Rainbow perfect matchings and Hamilton cycles in the random geometric graph

Rainbow perfect matchings and Hamilton cycles in the random geometric graph

Abstract

Given a graph on vertices and an assignment of colours to the edges, a rainbow Hamilton cycle is a cycle of length visiting each vertex once and with pairwise different colours on the edges. Similarly (for even ) a rainbow perfect matching is a collection of independent edges with pairwise different colours. In this note we show that if we randomly colour the edges of a random geometric graph with sufficiently many colours, then a.a.s. the graph contains a rainbow perfect matching (rainbow Hamilton cycle) if and only if the minimum degree is at least (respectively, at least ). More precisely, consider points (i.e. vertices) chosen independently and uniformly at random from the unit -dimensional cube for any fixed . Form a sequence of graphs on these vertices by adding edges one by one between each possible pair of vertices. Edges are added in increasing order of lengths (measured with respect to the norm, for any fixed ). Each time a new edge is added, it receives a random colour chosen uniformly at random and with repetition from a set of colours, where is a sufficiently large fixed constant. Then, a.a.s. the first graph in the sequence with minimum degree at least must contain a rainbow perfect matching (for even ), and the first graph with minimum degree at least must contain a rainbow Hamilton cycle.

1 Introduction

Let be i.i.d. points in chosen with the uniform distribution, where is fixed. Fix . Unless otherwise stated, distances and lengths in are measured with respect to the norm. We construct the random geometric graph of radius as follows. The vertices of are indexed by , and each pair of different vertices are joined by an edge if and only if and are within (-normed) distance . The length of an edge is defined to be , and is always at most by construction. With probability all points in are different, and fall in general position. Therefore, we will often identify vertex and point (i.e. we regard as the vertex set), and assume all edges have different lengths. Random geometric graphs (or more precisely a slight variation of the model defined above) were first introduced by Gilbert [12], and have been widely investigated ever since. They provide a theoretical model for wireless ad-hoc networks, and have relevant applications in statistics. We refer the reader to Penrose’s monograph [16] and a more recent survey by Walters [17] for further details and references on the subject.

We consider the natural coupling in which all with share one common vertex set . We call this coupling the random geometric graph process, and denote it by . Intuitively, the process starts at time with an empty graph on vertex set (almost surely, assuming that all vertices are at different positions). Then, as we increase from to , we add edges one by one in increasing order of length. By construction, each snapshot of the process at a given time is distributed precisely as a copy of . Finally, is deterministically the complete graph for all , where is the distance between two opposite corners of .

A lot of work has been done to describe the connectivity properties of random geometric graphs in this process and the emergence of spanning subgraphs (such as perfect matchings and Hamilton cycles). A celebrated result of Penrose [15] asserts that a.a.s.2 the first edge added during the process that gives minimum degree at least also makes the graph -connected. More precisely, let

 ˆrδ≥k =ˆrδ≥k(\boldmathX)=min{r≥0:G(\boldmathX;r) has minimum degree at least k}and ˆrk-conn =ˆrk-conn(\boldmathX)=min{r≥0:G(\boldmathX;r) is k-% connected}.

Then, for every constant ,

 limn→∞Pr(ˆrδ≥k=ˆrk-conn)=1. (1)

In view of this, Penrose (cf. [16]) asked whether a.a.s. that first edge in the process that gives minimum degree at least (and ensures -connectivity) is also responsible for the emergence of a Hamilton cycle. A first step in this direction was achieved by Díaz, Mitsche and Pérez [6], who showed (for dimension ) that, given any constant , a.a.s.  contains a Hamilton cycle. Some of their ideas were recently extended by three research teams (Balogh, Bollobás and Walters; Krivelevich and Müller; and Pérez-Giménez and Wormald), who independently settled Penrose’s question in the affirmative (but only two papers [4] and [14] were finally published). In particular, a more general packing result in [14] implies that

 a.a.s.\ {G(\boldmathX;ˆrδ≥2) contains a Hamilton cycle, andG(\boldmathX;ˆrδ≥1) contains a% perfect matching (for even n). (2)

Clearly, this claim is the best possible, since any graph with minimum degree less than (less than ) cannot have a perfect matching (respectively, Hamilton cycle).

