Designing preference functions for de Bruijn sequences with forbidden words

Abstract

A preference function provides a method to build periodic sequences by specifying a set of rules that determine which symbols are to be attempted before others, when the sequence is constructed one symbol at a time. The well-known prefer-one, prefer-opposite, and prefer-same binary de Bruijn sequences are all constructed using appropriate preference functions. In this article we provide some fairly general results that give conditions for a pair of an initial word and a preference function on a q-ary alphabet to produce sequences that include every pattern of given size $n\ge 1$–except possibly some specified set of patterns. We provide several old and new constructions that showcase the flexibility of the results. Specifically, we give a construction for square-free and general separative de Bruijn sequences. The existence of these sequences was established more than a decade ago but nonconstructively. An important special case of these separative sequences produces universal cycles for permutations. We also build a preference function for binary de Bruijn sequences of patterns with a maximum density of ones. As for full de Bruijn sequences, the main result helps furnish a recursive construction from arbitrary cyclic permutations of q symbols. Finally, we build a preference function that extends a full de Bruijn sequence of order n into one of order $n+1$.

Appendix A: Proof of Proposition 2.10

By Lemma 2.9$I^y$ occurs in $S_x$. Identifying $S_x$ and $S_y$ with cycles on the de Bruijn digraph of order n, we will show that $S_y$ can be obtained from $S_x$ via a chain of cross-join operations (see e.g. Fredricksen [10] for an exact definition).

First we note that starting the cycle $S_x$ at the word $I^y$ can be regarded as the ‘cross’ part of a cross-join operation, when the predecessor of $I^y$ is redirected to $I^x$ and vice versa. This cuts $S_x$ into two cycles $C_x$ and $C_y$ holding the n-words $I^x$ and $I^y$ respectively. Let us also refer by $\mathbf {T}$ to the path in $S_x$ that starts with $I^x$ and ends with the predecessor of $I^y$ (this is the cycle $C_x$ with exactly one edge missing.)

Enumerate the consecutive n-words of $C_y$ with $I^y$ being first. Now let $\mathbf {v}_1$ be the first n-word, beginning at $I^y$ which has a potential successor, other than its own successor in $C_y$, that appeared already in the path $\mathbf {T}$. $\mathbf {v}_1$ exists for sure among the vertices of $C_y$ because, trivially, the predecessor of $I^y$ is one such (previously being the predecessor of $I^x$ in $S_x$.) If $\mathbf {v}_1$ coincides with the the latter then $S_y$ is just a rotation of $S_x$ and the claim of the proposition is clearly valid. Otherwise, denote by $\mathbf {w}_1$ the leftmost successor of $\mathbf {v}_1$ in $\mathbf {T}$. This means that the right successor of $\mathbf {v_1}$ in $S_y$ is $\mathbf {w}_1$. Moreover, redirecting $\mathbf {v_1}$ to $\mathbf {w}_1$, and the predecessor of $\mathbf {w}_1$ in $\mathbf {T}$ to the successor of $\mathbf {v_1}$ in $C_y$ rejoins $C_x$ and $C_y$ into one cycle.

The new cross-joined cycle, say $S_1$, agrees with $S_y$ on at least the path from $I^y$ to $\mathbf {w_1}$. To prepare for a second iteration of the argument, let $\mathbf {T}_1$ be the sub-path of $\mathbf {T}$ that starts at $I^x$ and ends with the predecessor of $\mathbf {w}_1$ and let $\mathbf {T}_2$ be the remaining part of $\mathbf {T}$. If no word in $\mathbf {T}_2$ has a potential successor in $\mathbf {T}_1$ then the cycle $S_1$ is already a cycle that respects P when it is started with $I^y$ so it must coincide with $S_y$. Otherwise, we find the first such word, $\mathbf {v_2}\in \mathbf {T}_2$ and its leftmost successor in $\mathbf {T}_1$ and repeat the previous logic. Evidently, the argument can be inductively repeated until we get a cycle (and therefore a sequence) that fully respects P, and so it must be $S_y$.