On greedy algorithms for binary de Bruijn sequences

Abstract

We propose a general greedy algorithm for binary de Bruijn sequences, called Generalized Prefer-Opposite 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 algorithm.

Proof of Theorem 3

To make the proof easier to follow we start with an illustration.

Let $m=3$ and $h(x_0,x_1,x_2)=1 + x_0 + x_1 \cdot x_2 + x_2$. Figure 6 is the state graph ${\mathscr {G}}_h$ corresponding to the FSR that produces the de Bruijn sequence ${\mathbf {S}}_3:=(1100\,0101)$. Figures 7 and 8 give, respectively, the state graphs corresponding to the functions $f_4=1+ x_1 + x_2 \cdot x_3 + x_3$ and $f_5=1 + x_2 + x_3 \cdot x_4 + x_4$.

Fig. 6

The state graph ${\mathscr {G}}_{h_3}$ for $h_3=1+ x_0 + x_1 \cdot x_2 + x_2$

Fig. 7

The state graph ${\mathscr {G}}_{f_4}$ for $f_4 = 1+ x_1 + x_2 \cdot x_3 + x_3$

Fig. 8

The state graph ${\mathscr {G}}_{f_5}$ for $f_5 = 1+ x_2 + x_3 \cdot x_4 + x_4$

Table 7 Order of appearance in a run of the GPO algorithm

For fixed n and m, with $n \ge m$ and $j \in \llbracket {2^m-1}\rrbracket$, let ${\mathbf {u}}_j$ denote the $2^m$ consecutive states that form a cycle ${\mathscr {C}}_f$ in ${\mathscr {G}}_f$. These states are typeset in gray in Figs. 6, 7 and 8. As j progresses from 0 to $2^m-1$, we follow the directed edges to traverse all of the states clockwise from top left. For example, the state ${\mathbf {u}}_2$ as shown in Fig. 8 is 00010. We underline the first $m=3$ bits of the states in ${\mathscr {C}}_{f_4}$ and ${\mathscr {C}}_{f_5}$ and put a line over their last $m=3$ bits.

For a fixed $n > m$, let ${\mathscr {R}}_f$ be the graph obtained by removing the edges in ${\mathscr {C}}_f$ from the state graph ${\mathscr {G}}_f$. Note that ${\mathscr {R}}_f$ has $2^m$ disjoint trees as components. Let $T_j$ denote the tree in ${\mathscr {R}}_f$ that contains ${\mathbf {u}}_{j}$ as its root, for $j \in \llbracket {2^m-1}\rrbracket$. In Fig. 8, for instance, the vertex set of $T_0$ is $\{10110,00110,01100,11000\}$.

The algorithm yields ${\mathbf {S}}_4:=(1100 \, 1101 \, 0000 \, 1011)$ on input $f_4$ and initial state ${\mathbf {u}}_0=1100$. Note that the states in ${\mathscr {C}}_{f_4} \subset {\mathscr {G}}_{f_4}$ occur in the exact same order as they do in ${\mathbf {S}}_4$. In one period the states appear in the order

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

The same holds for $n=5$ with initial state ${\mathbf {u}}_0=11000$. The states in ${\mathscr {C}}_{f_5} \subset {\mathscr {G}}_{f_5}$ follow the same order of appearance as they do in the resulting de Bruijn sequence ${\mathbf {S}}_5:=(11000001\,10100111\,11011001\,00010101)$. The states appear in the order listed in Table 7.

Let ${\mathbf {m}}_0,{\mathbf {m}}_1,\ldots ,{\mathbf {m}}_{2^m-1}$ be the consecutive states of ${\mathbf {S}}_m$. There is a natural bijection between ${\mathbf {u}}_k$ and ${\mathbf {m}}_k$ for $k \in \llbracket {2^m-1}\rrbracket$ given by ${\mathbf {u}}_k:=a,b,c,d,e \longleftrightarrow {\mathbf {m}}_k:=a,b,c$. Hence, ${\mathbf {u}}_0 = 11000 \longleftrightarrow {\mathbf {m}}_0=110$ and so on. Slightly abusing the notation, we use $T_{\widehat{{\mathbf {m}}_k}}$ to denote the tree in ${\mathscr {R}}_f$ whose root is $T_{\widetilde{{\mathbf {u}}_k}}$ with $\widehat{{\mathbf {m}}_k}$ being the first $m=3$ bits of $\widetilde{{\mathbf {u}}_k}$.

