the rule of assignment in Line 10 of Algorithm 1 since the respective preferred states designated by Line 8 have appeared earlier. The resulting de Bruijn sequence is $(1110~{}0001~{}0011~{}0101)$. $\square$

3.3 Family Three

For order $n\geq 4$ we have yet another family of de Bruijn sequences that follows from the GPO Algorithm.

Theorem 7**.**

Given $n\geq 4$, let the feedback function be

[TABLE]

and the initial state ${\mathbf{b}}$ be any $n$-stage state of the sequence $(1110)$. Then the GPO Algorithm generates the $\mathcal{F}_{3}(n)$ family of de Bruijn sequences.

Proof.

The state graph ${\mathcal{G}}_{f}$ of the FSR with feedback function $f$ in Equation 5 also satisfies the conditions in Theorem 1. The coefficient of $x_{0}$ in $f$ is [math], implying that each non-leaf state has two children. The graph ${\mathcal{G}}_{f}$ contains a cycle whose vertices are the $n$-stage states of the periodic sequence $(1,1,1,0)$. All other $n$-stage states are divided into disjoint trees whose respective roots are the vertices of the cycle. ∎

Example 3**.**

The state graph ${\mathcal{G}}_{f}$ for $n=4$ and $f$ taken from Equation (5) is in Figure 2.

Starting from ${\mathbf{b}}=1101$ yields the sequence of states

[TABLE]

As before, the states in gray are governed by Line 10 of Algorithm 1 since the respective preferred states have already appeared. The output is $(1101~{}0011~{}0000~{}1011)$. $\square$

We have discussed three families of de Bruijn sequences that the GPO Algorithm can construct by carefully selecting the feedback functions and the initial states that combine to satisfy the requirements in Theorem 1. Table 2 lists four more verified families of feedback functions with ${\mathbf{b}}={\mathbf{0}}^{n}$ for $n\geq 5$. The proof, for each family, follows the usual line of arguments to check that the two conditions in Theorem 1 are met. The details are omitted here for brevity. Many more combinations of feedback functions and initial states ${\mathbf{b}}$ are still there to be discovered. We leave this as an open direction to pursue.

4 Modifications of the GPO Algorithm

The GPO Algorithm is not guaranteed to generate de Bruijn sequences when the pair $(f,{\mathbf{b}})$ fails to satisfy the conditions in Theorem 1. This section shows that some modifications to the GPO Algorithm may turn such a pair into a viable choice.

4.1 Prefer No

For a given $n$ and t\in\left\llbracket{n-1}\right\rrbracket let the feedback function be

[TABLE]

Note that ${\mathcal{G}}_{f}$ with $f$ in Equation (6) does not satisfy the conditions in Theorem 1 since ${\mathcal{G}}_{f}$ is not connected, being divided into two trees with respective roots ${\mathbf{0}}^{n}$ and ${\mathbf{1}}^{n}$. We modify the GPO Algorithm into the Prefer-No Algorithm, presented here as Algorithm 2, and prove that it generates a de Bruijn sequence for each $0\leq t<n$. When $t=n-1$, the resulting sequence is the same with the one produced by Prefer-Opposite in [15]. Taking $t\in\{0,1\}$ produces the same sequence as the output of Prefer-One.

Theorem 8**.**

The Prefer-No Algorithm generates a de Bruijn sequence for each tuple $(t,n)$ satisfying $n\geq 2$ and $1\leq t<n$. We call the resulting set of sequences the family $\mathcal{F}_{4}(n)$.

Proof.

Using the same method as in the proof of Theorem 1 we can confirm that the first state that Algorithm 2 visits twice must be ${\mathbf{0}}^{n}$.

Now, note that Algorithm 2 does not terminate at $1{\mathbf{0}}^{n-1}$ since it can still proceed to ${\mathbf{0}}^{n}$. For the same reason, it does not terminate at $0{\mathbf{1}}^{n-1}$ or ${\mathbf{1}}^{n}$. Any other state, say ${\mathbf{u}}$, has two possible successors. At least one of them must have not been visited yet since their two possible predecessors include ${\mathbf{u}}$. Thus, Algorithm 2 will not stop until it reaches ${\mathbf{0}}^{n}$ the second time around.

