3.1. Sequential editing

We begin with an $n+t$ long bit sequence

\begin{equation*} C_{0}, C_1, \dots, C_{t+n-1}. \end{equation*}

The new bit sequence of the same length,

\begin{equation*} b_{0}, b_1, \dots, b_{t+n-1}, \end{equation*}

is produced by following two rules:

1. Rule 1:

   \begin{equation*} b_i = C_i, 0 \le i \le n-1; \end{equation*}

2. Rule 2: For $i\ge 0$ determine bit $b_{i+n}$ by first asking if the feedback logic bit $f(b_{i+1},\dots,b_{i+n-1})$ has been previously defined; if so, set $b_{i+n}$ accordingly:

   \begin{equation*} b_{i+n} = b_{i} \oplus f(b_{i+1},\dots,b_{i+n-1}); \end{equation*}
   otherwise, define (and remember) the feedback logic bit in such a way that $b_{i+n}$ and $C_{i+n}$ agree.

Here, we give some terminology and indexing practice. We say that the sequence $b$ is obtained from the coin toss sequence $C$ by sequential editing. Each time a $b_{i+n}$ has freedom – because the necessary feedback bit has not yet been set – we set the feedback bit so that $b_{i+n}=C_{i+n}$ ; but at any time the bit $b_{i+n}$ ‘has no choice’, we assign it the forced value. Such a time $i$ is a time of a potential edit; if it turns out (by chance) that $b_{i+n}$ and $C_{i+n}$ agree, then no actual edit has taken place; if it is forced to take $b_{i+n}$ equal to $\overline{C_{i+n}}$ then an actual edit has taken place, and we label the time of this actual edit as $i$ rather than $i+n$ . The sequence $b$ obtained by this process always has the de Bruijn property. In terms of the de Bruijn graph with all vertices resolved, a potential edit occurs at time $i$ when $e_{i}$ , the edge from $v_{i}$ to $v_{i+1}$ , is going in to a vertex $v=v_{i+1}$ where $f(v)$ , the resolution of that vertex, is already known, so that the successor edge, $e_{i+1}=\pi _f(e_i)$ is determined — this is equivalent to determining $b_{i+n}$ , the rightmost bit of $e_{i+1}$ .

3.2. Shotgun editing

Now we define a second, generally different, way to edit the coin toss sequence $C_i$ to produce a sequence $a_i$ . We call this the shotgun edit. Unlike $b_i$ obtained by sequential editing, the sequence $a_i$ obtained by shotgun editing may not have the de Bruijn property.

The symbols $I$ , $J$ , $I_k$ , $J_k$ denote intervals of integers contained in the $(n+t)$ -long interval $[0,1,2,\dots,t+n-1]$ . We use $\ell (I)$ and $r(I)$ to denote the left- and right- endpoints of the interval $I$ . A binary sequence

(9) \begin{equation} C_{0},C_1, \dots, C_{t+n-1}, \end{equation}

has an $m$ -long repeat at $(I,J)$ if $\ell (I)\lt \ell (J)$ , $|I|=m=|J|$ and the two ordered $m$ -tuples $(C_i\;:\;i\in I)$ and $(C_j\;:\;j\in J)$ are equal. We say that (9) has a leftmost Footnote ⁴ $m$ -long repeat at $(I,J)$ if, in addition, either $\ell (I)=0$ or

\begin{equation*} C_{\ell (I)-1} \neq C_{\ell (J)-1}. \end{equation*}

This given, the shotgun edit of coin toss (9) is readily defined: make a list $(I_1,J_1),(I_2,J_2),\dots$ of all the leftmost $n$ -long repeats found in (9). Let

\begin{equation*} a_i = \begin {cases} \overline {C_i} & \textit {if} i=r(J_k) \textit {some} k\\[5pt] C_i & \textit {otherwise.} \end {cases} \end{equation*}

3.3. Zero and first generation words

Let

\begin{equation*} C_{0},C_1, \dots, C_{t+n-1}, \end{equation*}

be a coin toss sequence whose leftmost $n$ -tuple repeats occur at $(I_1,J_1),(I_2,J_2),\dots$ . The zero-generation words of length $h$ are simply words of the form:

\begin{equation*} (C_i,C_{i+1},\dots,C_{i+h-1}). \end{equation*}

A first-generation word is a zero-generation word with exactly one bit complemented, with the index of the complemented bit required to be $r(J_k)$ for some $k$ :

\begin{equation*} (C_i,C_{i+1},\dots,\overline {C_{i+j}}, \dots, C_{i+h-1}), i+j = r(J_k). \end{equation*}

3.4. The good event $G_{(t)}$

We always consider $n$ to be understood, but sometimes we will not want to emphasise the role of $t$ , hence writing $G \equiv{G_{(t)}}$ . Henceforth we shall always assume that $t$ is at most $N=2^n$ , since we are interested in cycle lengths for permutations on a set of size $N$ . Let

\begin{equation*} C_{0},C_1, \dots, C_{t+n-1}, \end{equation*}

be a length $n+t$ coin toss sequence whose leftmost $n$ -repeats occur at $(I_1,J_1),(I_2,J_2),\dots$ . Then the good event $G_{(t)}$ is defined to be the conjunction of these six conditions:

1. (a) neither the initial $n$ -long word of the coin toss sequence, nor any of its $1$ -offsFootnote ⁵ is repeated (probability of failure $O(tn/N)$ );

2. (b) all intersections of the form $I_k\cap J_{k^{\prime }}$ are empty (probability of failure $O(t^3n/N^2+tn^3/N)$ );

3. (c) the sets $I_1,I_2,\dots$ are pairwise disjoint; likewise $J_1,J_2,\dots$ (probability of failure $O(t^2n^2/N^2+t^3n/N^2+tn^3/N)$ );

4. (d) no first-generation word of length $n-1$ equals a zero-generation word of length $n-1$ , or another first-generation word of length $n-1$ (probability of failure $O(t^4n^2/N^3+t^3n^3/N^2+tn^3/N)$ );

5. (e) for every leftmost $(n-1)$ -repeat $(I,J)$ we have

   \begin{equation*} r(J_k)\notin I \cup J \cup \{\ell (I)-1,\ell (J)-1\}, \end{equation*}
   for all $k$ (probability of failure $O(t^3n/N^2+t^2n^2/N^2+tn^3/N)$ );

6. (f) there is no $(2n-1)$ -repeat (probability of failure $O(t^2/N^2)$ ).

The indicated probabilities of failure will be proven below in Theorem 4. First, though, we will prove a theorem that explains why $G$ is called the ‘good event’.

Theorem 2. If the coin toss sequence

\begin{equation*} C_{0},C_1, \dots, C_{t+n-1}, \end{equation*}

belongs to the good event $G$ , then

These conclusions, along with Theorem 4 in the next subsection, will provide substantial control of the prevalence of $(n-1)$ -tuple repeats

Proof of Conclusion 1. Assume, to the contrary, that the $a$ and $b$ sequences differ; let $i$ be the first position of disagreement:

\begin{equation*} a_j = b_j, j\lt i;\; a_i\ne b_i. \end{equation*}

There are two possibilities: (1) $a_i\neq b_i$ and $b_i=C_i$ ; or (2) $a_i\neq b_i$ and $a_i=C_i$ .

Case (1). Since $a_i\neq C_i$ we have $i=r(J_k)$ for some $k$ , and there is a leftmost $n$ -repeat in the $C$ sequence at $(I_k,J_k)$ . But $a_j=C_j$ for $j\in I_k$ (condition(b)); and $a_j=C_j$ for $j\in J_k\setminus \{i\}$ (condition (c)). Hence $b_j=a_j=C_j$ for $j \in I_k\cup J_k\setminus \{i\}$ . But $b_i\neq a_i \neq C_i$ , so in fact the $b$ -sequence itself has an $n$ -repeat at $(I_k,J_k)$ . But the $b$ -sequence has the de Bruijn property, and so the $(I_k,J_k)$ repeat can be backed up $d=\ell (I_k)\gt 0$ steps to reveal

\begin{equation*} (b_0,\dots,b_{n-1}) = (b_{i-d-n+1},\dots,b_{i-d}). (d=\ell (I_k)\gt 0). \end{equation*}

Since $i-d\lt i$ ,

\begin{equation*} (a_0,\dots,a_{n-1}) = (a_{i-d-n+1},\dots,a_{i-d}), \end{equation*}

so in fact

(10) \begin{equation} (C_0,\dots,C_{n-1}) = (a_{i-d-n+1},\dots,a_{i-d}). \end{equation}

The word on the right side of the last equality is either a zero-generation or a first-generation word; either case contradicts condition (a).

