Avoidance Coupling

We examine the question of whether a collection of random walks on a graph can be coupled so that they never collide. In particular, we show that on the complete graph on n vertices, with or without loops, there is a Markovian coupling keeping apart Omega(n/log n) random walks, taking turns to move in discrete time.


Introduction
Coupling of Markov chains has proved to be a valuable tool, notably, in recent years, in proving rapid mixing.Our intent here is to isolate one very simple type of Markov chain (random walk, especially on a complete graph) and to explore one particular capability, that of avoiding collision.
As an application, one may envisage some anti-virus software moving from port to port in a computer system to check for incursions.It is natural to have such a program implement a random walk on the ports so as not to be predictable.If another program (possibly with a different purpose) also does a random walk on the ports, it may be desirable or even essential to prevent the programs from examining the same port at the same time.
If two random walks are independent, they will collide in polynomial time on any finite, connected, non-bipartite graph, even if a scheduler tries to keep them apart [CTW93,TW93].Only if the scheduler is clairvoyant-that is, knows the entire future of each walk-is there a possibility of avoiding collision forever, and that case rests on a complex proof [Gác11] for enormous graphs.
Coupling, on the other hand, is a much more powerful technique for keeping random walks apart.On the cycle C n , for example (where the clairvoyant scheduler has no chance), coupling can easily keep linearly many random walks apart, simply by having them either all move clockwise, or all counter-clockwise, at the same time.
Keeping random walks apart on a complete graph K n by coupling-especially Markovian coupling, which we define more formally below-appears to be a more difficult task.We apply a number of techniques to achieve such couplings, depending on number-theoretic properties of n.For infinitely many n there is a Markovian coupling which keeps apart (n − 1)/2 random walkers, and for all n there is a Markovian coupling which keeps apart Ω(n/ log n) random walkers.We have essentially the same results on the looped version of the complete graph K * n , and in this case we also have non-Markovian couplings of linearly many walkers, for all n.
The related problem of coupling two Brownian motions (on various domains) so as to keep them at least some positive distance apart has been studied in some depth-see [BBC07,Ken09,BBK10].

Preliminaries
We refer the reader to a modern text such as [AF02,LPW09] (both of which are accessible online), for background on discrete Markov chains.All of our Markov chains are time-homogeneous and have finite state spaces.
A coupling of Markov chains is nothing more than an implementation of the chains on a common probability space, in such a way that each chain, viewed separately, is faithful to its transition matrix.In what follows, X t and Y t for t = 0, 1, 2, . . .will represent simple discrete-time random walks on the loopless complete graph K n , or its looped counterpart K * n (in which each vertex has a single self-loop, and the walk stays where it is with probability 1/n at each step).Clearly a time t for which X t = Y t should constitute a collision, but what if X t+1 = Y t ?If in addition Y t+1 = X t we call such an event a "swap", otherwise a "shove".
Allowing swaps and shoves makes things easy-on K n , for example, we could couple n walks simply by choosing, at each turn, a uniformly random derangement (or a uniformly random cycle) σ , and having the walker at i move to σ(i).
Instead, we make the issue of swaps and shoves moot by having the walkers move alternatelyin the case of two walkers, in the order X 0 , Y 0 , X 1 , Y 1 , . . . .Then the events X t = Y t and X t+1 = Y t both constitute collisions.Multiple walkers are assumed to take turns in a fixed cyclic order, and again, a collision is deemed to occur exactly if a walker moves to a vertex currently occupied by another.We call a coupling that forbids collisions an avoidance coupling.
Note that a collection of random walkers who move in continuous time, that is, after independent exponential (mean 1) waiting times, can be coupled so as to take turns as above; if there are k walkers, we simply have walker j wait for a random time after walker j−1 (modulo k ) has moved, where the time is distributed according to a Gamma distribution with shape parameter 1/k and scale parameter 1. (The sum of k independent such random variables is an exponential random variable with mean 1.) Thus, any coupling of our alternating discrete-time walkers can be applied to the continuous-time case.
