The cover time of a biased random walk on a random regular graph of odd degree

We consider a random walk process which prefers to visit previously unvisited edges, on the random $r$-regular graph $G_r$ for any odd $r\geq 3$. We show that this random walk process has asymptotic vertex and edge cover times $\frac{1}{r-2}n\log n$ and $\frac{r}{2(r-2)}n\log n$, respectively, generalizing the result from Cooper, Frieze and Johansson from $r = 3$ to any larger odd $r$. This completes the study of the vertex cover time for fixed $r\geq 3$, with Berenbrink, Cooper and Friedetzky having previously shown that $G_r$ has vertex cover time asymptotic to $\frac{rn}{2}$ when $r\geq 4$ is even.


Introduction
We consider a biased random walk on the random r-regular n-vertex graph G r for any odd fixed r 5, i.e. a graph chosen uniformly at random from the set of r-regular graph on an even number n of vertices. This random walk was introduced by Orenshtein and Shinkar [10]. In short, the walker walks from its current vertex along a previously unvisited edge whenever possible, and otherwise along an already visited edge chosen uniformly at random. See Section 2 for a precise definition. In [5] it is shown that with high probability, G 3 is such that the expected vertex cover time C b V (G 3 ) and expected edge cover time C b E (G 3 ) of the biased random walk satisfy 1 We generalize this result as follows.
Theorem 1. Suppose 3 r = O(1) is odd, and let G r be chosen uniformly at random from the set of r-regular graphs on n vertices. Then with high probability, G r is such that n log n.
With this the asymptotic leading term of C b V (G r ) is known for all r 3, with Berenbrink, Cooper and Friedetzky [3] having previously shown that C b V (G r ) ∼ rn 2 for any even r 4. They also showed that for even r, C b E (G r ) = O(ωn) for any ω tending to infinity with n, with the ω factor owing to the w.h.p. 2 existence of cycles of length up to ω.
Cooper and Frieze [6] considered the simple random walk on G r , showing that for any r 3, C s V (G r ) ∼ r−1 r−2 n log n and C s E (G r ) ∼ r(r−1) 2(r−2) n log n, and we see that the biased random walk speeds up the cover time by a factor of 1/(r − 1) for odd r. Cooper and Frieze [7] also consider the non-backtracking random walk, i.e. the walk which at no point reuses the edge used in the previous step, showing that C nb V (G r ) ∼ n log n and C nb E (G r ) ∼ r 2 n log n. Here, the biased random walk gains a factor of 1/(r − 2) for odd r. Theorem 1 will follow from the following theorem. Let C b V (G; s) (C b E (G; t)) denote the expected time taken for the biased random walk to visit s vertices (t edges). Note that C b · (G; ·) is defined as an expectation over the space of random walks on the fixed graph G, and that E C b · (G r ; ·) takes the expectation of C b · (G; ·) when G is chosen uniformly at random from the set of r-regular graphs.
Theorem 2. Suppose r 3 is odd, and suppose G r is chosen uniformly at random from the set of r-regular graphs on an even number n of vertices. Let (1 − log −r n)n s n and (1 − log −2 n) rn 2 t rn/2, and let ε > 0. Then 1 ± ε r − 2 n log n n − s + 1 + o(n log n), n log rn rn − 2t + 1 + o(n log n).
We take a = b ± c to mean that b − c < a < b + c. The (1 − log −2 n) factor in the lower bound for t is a fairly arbitrary choice, and may be replaced by (1−1/ω) for any ω tending to infinity sufficiently slowly. The specific choice of log −2 n is made to aid readability.
Applying Theorem 2 with s = n and t = rn/2 gives E C b V (G r ) ∼ 1 r−2 n log n and E C b E (G r ) ∼ r 2(r−2) n log n. Some more work is needed to conclude that w.h.p. G r is such that C b V (G r ), C b E (G r ) have the same asymptotic values, see Section 6.3.