Case (2). Because $i$ is a sequential edit point, ( $b_i\neq C_i$ ), it must be that the $(n-1)$ -long word $(b_{i-n+1},\dots, b_{i-1})$ is appearing for a second or later time, say

\begin{equation*} (b_{\ell -n+1},\dots, b_{\ell -1}) = (b_{i-n+1},\dots, b_{i-1}), \ell \lt i. \end{equation*}

We must have $b_{\ell }=C_{\ell }$ , since no sequential editing took place at time $\ell$ . (The relevant bit of the feedback logic had not yet been determined.) We know that $b_i\neq b_{\ell }$ , else the $b$ -sequence contains an $n$ -repeat which, as was explained in Case (1), backs up to yield the contradictory (10). So, $C_i\neq b_i \neq b_{\ell } = C_{\ell }$ ; that is, $C_i=C_{\ell }$ and

\begin{equation*} (b_{\ell -n+1},\dots, b_{\ell -1}, C_{\ell }) = (b_{i-n+1},\dots, b_{i-1}, C_i), \ell \lt i. \end{equation*}

Because $i$ is the first point at which the $b$ and $a$ sequences disagree,

(11) \begin{equation} (a_{\ell -n+1},\dots, a_{\ell -1}, C_{\ell }) = (a_{i-n+1},\dots, a_{i-1}, C_i), \ell \lt i. \end{equation}

Suppose (for the sake of a contradiction) that none of the $a$ bits appearing on either side of this last Equation (11) was edited by the shotgun process. Then we have

(12) \begin{equation} (C_{\ell -n+1},\dots, C_{\ell -1}, C_{\ell }) = (C_{i-n+1},\dots, C_{i-1}, C_i), \ell \lt i. \end{equation}

We have thus discovered an $n$ -long repeat in the coin toss sequence, but it might not be a leftmost $n$ -long repeat. So, we look left to determine the least $m\ge 1$ such that either $\ell -n+1-m\lt 0$ (i.e., you’ve gone ‘off the board’) or the run of equalities is broken:

\begin{equation*} C_{\ell -n+1-m} \neq C_{i-n+1-m}. \end{equation*}

One of these two will happen for $m\lt n$ or else the $C$ -sequence is found to contain a $2n$ -repeat, contradicting assumption (f). But then we have found a leftmost $n$ -repeat in the $C$ -sequence beginning at $\ell -n+1-m+1$ and $i-n+1-m+1$ ; shotgun editing would consequently modify the $C$ -bit at position $i-n+1-m+1+n-1=i-m+1$ . Since

\begin{equation*} i-n+1 \lt i-m+1 \le i, \end{equation*}

we have found that one of the $C$ -bits on the right side of equation (12), namely the one whose index is $i-m+1$ , is changed by shotgun editing, contrary to our earlier supposition that none of the $a$ bits appearing on either side of the equality (11) was edited by the shotgun process.

By condition (c), every $n$ -long word in the $a$ sequence either is a zero-generation word (matches exactly the corresponding $C$ -bits) or is a first-generation word (matches the corresponding $C$ -bits with exactly one change). Thus, at least one of the $n$ -long words appearing in (11) is a first-generation word, and this contradicts condition (d).

Proof of Conclusion 2. We will make use of the $a$ and $b$ sequences being equal. Suppose we have a leftmost $(n-1)$ repeat in the coin toss sequence,

(13) \begin{equation} C_{i+j} = C_{\ell +j}, 0\le j\lt (n-1);\; \textit{and}\; i=0 \;\textit{or } C_{i-1} \neq C_{\ell -1}. \end{equation}

By condition (e), none of these $2n$ bits (or $2n-2$ in case $i=0$ ) can be edited by the shotgun edit. Hence, we have a leftmost $(n-1)$ -repeat at the same place in the $a$ sequence, whence also the $b$ sequence.

On the other hand, suppose we have a leftmost $(n-1)$ -repeat in the $a$ sequence,

(14) \begin{equation} (a_i, a_{i+1}, \dots, a_{i+n-2}) = (a_\ell, a_{\ell +1}, \dots, a_{\ell +n-2}), \end{equation}

and

\begin{equation*} i=0, \quad \textit{or}\quad a_{i-1} \neq a_{\ell -1}. \end{equation*}

If

\begin{equation*} (a_{i},\dots,a_{i+n-2}) \neq (C_{i},\dots,C_{i+n-2}). \end{equation*}

then $(a_{i},\dots,a_{i+n-2})$ is a first-generation word of length $n-1$ which equals the first- or zero-generation word $(a_{\ell },\dots,a_{\ell +n-2})$ , which is forbidden by condition (d). So,

(15) \begin{equation} (a_{i},\dots,a_{i+n-2}) = (C_{i},\dots,C_{i+n-2}). \end{equation}

Similarly,

(16) \begin{equation} (a_{\ell },\dots,a_{\ell +n-2}) = (C_{\ell },\dots,C_{\ell +n-2}). \end{equation}

Altogether by (14),(15),(16) we have

(17) \begin{equation} (C_{i},\dots,C_{i+n-2}) = (C_{\ell },\dots,C_{\ell +n-2}). \end{equation}

If $i=0$ , then the last is a leftmost $(n-1)$ -repeat in the $C$ sequence, as asserted. So, to conclude, suppose for the sake of a contradiction that $i\gt 0$ and that $C_{i-1}= C_{\ell -1}$ . Then, we have an $n$ -long repeat

\begin{equation*} (C_{i-1},\dots,C_{i+n-2}) = (C_{\ell -1},\dots,C_{\ell +n-2}). \end{equation*}

Sliding left for $m$ steps, we will encounter a leftmost $n$ -repeat in the coin toss sequence

\begin{equation*} (C_{i-1-m},\dots,C_{i+n-2-m}) = (C_{\ell -1-m},\dots,C_{\ell +n-2-m}), \end{equation*}

with $0\le m\lt n-1$ by condition (f). But in such a case $a_{\ell +n-2-m} \neq C_{\ell +n-2-m}$ by the definition of shotgun editing. However, for $0\le m\lt n-1$

\begin{equation*} \ell \le \ell +n-2-m \le \ell +n-2, \end{equation*}

and by (16) $a_{\ell +n-2-m} = C_{\ell +n-2-m}$ The supposition that $i\gt 0$ and that $C_{i-1}= C_{\ell -1}$ has been contradicted, and so (17) is, indeed, a leftmost $(n-1)$ -repeat as needed.

3.5. Probability

In this section we bound the probability of failure of any one of the conditions (a)–(f) appearing in Theorem 2. Let $S$ be a set of relations, each of the form $C_i=C_j$ or $C_i \neq C_j$ with $i\lt j$ . We assume always that $S$ has at most one relation for a given $(i,j)$ ; that is, we don’t allow both $C_i=C_j$ and $C_i \neq C_j$ . What is the probability that a coin toss sequence $C$ will satisfy such a set of relations? The desired probability is $2^{-|S|}$ provided the graph associated with $S$ is cycle free. The graph we have in mind here is $(V,E)$ where $V$ is the set $0,1,2,\dots$ and $E$ is the set of pairs $\{i,j\}$ such that at least one (and by convention exactly one) of the relations $C_i=C_j$ or $C_i \neq C_j$ belongs to $S$ .

In particular, if the graph of $S$ consists of the $n$ pairs $(i,j),(i+1,j+1),\dots,(i+n-1,\, j+n-1)$ the probability is $2^{-n}=1/N$ . This is quite clear if $I=\{i,\dots,i+n-1\}$ and $J=\{j,\dots,j+n-1\}$ are disjoint, since then the underlying graph has no vertex of degree 2. It is also true if $I$ and $J$ overlap, (of course $I\neq J$ ): every vertex of degree $2$ in the graph (i.e., every element of $I\cap J$ ) has one larger neighbour and one smaller neighbour. But a cycle would require at least one vertex with two smaller neighbours.

We will have frequent occasion below, in the proof of Theorem 4, to consider sets $S$ whose graphs are the union of two such $n$ -sets of pairs $(i_1,j_1)$ , $(i_1+1,j_1+1), \dots, (i_1+n-1,j_1+n-1)$ and $(i_2,j_2)$ , $(i_2+1,j_2+1), \dots$ $(i_2+n-1,j_2+n-1)$ . We begin with a lemma which shows that in many situations which arise in these proofs the probability in question is $1/N^2$ .

Lemma 3. Let $\mathcal{G}$ be the graph whose edges consist of two sets of pairs

\begin{equation*}(i_1,j_1),(i_1+1,j_1+1),\dots,(i_1+m_1-1,j_1+m_1-1).\end{equation*}

and

\begin{equation*}(i_2,j_2), (i_2+1,j_2+1),\dots,(i_2+m_2-1,j_2+m_2-1).\end{equation*}

Then $\mathcal{G}$ is cycle free if any one of the following three conditions holds, where we assume $i_1\lt j_1$ and $i_2\lt j_2$ :

1. (i) $I_1\cap I_2 =\emptyset$

2. (ii) $J_1\cap J_2 =\emptyset$

3. (iii) $(I_1\cup I_2) \cap (J_1\cup J_2) = \emptyset$ and $j_2-i_2 \neq j_1-i_1$ .