In this paper we consider an edge-coloured version of the random geometric graph. (Throughout the manuscript, we will use the term edge colouring (and other terms alike) to denote an assignment of colours to the edges, not necessarily proper in a graph-theoretical sense.) Let be a random vector of colours, chosen independently and with replacement from a set of colours of size . We use this vector to colour the edges of the random geometric graph: each edge of () is assigned colour . We denote this model by . Similarly, we consider the coupled process in which, as we increase from to , we add new edges in increasing order of length, and each new edge is coloured according to .

Given a graph with colours assigned to the edges, we say the graph (or its edge set) is rainbow if all edges receive different colours. Recently, there have been many papers written on the subject of rainbow spanning structures in randomly edge coloured random graphs and digraphs (see e.g. [1, 2, 3, 5, 8, 9, 10, 11, 13]). Let denote the binomial random graph where each edge has independently been assigned a uniformly random colour from a set of size . For graphs with maximum degree and vertices, Ferber, Nenadov and Peter [10] showed that a.a.s.  contains a rainbow copy of , provided that and . Here, the number of colours is asymptotically optimal, whereas the bound on most likely is not. For Hamilton cycles tighter results are known. In [5], Cooper and Frieze determined that a.a.s.  contains a rainbow Hamilton cycle if and . This was later improved by Frieze and Loh [11] and recently even further improved by Ferber and Krivelevich [8] who showed that it holds when and . Here, the number of colours is asymptotically optimal and the bound on is optimal. Bal and Frieze [3] examined the case when the number of colours is exactly optimal, showing that a.a.s. contains a rainbow Hamilton cycle as long as .

In this manuscript, we investigate the emergence of rainbow spanning structures in the random geometric graph. Our main contribution is to extend (2) to a rainbow context. We show that a.a.s. the first edge in the edge-coloured random geometric graph process that gives minimum degree at least also creates a rainbow Hamilton cycle, provided that the number of colours is at least , where is a sufficiently large constant. Similarly, under the same assumptions (with even), the first edge in the process that ensures that the minimum degree is at least creates a rainbow perfect matching. To state the result more precisely, let

 ˆrRPM=ˆrRPM(\boldmathX,\boldmathZ)=inf{r≥0:G(\boldmathX;\boldmathZ;r) contains a rainbow perfect % matching}and ˆrRHC=ˆrRHC(\boldmathX,\boldmathZ)=inf{r≥0:G(\boldmathX;\boldmathZ;r) contains a rainbow Hamilton % cycle},

where we use the convention that . Note that whenever () the infimum in the above definition is actually a minimum, and it is precisely the length of the first edge in the process that creates a rainbow perfect matching (respectively, rainbow Hamilton cycle).

Theorem 1.

Given a fixed integer , there exists a sufficiently large constant satisfying the following. Let be i.i.d. points in chosen uniformly at random, and let be a random vector of colours, choosen independently and with replacement from a set of colours of size . Consider the random geometric graph process (for any fixed -normed distance, ) with a random colouring of the edges given by . Then,

 limn→∞Pr(ˆrRHC(% \boldmathX,\boldmathZ)=ˆrδ≥2(\boldmathX))=1,

and for even

 limn→∞(n even)Pr(ˆrRPM(\boldmathX,\boldmathZ)=ˆrδ≥1(% \boldmathX))=1.
Remark 2.

For , we can only claim that

 limn→∞Pr(ˆrRHC(% \boldmathX,\boldmathZ)=ˆr2-connconn(\boldmathX))=1andlimn→∞(n even)Pr(ˆrRPM(\boldmathX,\boldmathZ)=ˆr1-connconn(\boldmathX))=1, (3)

and in fact it is not known whether (1) holds (see [15] and [16]). We include a justification of (3) in Section 5 for completeness.