Next, we show that the algorithm visits every state. Before terminating, the algorithm must have visited $1{\mathbf{0}}^{n-1}$. Each non-leaf descendant of $1{\mathbf{0}}^{n-1}$ has two children. By a repeated application of Lemma 2 we confirm that the algorithm must have visited ${\mathbf{0}}^{n}$ and all of its descendants. These include ${\mathbf{1}}^{n-1}0$, which is the successor of ${\mathbf{1}}^{n}$. Hence, the algorithm must have visited both ${\mathbf{1}}^{n}$ and $0{\mathbf{1}}^{n-1}$ by following its rule. Next, each non-leaf descendant of $0{\mathbf{1}}^{n-1}$ has two predecessors. Again, we use Lemma 2 to verify that the algorithm visits all of the descendants of $0{\mathbf{1}}^{n-1}$. Thus, all possible states are covered and the resulting sequence is indeed de Bruijn. ∎

Example 4**.**

Figure 3 gives the state graph ${\mathcal{G}}_{f}$ with $f=x_{2}\cdot x_{3}$, obtained by letting $n=4$ and $t=2$ in Equation (6).

Running Algorithm 2 produces the sequence of states

[TABLE]

The underlined state $1111$ is based on the rule in Line 7. The entries in gray follows the rule in Line 13. Taking $n=6$, one obtains the $5$ distinct de Bruijn sequences in Table 3. $\square$

4.2 From Primitive Polynomials

We can use any primitive polynomial as the main ingredient. Given a primitive polynomial $g(x)=1+a_{1}x+\ldots+a_{m-1}x^{m-1}+x^{m}\in{\mathbb{F}}_{2}[x]$ of degree $m<n$, let

[TABLE]

The state graph ${\mathcal{G}}_{f}$ with $f$ given in Equation (7) is divided into two disjoint components. For example, the state graph for $f(x_{0},x_{1},x_{2},x_{3})=x_{1}+x_{2}$ is in Figure 4.

The Prim-Poly Algorithm, presented here as Algorithm 3, generalizes the example by modifying the GPO Algorithm. It produces another family, which we call $\mathcal{F}_{5}(n)$, of de Bruijn sequences by using the function in Equation 7 and suitable initial states when $g(x)$ ranges over all primitive polynomials of degrees $1$ to $n-1$. The construction in [17] is the special case when $g(x)=1+x+x^{2}$.

Theorem 9**.**

Let $g(x)$ be a given primitive polynomial of degree $m$. Algorithm 3 generates de Bruijn sequences for $1\leq m<n$ from the feedback function $f$ in Equation (7).

Proof.

Suppose that Algorithm 3 takes an initial state ${\mathbf{b}}$, which is a nonzero $n$-stage state of the $m$-sequence ${\mathbf{m}}$ with primitive characteristic polynomial $g(x)$ of degree $m<n$. We can use the similar argument to the proof of Theorem 1 to confirm that ${\mathbf{b}}$ must be the first state to be visited twice.

The algorithm’s rule of assignment forces ${\mathbf{0}}^{n-1}1$ to be the successor of ${\mathbf{0}}^{n}$, which in turn is the successor of $1{\mathbf{0}}^{n-1}$. Hence, the algorithm does not terminate at ${\mathbf{0}}^{n}$. Any other state ${\mathbf{v}}\neq{\mathbf{0}}^{n}$ has two possible successors. When the algorithm visits ${\mathbf{v}}$, there remains at least one of ${\mathbf{v}}$’s successors to visit, allowing the algorithm to proceed further. The algorithm does not stop until it reaches ${\mathbf{b}}$ the second time around.

