Short reachability networks

We first give an upper bound construction. Let $n\geq 2$ be given. Our sequence of transpositions starts with the following transpositions, in order.

1. 1.

   $(1,2)$.

2. 2.

   $(1,x)$ for all odd $3\leq x\leq n$.

3. 3.

   $(2,y)$ for all even $4\leq y\leq n$.

4. 4.

   $(x,x+1)$ for all odd $3\leq x\leq n-1$.

If $n$ is odd, we also add the transposition $(1,2)$ at the end, which will be needed to reach the position $(1,n)$. This defines a sequence of transpositions of length $\left\lceil 3n/2\right\rceil-2$, and it is straightforward to check that this sequence forms a 2-reachability network.

The main work in this proof is in showing the lower bound. Suppose that $\sigma=(\sigma_{1},\dots,\sigma_{\ell})$ is a shortest $2$-reachability network on $n\geq 2$ elements. Given two (distinct) counters on positions $1$ and $2$, a subsequence of $\sigma$ defines a permutation that moves the counters to new (distinct) positions $x,y\in\{1,\dots,n\}$. By the definition of $2$-reachability, it must be possible to reach any such pair of positions $(x,y)$.

We will consider a process on an auxiliary multigraph $G$ with vertex set $\{1,\dots,n\}$. An example is drawn in Figure 1 for the sequence of transpositions given earlier for the upper bound (for $n=9$). The process begins by growing two trees of black edges from the vertices 1 and 2, which we denote $T_{1}$ and $T_{2}$ respectively. We initialise $G$ as the empty graph and process the transpositions in order, adding an edge for each. We start with a set of active vertices $\{1,2\}$. Any other vertex becomes active when it first appears in a transposition with an active vertex. Note that no counter can be sitting on an inactive vertex, and (by minimality) no transposition joins two inactive vertices. When we process a transposition $(a,b)$, we add an edge $ab$ to $G$. We colour the edge black if either $a$ or $b$ was inactive, and red otherwise. The black edges always form a forest consisting of at most two trees each containing one of the starting positions.

We now modify our sequence to obtain a sequence with a nicer form (and at most the same length). Let $c_{1}$ and $c_{2}$ denote the counters that start on $1$ and $2$ respectively, and let $(a,b)$ where $a\in V(T_{1})$ and $b\in V(T_{2})$ be the transposition at the first time the two trees meet. Before this time, the counter $c_{1}$ is contained in $V(T_{1})$ and the counter $c_{2}$ is contained in $V(T_{2})$, so immediately after $(a,b)$ the first counter $c_{1}$ is contained in $V(T_{1})\cup\{b\}$ and the second counter $c_{2}$ is contained in $V(T_{2})\cup\{a\}$. We replace the initial sequence of transpositions, up to and including $(a,b)$, by the following sequence. Start with the transpositions $(1,v)$ for all $v\in V(T_{1})\setminus\{a\}$, followed by the transpositions $(1,u)$ for all $u\in V(T_{2})\setminus\{b\}$. Lastly, we do $(1,2)$, $(1,a)$ and $(2,b)$. The new sequence can reach every pair $(x,y)$ that the original sequence could reach, and uses at most as many transpositions. We colour the edge $(1,2)$ black, so the black edges form a spanning tree which we will view as rooted at the pair of vertices $\{1,2\}$.

We now prove that the sum of the red degrees (counted at the end of the process) is at least $n-2$, which shows that the number of edges in $G$ (and transpositions in our sequence) is at least $n-1+\left\lceil(n-2)/2\right\rceil=\left\lceil 3n/2\right\rceil-2$, as desired. We first need some more definitions. The subtree $T_{v}$ rooted at $v$ is the set of vertices $u$ (including $v$) such that the unique path from $u$ to $\{1,2\}$ in the black tree passes through $v$. Any vertex in the subtree rooted at $v$ is said to be above $v$. The deficit of a vertex $v$ is defined as

where $d_{R}(u)$ denotes the number of red edges incident to $u$. Note that the red edges may join vertices of $T_{v}$ to vertices outside $T_{v}$, so may not sit inside the tree. In order to bound the sum of the red degrees, we will inductively bound the deficit of vertices.

Each vertex $v\not\in\{1,2\}$ has a unique vertex adjacent to $v$ on the unique path in the black tree from $\{1,2\}$ to $v$. We call this the parent of $v$ and we denote it by $p(v)$ (which could equal $1$ or $2$). The children $\mathcal{C}(v)$ of a vertex $v$ are the vertices $u$ for which $p(u)=v$. We may rewrite the formula for the deficit of $v$ in the following inductive manner.

Note that the first edge that can carry a counter to $v\not\in\{1,2\}$ is the black edge to its parent; if $v$ is not incident to a red edge, then this is the only way to get a counter to $v$.