Combining Theorem 1 and Theorem 8.4 in [16], we immediately obtain the limiting probabilities of having a rainbow perfect matching and having a rainbow Hamilton cycle, assuming that the number of colours is sufficiently large.

Corollary 3.

Under the same assumptions of Theorem 1, put

 r=d√(2/d)logn+(3−d−2/d)loglogn+x22−dθn,

and let , where and are the volumes of the -dimensional and -dimensional unit -balls, respectively. Then

 limn→∞(n even)Pr(G(% \boldmathX;\boldmathZ;r) has a rainbow perfect matching)=⎧⎨⎩0x→−∞;exp(−e−α−f)x→α;1x→∞.
Corollary 4.

Under the same assumptions of Theorem 1, put

 r=d√(2/d)logn+(4−d−2/d)loglogn+y22−dθn,

and let , and be as in Corollary 3. Then if ,

 limn→∞Pr(G(\boldmathX;% \boldmathZ;r) has a rainbow Hamilton cycle)=⎧⎪⎨⎪⎩0y→−∞;exp(−2e−α−f/d)y→α;1y→∞.

Otherwise if ,

 Unknown environment '%'

In Section 2, we will discuss how we partition into a grid of small -dimensional cubic cells. Having this partition will simplify our task by allowing us to locally search for short rainbow paths or cycles within each cell or small cluster of cells. In Section 3, we will show how to find these paths or cycles. In Section 4, we will show how to connect the pieces together into one rainbow Hamilton cycle. A simple adaptation of the argument can be used to build a rainbow perfect matching. Finally, in Section 5 we will discuss the case , and pose some open questions in Section 6.

2 Tessellation and graph of cells

The main goal in this section is to prove Lemmas 13 and 14, which will be crucial in the construction of the rainbow Hamilton cycle and perfect matching. We will adapt and extend many ideas from [14]. Throughout the paper, and the -norm () in are fixed. Note that the volume of the unit -dimensional -ball satisfies

 2d/d!≤θ≤2d (4)

since the -ball contains the -ball and is contained in the -ball, which have volume and respectively. We will also make frequent use of the following inequality throughout the argument, often without explicitly mentioning it. For any ,

 ∥X∥∞≤∥X∥p≤d∥X∥∞. (5)

Moreover, we will pick a sufficiently small constant so that several requirements in the argument are met. Later we will choose sufficiently large with respect to this (recall is the number of colours). We remark that our choice of does not depend on or other parameters that may be introduced later in the statements.

We use the standard , , and asymptotic notation as with the following extra considerations. We do not assume any sign on a sequence satisfying or , but on the other hand a sequence satisfying or is assumed to be positive for all but finitely many . Furthermore, the constants involved in the bounds of the definitions of , and may depend on , but not on or . Whenever these constants depend on our choice of (in addition to ), we use the alternative notation , and instead. Our asymptotic statements are always uniform for all as a consequence of bounds (4) and (5).

Let be some function tending to infinity sufficiently slowly (in particular, ). We define and by

 θnr0d =(2d−1/d)logn+2d−2(3−d−2/d)loglogn−ωand θnr1d =(2d−1/d)logn+2d−2(4−d−2/d)loglogn+ω.

Then, by Theorem 8.4 in [16], the respective lengths and of the critical edges of the process that give minimum degree 1 and 2 satisfy

 r0≤ˆrδ≥1≤ˆrδ≥2≤r1∼r0a.a.s. (6)

Our argument will use edges of length at most to construct most of the rainbow perfect matching or Hamilton cycle, and only use a few longer edges of length up to or at some exceptional places.