Once the algorithm visits ${\mathbf{b}}$ the second time around, it must have visited both of ${\mathbf{b}}$’s children in ${\mathcal{G}}_{f}$ by Lemma 2. Any non-leaf state in the tree $T_{f,{\mathbf{b}}}$ has two children. Lemma 2 ensures that the algorithm has visited all such states as well as their descendants, including ${\mathbf{0}}^{n-1}1$. To visit ${\mathbf{0}}^{n-1}1$ the algorithm must have visited ${\mathbf{0}}^{n}$ and, prior to that, $1{\mathbf{0}}^{n-1}$. Since any non-leaf descendant of $1{\mathbf{0}}^{n-1}$ has two children, again by Lemma 2, the algorithm must have covered both ${\mathbf{0}}^{n}$ and $1{\mathbf{0}}^{n-1}$ as well as all of the descendants of $1{\mathbf{0}}^{n-1}$. Before reaching ${\mathbf{b}}$ for the second time, the algorithm must have visited all possible states. Thus, the resulting sequence is de Bruijn. ∎

Example 5**.**

Let $n=6$ and $1\leq m\leq 5$. There are $12$ choices for the primitive polynomial $g(x)$. There are $228$ distinct de Bruijn sequences of order $6$ that Algorithm 3 can construct. Table 4 presents only one example for each primitive polynomial $g(x)$ of degree $m$. To save space, $g(x)$ is represented by the coefficients of its monomials in descending degree, e.g., $1101$ represents $g(x)=x^{3}+x^{2}+1$. $\square$

Based on distinct initial states, almost all of the de Bruijn sequences generated by Algorithm 3 over all degree $m$ primitive polynomials for $1\leq m<n$ are distinct. Computational evidences suggest that the only exemption occurs for each primitive polynomial $g(x)$ of degree $m=n-1$ where there is exactly one pair of distinct states, namely $0{\mathbf{1}}^{n-1}$ and ${\mathbf{1}}^{n-1}0$, that yields two cyclically equivalent de Bruijn sequences.

Let $\varphi$ be Euler’s totient function. There are $\frac{\varphi\left(2^{m}-1\right)}{m}$ primitive binary polynomials of degree $m\geq 1$. For $1\leq m\leq n-2$ the number of distinct initial states from the $m$-sequence ${\mathbf{m}}$ is its period $2^{m}-1$. Distinct initial sequences yield distinct de Bruijn sequences. When $2<n=m+1$, there are only $2^{n-1}-2$ distinct de Bruijn sequences produced from the $2^{n-1}-1$ distinct initial states since ${\mathbf{1}}^{m}0$ and $0{\mathbf{1}}^{m}$ yield the same de Bruijn sequence. We formalize this observation and note that the proof follows the same reasoning as that of Theorem 5. The crux move is to show that there are no other collisions than the one already mentioned above. The details are omitted here.

Theorem 10**.**

Let $n>2$. Then

[TABLE]

Next, we describe the family $\mathcal{F}_{6}$, which is very closely related to $\mathcal{F}_{5}$. Again, let $g(x)$ ranges over all suitable primitive polynomials $1+a_{1}x+\ldots+a_{m-1}x^{m-1}+x^{m}\in{\mathbb{F}}_{2}[x]$ of degree $1\leq m<n$. Let $f(x_{0},x_{1},\ldots,x_{n-1})$ be the feedback function in Equation (7), defined by $g(x)$. We use the feedback function

[TABLE]

as an input in a slightly modified Algorithm 3. The initial state is now the complement of any string of length $n$ in the $m$-sequence ${\mathbf{m}}$. Consequently, in Line 8, we use ${\mathbf{c}}=0{\mathbf{1}}^{n-1}$ and, in Line 9 the assignment rule becomes ${\mathbf{c}}\leftarrow{\mathbf{1}}^{n}$. The respective analogues of Example 5 and Theorem 10 also hold for $\mathcal{F}_{6}(n)$.

4.3 Other Special Functions

A strategy that works well is to choose special feedback functions whose respective corresponding state graphs are similar to the successful ones above. Here we provide another example. Such functions are far too numerous to list. We invite the readers to come up with their own favourites.