Proof outline
The random r-regular graph G r is chosen according to the configuration model, introduced by Bollobás [4]. Each vertex v ∈ [n] is associated with a set P(v) of r configuration points, and we let P = ∪ v P(v). We choose u.a.r. (uniformly at random) a perfect matching µ of the points in P. Each µ induces a multigraph G on [n] in which u is adjacent to v if and only if µ(x) ∈ P(v) for some x ∈ P(u), allowing parallel edges and self-loops. Any simple r-regular graph is equally likely to be chosen under this model. We study a biased random walk. On a fixed graph G, this process is defined as follows. Initially, all edges are declared unvisited, and we choose some vertex v 0 as the active vertex. At any point of the walk, the walk moves from the active vertex v along an edge chosen uniformly at random from the unvisited edges incident to v, after which the edge is permanently declared visited. If there are no unvisited edges incident to v, the walk moves along a visited edge chosen uniformly at random. The other endpoint of the chosen edge is declared active, and the process is repeated.
A biased random walk on the random r-regular graph can be seen as a random walk on the configuration model, where we expose µ along with the walk as follows. Initially choosing some point x 0 ∈ P, we walk to x 1 = µ(x 0 ), chosen u.a.r. from P \ {x 0 }. Suppose x 1 ∈ P(v 1 ). From x 1 , the walk moves to some unvisited x 2 ∈ P(v 1 ). In general, if W k = (x 0 , x 1 , . . . , x k ) then (i) if k is odd, the walk moves to x k+1 = µ(x k ) (chosen u.a.r. from P \ {x 0 , . . . , x k } if x k is previously unvisited), and (ii) if k is even, the walk moves from x k ∈ P(v k ) to x k+1 ∈ P(v k ), chosen u.a.r. from the unvisited points of P(v k ) if such exist, otherwise chosen u.a.r. from all of P(v k ).
We define C(t) to be the number of steps taken immediately before the walk exposes its t-th distinct edge. To be precise, if W k = (x 0 , . . . , x k ) denotes the walk after k steps, then Note that the set in this definition always will consist of exactly one k, which is even. We divide by 2 because the sequence (x 0 , x 1 , . . . , x k ) consists of k/2 edge traversals and k/2 steps between configuration points within one vertex, and the quantity we are interested in is the number of edge traversals. We also let W (t) = W 2C(t) . Note that C(t) is a random variable over the combined probability space of random graphs and random walks, as opposed to the classical cover times C b V (G r ) and C b E (G r ) which are variables over the space of random graphs only.
We will show (Lemma 13) that if which does not contribute significantly to the cover time. The main part of the proof is devoted to calculating E (C(t + 1) − C(t)) when t t 1 . We define the random graph G(t) ⊆ G r as the graph spanned by the first t distinct edges visited by the walk, i.e. the trace of W (t). If, immediately after discovering its t-th edge, the biased random walk inhabits a vertex incident to no unvisited edges, then a simple random walk commences on G(t), and C(t + 1) − C(t) is the number of steps taken for this simple random walk to hit a vertex incident to an unvisited edge. We construct from G(t) a graph G * (t) by contracting all vertices incident to at least one unvisited edge into one "supervertex" x. Thus, conditioning on W (t), the graph G * (t) is a fixed graph, i.e. one with no random edges. We will show that when t t 1 , w.h.p.
x lies on "few" cycles of "short" length and has the appropriate number of self-loops (to be made precise in Section 4), which will imply that the expected hitting time of x for a simple random walk on G * (t) is Readers familiar with the proof for the cubic graph [5] will recognize the general idea of this outline. In the case of cubic graphs, the set of vertices visited exactly once coincides with the set of vertices incident to one unvisited edge, modulo the starting vertex of the walk. This is no longer true when r 5, which forces a more detailed study of the edges not visited by the walk.
The paper is laid out as follows. Sections 3, 4 and 5 respectively discuss properties of the random regular graph, hitting times of simple random walks, and a uniformity lemma for biased random walks, and may be read in any order. Section 6 contains the calculation of the cover time. Sections 7 and 8 are devoted to bounding the sizes of certain sets appearing in the calculations.

Properties of the random regular graph
Here we collect some properties of random r-regular graphs, chosen according to the configuration model.

Lemma 3. Let r
3. Let G r denote the random r-regular graph on vertex set [n], chosen according to the configuration model. Let ω tend to infinity arbitrarily slowly with n. Its value will always be small enough so that where necessary, it is dominated by other quantities that also go to infinity with n.
(i) With high probability, the second largest in absolute value of the eigenvalues of the transition matrix for a simple random walk on G r is at most 0.99.
(ii) With high probability, G r contains at most ωr ω cycles of length at most ω, (iii) The probability that G r is simple is Ω(1).
Friedman [8] showed that for any ε > 0, the second eigenvalue of the transition matrix is at most 2 √ r − 1/r + ε w.h.p., which gives (i). Property (ii) follows from the Markov inequality, given that the expected number of cycles of length k ω can be bounded by O(r k ). For the proof of (iii) see Frieze and Karoński [9], Theorem 10.3. Note that (iii) implies that any property which holds w.h.p. for the configuration multigraph holds w.h.p. for simple r-regular graphs chosen uniformly at random.
Let G(t) denote the random graph formed by the edges visited by W (t). Let X i (t) denote the set of vertices incident to i unvisited edges in G(t) for i = 0, 1, . . . , r. Let X(t) = X 1 (t) ∪ · · · ∪ X r (t) denote the set of vertices incident to at least one unvisited edge. Let G * (t) denote the graph obtained from G(t) by contracting the set X(t) into a single vertex, retaining all edges. Define λ * (t) to be the second largest eigenvalue of the transition matrix for a simple random walk on G * (t).
We note that by [2,Corollary 3.27], if Γ is a graph obtained from G by contracting a set of vertices, retaining all edges, then λ(Γ) λ(G). This implies that λ * (t) = λ(G * (t)) λ(G) 0.99 for all t. Initially, for small t, we find that w.h.p. G * (t) consists of a single vertex. In this case there is no second eigenvalue and we take λ * (t) = 0. This is in line with the fact that a random walk on a one vertex graph is always in the steady state.
We define C(t) to be the number of steps the biased random walk takes to traverse t distinct edges of G r . Of course, if G r is disconnected and the random walk starts in a connected component of less than t edges, then C(t) = ∞. We resolve this by defining a stopping time T * = min{t : λ * (t) > 0.99}, and setting C * (t) = C(min{t, T }). Strictly speaking, the estimates of C(t) in the upcoming sections are estimates of C * (t), but we do not make any explicit distinction between the two, noting that by Lemma 3 (i), w.h.p. T * = ∞ which implies that C * (t) = C(t) for all t.