A coupling is Markovian if it is itself a Markov chain, meaning, in the two-walker case, that X t+1 depends only on X t and Y t , while Y t+1 depends only on Y t and X t+1 .(Brownian couplings with the analogous property are referred to as "co-adapted" in [Ken09].)To allow the walkers to alternate, we tacitly assume that the state of the coupled chain includes the information of whether it is the first or second player's turn to move.For multiple walkers, the dependence is, similarly, only on the current locations of all the walkers, and on whose turn it is to move.
Note that the transformation above from discrete-time to continuous-time chains does not preserve the Markovian property.Indeed, it is rarely possible to get a Markovian avoidance coupling for continuous-time chains: Theorem 2.1.Let M be an irreducible, continuous-time, finite-state Markov chain.Then there is no Markovian coupling of two or more copies of M, without simultaneous transitions, that avoids all collisions.
Proof.Suppose X and Y are copies of M that are coupled in this way.Since M is irreducible, we may fix some tour v 0 , v 1 , . . ., v k of all the states that has a positive probability.Let r i > 0 be the rate at which the transition v i−1 → v i occurs, and consider the probability p that when started in state v 0 , the single chain X follows the tour exactly and completes it in time less than ε.Then where the constants implied by the Θ notation depend on the Markov chain and the tour, but not on ε.
Next we start the coupled chain (X, Y ) in state (v 0 , s) for some s, and consider the probability q(s) that its projection onto the first chain takes the tour and completes it in time less than ε.Since the coupled chain is collision-avoiding, it must take at least one additional step in order to move the second walker out of the way.But then at least k + 1 transitions must take place within time ε, thus where R is the maximum, over all states of the coupled chain, of the rate of transition out of that state.
For the coupling to be faithful, however, we must have q(s) ≥ p for some s.Since p = Θ(ε k ) and q(s) = O(ε k+1 ), this is impossible for small enough ε.

Two walkers on three vertices
No avoidance coupling is possible for two walkers on K 3 , since there is no choice of where to move, hence no room for randomness.On the looped graph K * 3 , however, a walker stays where she is with probability 1/3.We shall see that this is enough to permit an avoidance coupling, but not a Markovian one.In fact, we can completely analyze the more general walk on K * 3 in which a walker stays in place with some arbitrary probability s, and moves to each of the other two vertices with probability (1−s)/2.Theorem 3.1.Consider two walkers on K * 3 , each with looping probability s ∈ [0, 1).There exists an avoidance coupling if and only if s ≥ 1 3 , and there exists a Markovian avoidance coupling if and only if s ≥ 1 2 .
Proof of Theorem 3.1, non-Markovian case.We first show that an avoidance coupling exists in the case s = 1/3 (i.e., ordinary random walk on K * 3 ).Given X t and Y t , the pair (X t+1 , Y t+1 ) is chosen uniformly at random among the allowed pairs, except (X t , Y t ) itself.
Thus, for example, if Alice is at 0 and Bob at 1, their new positions will be (0, 2), (2, 0), or (2, 1) each with probability 1/3.Notice that this coupling is not quite Markovian, as Bob's move depends on Alice's previous position-he is not permitted to stay put when Alice has just done so.
Using this fact it easily follows that Alice's sequence is i.i.d.uniform; using the fact again, it follows that Bob's sequence is also i.i.d.uniform, as required.(We remark that this coupling is invariant under time reversal, except that Alice and Bob exchange roles.) Turning now to the case s ≥ 1/3, we can modify the above coupling as follows.At each round, with a suitable probability let both walkers stay in place.Otherwise they proceed to the next round.This clearly increases the probability that each walker stays in place at any step, without otherwise changing their trajectories.
Finally we must show that no avoidance coupling is possible if s < 1/3.Consider the event that X 0 , . . ., X n alternate between two (unspecified) states of K * 3 .This has probability 2 ( 1−s 2 ) n , since each jump has probability 1−s 2 .However, this event forces Proof of Theorem 3.1, Markovian case.Suppose first that there is a Markovian avoidance coupling.Let p ab be the probability that Alice stays at a given that it is her move, that she is at a, and that Bob is at b. Let q ab be the probability that Bob stays at b, given that it is Bob's move, and again that Alice is at a and Bob at b.That these quantities may only be defined for certain pairs a, b will not interfere with our arguments.
Suppose Alice has just moved from 0 to 1. Her conditional probability of next moving back to 0 is (1−s)/2.Bob must have been at 2 and will stay there with probability q 12 , after which Alice moves to 0 with probability 1−p 12 .We conclude that (1−s)/2 = q 12 (1−p 12 ).