Let . We tessellate into -dimensional cubic cells of side length

 s=⌈(s′)−1⌉−1∼s′=Θ(ε1/dr0),

that is, of volume

 sd∼εd21−dθr0d∼εlogn/n,

arranged in a grid fashion. Let denote the set of cells. There are cells in , where we recall that the constant hidden in the notation depends on (and ). Clearly (assuming that is sufficiently small given and by (4)), the vertices inside each cell induce a clique in , and in fact a stronger property holds in view of the following definition.

Definition 1.

The graph of cells is a graph with vertex set (i.e. the set of cells of the tessellation), and two cells are adjacent in if they are at (-normed) distance at most .

This implies that for any pair of adjacent cells and any pair of points that belong to these cells, and must be adjacent in the graph . (This is true regardless of the -norm being used, in view of (5).) The degree of a cell in the graph of cells is at most the number of cells contained in a ball of radius centered at the center of . As each cell has volume and the ball of radius has volume , we deduce that

 the maximum degree of the graph of cells is Δ(GC)=O(1/ε). (7)

Note that the number of points of that fall into each cell is distributed as with expectation . Then, we can easily bound the maximum number of points in a cell.

Lemma 5.

A.a.s. no cell in contains more than vertices of .

Proof.
 Pr[Bin(n,sd)≥logn]≤(n⌈logn⌉)sd⌈logn⌉≤(eε+o(1))⌈logn⌉=o(1/n),

since (assuming is sufficiently small). Thus, a union bound over all cells shows that a.a.s. none has more than many points. ∎

Similarly, we can perform analogous calculations to bound the number of vertices of that fall inside of a ball of radius centered around a vertex , take a union bound over all choices of , and conclude the following.

Lemma 6.

Given any constant , the maximum degree of the power graph is a.a.s. .

Definition 2.

We say a cell is dense if it has at least points of in it, and is otherwise sparse.

Note that this definition is different from the corresponding notions in [14] and [4], which only require dense cells to contain a large but constant number of points of . We will show we cannot have too many sparse cells or too large connected sets of sparse cells in the graph of cells (or in its -th power , for a fixed ). We shall also take into account whether these cells are “close” to the boundary of the cube . To make this precise, define for and . These are the facets (i.e. -dimensional faces) of the boundary of .

Lemma 7.
1. A.a.s. the number of sparse cells is at most .

2. Moreover, for any arbitrary constants and , a.a.s. the power graph

1. has no connected set of at least cells which are all sparse;

2. has no connected set of at least cells which are all sparse and such that some cell in lies within distance from at least facets of (for );

3. has no sparse cell within distance from facets of .

Remark 8.
1. The property “of at least cells” in the statement can be replaced by “of total volume at least ”, and the claim is still valid.

2. Note that this lemma provides an analogue of Lemma 4 in [14]. However, the latter gives a bound on the size of , for arbitrarily small (possibly much smaller than any fixed function of ). Here, we cannot achieve that, given our more restrictive definition of dense cell (i.e. sparse cells are more abundant). However, the current statement will suffice for our purposes. In particular, Lemma 4 in [14] is used in the proof of Lemma 5 in [14] with , which is greater than (if is sufficiently small), and thus this situation is covered by our present statement. Hence, Lemma 5 in [14] is still valid with our definition of dense cells, since we can replace all uses of Lemma 4 [14] in the proof by its counterpart in this manuscript.

Proof of Lemma 7.

Recalling that the number of points in any fixed cell is distributed as , the probability that a cell is sparse is (with the convention that for all )

 ⌈ε3logn⌉−1∑k=0(nk)sdk(1−sd)n−k ≤⌈ε3logn⌉−1∑k=0(ensdk(1−sd))ke−sdn ≤⌈ε3logn⌉(e+o(1)ε2)ε3logne−(ε+o(1))logn =n−ε+ε3log(e/ε2)+o(1) ≤n−ε(1−ε/2)(for large n),

provided that is chosen so that . Note that this is possible since as . In particular, this condition implies that and so the number of sparse cells is a.a.s. at most by Markov’s inequality. This proves part 1.