Simple random walks
We are interested in calculating C(t + 1) − C(t), i.e. the time taken between discovering the t-th and the (t+1)-th edge. Between the two discoveries, the biased random walk can be coupled to a simple random walk on the graph induced by W (t), and in this section we derive the hitting time of a certain type of expanding vertex set.
Implicitly, when we state results about a vertex set S in a graph G we are considering a sequence of vertex sets (S n ) in a graph sequence (G n ). We say that G has positive eigenvalue gap if the second largest eigenvalue λ n of the transition matrix for G n satisfies lim sup λ n < 1.
Consider a simple random walk on an r-regular graph G = (V, E) with eigenvalue gap 1 − λ > 0. For a set S of vertices and a probability measure ρ on V , let E ρ (H(S)) denote the expected hitting time of the set S, i.e. the number of steps the walk takes until it reaches S, when the initial vertex is chosen according to ρ. Let π denote the stationary distribution of the random walk, uniform in the case of a regular graph and proportional to degrees in general. Let P (t) u (v) denote the probability that a simple random walk starting at u occupies vertex v at time t.

Lemma 4.
Suppose v is a vertex of a graph. Then the hitting time of v, starting from the stationary distribution π, is given by Lemma 4 can be found in [2] (Lemma 2.11), and can be applied to hitting times of sets by contracting a set of vertices to a single vertex, retaining all edges. The following bound will be frequently used. Suppose G is a graph with eigenvalue gap 1 − λ(G), and S is a set of vertices in G. Then if G S is the graph obtained by contracting S into a single vertex, retaining all edges, we have equal hitting times for S in G and G S and t in a graph with eigenvalue gap 1 − λ (see for example Jerrum and Sinclair [11]), and λ(Γ) λ(G) for any Γ obtained from G by contracting a set of vertices (see [2,Corollary 3.27]). From an arbitrary starting distribution ρ, we have for any constant ε > 0, which follows from the fact that if the graph has positive eigenvalue gap, after O(log n) steps the walk has mixed to within ε distance in total variation from the stationary distribution.
In the following lemma we define N d (S) to be the set of vertices at distance exactly d from the vertex set S.

Lemma 5.
Suppose G is an r-regular graph on n vertices with positive eigenvalue gap. Let ω tend to infinity arbitrarily slowly with n. Suppose S is a set of vertices with |S| = o(n) even such that Proof. We first note that the set S contains exactly |S|/2 edges. Indeed, as |N (S)| = (r − 1)|S| and the total degree of S is r|S|, S contains at most |S|/2 edges. As |N 2 (S)| = (r −1) 2 |S|, each vertex of N (S) must have exactly one edge to S, implying that S contains at least |S|/2 edges. Consider the graph G S obtained by contracting S into a single node s, retaining all edges. In the graph G S , s has degree r|S|. Then s is a node with exactly |S|/2 self-loops, and is otherwise contained in no cycle of length at most ω, as |N d (S)| = (r −1) d |S| ensures that G S is locally a tree up to distance ω from s. Since π s = |S|/n = o(1) we may choose ω tending to infinity with ωπ s = o(1). We have Repeating the argument following (1), We now argue that It is argued in Cooper and Frieze [6], Lemma 7, that with no loop at vertex s, the expected number of returns to s within ω steps is (r − 1)/(r − 2) + o(1). With |S|/2 loops, when at s, there is a 1/r chance of using the loop and so each visit to s yields 1/(r − 1) expected returns. I.e. the 2 of [6] becomes The set S in Lemma 5 can be thought of as a "pure" root set, and we now show that the hitting time remains similar when we allow some impurities in both the structure of the set and in the starting distribution.