Theorem 4. Let $G$ be the good event. Then,

\begin{equation*} \mathbb {P}(G) \ge 1 - O\left (t^4n^2/N^3+t^3n^3/N^2+tn^3/N\right ). \end{equation*}

Proof. We shall show that the probability that a random coin toss sequence of length $t+n$ fails any one of the conditions (a) through (f) in the definition of $G$ is $O(t^4n^2/N^3+t^3n^3/N^2+tn^3/N)$ . (More explicitly, each will be shown to fail with the probability indicated in the definition of $G$ .) We invoke the above Lemma during the proof by citing Lemma (i), Lemma (ii) and Lemma (iii).

Condition (a): [neither the initial $n$ -long word of the coin toss sequence, nor any of its $1$ -offs, is repeated.] Consider first an exact repetition. There are $t-1$ places where the repeated sequence can start, and by earlier remarks the probability that the second sequence repeats the first is $1/N$ . The same argument applies to the $1$ -offs of the initial pattern, and we conclude that the probability for condition (a) to fail is less than $t(n+1)/N$ .

Condition (b): [all intersections $I_k\cap J_{k^{\prime }}$ are empty.] For $k=k^{\prime }$ we bound the probability of failure by $tn/N$ using the same technique as in case (a). Suppose that $I_1\cap J_2 \neq \emptyset$ . By the $k=k^{\prime }$ case of the proof we may assume $J_1$ disjoint from $I_1$ and to its right; and $I_2$ disjoint from $J_2$ and to its left. If $I_1\cap I_2=\emptyset$ then Lemma (i) yields the upper bound $O(t^3n/N^2)$ . If $J_1\cap J_2=\emptyset$ then Lemma (ii) yields $O(t^3n/N^2)$ . In the remaining case $I_1$ meets $I_2$ , $J_1$ meets $J_2$ , and $I_1$ meets $J_2$ . Thus the union $I_1\cup I_2\cup J_1\cup J_2$ is an interval, and a bound of $O(tn^3/N)$ results.

Condition (c): [the sets $I_1,I_2,\dots$ are pairwise disjoint; likewise $J_1,J_2,\dots$ .] We will prove the assertion regarding $I_1,I_2,\dots$ ; the other assertion is proven in an entirely similar manner. Suppose $I_1\cap I_2\neq \emptyset$ . We may assume $J_1\cap J_2\neq \emptyset$ ; otherwise Lemma (ii) implies an upper bound of $O(t^3n/N^2)$ . So now, both intersections $I_1\cap I_2$ and $J_1\cap J_2$ are nonempty. If any one of the four intersections $I_a\cap J_b$ is nonempty, then again the union $I_1\cup I_2\cup J_1 \cup J_2$ is an interval, and we have the upper bound $O(tn^3/N)$ . So assume $(I_1\cup I_2) \cap (J_1 \cup J_2) = \emptyset$ . Assume, for the sake of a contradiction, that $r(J_2)-r(I_2)=d=r(J_1)-r(I_1)$ . Then we have $I_1\neq I_2$ and $J_1\neq J_2$ . Without loss, let us say $I_1$ is left of $I_2$ and $J_1$ is left of $J_2$ . We have $C_{\ell (I_2)-1} \neq C_{\ell (J_2)-1}$ because $(I_2,J_2)$ is assumed to be a leftmost $n$ -repeat. Since $I_1$ is left of $I_2$ , $\ell (I_2)-1\in I_1$ ; but then $C_{\ell (I_2)-1} = C_{\ell (J_2)-1}$ , the contradiction. So, $r(J_2)-r(I_2)=d=r(J_1)-r(I_1)$ is untenable and now Lemma (iii) implies an upper bound of $O(t^2n^2/N^2)$ .

Condition (d): [no First-generation word of length $n-1$ equals a Zero-generation word of length $n-1$ , or another First-generation word of length $n-1$ ]. Suppose the first assertion is violated. Then we have, for some $i$ , some $d\gt 0$ and some $j\in \{i,i+1,\dots,i+n-2\}$ ,

(18) \begin{equation} C_{\ell }=C_{\ell +d}\; \textit{for}\; \ell \in \{i,i+1,\dots,i+n-2\}\setminus \{j\}, \textit{and}\; C_{j}\neq C_{j+d}, \end{equation}

with $r(J_k) \in \{j,j+d\}$ and with $(I_k,J_k)$ a leftmost $n$ -repeat. Let $I = \{i,i+1,\ldots,i+n-2\}$ and let $J = \{i+d,i+d+1,\ldots,i+d+n-2\}$ . If $I$ and $I_k$ are disjoint then using Lemma (i) the probability is bounded by $O(t^3 \, n/N^2)$ . So assume they intersect, so their union is an interval. Similarly we can assume, using Lemma (ii), that $J$ and $J_k$ also intersect so their union is another interval. If these two intervals intersect, forming another interval, we have a probability bound of $O(t \, n^3/N)$ . Otherwise $( I \cup I_k) \cap (J \cup J_k)$ is empty. If $r(J_k) - r(I_k) = d$ then $C_j = C_{j+d}$ , contradicting (18). So Lemma (iii) implies $O(t^2 \, n^2/N^2)$ for the probability. This gives an overall bound of $O(t \, n^3/N +t^2 \, n^2/N^2 +t^3 \,n/N^2)$ .

Next, suppose the second assertion of (d) is violated. Then we have, for some $i$ , some $d\gt 0$ and some $j_1,j_2\in \{i,i+1,\dots,i+n-2\}$ ,

(19) \begin{equation} (C_i, C_{i+1},\dots, \overline{C_{j_1}}, \dots, C_{i+n-2}) = (C_{i+d}, C_{i+d+1},\dots, \overline{C_{j_2+d}}, \dots, C_{i+d+n-2}), \end{equation}

with $j_1=r(J_1)$ , $j_2+d=r(J_2)$ and $(I_1,J_1), (I_2,J_2)$ leftmost $n$ -repeats. As reasoned before we have $I\cap J$ , $I_1\cap J_1$ and $I_2\cap J_2$ all empty with probability at least $1-O(tn^2/N)$ . It follows, from the sheer geometry of the situation, that $I\cap I_1=\emptyset$ . We may assume that $I_1\cap I_2=\emptyset$ , since (as proven in (c) above) the probability of failure is $O(t^3n/N^2+t^2n^2/N^2+tn^3/N)$ . We may assume that $I_1\cap I=\emptyset$ , for otherwise the union of $I_1$ and $I$ and $J_1$ is a connected interval, and then by reasoning as above a bound of $O( t^3n^3/N^2)$ results. We now have all three intersections $I\cap I_1$ , $I\cap I_2$ and $I_1\cap I_2$ being empty; and by an obvious embellishment of Lemma (i) the probability of the remaining case is $O( t^4n^2/N^3)$ .

Condition (e): [for every leftmost $(n-1)$ -repeat $(I,J)$ we have

\begin{equation*} r(J_k)\notin I \cup J \cup \{\ell (I)-1,\ell (J)-1\}, \end{equation*}

for all $k$ .] Say $I=\{i,i+1,\dots,i+n-2\}$ and $J=\{i+d,i+d+1,\dots,i+d+n-2\}$ . The probability that $I_k\cap J_k$ is not empty is $O(tn/N)$ , so assume $I_k\cap J_k=\emptyset$ . If $r(J_k) \in I\cup \{i-1\}$ , then, since $I_k$ lies entirely to the left of $J_k$ , $I_k\cap I=\emptyset$ . By Lemma (i) the probability of this is $O(t^3n/N^2)$ .

The probability that $I\cap J$ is not empty is $O(tn/N)$ , so assume both $I_k\cap J_k$ and $I\cap J$ are empty. The probability that $I_k\cap I$ is empty is, by Lemma (i), $O(t^3n/N^2)$ , so assume $I_k\cap I\neq \emptyset$ . If $I\cap J_k$ or $I_k\cap J$ is nonempty then $I_k\cup I \cup J_k \cup J$ is an interval, and the probability of this is $O(tn^3/N)$ . So, assume $(I\cup I_k) \cap (I_k\cup J) = \emptyset$ . If $r(J_k)-r(I_k)=r(J)-r(I)$ , then

\begin{align*} C_{\ell (I)-1} = C_{\ell (J)-1} & \quad \textit{by}\; \ell (I)-1 \in I_k \\[5pt] C_{\ell (I)-1} \neq C_{\ell (J)-1} & \quad \textit{by}\; (I,J)\; \textit{being a leftmost }\; (n-1)-\textit{repeat}, \end{align*}

an impossibility. So, $r(J_k)-r(I_k)\neq r(J)-r(I)$ and Lemma (iii) gives the bound $O(t^2n^2/N^2)$ for this final scenario.