Similarly, suppose Bob has just moved from 0 to 2. His conditional probability of next moving back to 0 is (1−s)/2.Alice must have been at 1 and will stay there with probability p 12 , after which Bob moves to 0 with probability 1−q 12 .So we get (1−s)/2 = p 12 (1−q 12 ).Combined with the conclusion of the previous paragraph, this gives p 12 = q 12 , and similarly p ab = q ab for all a = b.
Since the equation p 12 (1−p 12 ) = (1−s)/2 has no real solutions for s < 1 2 , the presumed coupling cannot exist in this case.
We now demonstrate that, conversely, when 1 2 ≤ s < 1 there is a Markovian avoidance coupling for two walkers.Let p and 1−p be the two (possibly equal) values of x satisfying x(1 − x) = (1 − s)/2, and note that then p 2 + (1−p) 2 = s.Letting i ′ stand for i+1 mod 3, put p ii ′ = p, p i ′ i = 1−p, and q ij = p ij .We claim that these values are the holding probabilities p ij , q ij (as defined earlier in the proof) of a Markovian avoidance coupling.
To show this, condition on the event that Alice is at i at time 1.We will show that, conditioned also on Bob's position at time 0, Alice's next step is to i (respectively i ′ ) with the correct probability s (respectively (s−1)/2); hence the probability she moves to i ′′ is correct also.Since the coupling is Markovian, Alice's future depends on her past only through (X 1 , Y 0 ), so this will suffice to prove that Alice's trajectory has the correct law.By the symmetry of our construction, the same will then apply to Bob.
Suppose first that Y 0 = i ′ , that is, that Bob was at i ′ one move ago.Then, with all probabilities Observe that the coupling is invariant under replacing state i with −i mod 3, swapping ′ and ′′ , and substituting 1 − p for p.Since the above conditional probabilities are symmetric in p and 1 − p, it follows that the distribution of X 2 conditional on {X 1 = i, Y 0 = i ′′ } is also correct.
4 Two walkers for composite n Theorem 4.1.For any composite n = ab, where a, b > 1, there exist Markovian avoidance couplings for two walkers on K n and on K * n .Proof.We partition [n] := {0, 1, . . ., n−1} into b "clusters" S 1 , . . ., S b each of size a.We construct a coupling so that, when it is Alice's turn to move, she and Bob are in different clusters.(This is where we use b > 1.) For the coupling on K n , Alice's protocol is to move with probability a(b−1) ab−1 to a random vertex in Bob's cluster (other than Bob's vertex), and move with probability a−1 ab−1 to a random vertex in her own cluster (other than her current vertex).(This is where we use a > 1.) Bob's protocol is to move to a random new vertex in his current cluster, unless Alice is also in his cluster, in which case he moves to a uniformly random vertex in a uniformly random unoccupied cluster.(After Bob moves, he and Alice are once again in different clusters.) The coupling for K * n is essentially the same, except that Alice's probability of moving to Bob's cluster is b−1 b , and when either Alice or Bob move within their own cluster, the new vertex may be the same as the current vertex.
Regardless of how Alice and Bob start, after Alice and then Bob move, Bob is at a uniformly random new vertex (for the coupling on K n ) or a uniformly random vertex (for the coupling on K * n ).Thus Bob's walk has the correct distribution.When the coupled walks are viewed backwards in time, the protocol is the same but with the roles of Alice and Bob reversed.Thus Alice's walk also has the correct distribution.
Note that when a = 2, the above coupling has minimum entropy, meaning that the entropy of the coupling is equal to the entropy of a single walker.The coupling does not have minimum entropy for a > 2; it can, however, be modified to have minimum entropy, at the cost of giving up the Markovian property.Specifically, we can give each cluster the structure of a directed cycle and insist that when Alice moves to Bob's cluster, she chooses the site after Bob's; and if she stays in her own cluster, Bob copies Alice's movement in his own cluster.To copy Alice's movement, Bob needs to remember where Alice was on her previous turn, which is why this modified coupling is not Markovian.