We show the following claim inductively, which we will extend to $v\in\{1,2\}$ to finish the proof.

We now show that the vertices $1$ and $2$ have deficit at most $1$ as well. The difficulty in this case is that it is possible to put a counter on $1$ (say) using the black edge $(1,2)$. This means that we can put a counter on $\ell(x)$ and $\ell(y)$ without requiring a red edge between them. However, this does not take into account the fact that the counters are distinguishable, and we only need to modify the argument from the claim slightly to make use of this.

Let the counters starting in positions 1 and 2 be $c_{1}$ and $c_{2}$ respectively, and suppose there is a child $x$ of $1$ which has deficit 1. Then there is a vertex $\ell(x)$ such that the only way for a counter to reach $\ell(x)$ is to follow the path of black edges from $1$ to $x$ to $\ell(x)$. If the transposition $(1,x)$ occurs before the black edge $(1,2)$, then there is no way to move the counter $c_{2}$ to $\ell(x)$, and $x$ must have deficit $0$. The black edge $(1,2)$ is the only way to put a counter on $1$ without using a red edge, so after this transposition we can apply a proof similar to that of the claim above to see the deficit of 1 is at most 1. Indeed, if $x$ and $y$ are children of $1$ with deficit 1, then there has to be some transposition that can place a counter on $1$ in between $(1,x)$ and $(1,y)$ in order to reach $(\ell(x),\ell(y))$. We have just argued that this is not the transposition $(1,2)$, and so it must be a red transposition. Likewise, there must be a red transposition after the last black edge to a vertex $z$ of deficit 1 in order to end with the counters in $1$ and $\ell(z)$.

Since every vertex is in either the subtree rooted at 1 or the subtree rooted at 2 and the deficit of these two subtrees is at most $1$, the sum of the red degrees must be at least $n-2$. Hence, there are at least $\left\lceil(n-2)/2\right\rceil$ red edges and at least $\left\lceil 3n/2\right\rceil-2$ transpositions in the sequence. ∎

Let $t\geq 3$ be a natural number and choose $\varepsilon\in(0,\frac{1}{t+1})$. Let $n>t$ and set $L=\left\lfloor n^{1-\varepsilon}\right\rfloor$. Recall that the counters start in positions $\{1,\dots,t\}$.

Let $G$ be a random bipartite graph with vertices $A\cup B=\{a_{t+1},\dots,a_{n}\}\cup\{b_{1},\dots,b_{L}\}$ constructed by adding two uniformly random edges from $a_{j}$ to $\{b_{1},\dots,b_{L}\}$ for each $j$. The random transposition sequence begins with $L$ phases. In the $i$th phase, we add the transpositions $(1,j)$ for each $j\in\{2,\dots,t\}$ followed by $(1,j)$ for the $j\in\{t+1,\dots n\}$ such that $a_{j}$ is adjacent to $b_{i}$. Finally, the sequence ends with a $t$-reachable network over the first $t$ positions.

Fix positions $x_{1},\dots,x_{t}\in\{1,\dots,n\}$ for which we need to find a subsequence that puts our $t$ counters into those positions. Let the $x_{i}$ which are in $\{t+1,\dots,n\}$ be $x_{i_{1}},\dots,x_{i_{m}}=y_{1},\dots,y_{m}$.

Suppose that there is a matching $\{a_{y_{j}}b_{s_{j}}:j\in[m]\}$ in $G$ containing the vertices $a_{y_{1}},\dots,a_{y_{m}}$. For $j\in[t]$, we can put the $j$th counter in position $y_{j}=x_{i_{j}}$ during phase $s_{j}$ using the transpositions $(1,j)$ and $(1,y_{j})$. The remaining counters can easily be positioned using the $t$-reachable sequence at the end.

It remains to show that there is some choice for $G$ such that every set of $s\leq t$ vertices from $A$ is in a matching. The probability that a given set of $s$ vertices from $A$ has at most $s-1$ neighbours in $B$ is at most

Hence, by the union bound, the probability that there is a set $A^{\prime}\subseteq A$ of size at most $t$ and with at most $|A^{\prime}|-1$ neighbours is at most

Since $\varepsilon<1/(t+1)$, this probability is less than 1 for $n$ sufficiently large. This implies there exists a suitable choice for the graph $G$ (for sufficiently large $n$).

Note that for $j\in\{t+1,\dots,n\}$, the transposition $(1,j)$ appears exactly twice in the sequence since the vertex $a_{j}$ has degree exactly 2. We can create a $t$-reachable network on the first $t$ positions using $O(t\log t)$ star transpositions. Hence, there exists a $t$-reachability network using at most

transpositions. ∎

The upper bound is given in the discussion above, so we only need to prove a matching lower bound. For this we proceed much as in the proof of Theorem 2. We colour the transposition $(1,x)$ black if it is the only occurrence, red if it has occurred previously and blue otherwise (i.e. if it is a first occurrence, but will occur later as well). There are $n-1$ transpositions that are either black or blue, and it suffices to show that there are at least $(n-1)/2$ red transpositions.