Lemma 7.
Let ω tend to infinity arbitrarily slowly with n. Suppose G is an r-regular graph on n vertices whose transition matrix has second largest eigenvalue λ 0.99, containing at most ωr ω cycles of length at most ω. Suppose S is a root set of order ω, and a simple random walk is initated at a vertex y chosen according to a distribution µ satisfying µ(y) r/|S| for y ∈ S and µ(y) = 0 for y / ∈ S. The expected time to reach a vertex in S at distance at least 2 from y is then for any constant ε > 0.
Proof. We first argue that Consider the contracted graph G S , and let s denote the contracted node. Then s has degree r|S|, and s has at most (1/2 + ω −3 )|S| self-loops. Apart from the self-loops, s lies on at most |S|/ω 3 cycles of length at most ω, as any cycle of G S containing s corresponds to a path between members of S in G. Let R = N (S), and note that |R| = Ω(|S|). Consider the graph Γ, defined as G S induced on the set of vertices at distance 1, 2, . . . , ω from s. Note that s is not included in Γ. The graph Γ contains all of R, and as s lies on at most |S|/ω 3 cycles of length at most ω in G S , the number of components in Γ containing more than one member of As G contains at most ωr ω short cycles, the number of components of Γ containing a cycle is at most ωr ω = O(|R|/ω 3 ) (choosing ω tending to infinity sufficiently slowly). This leaves (1 − o(1))|R| connected components in Γ which are all complete (r −1)-ary trees of height ω, each rooted at a member of R and containing no other member of R. Let T denote the set of vertices on such components.
Arbitrarily choose |S|/2 of the self-loops of s in G S , and designate them as good. Also say that an edge is good if it has both endpoints in T ∪ {s}. All other edges are bad.
Consider a simple random walk Z(τ ) of length ω on G S , starting at s. Let B τ denote the event that Z(τ ) traverses a bad edge to reach Z(τ + 1). Whenever the walk visits s, the probability that it chooses a bad edge is O(ω −3 ). If the walk is inside T , there are no bad edges to choose. So for any τ 0 we have If B r does not occur for any r τ − 1, then the walk (Z(0), . . . , Z(τ − 1)) can be viewed as the same Markov chain as considered in Lemma 5. So, by Lemma 5, This shows that E π (H(S)) ∼ r r−2 n |S| . We show (3), noting that we only need to consider |S| = o(n/ log n) due to the O(log n) term. Suppose y ∈ S is chosen according to a distribution µ with µ(y) r/|S| for any y ∈ S. Let S(y) denote the set of vertices in S at distance at least 2 from y, and let H be the random time at which S(y) is hit. The probability of H ω steps is then bounded by r/ω 3 , as there are at most |S|/ω 3 paths of length at most ω between vertices of S which use no edges induced by S. For any t we have so if we let p t denote the probability that the walk is in S at time t, Since |S| = o(n/ log n), we conclude that Pr {H C log n} = o(1) for any constant C.
With t = C log n for C large enough, we have P (t) S (·) within ε of π in total variation distance. If H denotes the first time after time t that S(y) is hit, we then have Since H = H with probability 1 − o(1), the lemma follows.
The next lemma is needed in the study of the sizes of X i (t), and will be applied with R = X 1 (t), S = X(t).
Lemma 8. Let G be an r-regular graph with positive eigenvalue gap. Let R ⊆ S ⊆ V be vertex sets. Suppose a simple random walk is initiated at a vertex y chosen according to a distribution µ with µ(y) r |R| for y ∈ R and µ(y) = 0 for y / ∈ R. Suppose the walk ends as soon as it hits S \ {y}. Then there is a constant B > 0 such that for any x ∈ S, the probability that the walk ends at x is at most B/|R|.
Proof. We will in general write (w 0 , w 1 , . . . ) for a simple random walk. Define S(x, y) = S \ {x, y}. The following reversibility property is central to the proof. If T S(x,y) denotes the first time at which S(x, y) is visited, then as the graph is regular, Let µ denote the uniform probability measure on R. Then, writing · or y for the starting vertex chosen by µ, the probability that x is the first vertex of S \ {y} to be hit is Write Q xy (t) = P x (w t = y and w 1 , . . . , w t−1 = y and T S(x,y) > t), so that Q xy (t) denotes the probability that the walk avoids S(x, y) and first visits y is at time t (or returns to y if x = y). Then for all t 1, the electronic journal of combinatorics 27(4) (2020), #P4.11 so, letting 1 x=y = 1 if x = y and 0 otherwise, As hitting times are almost surely finite, writing T + y = min{t 1 : w t = y}, we have Letting x = y and plugging (6) into (5) we obtain Now let x = y and note that in this case T + y = T y . Plug (6) and (7) into (5) to obtain t 0 P x (w t = y and T S(x,y) > t) = P x (T y < T S(x,y) ) P y (T S(x,y) < T + y ) .
The identity (8) also holds for x = y, as P y (T y < T S(x,y) ) = 1, matching the expression in (7). Plugging (8) into (4) yields Claim 9. P y (T S(x,y) < T + y ) Proof of Claim 9. Corollary 2.8 of [2] states that for any states i, j in a Markov chain, .
We consider running a simple random walk on the graph obtained by contracting S(x, y) to a single vertex, and set i = y, j = S(x, y). As π y = 1/n, we have .
For any two states k, in a Markov chain with eigenvalue gap 1 − λ we have the electronic journal of combinatorics 27(4) (2020), #P4.11 Applying Claim 9 to (9) and choosing B = 2r 1−λ finishes the proof, as the events {T y < T S(x,y) } are mutually exclusive for y ∈ R, so