Monotonicity
The purpose of this section is to show that existence of avoidance couplings for k walkers on K * n is monotone in n.We do not know whether the corresponding statement holds for the unlooped case K n , nor if we impose the Markovian condition or the minimum entropy condition.
Theorem 5.1.If there is an avoidance coupling of k walkers on K * n , then there is an avoidance coupling of k walkers on K * n+1 .
The following concept will be useful for the proof.Suppose that k walkers walk on K * 2 , taking turns in cyclic order as usual, in such a way that no two walkers are simultaneously at vertex 1 (but several walkers can be at 2), and so that the trajectory of any given walker is an i.i.d.Bernoulli sequence that is 1 with probability p at each step.We call such a coupling a 1-avoidance coupling of Bernoulli(p) walkers.
Lemma 5.2.If there is an avoidance coupling of k walkers on K * n , then there is a 1-avoidance coupling of k Bernoulli(1/n) walkers.
Proof.Let any given Bernoulli walker be at 1 exactly when the corresponding walker on K * n is at vertex 1.
Lemma 5.3.If there is a 1-avoidance coupling of k Bernoulli(p) walkers, then there is a 1avoidance coupling of k Bernoulli(q) walkers for all q < p.
Proof.We simply thin the process of 1s.Suppose we have a Bernoulli(p) coupling, and take an independent process of i.i.d.coin flips, heads with probability q/p, indexed by turns (i.e. times at which any walker is allowed to move).To get the Bernoulli(q) coupling, take a walker to be at 1 whenever the original walker is at 1 and the corresponding coin flip is heads.
Proof of Theorem 5.1.Suppose we have an avoidance coupling of k walkers on K * n .By Lemmas 5.2 and 5.3, there exists a 1-avoidance coupling of k Bernoulli(1/(n+1)) walkers.Take such a coupling, independent of the original coupling on K * n .To get a coupling on K * n+1 , take a given walker to be at the same vertex as the corresponding walker on K * n , unless the corresponding Bernoulli walker is at 1, in which case take it to be at vertex n + 1.
6 Linear number of walkers for special n Theorem 6.1.There exists a minimum-entropy Markovian avoidance coupling for k walkers on K * n for any k ≤ 2 d and any n = 2 d+1 or 2 d+1 + 1, as well as on K n for n = 2 d+1 + 1.
Corollary 6.2.There exists an avoidance coupling for k walkers on K * n for any k ≤ n/4.
Proof.This is immediate from Theorem 6.1 and the monotonicity result, Theorem 5.1.
Proof of Theorem 6.1.We begin with the case of 2 d walkers, labeled 0, . . ., 2 d −1 = ω on K * n for n = 2 d+1 or 2 d+1 + 1.Let ε i t be independent uniform {±1} random variables.Let δ t be independent uniform {0, 1} random variables.Let the trajectory of walker j be denoted {X j t }.Let d−1 i=0 j i 2 i be the binary representation of j , for 0 ≤ j ≤ ω .Given X 0 t , the positions of the other walkers are given by X j t = X 0 t + j i ε i t 2 i where all positions are understood modulo n.We then define inductively We first show that this is indeed a coupling of random walkers.For all j , we have X ω t = X j t + (1−j i )ε i t 2 i and so Since (1−j i )ε i t + j i ε i t+1 is ±1 we find that the sum is uniform on the odd numbers in [−2 d , 2 d ], and so X j t+1 − X j t is uniform on [1, 2 d+1 ], as needed for the walk on K * 2 d+1 or on K 2 d+1 +1 .The process (X j t ) is Markov since the ε's and δ 's used to define X j t+1 in terms of X j t in (1) are disjoint from those used in any other time step.Note that the j th trajectory determines all the bits, so this is also a minimum entropy coupling.
Next we establish avoidance.Let j < j ′ be two walkers.We have Note that |j ′ i − j i | ≤ 1, hence |∆| < 2 d , and so ∆ = 0 mod n implies ∆ = 0.If i 0 is minimal such that j i 0 = j ′ i 0 then ∆ is divisible by 2 i 0 but not by 2 i 0 +1 , and so is non-zero.Thus there are no collisions within any round.Between consecutive rounds we have Let i 1 be maximal such that j i 1 = j ′ i 1 .Since j < j ′ this implies j i 1 = 0 and j ′ i 1 = 1.We have Terms for i > i 1 contribute at most 2 d − 2 i 1 +1 to the sum in (2).Terms for i < i 0 contribute at most 2(2 and so ∆ = 0 mod n.