Let $V_{\ell }:=\{{\mathbf {u}}_{\ell -1}\} \cup \{{\mathbf {v}}\in T_{\ell } : {\mathbf {v}}\ne {\mathbf {u}}_{\ell }\}$, for $\ell \in \llbracket {i+1,j}\rrbracket$, be the set that contains ${\mathbf {u}}_{\ell -1}$ and the vertices of $T_{\ell }$ except for its root ${\mathbf {u}}_{\ell }$. Notice that elements in $V_{\ell }$ always come in pairs as conjugate states. The first two states are conjugate as are the last two states. Their last $m=3$ bits are ${\mathbf {m}}_{\ell }$ and ${\mathbf {m}}_{\ell +1}$. This fact follows from how f is defined in Eq. (1) where h is modified by shifting the focus to the last m entries, instead of the first m entries. In Fig. 8 we have

$$\begin{aligned} V_1&=\{10100,00100,01000,11000\} \rightarrow {\mathbf {m}}_1=100,{\mathbf {m}}_2=000, \\ V_2&=\{10001,00001,10000,00000\} \rightarrow {\mathbf {m}}_2=000,{\mathbf {m}}_3=001, \\ V_3&=\{00010,10010,01001,11001\} \rightarrow {\mathbf {m}}_3=001,{\mathbf {m}}_4=010. \end{aligned}$$

One of the two states in each conjugate pair belonging to $V_{\ell }$ has a successor which is a leaf in either $T_{\widehat{{\mathbf {m}}_{\ell }}}:=T_{\widetilde{{\mathbf {u}}_{\ell }}}$ or $T_{\widehat{{\mathbf {m}}_{\ell +1}}}:=T_{\widetilde{{\mathbf {u}}_{\ell +1}}}$. The successor of 10100 is 01001, which is a leaf in $T_{\widehat{000}}=T_{\widetilde{{\mathbf {u}}_2}}=T_3$. The successor of 11000 is 10000, which is a leaf in $T_{\widehat{001}}=T_{\widetilde{{\mathbf {u}}_3}}=T_2$.

We will use the enumeration of the trees that contribute some leaves to a well-chosen subsequence of a de Bruijn sequence as a tool in the proof of the next theorem. Figure 7 will be useful to confirm the only two cases where a collision in the output occurs.

Proof of Theorem 3

Let $n > m \ge 2$ be fixed. Let ${\mathbf {S}}_m$ be the de Bruijn sequence of order m produced by an FSR with feedback function h. Let f be defined based on h as in Eq. (1). From the state graph ${\mathscr {G}}_f$ we let ${\mathscr {C}}_f$ and ${\mathscr {R}}_f$ be the subgraphs defined earlier. Arithmetic operations on the indices are taken modulo $2^m$ in this proof.

Let ${\mathbf {u}}_j : j \in \llbracket {2^m-1}\rrbracket$ be the consecutive states of length n in the directed cycle ${\mathscr {C}}_f \subset {\mathscr {G}}_f$. Choose one of the vertices, which are in a one-to-one correspondence with the m-stage states of ${\mathbf {S}}_m$, arbitrarily as ${\mathbf {u}}_0$. Recall that $T_j$ is the largest tree in ${\mathscr {R}}_f$ whose root is ${\mathbf {u}}_{j}$. The vertex sets of the trees $T_j : j \in \llbracket {2^m-1}\rrbracket$ partition the set of vertices of ${\mathscr {G}}_f$.

We choose an arbitrary index $i \in \llbracket {2^m-1}\rrbracket$ and let ${\mathbf {S}}_n^i$ be the de Bruijn sequence generated by the algorithm on initial state ${\mathbf {u}}_i$. Lemma 1 implies that, among the states ${\mathbf {u}}_0, {\mathbf {u}}_1, \ldots , {\mathbf {u}}_{2^m-1}$, the second state that the algorithm visits must be ${\mathbf {u}}_{i+1}$. For a contradiction, let ${\mathbf {u}}_k$ with $k > i+1$ be the second state visited. Then, both of its children in ${\mathscr {G}}_f$ must have already been visited. This is impossible since one of the children, which is ${\mathbf {u}}_{k-1} \ne {\mathbf {u}}_i$, has not been visited yet. Thus, by their order of appearance in ${\mathbf {S}}_n$, the states are

$$\begin{aligned} {\mathbf {u}}_i, {\mathbf {u}}_{i+1}, \ldots , {\mathbf {u}}_{2^m-1}, {\mathbf {u}}_0, \ldots , {\mathbf {u}}_{i-1}. \end{aligned}$$

Combining this fact and Lemma 1 makes it clear that by the time ${\mathbf {u}}_j$ is visited, all states in $T_{\ell }$ for all $\ell \in \llbracket {i+1,j}\rrbracket$ must have been visited. So in the remaining run of the algorithm, each state visited after ${\mathbf {u}}_j$ must belong only to $T_{t}$ for some $t \in \llbracket {j+1,i}\rrbracket$. Similarly, suppose that we run the algorithm with initial state ${\mathbf {u}}_j$ for some $j \ne i$ to generate ${\mathbf {S}}^j_n$. By the time ${\mathbf {u}}_i$ is visited, all states in $T_{\ell }$ for $\ell \in \llbracket {j+1,i}\rrbracket$ must have been visited. Each of the remaining states to visit belongs only to $T_{s}$ for some $s \in \llbracket {i+1,j}\rrbracket$.