On vertices with at least one unvisited edge
We now move from considering simple random walks, to the biased random walk W k .
Recall that W (t) = W 2C(t) denote the state of the walk immedately after it visits its (2t − 1)-th distinct configuration point. See Section 2 for details. The walk W (t) induces a colouring on the edges and vertices of G r as follows. An edge is coloured red, green or blue if it has been visited zero, one or at least two time(s), respectively. A vertex is (i) green if it is incident to exactly r − 1 green edges and one red edge, (ii) red if it is incident to red edges only, and (iii) blue otherwise.
Recall that X i (t) denotes the set of vertices incident to exactly i red edges in W (t). We let X g 1 (t), X b 1 (t) denote the green and blue vertices of X 1 (t), respectively, and set We have X(t) = X g 1 (t) ∪ Z(t). Note that the number of configuration points not visited by W (t) is exactly rn − 2t + 1. We will eventually show that |X g . In this section we present a "sprinkling" tool used to show that X g 1 (t) is a root set of order ω, which will imply that X(t) is also a root set of order ω. Before this, we state our results on the necessary set sizes.
Parts (i), (iii) and (iv) are proved in Section 7, and part (ii) in Section 8.

A sprinkling technique
The green edges and vertices are the focus of this section. Suppose W = W (t) = (x 0 , x 1 , . . . , x k ). Recall that a vertex is green if r − 1 of its configuration points appear exactly once in W , and its remaining configuration point does not. Suppose x 2i , x 2i+1 , x 2i+2 , x 2i+3 are consecutive configuration points visited exactly once by W , where x 2i+1 , x 2i+2 belong to a green vertex v. In this situation we call the pair (x 2i+1 , x 2i+2 ) a green link. Let L(W ) ⊆ P × P be the set of green links in W . For i < j, say that (x 2i , . . . , x 2j+1 ) is a green segment if (x 2i+1 , x 2i+2 ), (x 2i+3 , x 2i+4 ), . . . , (x 2j−1 , x 2j ) ∈ L(W ), and a maximal green segment if (x 2i−1 , x 2i ), (x 2j+1 , x 2j+2 ) / ∈ L(W ) (or if i = 0 or 2j + 1 = k, respectively). Form the contracted walk W by replacing each maximal green segment (x 2i , . . . , x 2j+1 ) by the single edge (x 2i , x 2j+1 ). Two walks W 1 , W 2 are said to be equivalent if W 1 = W 2 and L(W 1 ) = L(W 2 ), and we let [W ] = ( W , L) denote the equivalence class of the walk W . Note that X i (W 1 ) = X i (W 2 ) for any i and W 1 ∼ W 2 , and that the two walks induce the same vertex colouring. Proof. Recall the distinction between W k and W (t) from Section 2. Let w = (x 0 , . . . , x k ) be a walk with Pr {W (t) = w} > 0, i.e. a walk with Pr {W k = w} > 0, and such that |{x 0 , . . . , x k }| = 2t − 1. We calculate Pr {W k = w} exactly. For 0 i < k, let w i = (x 0 , . . . , x i ) and let s i denote the number of distinct edges traversed by w i .
If i is even, the walk proceeds to µ(x i ), i.e. the other endpoint of the edge attached to ∈ w i is guaranteed by the assumption that Pr {W k = w} > 0. If this is not the first time x i is visited, then µ(x i ) is uniquely determined by w i , and Pr {W i+1 = w i+1 | W i = w i } = 1, again using the assumption that Pr {W k = w} > 0.
If i is odd, suppose the configuration point x i belongs to a vertex v such that 0 j < r of the points in P(v) have not been visited by w i . If j > 0 then we have Pr {W i+1 = w i+1 | W i = w i } = 1/j, while if j = 0 then this probability is 1/r. Say that the step has type a if the probability in question is 1/a (note that the j = 0 case is type r, which will be convenient notationally).
All in all, letting e(w) = t − 1 denote the number of distinct edges visited by w, for some integers i 1 , i 2 , . . . , i r 0, counting the number of steps of the different types. The 1/rn factor accounts for the starting point of the walk. We argue that (11) can be expressed as a function of [w].
Suppose w has w = (x 0 , . . . , x ) and green link set L(w). If e(w) denotes the number of distinct edges visited by w, we have e(w) = e( w ) + |L(w)|.
It remains to argue that r a=1 a −ia is determined by w and L(w). For a single vertex v let i a (v) denote the number of steps (x 2i+1 , x 2i+2 ) of type a with x 2i+1 ∈ P(v). If v is green in the colouring induced by w, then i a (v) = 1 for a ∈ {2, 4, . . . , r − 1} and i a (v) = 0 otherwise. Indeed, w makes exactly (r − 1)/2 visits to v, and the i-th visit involves a step of type r − 2i + 1. If v is not green, then every visit to a configuration point of v is recorded by w . Indeed, w is exactly the subsequence of w of configuration points belonging to non-green vertices. So i a (v) is determined by L(w) if v is green and by w otherwise. This shows that the right-hand side of (11) can be expressed as a function of [w], which finishes the proof.
Form F 1 from F by replacing (x 2i , x 2i+1 ) by (x 2i , p 1 ) and (q 1 , x 2i+1 ). We repeat the above using F 1 with the next (p 2 , q 2 ) ∈ L, forming W 2 , W 3 , . . . , until all links have been placed in the walk. The final walk is W (t). We refer to this as sprinkling the links into the green edges of the contracted walk. Note that this can be viewed as a Pólya urn process; the urns are the edges of F , and the number of balls in urn (x 2i , x 2i+1 ) ∈ F is the length of the path between the two points in W (t). Indeed, this value is initially 1 for each member of F , and if a member of F is picked then it is replaced by two members, and so on.
the electronic journal of combinatorics 27(4) (2020), #P4.11 Proof. Say that v ∈ X g 1 is bad if its distance to the rest of X is at most ω, or if it lies on a cycle of length at most ω. Otherwise it is good. As X g 1 makes up almost all of X = X g 1 ∪ Z, to show that X is a root set of order ω it is enough to show that the number of bad vertices in X g 1 is at most X g 1 /ω 3 . By Lemma 3, at most ωr ω = o(X g 1 (t)) vertices lie on cycles of length at most ω. Let 1 , 2 ∈ L be links. As they are sprinkled onto the φ = Ω(nδ 1/2 ) green edges of [W (t)], the probability that they are placed within distance ω of each other is O(r ω /φ(t)). Indeed, any green edge in [W (t)] is within graph distance at most ω of at most r ω other green edges. The expected number of pairs 1 , 2 within distance ω of each other is bounded by As X g 1 = Ω(nδ), choosing ω small enough this shows that all but X(t)/ω 3 of the links are at distance at least ω from other links. We conclude that X(t) is a root set of order ω.