To see that this coupling is Markovian, note that X 0 t is determined by X ω t−1 and δ t .Similarly, X 2 i t is determined by X 0 t and ε t i , and all other walkers' positions are determined by the positions of the previous walkers in the round.
We can reduce the number of walkers to any value between 2 and 2 d by simply removing walkers other than 0 and ω .The Markovian property is preserved if we first remove walkers whose indices are not powers of 2.
Finally we turn to the case of k walkers on K * n for n = 2 d+1 + 1.To do this we simply add to the coupling on K n rounds in which all walkers rest, beginning with walker 0. For the Markovian property, we need to ensure that each walker j = 0 can detect when walker 0 has decided to rest.This is so because on K n , given X ω t , no vertex is a possible value for both X 0 t and X 0 t+1 (otherwise X 0 t+1 , which depends only on X ω t and δ t+1 but not on X 0 t , might stay in place).
We say that an avoidance coupling of k walkers stays in waves if, for some distinguished walker w , whenever w stays in place, all the other walkers do likewise at the following k−1 turns, while if w moves, all others do so too.(The coupling on K * 2 d+1 +1 in the last proof stays in waves.)Note that any Markovian avoidance coupling that stays in waves on K * n may be modified to obtain a Markovian avoidance coupling on K n by removing all the looping rounds.
7 Many walkers for general n Theorem 7.1.There exists a Markovian avoidance coupling of k walkers on K * n for any k ≤ n/(8 log 2 n), and on K n for any k ≤ n/(56 log 2 n).
The constants in this theorem can easily be improved.However, as noted below, our methods will not go beyond n/(log 2 n).To prove the theorem, we make use of two lemmas which allow us to combine avoidance couplings.
Lemma 7.2.Suppose that we have avoidance couplings of r walkers on K * m and of s walkers on K * n .Then there is an avoidance coupling of k walkers on K * mn , for any k satisfying r + s − 1 ≤ k ≤ rs.If the given couplings are Markovian, then so is the new coupling.If the given couplings stay in waves, then so does the new coupling.If the given couplings are minimum-entropy, then the new coupling is too.
Proof.We identify K * mn with K * m × K * n , and note that if X t and Y t are independent random walks on K * m and K * n respectively, then (X t , Y t ) is a random walk on K * mn .Given an avoidance coupling {X i t } of r walkers on K * m and an independent avoidance coupling {Y j t } of s walkers on K * n , we construct a coupling on K * mn of rs walkers with labels (i, j), for 1 ≤ i ≤ r and 1 ≤ j ≤ s.The walkers move in lexicographic order.The trajectory of walker (i, j) is given by (X i t , Y j t ), which as noted above is a random walk on K * mn .That the walkers avoid collisions follows from the product construction and the collision avoidance of the given couplings.If the given couplings are Markovian, then since the walkers on K * mn move in lexicographic order, the resulting coupling is also Markovian.It is clear that the coupling stays in waves provided both original couplings do.Finally, no randomness is required beyond that in the couplings on K * m and on K * n , so if they are minimum entropy, so is the resulting coupling.
To construct a coupling of fewer walkers, just eliminate some of the walkers, as long as walkers (i, 1) and (1, j) (for each 1 ≤ i ≤ r and 1 ≤ j ≤ s) are kept.All other trajectories are determined by those, so the Markov property is maintained.
We remark that a variant of the above construction can be used to combine an avoidance coupling of r walkers on K m and an avoidance coupling of s walkers on K * n that stays in waves to produce an avoidance coupling of rs walkers on K mn .Lemma 7.3.Suppose we have avoidance couplings for k walkers on K m and on K n (respectively, K * m and K * n ).Then we have an avoidance coupling for k walkers on K m+n (respectively, K * m+n ).If the original couplings are Markovian then so is the resulting coupling.