Proposition 11**.**

Let $n\geq 4$. We run Algorithm 4 on

[TABLE]

The initial state can be any of the $n$-stage state of the periodic sequence $(0111)$. Then the resulting sequence is de Bruijn of order $n$.

The proof of Proposition 11 follows the argument in the proof of Theorem 9 since the state graph for $f$ in Equation (9) is similar with the one in Theorem 9.

Example 6**.**

Figure 5 presents ${\mathcal{G}}_{f}$ with $n=4$ and $f$ in (9).

Starting from ${\mathbf{b}}=1011$ we obtain the sequence of states

[TABLE]

$\square$

5 Conclusion

We put many known greedy algorithms for binary de Bruijn sequences into the framework of GPO Algorithm. We establish easy-to-check conditions that guarantee the efficacy of this general algorithm and its variants. The Boolean maps, i.e., the truth table in its algebraic normal form (ANF) is used to output the resulting de Bruijn sequence. Efficiency in either time or memory complexity as $n$ grows is not a strong aspect of most greedy algorithms. Here, the bottleneck clearly lies in checking if a particular $n$-string has appeared. In our basic python implementation, the search is iterated through the state graph. A typical modern laptop with $16$ GB of memory has the required resources to comfortably handle $n\leq 15$. Devising a better general search approach is interesting but lies outside the scope of this paper.

A complete list of suitable feedback functions and initial states are easy to generate for small $n$. Table 5 is for $n=3$. The combined Tables 7 and 8 is for $n=4$. They lead to the following assertion. For any $n>2$, let ${\mathbf{S}}$ be any de Bruijn sequence of order $n$ and let ${\mathbf{b}}$ be any $n$-string. Then there exists $f(x_{0},x_{1},\ldots,x_{n-1})$ such that ${\mathbf{S}}$ is the output of the GPO Algorithm on input $(f,{\mathbf{b}})$.

One defines a function $g(x_{0},x_{1},\ldots,x_{n-2})$ while traversing ${\mathbf{S}}$, starting from the initial state ${\mathbf{b}}=b_{0},b_{1},\ldots,b_{n-1}$. Since the root is special, $g(b_{0},b_{1},\ldots,b_{n-2}):=b_{n-1}$. If $y$ is the next bit after the first occurance of any other $n-1$ string $(x_{0},x_{1},\ldots,x_{n-2})$, then $g(x_{0},x_{1},\ldots,x_{n-2}):=\overline{y}$. This function $g$ defines a suitable feedback function $f(x_{0},x_{1},\ldots,x_{n-1}):=g(x_{1},\ldots,x_{n-1})$ to use in the GPO Algorithm.

Example 7**.**

Let $n=3$, ${\mathbf{S}}=(0001~{}0111)$, and ${\mathbf{b}}=101$. We have $g(1,0)=1$, $g(0,1)=0$, $g(1,1)=0$, and $g(0,0)=1$. Hence, $g(x_{0},x_{1})=1+x_{1}$ and $f(x_{0},x_{1},x_{2}):=g(x_{1},x_{2})=1+x_{2}$. On input ${\mathbf{b}}=101$ and $f(x_{0},x_{1},x_{2})=1+x_{2}$, the algorithm generates ${\mathbf{S}}$.

Let $n=4$, ${\mathbf{S}}=(1111~{}0110~{}0101~{}0000)$, and ${\mathbf{b}}=1111$. We have

[TABLE]

Hence, $g(x_{0},x_{1},x_{2})=x_{0}\cdot x_{1}+x_{0}+x_{1}\cdot x_{2}+x_{1}+1$ and, therefore, $f(x_{0},x_{1},x_{2},x_{3})=g(x_{1},x_{2},x_{3})=x_{1}\cdot x_{2}+x_{1}+x_{2}\cdot x_{3}+x_{2}+1$. $\square$

Acknowledgements

The work of Z. Chang is supported by the National Natural Science Foundation of China under Grant 61772476 and the Key Scientific Research Projects of Colleges and Universities in Henan Province under Grant 18A110029. Nanyang Technological University Grant Number M4080456 supports the research carried out by M. F. Ezerman and A. A. Fahreza.