Calculating the cover time
Recall the definitions (10) of δ i and t i , 0 i 4. We begin by showing that the time taken to find the first t 1 edges contributes insignificantly to the cover time.
This is proved in Section 6.1. We then move on to estimating the expected cover time increment for larger t.
Lemma 14. For t 1 t t 4 and any ε > 0, Lemma 14 is proved in Section 6.2. The time to discover the final O(log n) edges can be bounded using (2). The increment C(t + 1) − C(t) is the number of steps taken by a simple random walk to reach a set of configuration points of size rn − 2t + 1, which corresponds to a vertex S of size at least (rn − 2t + 1)/r. By (2), we then have This shows that for t t 1 , proving the edge cover time statement of Theorem 2. The vertex cover time follows from a simple argument using Lemma 10 (iv): the walk is expected to cover all but s vertices at time δ = ((s + 1)/n) 2/r , which accounts for the r/2 factor separating the vertex and edge cover times. The details are in Section 6.3.
To prove Theorem 1 it remains to show that E G (C(t)) ∼ E (C(t)) for almost all fixed r-regular graphs G. This is done in Appendix A.
Theorem 15. Let G = (V, E) be an r-regular graph on n vertices, and suppose that each of the eigenvalues of the adjacency matrix with the exception of the first eigenvalue are at most λ G (in absolute value). Let A be a set of cn vertices of G. Then for every , the number of walks of length in G which avoid A does not exceed (1 − c)n((1 − c)r + cλ G ) .
The set A of Theorem 15 is fixed. In our case we choose a point x k+1 uniformly at random from P(X 1 (t)), so we consider a simple random walk initiated at a uniformly random vertex u ∈ X 1 (t). The subsequent walk now begins at vertex u and continues until it hits a vertex of Y u = X(t) \ {u}. Because the vertex u is random, the set Y u differs for each possible exit vertex u ∈ X 1 (t). To apply Theorem 15, we split X 1 (t) into two disjoint sets A, A of (almost) equal size. For u ∈ A, instead of considering the number of steps needed to hit Y u , we can upper bound this by the number of steps needed to hit B = A ∪ X 2 ∪ · · · ∪ X r , and vice versa. Suppose without loss of generality that u ∈ A.
Let S( ) be a simple random walk of length starting from a uniformly chosen vertex of A. Thus S( ) could be any of |A|r uniformly chosen random walks. Let c = |B |/n. The probability p that a randomly chosen walk of length starting from A has avoided B is, by Theorem 15, at most where λ .99 (see Lemma 3) is the absolute value of the second largest eigenvalue of the transition matrix of S. Thus So, Now, for any t we have r −1 (rn − 2t) |B | rn − 2t, and the probability that x k+1 ∈ X 1 is |X 1 |/(rn − 2t + 1). Summing over 0 t t 1 , (13) gives = o(n log n).