Proof.Partition the vertex set of K m+n into two clusters U and V of sizes m and n, respectively.We will ensure that when it is the first walker's turn, all of the walkers are in the same cluster.At each of her turns, the first walker flips an appropriately biased coin to decide whether to move within her current cluster or to switch to the other cluster.If she stays in her current cluster she moves according to that cluster's coupling rules, and so do the rest of the walkers.If she switches to the other cluster, she moves to a uniformly random vertex therein.Each subsequent walker now chooses a random k -walker configuration in the new cluster (say, V ) consistent with the walkers that are already in V , in accordance with the stationary distribution on configurations of the K n (or K * n ) coupling arising just before a move of the first walker.He then moves to his allotted space in this configuration.
For general n = n i 2 i , n i ∈ {0, 1}, we write where 2 d+1 |r .If r ≥ 0 and n = 0, then Lemma 7.3 provides a Markovian avoidance coupling for K * n ; this inequality indeed holds whenever Now any n ≥ 8 satisfies (d+2)2 d+2 ≤ n < (d+3)2 d+3 for some integer d ≥ 0. Thus the above gives a Markovian avoidance coupling for any number of walkers up to 2 d .By the first inequality, 8 × 2 d ≤ n, so d+3 ≤ log 2 n, which combined with the second inequality gives 2 d > 1 8 n/(d+3) ≥ n/(8 log 2 n), which proves the theorem for K * n for n ≥ 8.The claim of the theorem is trivial for n < 8.
We now turn to the case of K n (without loops).Recall that if we have a Markovian avoidance coupling on K * n that stays in waves, then removing the looping rounds yields such a coupling on K n .
Fix d ≥ 1, and let S be the set of values of n for which Markovian avoidance couplings exist on K * n for every number of walkers up to 2 2d−1 , all of them staying in waves.By Lemma 7.3, S is closed under addition.From Theorem 6.1, we see that S contains 2 c +1 for all c ≥ 2d.Using Theorem 6.1 and Lemma 7.2, when a ≥ 1, b ≥ 1, and a+b ≥ 2d+1, there is a Markovian avoidance coupling for n = (2 a +1)(2 b +1) with x 2 b−1 − y walkers, where 1 ≤ x ≤ 2 a−1 and 0 ≤ y < 2 b−1 .In particular, S contains (2 a +1)(2 b +1), and specifically S contains 2 2d+1 + 1 + 2 i + 2 2d+1−i for all 1 ≤ i ≤ d (and also for i = 0 using Theorem 6.1 and Lemma 7.3).
We combined the number-theoretic avoidance coupling from Theorem 6.1 with the sum and product lemmas to obtain an avoidance coupling with Ω(n/ log n) walkers for any n.Given these three ingredients, this general-n construction is in a sense best possible up to constants.More precisely, we argue below that these three ingredients cannot be combined to obtain a coupling of more than n/ n walkers on K n or K * n , where n is the Hamming weight of n.
By the distributive law, any coupling that can be constructed using the sum and product lemmas 7.3 and 7.2 can be done by taking sums of products of basic constructions.Consider the product of s basic couplings of 2 d j walkers on either 2 d j +1 + 1 or 2 d j +1 vertices.Note that ab ≤ a b .The product lemma gives a coupling of 2 d walkers on K m or K * m , where d = d j and m ≥ 2 d+s and m ≤ 2 s .In particular m ≥ 2 d m .Next suppose that n is the sum of several such product terms, say n = i m i , each corresponding to the same d.Then n ≥ 2 d i m i ≥ 2 d n .In particular the number of walkers is at most n/ n .Thus, to improve on the Ω(n/ log n) bound for general n, more ingredients would be needed.
couplings, since the k th walker can be imagined.The answer is "yes" for the Markovian couplings that we exhibited, but it is not clear if this holds in general.
4. Monotonicity in loop weights.Suppose that K n is equipped with loops of weight w , so that a walker loops with probability w/(w+n−1).If there is an avoidance coupling for k walkers on K n with loops of weight w , must there be one with loops of weight w ′ , where w ′ > w ?The answer is "yes" for non-Markovian couplings, but what about the Markovian case?In particular, is the maximum number of Markovian avoiding walkers always at least as great on K * n as it is on K n ? 5. Minimum entropy couplings.Does existence of an avoidance coupling imply existence of a stationary avoidance coupling whose entropy equals that of a single random walk?
6. 1-avoidance.What is the largest p for which k i.i.d.Bernoulli(p) sequences can be coupled, taking turns to move as usual, so that no two simultaneously take the value 1?