Condition (f): [there is no $(2n-1)$ -repeat.] Easily, the failure probability is at most $2t^2/N^2$ .

3.6. Coin tossing versus paths in the de Bruijn graph

Proof. Without loss of generality we assume the given string has the de Bruijn property. (Else, the two probabilities are both zero.) First, let’s compute the probability that $b$ arose by sequential editing of a $(t+n)$ -long coin toss. The probability of the coin toss yielding $b_0,\dots,b_{n-1}$ is $(1/2)^n$ . Consider $b_i$ , with $i\ge n$ . If $(b_{i-n+1},\dots,b_{i-1})$ is equal to $(b_{j-n+1},\dots,b_{j-1})$ for some $j$ in the range $n\le j\lt i$ , then sequential editing says to let $b_i$ be what it ought to be: $b_{i-n}\oplus f(b_{i-n+1},\dots,b_{i-1})$ . In which case, it does not matter what value $C_i$ has. But if $i\ge n$ and $(b_{i-n+1},\dots,b_{i-1})$ has not been seen before (among $(n-1)$ -long words ending at a position greater then or equal to $n$ ), then $C_i$ must be equal to $b_i$ . (And, we remember henceforward the value of $f(b_{i-n+1},\dots,b_{i-1})$ is $b_i\oplus b_{i-n}$ .) Altogether, then, the probability that a length $t+n$ coin toss will yield a given sequence $b_0,b_1,\dots,b_{t+n-1}$ by sequential editing is $2^{-r}$ where

\begin{equation*} r = n-1+ \#\textit {distinct } (n-1)\textit {-long subwords ending at position } n-1 \textit { or later}. \end{equation*}

Now let’s compute the probability that $b$ arose by choosing a starting position and logic at random. Classify each position $i$ , $0\le i\le t+n-1$ , as Type I or Type II. The position is Type I if $i\ge n$ and the preceding $(n-1)$ long word $(b_{i-n+1},\dots,b_{i-1})$ is appearing for the first time in the $b$ -sequence. The position is Type II otherwise: either $i\lt n$ , or the preceding $(n-1)$ long word $(b_{i-n+1},\dots,b_{i-1})$ is appearing for the second or later time. It should be clear that the probability in question is

\begin{equation*} \left (\frac {1}{2}\right )^{n+\#\textit {Type I}}. \end{equation*}

The two probabilities just calculated agree.Footnote ⁶

3.7. Notation for paths starting at $k$ random $n$ -tuples

We now fix $k \ge 1$ and use the notation $e_1,\ldots,e_k$ to name $k$ random $n$ -tuples. Collectively, these $k$ edges of $D_{n-1}$ are denoted

(20) \begin{equation} \boldsymbol{{e}} = (e_1,e_2,\ldots,e_k) \in (\mathbb{F}_2^n)^k. \end{equation}

Picking a random feedback $f$ , and $k$ random $n$ -tuples, independent of $f$ , is equivalent to picking one element, uniformly at random from the space

(21) \begin{equation} S_{n,k} \;:\!=\; \{ (f,\boldsymbol{{e}})\,: \, \ f\;:\; \mathbb{F}_2^{n-1} \to \mathbb{F}_2, \boldsymbol{{e}} \in (\mathbb{F}_2^n)^k \}, \textrm{ with } |S_{n,k}|=2^{2^{n-1}+kn}. \end{equation}

The choice of $(f,\boldsymbol{{e}})$ from $S_{n,k}$ determines $k$ infinite periodic sequences of edges: for $a=1$ to $k$ ,

(22) \begin{equation} {\textrm{Seg}}(f,e_a) \;:\!=\; (e_{a,0} e_{a,1} e_{a,2}\cdots ) \textit{ where } e_{a,0}=e_a, \textrm{ and for } i \ge 0, e_{a,i+1}= \pi _f(e_{a,i}). \end{equation}

For the sake of comparison with coin tossing, we often look at such paths only up to time $t$ (this is what motivated our terminology segment):

(23) \begin{equation} \textrm{ for } a=1 \textrm{ to } k, \ \ {\textrm{Seg}}(f,e_a,t) = (e_{a,0} e_{a,1} \cdots e_{a,t}). \end{equation}

3.8. $(k,t)$ -sequential editing

Now we will define a modification of the sequential editing process that was discussed earlier in Section 3.1. The reader should bear in mind our ultimate goal. We wish to study what happens when a feedback logic $f$ is chosen at random; $k$ different starting $n$ -tuples $e_1,\dots,e_k$ are chosen at random; and $k$ walks of length $t$ are generated, the first starting from $e_1$ and using the logic $f$ to continue for $t$ steps; the second starting from $e_2$ , etc. As in Section 3.1, we wish to generate these walks using $k(n+t)$ coin tosses, and we would like to have an analogue to Theorem 5 saying that our procedure for passing from the coin toss to the $k$ walks perfectly simulates the process of choosing a logic and starting points at random. The reader can almost certainly envision the natural way to achieve this, but we will write out the details.

The first $n+t$ coins are used exactly as in Section 3.1: Rule 1 is applied to the first $n$ coin tosses to yield starting point $e_1$ , and then Rule 2 is applied $t$ times to get the overlapping $n$ -tuples $e_1=e_{1,0}, e_{1,1}, \ldots, e_{1,t}$ that form the first walk. Equivalently, this segment is spelled out by the $(n+t)$ de Bruijn bits $b_0 \ldots b_{t+n-1}$ , and along the way, some feedback logic bits have been defined.

Then, for the next $n$ coin tosses, $C_i$ for $i=t+n$ to $i=t+2n-1$ inclusive, sequential editing is suspended; again Rule 1 is applied, to give

\begin{equation*} e_2 \;:\!=\; (b_{t+n},\ldots,b_{t+2n-1}) \;:\!=\; (C_{t+n},\ldots,C_{t+2n-1}), \end{equation*}

with no new feedback logic bits learned. Then, Rule 2 is applied for the next $t$ input bits, $C_i$ for $i=t+2n$ to $i=2t+2n-1$ to create the second walk of length $t$ , ${\textrm{Seg}}(f,e_2,t)$ — remembering of course those feedback logic bits that were learned during the creation of ${\textrm{Seg}}(f,e_1,t)$ , and (most likely) learning some new feedback logic bits in the process. (It might be the case that $e_2=e_1$ , or that $e_2$ appears in the first walk, in which case, we don’t learn any new feedback logic bits.) If $k\gt 2$ , we continue in a similar fashion, first suspending editing for time $n$ , during which time we learn no new feedback logic bits and we form $e_a \;:\!=\; (b_{(a-1)(t+n)},\ldots,b_{(a-1)t+an-1}) \;:\!=\; (C_{(a-1)(t+n)},\ldots,C_{(a-1)t+an-1})$ , then returning to Rule 2 for the next $t$ bits, to fill out ${\textrm{Seg}}(f,e_a,t)$ .

For $k,t \ge 1$ we define

(24) \begin{align} {{Q-EDIT_{k,t}}}\;:\;\{0,1\}^{k(n+t)} \;\;\;\to \;\;\; (\mathbb{F}_2^{t+n})^k \end{align}

(25) \begin{align} (C_0,C_1,\ldots,C_{k(n+t)-1}) \;\;\; \mapsto \;\;\; ({\textrm{Seg}}(f,e_1,t),\ldots,{\textrm{Seg}}(f,e_k,t) ) \end{align}

as given by the above procedure.

It may, or should, seem intuitively obvious that ${Q-EDIT_{k,t}}$ , applied to an input uniformly chosen from $ \{0,1\}^{k(n+t)}$ , induces the same distribution on the $k$ segments of length $t$ in (22), as does a uniform pick from $S_{n,k}$ and iteration of $\pi _f$ from each of $e_1,\ldots,e_k$ . We claim that the argument given in the proof of Theorem 5 can be adapted to show this.