Phase two: Proof of Lemma 14
Lemma 16. Let ε > 0. For t 1 t t 4 , Proof. The proof of Lemma 14 is based on the following calculation. Define events is a root set of order ω}, and set E(t) = A(t) ∩ B(t). Then for any ε > 0, Lemma 7 shows that E (C(t + 1) − C(t)) can be calculated as Indeed, suppose E(t) holds. As X 1 (t) contains almost all unvisited configuration points, edge t is attached to some v ∈ X 1 (t) w.h.p., and a simple random walk commences at v, ending once it hits X \ {v}. As A(t) holds we have |X| ∼ rn − 2t. If E(t) does not hold, then we use the bound (1), stating that the hitting time is O(n/|X|) = O(n/(rn − 2t)) (as |X| (rn − 2t)/r) as long as the graph has a positive eigenvalue gap. We refer to the discussion in Section 3 justifying our assumption that the second largest eigenvalue stays at most 0.99 throughout the process. Lemmas 10 and 12 show that indeed, Pr we have Z(t) ⊆ Z(t 3 ) = ∅ w.h.p. by Markov's inequality. Note that this implies X g 1 (t) ⊆ X g 1 (t 3 ) and |X g 1 (t)| = rn − 2t. We also have Thus, repeating the calculation (12) of Lemma 12, we have that no two vertices of X g 1 (t 3 ) are placed within distance ω of each other. As X g 1 (t) ⊆ X g 1 (t 3 ), the same must be true of X g 1 (t). Thus the only vertices of X g 1 (t) which violate the root set constraints are those placed on the ωr ω short cycles of G r , and choosing ω small enough we have ωr ω (rn − 2t)/ω 3 for all t 3 t t 4 , so w.h.p. X(t) = X g 1 (t) is a root set of order ω.

The vertex cover time
Using Lemma 10 (iv) we can move between partial edge cover time C(t) = C E (t) and partial vertex cover time C V (t). The following lemma is almost exactly as the one in [5] for the case r = 3.
Lemma 17. Let (1 − log −r n)n s n. Then Proof. The set of undiscovered vertices at time t is X r (t). We write C V (s) = C(τ s ), where τ s = min{t : |X r (t)| = n − s}.
Suppose n − s tends to infinity with n. Let δ s = (1 − (s − 1)/n) 2/r , and let ω tend to infinity arbitrarily slowly with n. Define τ − s = (1 − δ s ω) rn 2 and τ + s = (1 − δ s ω −1 ) rn 2 . By Lemma 10 (iv), w.h.p., and similarly, w.h.p. X r (τ + s ) n − s. In particular, w.h.p. τ − s < τ s < τ + s . We have where T = 1 − n−s n rn 2 is an upper bound for τ s . Indeed, at time T there are rn − 2T < r(n − s) unvisited configuration points, so τ s < T . Conditional on the event {τ s > τ + s }, for t τ + s we can apply (2) to obtain We then have We calculate E (C(τ + s )) using Theorem 2 to obtain For a lower bound, we use and using the same argument as above it is enough to note that for t τ − s ,