We now examine what may allow ${\mathbf {S}}_n^i = {\mathbf {S}}_n^j$ while $i \ne j$. Without lost of generality, let $i < j$. If ${\mathbf {S}}_n^i = {\mathbf {S}}_n^j$, then we can partition the set of consecutive states visited by the algorithm on initial state ${\mathbf {u}}_i$ into two parts. Part A contains the sequence of states starting from the successor of ${\mathbf {u}}_i$ in ${\mathbf {S}}^i_n$ up until the state ${\mathbf {u}}_j$. This includes all states in $T_{\ell }$ for $\ell \in \llbracket {i+1,j}\rrbracket$. Part B hosts the sequence of states starting from the successor of ${\mathbf {u}}_j$ in ${\mathbf {S}}_n$ until the state ${\mathbf {u}}_i$. This includes all states in $T_{\ell }$ for $\ell \in \llbracket {j+1,i}\rrbracket$. We say that the two parts are self-closed since all of the successors of each state in Part A, except for ${\mathbf {u}}_j$, are also contained in Part A. Similarly with Part B.

Going back to ${\mathscr {G}}_f$, we consider each state ${\mathbf {a}}$ in Part A, except for ${\mathbf {u}}_j$, and check if the successor of ${\mathbf {a}}$ is a leaf. If yes, then we identify the corresponding tree in ${\mathscr {R}}_f$ by its root. Because Part A is self-closed, we claim that all trees corresponding to elements in Part A must also be in Part A. To confirm this claim, we use the bijection between ${\mathbf {u}}_{\ell }$ and ${\mathbf {m}}_{\ell }$ for $\ell \in \llbracket {2^m-1}\rrbracket$ to associate

$$\begin{aligned} {\mathbf {u}}_{\ell }:=u_0,u_1,\ldots ,u_{m-1}, \ldots ,u_{n-1} \longleftrightarrow {\mathbf {m}}_{\ell }:=u_0,u_1,\ldots ,u_{m-1}. \end{aligned}$$

(10)

and let $V_{\ell }:=\{{\mathbf {u}}_{\ell -1}\} \cup \{{\mathbf {v}}\in T_{\ell } : {\mathbf {v}}\ne {\mathbf {u}}_{\ell }\}$. This allows us to identify $T_{\ell }:=T_{{\mathbf {u}}_{\ell }}$ as $T_{{\mathbf {m}}_{\ell }}$ and vice versa.

The algorithm’s assignment rule requires that the respective successors of the conjugate states

$$\begin{aligned} {\mathbf {v}}:=v_0,v_1,\ldots ,v_{n-1}\quad \text{ and }\quad \overline{{\mathbf {v}}}:=v_0+1, v_1, \ldots , v_{n-1} \end{aligned}$$

must be either one of the two companion states

$$\begin{aligned} {\mathbf {w}}&:=v_1,\ldots ,v_{n-1},h(v_{n-m},v_{n-m+1},\ldots ,v_{n-1}) \text{ and } \\ \widehat{{\mathbf {w}}}&:= v_1,\ldots ,v_{n-1},h(v_{n-m},v_{n-m+1},\ldots ,v_{n-1})+1. \end{aligned}$$

Notice that $\widehat{{\mathbf {w}}}$ must be a leaf in the tree whose root has, as its first m bits,

$$\begin{aligned} v_{n-m+1}, \ldots , v_{n-1}, h(v_{n-m},v_{n-m+1},\ldots ,v_{n-1})+1. \end{aligned}$$

The latter is the companion state of the child of $(v_{n-m},\ldots ,v_{n-1})$ in ${\mathscr {G}}_h$.

Let a pair of conjugate states ${\mathbf {v}}$ and $\overline{{\mathbf {v}}}$ whose common last m bits is ${\mathbf {m}}_{\ell }$ be given. One of their two possible successors must be a leaf in $T_{\widehat{{\mathbf {m}}_{\ell +1}}}$ whose root is the companion state $\widehat{{\mathbf {m}}_{\ell +1}}$ of ${\mathbf {m}}_{\ell +1}$. We generalize this observation to vertices in $V_{i+1}=\{{\mathbf {u}}_i\} \cup \{{\mathbf {v}}\in T_{i+1}: {\mathbf {v}}\ne {\mathbf {u}}_{i+1} \}$. All of the states come in conjugate pairs whose respective last m bits are ${\mathbf {m}}_{i+1}, {\mathbf {m}}_{i+2}, \ldots , {\mathbf {m}}_{i+n-m}$. Each pair has a state whose successor is a leaf in ${\mathscr {G}}_f$. Enumerating the corresponding trees, we obtain

$$\begin{aligned} T_{\widehat{{\mathbf {m}}_{i+2}}}, T_{\widehat{{\mathbf {m}}_{i+3}}}, \ldots , T_{\widehat{{\mathbf {m}}_{i+1+n-m}}}. \end{aligned}$$