3.9. The good event $G_{(k,t)}$ for $(k,t)$ -sequential editing

There are two different ways to produce $k$ walks each of length $t$ out of a sequence of $k(n+t)$ coin tosses. The first, with $t'=(k-1)n+kt$ playing the role of $t$ , is simple sequential edit, to determine a starting $n$ -tuple $e$ , and one path $e_0,e_1,\ldots,e_{t'}$ corresponding to $t'=(k-1)n+kt$ iterates of $\pi _f$ starting from $e$ . The good event, regarding this first procedure, is really $G \equiv G_{((k-1)n+kt)}$ . We can then cut the path of length $t'$ to produce $k$ paths of length $t$ ; see (31) to see the natural notation associated with such cutting. The second procedure is is to apply ${Q-EDIT_{k,t}}$ , defined in the previous section, to produce a $k$ -tuple of starting edges, $\boldsymbol{{e}}$ , and $k$ segments of length $t$ , as in (23). The good event, regarding this second procedure, to be called $G_{(k,t)}$ , is designed so that the two procedures agree. We simply take all of the demands of the good event for simple editing on $k(n+t)$ coins, and throw in additional requirements to ensure the suspensions of editing involved in the definition of ${Q-EDIT_{k,t}}$ . Informally, these additional requirements are that every $(n-1)$ tuple which appears at some time $j$ involved in suspension occurs at no other time $i$ in the coin toss sequence. Formally, given $n,k,t$ , the bad event $B$ is given by

(26) \begin{equation} B = \bigcup _{ i \in [0, k(n+t)-n+1]} \ \ \bigcup _{j \in \cup _{a=1}^{k-1} [a(n+t)-n+2,a(n+t)]} M_{ij}, \end{equation}

where the event $M_{ij} = \emptyset$ if $i=j$ , and otherwise

\begin{equation*} M_{ij} = \{d_\textit {HAMMING}(C_iC_{i+1}\cdots C_{i+n-2},C_jC_{j+1}\cdots C_{j+n-2}) \le 1 \}, \end{equation*}

and the good event is then

(27) \begin{equation} {G_{(k,t)}} = G_{((k-1)n+kt)} \setminus B. \end{equation}

Since a word of length $n-1$ has $n$ neighbours at Hamming distance 1 or less, ${\mathbb{P}}(M_{ij}) = n/2^{n-1}$ for $i \ne j$ , so that ${\mathbb{P}}(B) \le (n+t)k^2n^2 \times 2/N$ , for the sake of extending Theorem 4.

We now consider the following to have been proved; it is a single theorem, to give the extensions of Theorems 2 and 4 and 5, appropriate to $k,t$ sequential editing. Note that in the final conclusion of Theorem 6 we treat $k$ as fixed while $n,t \to \infty$ , so that $t$ and $kt$ are of the same order, and we take the assumption $t/\sqrt{N} \to \infty$ so that the three terms in the bound from Theorem 4 are covered by a single term.

We summarise: there is an exact operation, sequential editing of $n+t$ coin tosses, which achieves the exact distribution of ${\textrm{Seg}}(f,e,t)$ , as induced by a uniform choice of $(f,e)$ from its $2^{2^{n-1}} 2^n$ possible values, followed by starting at $e$ and taking $t$ iterates of the permutation $\pi _f$ . There is a good event $G \equiv G_{t}$ , with ${\mathbb{P}}(G) \to 1$ provided that $t^3n^3/N^2 \to 0$ , for which the sequential edit agrees with the shotgun edit, and $v_i=v_j$ iff the coins have a leftmost $(n-1)$ -tuple repeat at $(i,j)$ . This sequential edit can be used with $k(n+t)$ in place of $n+t$ , to create one long segment; there is the corresponding good event $G_{t^{\prime }}$ , $t^{\prime }=(k-1)n+kt$ . There is a second, distinct operation, ${Q-EDIT_{k,t}}$ , for editing $k(n+t)$ coin tosses, to yield the exact distribution of $k$ segments of length $t$ under a single logic $f$ and $k$ starting $n$ -tuples, $\boldsymbol{{e}} = (e_1,\ldots,e_k)$ ; that is, the distribution of $({\textrm{Seg}}(f,e_1,t),\ldots,{\textrm{Seg}}(f,e_k,t))$ as induced by a uniform choice of $(f,{\mathbb{E}\,})$ from its $2^{2^{n-1}} 2^{kn}$ possible values. And there is a corresponding good event ${G_{(k,t)}} \subset G_{t^{\prime }}$ , with

\begin{equation*} {\mathbb {P}}(G_{t^{\prime }} \setminus {G_{(k,t)}}) \le 2k^2n^2(n+t)/N, \end{equation*}

formed by adding the constraint that $i$ or $j \in \cup _{0 \le a \lt k} [a(n+t)-n+2,a(n+t)]$ implies that there is not an $(n-1)$ tuple repeat at $(i,j)$ . On the event $G_{(k,t)}$ , the $k$ -sequential edit agrees exactly with the cutting of ${\textrm{Seg}}(f,e_1,k(n+t)-n)$ .

Figure 1. An example, one segment of length 290, where there are three leftmost $(n-1)$ -tuple repeats, at (56,153), (120,260) and (135,175).

Figure 2. The same example: one segment of length 290, where there are three leftmost $(n-1)$ -tuple repeats, at locations (56,153), (120,260) and (135,175). Now, the locations are plotted in standard Cartesian coordinates.

Figure 3. Coloring. An example, with $k=3$ , $n=10,t=90$ . The same one segment of length 290, as in Figure 1, where there are three leftmost $(n-1)$ -tuple repeats, at (56,153), (120,260) and (135,175). Now the first segment is coloured red, the second yellow and the third blue.

Figure 4. Coloring and cutting; a succinct way to visualise both. The $k$ segments of length $t$ are still shown as they appear along the single segment of length $k(t+n)-n$ . We also show the $\left(\substack{k \\[3pt] 2}\right)$ $t$ by $t$ squares where matches may occur between two differently coloured length $t$ segments. Note the repeat at (56,153) is a vertex coloured both red and yellow, hence orange. The repeat at (120,260) is a vertex coloured both yellow and blue, hence green. The vertex at (135,175) is coloured yellow twice - we could show it as an extra-saturated yellow but did not. The significance of the diagonals of the small squares is explained in Section 4.3.

3.10. A cutting example

We now illustrate some of the concepts just introduced, with an example and with Figures 1, 2, 3 and 4. Take $n=10,t=90,k=3$ . So, to generate $k=3$ segments of length $t=90$ , we start with $k(n+t)=300$ coin tosses, used to generate one segment of length $k(n+t)-n=290$ . When we have in mind a single segment of length $t$ , we will use a single subscript to label the edges, so that with $e=e_0$ , the segment is a list of $t+1$ edges

(28) \begin{equation} {\textrm{Seg}}(f,e,t)= e_0 e_1 \cdots e_t. \end{equation}

The coin tosses, indexed from $i=0$ to $i=299$ , are labelled $C_i$ , the de Bruijn bits formed by sequential edit are labelled $b_i$ , and the bits formed by shotgun edit are labelled $a_i$ . On the good event $G$ , we will have $a_i=b_i$ for all $i$ . The vertex $v_i$ is the $(n-1)$ - tuple of bits starting with $b_i$ , the edge $e_i$ is the $n$ -tuple of bits starting with $b_i$ and edge $e_i$ at time $i$ goes from vertex $v_{i}$ to $v_{i+1}$ :

\begin{equation*} v_{i} = b_{i}b_{i+1}\cdots b_{i+n-2}, \ e_i = b_{i}b_{i+1}\cdots b_{i+n-1}, \ e_i= (v_{i},v_{i+1}) . \end{equation*}

We also view the segment in (28) as a list of $t+2$ vertices, or as a list of $t+n$ bits, and abuse notation by writing equality, so that

(29) \begin{equation} {\textrm{Seg}}(f,e,t)= v_0 v_1 \cdots v_t v_{t+1}. \end{equation}

(30) \begin{equation} {\textrm{Seg}}(f,e,t)= b_0 b_1\cdots b_{n-1}b_n \cdots b_tb_{t+1}\cdots b_{t+n-1}. \end{equation}

Since we are particularly interested in leftmost $(n-1)$ -tuple repeats, we shall suppose that we are in the good event $G$ , and the leftmost $(n-1)$ -tuple repeats in the coin toss sequence are at (56,153), (120,260) and (135,175). Thanks to $G$ occurring, we know that all 291 edges $e_0$ to $e_{290}$ are distinct, and the only vertex repetitions are $v_{56}=v_{153}, v_{120}=v_{260}$ and $v_{135}=v_{175}$ . One way of indicating where these vertex repeats occur is to draw some auxiliary lines pointing to the locations, as in Figure 1. Figure 2 gives a two-dimensional (‘spatial’) view of the same situation.

When we cut the single long segment in (28) into $k=3$ segments, we use two indices; the first runs from 1 to $k$ , and the second runs from 0 to $t$ . Including the relation with (28), for Example 1, but with the labels $e_1,\ldots,e_k$ overloaded — since they also appear on the left side, naming the starting edges for the $k$ segments — this will give

\begin{equation*} \begin {array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} {\textrm {Seg}}(f,e_1,t) & = & e_{1,0}\cdots e_{1,90} & = & e_0\cdots e_{90} \\[5pt] {\textrm {Seg}}(f,e_2,t) & = & e_{2,0}\cdots e_{2,90} & = & e_{100}\cdots e_{190} \\[5pt] {\textrm {Seg}}(f,e_3,t) & = & e_{3,0}\cdots e_{3,90} & = & e_{200}\cdots e_{290}. \end {array} \end{equation*}

The same $k=3$ segments of length $t=90$ , presented as lists of vertices (which here are 9-tuples) are notated as

(31) \begin{equation} \begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}{\textrm{Seg}}(f,e_1,t) & = & v_{1,0}v_{1,1}\cdots v_{1,90}v_{1,91} & = & v_0v_1\cdots v_{90}v_{91} \\[5pt] {\textrm{Seg}}(f,e_2,t) & = & v_{2,0}v_{2,1}\cdots v_{2,90}v_{2,91} & = & v_{100}v_{101}\cdots v_{190}v_{191} \\[5pt] {\textrm{Seg}}(f,e_3,t) & = & v_{3,0}v_{3,1}\cdots v_{3,90} v_{3,91} & = & v_{200}v_{201}\cdots v_{290}v_{291}. \end{array} \end{equation}

Collectively, these $k$ segments are given by a deterministic function of $(f,\boldsymbol{{e}})$ , where $\boldsymbol{{e}}=(e_1,e_2,\ldots,e_k)$ names all $k$ starting points.