Set sizes
Recall the definition where X i denotes the set of vertices incident to i unvisited edges, and X b 1 is the set of vertices in X 1 which are incident to at least one edge which has been visited more than once.
Proof. We show that there exists a B > 0 such that for any m 1, beginning with m = 1 before the general statement. Note that any vertex v ∈ Z(t) has been visited some number 0 k < r of times. To be precise, there are k indices i 1 , . . . , i k such that x 2i j +1 ∈ P(v). Let j ∈ {1, 2} denote the number of new edges incident to v in the segment (x 2i j −1 , x 2i j , x 2i j +1 , x 2i j +2 ); the edge (x 2i j +1 , x 2i j +2 ) must be new by design of the random walk, while (x 2i j −1 , x 2i j ) may or may not appear earlier in the walk. Let L = L(r) denote the set of vectors ( 1 , 2 , . . . , k ) with i ∈ {1, 2} such that i r − 1, including in L the empty vector ∅ (for unvisited vertices), excluding the vector (2, 2, . . . , 2) consisting of (r − 1)/2 copies of 2 (for green vertices). We partition where v ∈ Z (t) for = ( 1 , . . . , k ) if and only if there exists a sequence 0 < s 1 < s 2 < · · · < s k t such that v moves from X r− 1 −···− j−1 to X r− 1 −···− j at time s j for j = 1, . . . , k, and is in X r− 1 −···− k at time t. If v ∈ X i at time s, the probability that v is chosen by random assignment is i/(rn − 2s), while Lemma 8 shows that the probability that v is at the end of a blue walk is O (1/(rn − 2s)). In either case, the probability that v moves from one set to another is at most B/(rn − 2s) for some B > 0. For a fixed = ( 1 , . . . , k ) ∈ L, with s 0 = 1, For b 1 we use the bound Combining (15) and (16), the probability that 1 ∈ Z (t) is bounded above by Collecting powers of rn − 2s j for j = 1, . . . , k, we have the electronic journal of combinatorics 27(4) (2020), #P4.11 Let N denote the number of indices j ∈ {1, . . . , k} with j = 1. Then We now argue that r/2 − k + N 3/2, or equivalently 2(k − N ) r − 3, for all ∈ L. Firstly, if 1 + · · · + k r − 3 then we have 2(k − N ) 2k − N = 1 + · · · + k r − 3. Secondly, if 1 + · · · + k = r − 2 then as r − 2 is odd we have As |L(r)| is a function of r only, and therefore constant with respect to n, it follows that Pr {1 ∈ Z(t)} =
We turn to bounding the probability that [m] ⊆ Z(t). We fix (1) , . . . , (m) ∈ L and bound the probability that i ∈ Z (i) (t) for i = 1, . . . , m. Let k(i) = dim (i) denote the number of components of (i) . Then, summing over all choices s (i) j for 1 i m and 1 j k(i), for some constant C > 0. By symmetry the same bound holds for any vertex set of size m. It follows that for any m, writing (n) m = n(n − 1) . . . (n − m + 1), For s > 1 we apply the binomial theorem to obtain We set s = e θ 1 + 2θ (as θ = o(1)) to obtain Corollary 19. For t = (1 − δ) rn 2 with δ = o(1), and 0 < θ = o (1), Proof. The number of free configuration points at time t is rn − 2t, so By Lemma 18 we have The technique used to prove Lemma 18 can be strengthened to obtain concentration for the number of unvisited vertices X r (t). This is the topic of the next lemma.
the electronic journal of combinatorics 27(4) (2020), #P4.11 Proof. Suppose a biased random walk is initiated at a vertex chosen uniformly at random. The probability that no vertex in [m] is not incident to any of the first t distinct edges visited by the walk is bounded by The factor 1 − m/n is the probability that the initial vertex is in [m], and the bound otherwise is exactly as in (16). The inequality (17) follows from the arguments used in the proof of Lemma 18. For m = 1 we need the converse inequality to (16). From 1 − x e −x/(1−x) , valid for (1)).
Lemma 20 relates the number of unvisited edges to the number of unvisited vertices: we expect |X r (t)| = n − s to occur when t ∼ 1 − s n 2/r . This heuristically explains why . See Section 6.3 for details.

The green edges
Let Φ(t) denote the number of green edges in W (t).
Proof of Lemma 21. Firstly, let us see how Φ(t) changes with time. Fix ε 1 > 0 such that let X (t) = {X g 1 (t) (1 − ε 1 )(rn − 2t)} and let 1 t denote the indicator variable for X (t). We note that with λ = 1/ log n, by Corollary 19 Pr X (t) E e −λX g for any t t ε .
Proof. Condition on a [W (t)] such that X g 1 (t) (1 − ε 1 )(rn − 2t), recalling that the set X g 1 (t) is the same for all walks in [W (t)], as mentioned in Section 5.2. If the next edge is added without entering a blue walk, then Φ(t + 1) = Φ(t) + 1. So, Suppose the new edge chooses a vertex of X 1 (t), thus entering a blue walk. We may view this as a walk on [W (t)], and any time a green edge is traversed, we ask if the green edge in [W (t)] contains a green link in W (t), in which case the blue walk ends. If not, the green edge turns blue and Φ decreases by one. There are L(t) = r−1 2 X g 1 (t) green links, distributed into the Φ(t) green edges by a Pólya urn process as discussed in Section 5. Suppose e 1 , e 2 , . . . , e are green edges in [W (t)], and let K 1 , K 2 , . . . , K be the lengths of the corresponding paths in W (t), corresponding to the first entries of a vector (k 1 , . . . , k φ(t) ) drawn uniformly at random from all vectors with k i 1 and φ(t) i=1 k i = Φ(t). The number of such vectors is Φ(t)−1 φ(t)−1 , and noting that Φ(t) = φ(t) + L(t), the probability that none of the edges contain a green link is exactly Define for 0 < θ = o(1), f t (θ) = E e −θΦ(t) 1 t .

It follows by induction and from (21) that
We bound the two terms in (22) separately. Firstly, ne θΛ(t) η = n exp δ ε log 2 n nδ