Fix , and . In order to prove 2(b), it is enough to show that contains no connected set of exactly sparse cells within distance from facets of . Observe that the events that two or more cells are sparse are negatively correlated. Therefore,

 the probability that k given cells are sparse is at most n−kε(1−ε/2). (8)

If we also choose small enough so that , the probability that a given set of cells are all sparse is . Since there are only possible connected sets of cells in the power graph lying within distance from facets of , we can take the union bound and complete the proof of part 2(b). Part 2(a) follows as a particular case of part 2(b) taking . Finally, the expected number of sparse cells within distance from facets of is at most times , which is . This immediately yields part 2(c) and completes the proof. ∎

Given a set of cells , we denote by the subgraph of the graph of cells induced by .

Definition 3.

Let be the set of dense cells, and let be the set of cells in the largest component of (i.e. the subgraph of induced by dense cells). If there are two or more such largest components, pick one according to any arbitrary deterministic rule (we will see that a.a.s.  is very large, so the choice of is unique). Call cells in good. Cells that are not good, but are adjacent in the graph of cells to some good cell are called bad. Bad cells must be sparse by construction. Cells that are not adjacent to good cells are called ugly. Note that ugly cells may be dense or sparse. Let and denote the set of bad and ugly cells, respectively.

As a crucial ingredient in our argument, we will use Lemma 5 in [14], which shows that a.a.s. ugly cells (which are called “bad” in that paper) appear in small clusters far enough from each other. (Note that this result is stated for dimension , and then extended to general in Section 4 of [14].) Unfortunately, the definition of dense cell we use in the present manuscript is more restrictive than the one in [14] (they only require a dense cell to contain at least points, for a large constant ; whilst here we require at least points). In order to overcome this minor obstacle, we simply observe that our Lemma 7 extends Lemma 4 in [14] to a less restrictive notion of sparse cell (although with a slightly weaker bound). In view of Remark 8(2), the proof of Lemma 5 in [14] is also valid in our setting by trivially replacing Lemma 4 in [14] by Lemma 7 of the present paper. Hence, adapting Lemma 5 in [14] to our current notation, we obtain the following statement.

Lemma 9 ([14]).

A.a.s. all connected components of have -diameter at most .

Next, we obtain useful bounds on the number of bad and ugly cells.

Lemma 10.

A.a.s. there are at most bad cells and at most ugly cells.

In particular, this implies that a.a.s. our choice of in Definition 3 was unique.

Proof.

Since bad cells are sparse by definition, the first part of the statement follows trivially from Lemma 7(1).

To prove the second part, consider a cell which is at distance at most from exactly facets of the cube , for some . Without loss of generality, assume these facets are precisely . Let (i.e.  is the point in with largest coordinates), and let denote the ball with centre and radius . By construction, the set

 S=P+(B∩([0,∞)i×Rd−i))

contains precisely those points within distance of and “above” with respect to the first coordinates. Moreover, is fully contained in the cube , and every cell intersecting must belong to the set of cells that are adjacent to in the graph of cells . Therefore, the number of cells in satisfies

 |N|+1 ≥vol(S)/sd =2−iθr0d(1−4ds/r0)d/sd ≥(2d−i−1/d+o(1))ε−1(1−4d2s/r0) ≥(2d−i−1/d)ε−1(1−αε1/d)(eventually) (9)

for some constant , where we used and the fact that for . Furthermore, let be the event that is ugly and is adjacent in to at most other ugly cells. This event implies that there must be a set of at least

 |N|−(4d2)d≥(2d−i−1/d)ε−1(1−2αε1/d) (10)

sparse cells adjacent to . Note that the last line follows since we can choose sufficiently small, given (and ). Thus, by (8) and summing over all possible choices of such , we obtain that the probability of is at most

 Oε(1)⋅n−(2d−i−1/d)(1−2αε1/d)(1−ε/2)≤n−(2d−i−1/d)(1−Θ(ε1/d))(eventually),

assuming that is sufficiently small. Hence, summing over and over the possible choices of , we show that the expected number of cells that are ugly and are adjacent in to at most other ugly cells is

 d∑i=0Oε((n/logn)(d−i)/dn−(2d−i−1/d)(1−Θ(ε1/d)))=nO(ε1/d),

since . By Markov’s inequality and in view of Lemma 9 (which implies that a.a.s. there are no ugly cells that are adjacent to more than other ugly cells), we conclude that a.a.s. there are at most ugly cells. This finishes the proof of the second statement. ∎

Lemma 11.