3.11. Coloring

Imagine the $k$ segments of length $t$ as pieces of (directed) yarn, with $k$ different ‘primary’ colours. Vertices that appear only once get the primary colour of the segment they come from; vertices that appear twice on the same segment might be visualised as having a more saturated version of the primary colour of that segment. The interesting case occurs when a vertex appears on two different segments; such a vertex, call it $v^{\#}$ , gets each of two primary colours — and its secondary colour shows which two segments this vertex lies on; for example imagine that the two strands are red and yellow, so that $v^{\#}$ is coloured orange. Figure 3 on page 31 and Figure 4 illustrate this colouring.

4.1. Big picture perspective: $k$ coloured segments, $m$ toggle points

We will have $k$ segments each of length $t=N^{.6}$ . The expected number of leftmost $(n-1)$ -tuple repeats within a single segment is about $\left(\substack{t \\[3pt] 2}\right)/N\doteq .5 N^{.2}$ . The expected number of repeats between two different given segments is about $t^2/ N = N^{.2}$ , so the expected number of repeats between two different segments, combined over all $\left(\substack{k \\[3pt] 2}\right)$ choices for which two segments, is about $\left(\substack{k \\[3pt] 2}\right) \times N^{.2}$ . This is a huge number of repeats (each based on one vertex having a secondary colouring), and we intend to find $m$ such repeats, say at $v^{\#}_1,\ldots,v^{\#}_m$ in narrowly constrained spatial positions. The goal is to show that, with high probability, for all $2^m$ choices of how to change $f$ by toggling the values of $f(v^{\#}_i)$ for $i \in I \subset \{1,2,\ldots,m\}$ , the same $m$ vertices will be picked out by the narrow two-dimensional spatial constraints.

In this section we present further figures intended to assist the reader’s intuition. We also give an algorithm which for a given logic $f$ and starting edges $e_i$ finds $m$ vertices $v^{\#}_1$ , …, $v^{\#}_m$ . These points—we call them toggle points—give rise to a family of $2^m$ functions. We also define (in Section 4.10) a process called relativisation which associates with $\pi _f$ in $S_N$ a permutation $\sigma$ in $S_k$ , $k$ being fixed and $N \to \infty$ . It will be shown that as $f$ varies over the $2^m$ functions in a ‘toggle class’, the resulting $\sigma ’s$ cover $S_k$ almost uniformly. In Section 5, it will be shown that this uniform coverage of $S_k$ (for each fixed $k$ ) is a sufficient condition to prove Theorem 1.

A critical issue is that the algorithm for choosing the toggle points must be such that, if the feedback $f$ is replaced by any one of the $2^m$ functions in the toggle class, the algorithm would find the same toggle points and the same class.Footnote ⁷ However this is not necessarily the case and the probability of success for the algorithm must be estimated; this leads to the definition of an event $H$ .

4.2. Toggling: The case $k=2$ and $m=1$

We show what can happen when we toggle one bit of a logic $f$ . We have two segments of length $t$ , which share a vertex $v^{\#}$ . Toggling changes the value of $f$ , only at $v^{\#}$ , and gives a new logic ${f^{*}}$ . Suppose that the segments under $f$ were red and yellow, and that $v^{\#}$ appears at position $i$ on the red segment, and position $j$ on the yellow segment. Overall, this repeat has spatial location $(i,j)$ and colour orange. Exactly one such repeat was visualised in Figures 2 and 3; it occurred with $(i,j)=(56,53)$ . The displacement is $i-j$ — we have a preferred sequence of colours, (derived from the rainbow ROY G. BIV) where red comes before yellow — hence the displacement is 3, rather than $-3$ , in this example.

4.3. Picking the ‘earliest’ toggle with a small displacement

Consider the case where we have $k=2$ segments and want to find a single vertex $v^{\#}$ via a recipe which, when applied to the segments under the toggled logic ${f^{*}}$ , still picks out the same vertex. A good recipe involves naming a small bound $d$ on the absolute displacement $|i-j|$ (thus staying close to the ‘diagonal’), and then picking the ‘earliest’ pair $(i,j)$ that satisfies the displacement bound. This was the key to overcoming the ‘fallacy’ described in Footnote 7.

The specific choice of how to define earliest is somewhat arbitrary; we will take smallest $(i+j)$ as the first criterion for earliest, with ties to be broken according to smallest value of $\max\!(i,j)$ — given that $i+j = i'+j'$ , this is equivalent to taking smallest absolute displacement for the tie-break criterion. For use in the case of $k$ colours and $\left(\substack{k \\[3pt] 2}\right)$ colour pairs $\alpha =(a,b)$ , break further ties according to $\min\!(a,b)$ and then $\max\!(a,b)$ .

The logic $f$ , with value at $v^{\#}$ complemented, gives a new logic ${{f^{*}}}={\textrm{Toggle}}(f,\{v^{\#} \})$ , so that ${{f^{*}}}(v^{\#})=1-f(v^{\#})$ , while ${{f^{*}}}(v)=f(v) \ \ \forall v \ne v^{\#}$ . It is possible that changing the logic bit at $v^{\#}$ , will cause an earlier pair to become available as the location of a match between the two segments; so that the recipe for picking the earliest small displacement match, applied to ${f^{*}}$ , picks out a different vertex instead of $v^{\#}$ . In this case, the word toggle is very misleading: the overall operation (find $v^{\#}$ , then complement the logic at that vertex) is not an involution. Our programme is to specify a displacement bound $d$ that varies with $n$ , in such a way that 1) with high probability, at least one small displacement match can be found, and 2) with high probability, the vertex for the earliest small displacement match is the same in the logic ${{f^{*}}} ={\textrm{Toggle}}(f,\{v^{\#} \})$ at the vertex selected for $f$ . The example in Figure 8, viewed with any $d \ge 3$ , illustrates what might go wrong with respect to 2).

Figure 6. About 333 of the 500 occurrences of repeats from Figure 5, but now viewed as among $k=3$ segments of length $t=N^{.6}\doteq 1.38 \times 10^6$ . The $\left(\substack{k \\[3pt] 2}\right)$ $t$ by $t$ above-diagonal squares from Figure 5 are superimposed, so the expected number of points is about $\left(\substack{k \\[3pt] 2}\right) \times N^{.2}\doteq 334.3$ . The approximately 167.1 repeats where both occurrences lie in the same segment, corresponding to the $k$ right triangles hugging the diagonal in Figure 5, are not shown. In Section 4.5 we discuss this picture, suggesting scaling for the axes, so that in each colour, the picture is approximately a standard (rate 1 per unit area) two-dimensional Poisson process. The colour scheme is intended to be purple, green, orange.

Figure 7. Toggling. An example with $t=90$ and displacement $d=3$ . The same repeat as shown in Figure 3 with location (56,153) and shown by the orange dot in Figure 4. When all $\left(\substack{k \\[3pt] 2}\right)$ squares are superimposed, as in Figure 6, the spatial location becomes $(i,j)=(56,53)$ . Before the toggle, we have two segments of length $t=90$ ; after the toggle, the segments have length $t \pm d$ , that is, 93 and 87.

Figure 8. Toggling. This is a continuation of the example in Figure 7, with one repeat with location (56,153), shown by the orange dot in Figure 4. When all $\left(\substack{k \\[3pt] 2}\right)$ squares are superimposed, as in Figure 6, the spatial location becomes $(i,j)=(56,53)$ . Now suppose there were an additional repeat, (which would have been shown by a red dot at (53,58) in Figure 4,) shown here in Figure 8 by the pair of red dots for $f$ . After the toggle at the orange vertex, vertex 53, along the segment that starts red and finishes yellow, is the same as vertex 55, along the segment that start yellow and finishes red. So, in the logic ${f^{*}}$ , we have two matches between the two segments: the original, at (56,53), shown by the orange dots, and a new one, at (53,55), shown by the red dots.