Hence, going through each conjugate pair in $\displaystyle {\cup _{\ell \in \llbracket {i+1,j}\rrbracket } V_{\ell }}$ and identifying the tree that contains a successor which is a leaf gives us the following list of trees:

$$\begin{aligned} \bigcup _{\ell \in \llbracket {i+1,j}\rrbracket } \{T_{\widehat{{\mathbf {m}}_{\ell +1}}}, T_{\widehat{{\mathbf {m}}_{\ell +2}}}, \ldots , T_{\widehat{{\mathbf {m}}_{\ell +n-m}}}\}. \end{aligned}$$

Part A must then have the following property. Because it is self-closed, it contains not only all of the states of the trees $T_{\ell }$ for $\ell \in \llbracket {i+1,j}\rrbracket$, but also all states of the trees with respective roots $\displaystyle {\widehat{{\mathbf {m}}_{i+2}}, \widehat{{\mathbf {m}}_{i+3}},\ldots , \widehat{{\mathbf {m}}_{j+n-m}}}$. In total, the states in Part A come from

$$\begin{aligned} j+n-m-(i+2)+1=j-i+(n-m-1) \end{aligned}$$

trees. The assumption that ${\mathbf {S}}_n^i = {\mathbf {S}}_n^j$, however, implies that there are $j-i$ distinct trees that contribute their states to Part A. This is impossible if $n > m+1$. Thus, whenever $n > m+1$, distinct initial states ${\mathbf {u}}_i$ and ${\mathbf {u}}_j$ generate distinct de Bruijn sequences.

Now, let $n=m+1$ and ${\mathbf {u}}_i$ be the initial state. Suppose that Part A contains states contributed by only one tree and let ${\mathbf {u}}_i:=u_0,u_1,\ldots ,u_{n-1}$. Then ${\mathbf {u}}_{i+1}:=u_1,\ldots ,u_{n-1},u_n$ and the only relevant leaf (in $T_{i+1}$) is $\overline{{\mathbf {u}}_i}:=u_0+1,u_1,\ldots ,u_{n-1}$, which is the successor of ${\mathbf {u}}_i$ in ${\mathbf {S}}_n^i$. Thus, we have $u_1,u_2,\ldots ,u_{n-1},u_n+1=u_0+1,u_1,\ldots ,u_{n-1}$, i.e.,

$$\begin{aligned} u_0\ne u_1=u_2=\cdots =u_{n-1}\ne u_n. \end{aligned}$$

There are only two cases that satisfy this constraint.

• 1. Part A includes $T_{{\mathbf {0}}^m1}$, containing ${\mathbf {0}}^{m+1}$ and ${\mathbf {0}}^{m}1$, while Part B contains all of the other trees.

• 2. Part A consists of $T_{{\mathbf {1}}^m0}$, containing ${\mathbf {1}}^{m+1}$ and ${\mathbf {1}}^{m}0$, while Part B contains all of the other trees.

Starting with ${\mathbf {b}}={\mathbf {0}}^m1$, the last state of the generated de Bruijn sequence must be ${\mathbf {0}}^{m+1}$. If we begin with ${\mathbf {b}}=1{\mathbf {0}}^m$, then the second and third states of the resulting de Bruijn sequence must be ${\mathbf {0}}^{m+1}$ and ${\mathbf {0}}^m1$, respectively. These two de Bruijn sequences are shift equivalent. A similar argument can be made for the initial states ${\mathbf {1}}^m0$ and $0{\mathbf {1}}^m$.

If Part A fully contains the trees $T_i$ and $T_{i+1}$, then the above analysis confirms that it also contains leaves belonging to the trees $T_{\widehat{{\mathbf {m}}_{i+1}}}$ and $T_{\widehat{{\mathbf {m}}_{i+2}}}$. Because ${\mathbf {m}}_{i+1} \ne \widehat{{\mathbf {m}}_{i+1}}$ we have

$$\begin{aligned} {\mathbf {m}}_i=\widehat{{\mathbf {m}}_{i+1}} \text{ and } {\mathbf {m}}_{i+1} = \widehat{{\mathbf {m}}_{i+2}} \implies {\mathbf {m}}_i = {\mathbf {m}}_{i+2}, \end{aligned}$$

which is impossible since ${\mathbf {S}}_m$ is de Bruijn. One can proceed inductively to come to the same conclusion for the cases where there are more than 2 trees that contribute their states to Part A. This completes the proof. $\square$