We claim that the number of red transpositions is at least the number of black transpositions. Let the two counters be $c_{1}$ and $c_{2}$. Suppose $(1,x)$ is the last black transposition where $x\geq 3$. Then there needs to be a red transposition after $(1,x)$ in order to leave the counters $c_{1}$ and $c_{2}$ in $1$ and $x$ respectively. Similarly, in between any pair of black transpositions $(1,x)$ and $(1,y)$ where $x,y\geq 3$ there has to be a red transposition in order to leave the counters $c_{1}$ and $c_{2}$ in $y$ and $x$ respectively. This shows the claim when $(1,2)$ is used multiple times, and there is no black $(1,2)$.

If $(1,2)$ is used exactly once, then it must be the first black transposition. Indeed, if $(1,x)$ is a black transposition where $x\geq 3$, then there is no way to put the counter $c_{2}$ in $x$ unless $(1,2)$ has already been. If $(1,2)$ were to be the only black transposition, then $(1,x)$ must appear twice for every $x\geq 3$ and there are $2(n-2)+1$ transpositions. This is at least $\left\lceil 3(n-1)/2\right\rceil$ for $n\geq 3$ and we would therefore be done. Hence, if $(1,2)$ is used exactly once, it must be the first but not the last black transposition. The above argument shows there is a red transposition after the last black transposition and in between every pair of black transpositions $(1,x)$ and $(1,y)$ where $x,y\geq 3$. It is enough to show there is a red transposition between $(1,2)$ and the first black transposition $(1,x)$ where $x\geq 3$, and this is easy to argue: there must be a red transposition between $(1,2)$ and $(1,x)$ in order to leave $c_{1}$ in $x$ and $c_{2}$ in $1$.

By definition, the number of red transpositions is at least the number of blue transpositions. Since there is a total of $n-1$ black and blue transpositions, there must be at least $(n-1)/2$ of one of them and there are at least $(n-1)/2$ red transpositions. ∎

We use a discharging argument to show that, after disregarding a constant number of the transpositions, 1.6 is a lower bound on the average number of times that a star transposition is used.

Let $T=(T_{1},\dots,T_{\ell})$ be a sequence of star transpositions that forms a 2-uniformity network on $n$ elements. We assign each transposition $(1,a)$ a weight equal to the number of times $(1,a)$ is used in $T$. Let $\sigma_{1},\dots,\sigma_{m}$ be the transpositions of $T$ which are used exactly once and are of the form $(1,x)$ for $x\geq 3$. We transfer weight from the transpositions which are used multiple times to the transpositions which are used exactly once according to the following rules.

• •

  If the last appearance of $(1,a)$ and $(1,b)$ is between $\sigma_{i}$ and $\sigma_{i+1}$, then they each transfer $0.3$ to $\sigma_{i}$ and $0.1$ to $\sigma_{i+1}$.

• •

  If there is only one transposition which appears for the last time between $\sigma_{i}$ and $\sigma_{i+1}$, it transfers $0.4$ to $\sigma_{i}$.

• •

  Each transposition which appears between $\sigma_{i}$ and $\sigma_{i+1}$ for neither the first time nor the last time, transfers $0.6$ to $\sigma_{i}$ and $0.2$ to $\sigma_{i+1}$.

If $(1,a)$ is used exactly twice, then it transfers $0.4$ and ends with weight $1.6$. A transposition used more than twice transfers $0.8(i-2)+0.4$ so ends with weight $1.6+0.2(i-2)$. Hence, we only need to show that all but a constant number of the transpositions that are used once (the $\sigma_{i}$) end with weight at least $1.6$.

Let $\sigma_{i-1}=(1,a)$ and $\sigma_{i}=(1,b)$. In order to end with counters in both $a$ and $b$, there must a transposition between $\sigma_{i-1}$ and $\sigma_{i}$ which can place a counter in position 1. In particular, there is either a transposition which is not appearing for the first time, or $(1,2)$ appearing for the first time. This immediately shows that all but one of $\sigma_{1},\dots,\sigma_{m-1}$ have weight at least 1.4. We claim that in between either $\sigma_{i-1}$ and $\sigma_{i}$ or between $\sigma_{i}$ and $\sigma_{i+1}$ one of the following must hold:

1. 1.

   there are at least two transpositions appearing for the last time,

2. 2.

   there is a transposition appearing for neither the first nor the last time,

3. 3.

   there is the transposition $(1,2)$ appearing for the first time.