Recall, from Section 3, that $t$ is the length of our segments. To get high probability in 1), a necessary and sufficient condition is that

(32) \begin{equation} td/N \to \infty . \end{equation}

To get high probability in 2), a necessary and sufficient condition is that

(33) \begin{equation} d^2/N \to 0. \end{equation}

The argument that (33) suffices is somewhat delicate, akin to a stopping time argument; it is easier to prove — see (37) — that a sufficient condition is that

(34) \begin{equation} td^3/N^2 \to 0; \end{equation}

and then it will be easy to arrange for situations corresponding to pairs $(t,d)$ satisfying both (32) and (34).

4.4. Displacements caused by toggles

Suppose we have $k=3$ colours, as shown in Figure 9. There are three segments of length $t=90$ , with respect to $f$ . The segment with respect to $f$ , starting with $e_{1}$ , coloured red, has $v^{\#}_1$ in position 6 and $v^{\#}_2$ in position 35 — so the red segment, of length 90, is divided into an initial red path of length 6, followed by a red path of length 29, followed by a red path of length 55.

Figure 9. With starting edges $e_1,e_2,e_3$ , three segments under the logic $f$ are shown in the top part of the display; the red and yellow segments share a vertex $v^{\#}_1$ , coloured orange, early on, the red and blue segments share a vertex $v^{\#}_2$ , coloured purple, at a intermediate time, and the yellow and blue segments share a vertex $v^{\#}_3$ , coloured green, at a late time. We take ${{f^{*}}}={\textrm{Toggle}}(f,\{v^{\#}_1 \})$ and ${{f^{**}}}={\textrm{Toggle}}(f,\{v^{\#}_1,v^{\#}_2 \})$ to be the logics formed by toggling at $v^{\#}_1$ , and at both $v^{\#}_1$ and $v^{\#}_2$ . The middle part of the display shows the three segments under ${f^{*}}$ , and the bottom part of the display shows the three segments under ${f^{**}}$ .

The $f$ segment starting with $e_{2}$ , coloured yellow, has $v^{\#}_1$ in position 3 and $v^{\#}_3$ in position 75; hence, it is divided into yellow paths of lengths 3, 72, 15, in that order.

The $f$ segment starting with $e_{3}$ , coloured blue, has $v^{\#}_2$ in position 37 and $v^{\#}_3$ in position 71; hence, it is divided into blue paths of lengths 37, 34, 19, in that order.

Next, consider ${{f^{*}}} \;:\!=\; f$ , toggled at $v^{\#}_1$ . Its segment starting from $e_{1}$ has length 6 red followed by length (72 + 15) = 87, for a total length of 93. Its segment starting from $e_{2}$ has length 3 yellow, followed by length (29 + 55) = 84, for a total length of 87. The ${f^{*}}$ segment starting from $e_{3}$ is still length 90, all blue. More importantly, $v^{\#}_2$ has moved from position 35 on ${\textrm{Seg}}(f, e_1)$ to position 32 on ${\textrm{Seg}}(f,e_2)$ , and $v^{\#}_3$ has moved from position 75 on ${\textrm{Seg}}(f, e_2)$ to position 78 on ${\textrm{Seg}}(f,e_1)$ , so these have new positions under ${f^{*}}$ , i.e., have been displaced.

Now consider the full effect of changing from $f$ to ${f^{*}}$ , by toggling the logic at the bit $v^{\#}_1$ which appeared at positions $(i,j)=(i,i-d) = (6,3)$ , with $d=3$ , for the red and yellow segments: every red vertex later than 6 gets displaced by $-d$ , and every yellow vertex later than 3 gets displaced by $+d$ . If a match occurs at $(I,J)$ in the $f$ segments, and the colours involved are red, and some colour, call it $a$ , with $a$ not equal to yellow, then:

• Case 1. Color $a$ comes after red, in the list of $k$ colours: the ordered colour pair is (red, $a)$ . The index $I\gt i$ belongs to a red vertex in position $I$ under the logic $f$ , and this vertex has position $I-d$ under the logic ${f^{*}}$ . So the point at $(I,J)$ moves to position $(I-d,J)$ .Footnote ⁸

• Case 2. Color $a$ comes before red, in the list of $k$ colours; the colour pair is $(a,$ red). The index $J\gt i$ belongs to a red vertex, in position $J$ under the logic $f$ , but in position $J-d$ under the logic ${f^{*}}$ . So the point at $(I,J)$ moves to position $(I,J-d)$ .

If there is an orange match at $(I,J)$ for the $f$ segments, with $I\gt i$ and $J\gt j$ , this match will move to $(I-d,J+d)$ .

Similarly, a match between yellow, and some $a$ not equal to red, occurring at $(I,J)$ under the logic $f$ , moves to $(I+3,J)$ or $(I,J+3)$ under the logic ${f^{*}}$ , according to whether $a$ comes after or before yellow, in the list of all $k$ colours.

This effect can be seen in Figure 9: the orange dot is at (6,3) with displacement $d=3$ , the purple dot occurs at (35,37) under $f$ , but at (32,37) under ${f^{*}}$ and ${f^{**}}$ .

More cases can be seen in Figure 10.

Figure 10. Displacements caused by a single toggle. An example with $t=90$ , and three colours, red, yellow, and blue. Say the toggle is at $v^{\#}_2$ occurring at (red,blue) time $(35,40)$ , similar to the purple vertex at ( $35,37)$ in Figure 9, but with the displacement changed from −2 to −5, for the sake of being easier to see in the two-dimensional picture. We have thrown in several more matches between two different colours, at various earlier and later times, to show the resulting two-dimensional displacements. Red vertices at times greater than 35 have their time increased by 5, and blue vertices at times greater than 40 have their time decreased by 5. Two-dimensional match locations are indicated by a solid circle for the logic $f$ and an open circle for the logic ${f^{*}}$ .

4.5. The natural scale: by $1/\sqrt{N}$ for length, by $1/N$ for area

One gets an intuitive grasp of the process of spatial locations of places $(i,j)$ where two segments of different colours share a vertex, by looking at a picture such as that in Figure 6 — even though the axes are unlabelled.

One view would be that the square is $t$ by $t$ , with $n=34, N=2^n,t=N^{.6}$ , i.e., about 1.3 million by 1.3 million. The other natural view is that the square is about $t/\sqrt{N}$ by $t/\sqrt{N}$ , i.e., about 10.556 by 10.556, with area 111.43.

The latter point of view is natural, since at each $(i,j)$ , for each colour pair $(a,b)$ , $1 \le a \lt b \le k$ , with $\doteq$ to allow a small discrepancy for the failure of the good event, ${\mathbb{P}}($ an arrival Footnote ⁹ at $(i,j)$ in those colours) $ \;:\!=\;{\mathbb{P}}(v_{a,i}=v_{b,j}$ and $v_{a,i-1} \ne v_{b,j-1})$ $\doteq{\mathbb{P}}($ there is a leftmost $(n-1)$ -tuple repeat at a specific locationFootnote ¹⁰ in the coin tossing sequence) $= 1/N$ . Hence, scaling length by $1/\sqrt{N}$ , so that area is scaled by $1/N$ , leads to

\begin{equation*} \text { the expected number of arrivals per unit area } =1. \end{equation*}

The picture in Figure 6, viewed as occurring on a 10.556 by 10.556 square, closely resembles a (standard, rate $1 \, dy \, dx$ ) two-dimensional Poisson process, in each secondary colour pair. And overall, ignoring colour, the picture resembles the rate $\left(\substack{k \\[3pt] 2}\right) \, dy \, dx$ Poisson process on the $t/ \sqrt{N}$ by $t/ \sqrt{N}$ square.

There are additional requirements for the Poisson process, beyond having intensity $1 \, dy \, dx$ . Namely, probabilistic independence for the counts in disjoint regions. We do have a good Poisson process approximation, for a combination of two reasons. First, the good event $G={G_{(k,t)}}$ from Theorem 6 gives a high-probability coupling (since $t=N^{.6}$ entails $t^3n^3/N^2 \to 0$ ) between coin tossing and the $k$ de Bruijn segments of length $t$ . Second, the Chen–Stein method, Theorem 3 of [Reference Arratia, Goldstein and Gordon6], gives a total variation distance upper bound (tending to zero since $t=N^{.6}$ entails $t^3n/N^2 \to 0$ ) between the process of indicators of leftmost $(n-1)$ -tuple repeats for coin tossing, and a process with the same intensity, but mutually independent coordinates.

We get our intuition from the Poisson process. But for our proofs, we will work directly with the discrete, dependent processes.

4.6. Controlled regions for $m$ successive potential toggle vertices

4.6.2. The search regions

We divide the time interval $[0,t]$ into $m$ equal length pieces. On the earliest piece, with times in [0,t/m], we demand that we can find a match $(i,j)$ with $|i-j|$ very very very small, but no upper bound on $\max\!(i,j)$ other than $\max\!(i,j) \lt t/m$ . In the natural scale of Section 4.5 we are searching for matches in a very very very thin and very very long rectangle surrounding the diagonal line $i=j$ ; this rectangle has a large area. On the second piece, with times in $[t/m,2t/m]$ , we relax the notion of thin, expanding by a large factor $r^2$ , relax the notion of long, dividing by the factor $r^2$ , thus keeping the area constant. We continue this pair of geometric progressions, so the $m$ th region is a thin long rectangle — but still with the same area.