If one of the first two cases occurs, then $\sigma_{i}$ ends with weight at least 1.6, while the last case can only occur twice. This analysis does not apply to $\sigma_{1}$ and $\sigma_{m}$, showing that the number of $\sigma_{i}$ which end up with weight less than 1.6 is at most four.

We now prove the above claim. Let $\sigma_{i-1}=(1,a)$, $\sigma_{i}=(1,b)$ and $\sigma_{i+1}=(1,c)$ and assume that none of the conditions above hold. In between $\sigma_{i-1}$ and $\sigma_{i}$, there can be only transpositions used for the first time and a single transposition $(1,\ell)$ used for the last time. We may write the sequence as

where the $f_{j}$ are distinct, not equal to 2 and are appearing for the first time.

The only way to end the sequence of lazy transpositions with the counters in $\{a,b\}$, $\{a,\ell\}$ or $\{b,\ell\}$ is to start the sequence (1) with the two counters in $\{1,\ell\}$. Let $p_{j}$ be the probability associated with the transposition $(1,j)$ in (1). The probability that this sequence ends with counters in $\{a,b\}$ and $\{a,\ell\}$ must be equal, so

In particular, if $q=(1-p_{f_{k+1}})\cdots(1-p_{f_{s}})$, then $1-p_{\ell}=p_{\ell}qp_{b}$ which implies $p_{\ell}=1/(qp_{b}+1)\geq 1/2$. We now claim that the probability a counter ends in $\ell$ is strictly higher than the probability a counter ends in $b$ unless $qp_{b}=1$. We condition on which of $1$ and $\ell$ contain a counter before the transposition $(1,\ell)$ and calculate the probability a counter ends in each of the positions $\ell$ and $b$.

In every case, the probability that a counter ends in $\ell$ is at least the probability that a counter ends in $b$, and the first one is strict unless $qp_{b}=1$.

Now we consider the transpositions between $(1,b)$ and $(1,c)$ below.

Since the transposition $(1,b)$ fires with probability 1, there is no way for the counters to end in both $\ell^{\prime}$ and $c$, giving a contradiction and proving the claim. ∎

We will again apply a discharging argument. Let $T$ be a given $t$-reachability network consisting of star transpositions. As before, we assign the transposition $(1,a)$ a weight equal to the number of times $(1,a)$ is used and let $\sigma_{1},\dots,\sigma_{m}$ be the subsequence of transpositions that are used exactly once and are of the form $(1,x)$ for $x\not\in[t]$. We transfer weight from a transposition used multiple times according to the $\sigma_{i}$ using the following simple rules.

• •

  For each transposition between $\sigma_{i}$ and $\sigma_{i+1}$ used for the last time (but not the first), transfer $1/t$ to $\sigma_{i}$.

• •

  For each transposition between $\sigma_{i}$ and $\sigma_{i+1}$ used for neither the first time nor the last time, transfer 1 to $\sigma_{i}$.

We will show all but at most $2t$ transpositions end up with weight at least $2-1/t$. Clearly, any transposition which is used multiple times ends with weight $2-1/t$ as required. We will ignore any $\sigma_{i},\sigma_{i+1}$ for which some transposition $(1,x)$ with $x\in[t]$ occurs for the first time between them. This disregards at most $2(t-1)$ of the $\sigma_{i}$.

We show all the other $\sigma_{i}$ (except $\sigma_{m}$) get weight at least $2-1/t$. Consider the section between $\sigma_{i}=(1,a)$ and $\sigma_{i+1}=(1,b)$

If there is a transposition between $\sigma_{i}$ and $\sigma_{i+1}$ used for neither the first nor the last time, then $\sigma_{i}$ has weight at least 2. So we assume all $(1,s_{j})$ are used for either the first or last time. Let $s_{1}^{\prime},\dots,s_{\ell^{\prime}}^{\prime}$ be the transpositions used for the last time. If $\ell^{\prime}\geq t-1$, then $\sigma_{i}$ receives weight at least $2-1/t$ as desired, and we show that this is always the case.

We show this by proving that we cannot simultaneously leave counters in the positions $\{a,s_{1}^{\prime},\dots,s_{\ell^{\prime}}^{\prime},b\}$ (which we should be able to if the set has at most $t$ elements). The transposition $\sigma_{i}=(1,a)$ must be used (as this is the only way to put a counter on $a$) and then the only way for a counter to ‘reload’ position 1 before $\sigma_{i+1}$ is using a transposition $(1,s_{j}^{\prime})$ that has been used before. (Here we use our assumption that no $(1,x)$ appears for the first time in our segment for $x\in[t]$.) When we use $(1,s_{j}^{\prime})$ to ‘reload’, then no counter can end in position $s_{j}^{\prime}$ since this is the last use of this transposition. Hence, we cannot ‘reload’ position $1$ to leave a counter in $b$. ∎