Here is a concrete way to accomplish the above, together with $t^3n^3/N^2 \to 0$ and with $k$ fixed. Let

\begin{equation*} t \;:\!=\; m \, N^{.6}, a \;:\!=\; N^{.1}, \text { so that } t/m = a \sqrt {N}. \end{equation*}

The last condition should be understood as ‘in the natural scale from Section 4.5, the $t$ by $t$ rectangle is $ma$ by $ma$ , and length $t/m$ for the discrete $i$ and $j$ corresponds to length $a$ ’. Let

\begin{equation*} r \;:\!=\; a^{1/(2m+1)}, \text { so that } r^{2m+1} = a, \end{equation*}

and, ignoring the factors of $\sqrt{2}$ involved in the 45 degree rotation, take the thin long rectangles to have shapes

(35) \begin{eqnarray} d_1 = \frac{1}{r^{2m}} & \textrm{ by } & w_1 = r^{2m+1} = (t/m)/\sqrt{N} \nonumber \\[5pt] d_2 = \frac{1}{r^{2m-2}} & \textrm{ by } & w_2 = r^{2m-1} \\[5pt] & \vdots & \nonumber \\[5pt] d_m = \frac{1}{r^2} & \textrm{ by } & w_m = r^{3} \nonumber \end{eqnarray}

Indexing by $\ell =1$ to $m$ , the $\ell$ th rectangle is $d_\ell \;:\!=\; r^{2\ell -2m-2}$ by $w_\ell \;:\!=\; r^{2m-2\ell +3}$ on the natural scale. Directly in terms of the discrete $i$ and $j$ , we define

(36) \begin{eqnarray}{\textit{Region}}_\ell =\left \{ (i,j)\;:\;\frac{|i-j|}{\sqrt{N}} \lt r^{2\ell -2m-2} \right . & \textrm{ and } & \\[5pt] \left . \frac{(\ell -1)t}{m} \le \min\!(i,j) \le \max\!(i,j) \le \frac{(\ell -1)t}{m} + \frac{t/m}{r^{2(\ell -1)}} \right \}, & & \nonumber \end{eqnarray}

so one checks that 1) as $\ell$ increases by 1, the thinness constraint relaxes by a factor of $r^2$ , while the width constraint becomes more severe by a factor of $r^2$ , so the area stays constant, 2) the first region, with $\ell =1$ , allows $i,j \in [0,t/m]$ and 3) the last region, with $\ell =m$ , has $|i-j|/\sqrt{N} \le 1/r^2 = o(1)$ as $n \to \infty$ .

Consider the possibility discussed in Section 4.3, where a toggle at a vertex appearing in two differently coloured segment enables a match within a single segment to become, after the toggle, an earlier match between two different segments. For each $\ell =1$ to $m$ , with the $(t,d)$ in (34) given by $t=w_\ell \sqrt{N}, d= d_\ell \sqrt{N}$ , the condition in (34) is indeed satisfied by our specific choice in (35). On the natural scale, and ignoring rotation, we are searching for a match in a $\delta = d_\ell$ by $W=w_\ell$ rectangle, thin and long, with $\delta \to 0$ and area $\delta W \to \infty$ . The condition (34), on the natural scale, means that $\delta ^3 W \to 0$ . It implies that, with high probability, we do not find a match between two differently coloured segments (at $(i,j)$ in the rectangle, with $|i-j|/\sqrt{N}\lt \delta$ ,) and simultaneously a nearby match within a single segment. Here, nearby means with both indices within distance $\delta \sqrt{N}$ from $i$ or $j$ . Now, the $\delta$ by $W$ rectangle can be covered by $W/\delta$ squares, each square of size $4 \delta$ by $4 \delta$ , and with each successive square being a translate, by $\delta$ , of the previous square. Ignoring constant factors,Footnote ¹¹ the expected number of arrivals in one square is order of $\delta ^2$ , and the chance of two or more arrivals in that one square is order of $\delta ^4.$ Thus the expected number of squares with two or more arrivals is order of

(37) \begin{equation} W/\delta \times \delta ^4 \ = \delta ^3 W \ \ \ \to 0. \end{equation}

4.7. Definition of the choice functions

Write $V = \mathbb{F}_2^{n-1}$ for the set of vertices in $D_{n-1}$ , and write ‘null’ for a special value, not in $V$ , used to encode ‘undefined’. Recall that we write $\boldsymbol{{e}} = (e_1,\ldots,e_k)$ for the starting $n$ -tuples for $k$ segments, and $S_{n,k} = \{(f,\boldsymbol{{e}}) \}$ for the space in which we make a uniform choice of logic and starting edges. Also recall our notation (31) for vertices along the $k$ segments. Note that we have both $k$ segments and $k$ colours; these are different concepts, and ultimately, colours will be labelled according to the segment labels under $f$ — but on the soon to be defined ‘happy’ event $H$ , finding $v^{\#}_i$ on two different segments of $f$ will be equivalent to finding $v^{\#}_i$ on two different colours. To keep track of the colours, let

(38) \begin{equation} {\mathcal{A}} \;:\!=\; \{\alpha = (a,b)\;:\;1 \le a \lt b \le k \} \end{equation}

For $\ell =1$ to $m$ , we define

(39) \begin{equation} {\textrm{Candidates}}_\ell \;:\;S_{n,k} \to [0,t]^2 \times{\mathcal{A}} \end{equation}

\begin{equation*} {\textrm {Candidates}}_\ell (f,\boldsymbol{{e}}) = \{ (i,j,a,b)\;:\;(i,j) \in {\textit {Region}}_\ell \text { and } v_{a,i} = v_{b,j} \} \end{equation*}

where ${\textit{Region}}_\ell$ is defined by (36).

For $\ell =1$ to $m$ , we define

(40) \begin{equation} {\textit{Choice}}_\ell \;:\;S_{n,k} \to V \cup \{{\textit{null}} \}, \ \ {\textit{Choice}}_\ell (f,\boldsymbol{{e}}) = v^{\#}_\ell \textrm{ or else }{\textit{null}} \end{equation}

where the value is $\textit{null}$ if the set of candidates is empty, and otherwise, picking the first $(i,j,a,b)$ in ${\textrm{Candidates}}_\ell (f,\boldsymbol{{e}})$ , $v^{\#}_\ell$ is the vertex with $v^{\#}_\ell =v_{a,i} = v_{b,j}$ . To be very careful, the order for first is the lex-first order on $(i+j, \max\!(i,j), a,b)$ .

4.8. The happy event $H = H(k,m,n)$

We now describe a subset of $S_{n,k}$ and refer to this subset as the happy event $H$ . One requirement for $(f,\boldsymbol{{e}}) \in H$ is that, for $\ell =1$ to $m$ , each of the values ${\textit{Choice}}_\ell (f,\boldsymbol{{e}}) \ne{\textit{null}}$ . Starting with such an $(f,\boldsymbol{{e}})$ , the choice functions pick out a set of $m$ distinct vertices; call them $v^{\#}_1,\ldots, v^{\#}_m$ , and name the set, ${{V^{\#}}} = \{ v^{\#}_1,\ldots, v^{\#}_m \}$ — we will use this notation in (42) below.

Given a set of vertices, $U \subset V$ , we denote the logic $f$ toggled at the vertices in $U$ as ${\textrm{Toggle}}(f,U)$ , defined by

(41) \begin{equation} {\textrm{Toggle}}(f,U) \;:\!=\;{{f^{*}}}, \textrm{ where }{{f^{*}}}(v) = \left \{ \begin{array}{l@{\quad}l} 1-f(v) & \mbox{if $v \in U$} \\[5pt] f(v) & \mbox{if $v \in V \setminus U$} \end{array} \right . . \end{equation}

We define $H$ as follows:

(42) \begin{equation} H = \{ (f,\boldsymbol{{e}})\;:\;\forall \ U \subset{{V^{\#}}}, \textrm{ with }{{f^{*}}} ={\textrm{Toggle}}(f,U), v^{\#}_\ell ={\textit{Choice}}_\ell ({{f^{*}}},\boldsymbol{{e}}) \end{equation}

\begin{equation*} { \textit {and } \text {the segments }} {\textrm {Seg}}(f,e_i,t) \text { collectively have } k(t+1) \text { distinct edges} \}. \end{equation*}

Informally, $(f,\boldsymbol{{e}})$ is in the happy event iff the $k$ segments involve no $n$ -repeats, and the choice recipes find $m$ potential toggle vertices, and all $2^m$ cousins ${f^{*}}$ , formed by toggling at a subset of those vertices, give rise to the same $v^{\#}_1,\ldots,v^{\#}_m$ .