Given any constant , a.a.s. every two ugly cells lying in different components of are at (-normed) distance at least apart.

Proof.

We will assume that the a.a.s. conclusions of Lemmas 7 and 9 hold, and deterministically prove that any pair of nonadjacent cells in at distance less than cannot both be ugly. Thus, consider any two different cells and that are not adjacent in the graph of cells , but are at distance less than from each other. Suppose that there are exactly facets of the -cube (for some ) at distance less than from or . Assume, without loss of generality, that these facets are precisely . In particular, both and must be at distance at most from these facets (for sufficiently small given ), and at distance at least from any other facet.

We will proceed in a similar fashion as in the proof of Lemma 10 in order to describe a large set of cells that are adjacent to or . Let and be respectively the points in and with largest coordinates. The hyperplane in orthogonal to vector (with respect to the Euclidean inner product) and passing through the origin splits into two halfspaces

 H={Q∈Rd:⟨Q,P′−P⟩≤0}andH′={Q∈Rd:⟨Q,P′−P⟩≥0}.

Consider the set

 Bi=B∩([0,∞)i×Rd−i)

where denotes the ball with centre and radius . Since has volume , at least one of the sets or has volume at least half of this. Assume it is (by otherwise reversing the roles of and ). Define the sets

 S=P+BiandS′=P′+Bi∩H′.

Since and are not adjacent in , and must be at distance greater than . In particular, . Moreover, by our choice of , any other point in is further away from than is, so and must be disjoint sets. Let be the set of cells different than and intersecting . By construction, every cell in must be adjacent in the graph of cells to or . Moreover, since , we can use (9) to infer

 |N|+2≥vol(S∪S′)/sd≥(3/2)(2d−i−1/d)ε−1(1−αε1/d)≥(4/3)d−idε−1,

for sufficiently small given , where we used that . In view of Lemma 9 (assuming its a.a.s. conclusion is true), if both and are ugly, all but at most cells in must be sparse. Then, we obtained a set of at least sparse cells within distance at most from each other (i.e. pairwise adjacent in the power graph for ). This contradicts the a.a.s. conclusion of Lemma 7. ∎

Lemma 12.

A.a.s. the following holds. Given any ugly cell , let be the set of good cells at -distance at most from . Then, the subgraph of induced by is connected and has diameter (as a graph) at most .

Proof.

Let . Tesselate by -dimensional cubes of side . We call these cubes boxes to distinguish them from the cells. Note that a cell may intersect a box and not be fully contained in it. Regardless of that, each box contains at least

 (t/s−2)d≥ε−1/(θ(2d)d+1) (11)

cells. We say that two different boxes and are adjacent if the set is topologically connected (i.e. they share at least one point). (Recall that the term adjacent has a different meaning for cells.) If two different cells and intersect the same box or two adjacent boxes, then and must be at -distance at most , and thus must be adjacent in the graph of cells .

Fix an ugly cell . We will assume that all the a.a.s. properties leading to the conclusions of Lemmas 7, 9, 10 and 11 hold, and deterministically prove the statement for cell . Let be the union of all boxes at -distance between and from cell . Since the side of a box is , then is topologically connected. Moreover, has two connected components and with, say, . (It is worth noting that these properties of hold regardless of how close cell is from some facets of , but this may cease to be true if we replaced by some other in the definition of .) Let and be the set of good cells contained in and (respectively), and let be the set of good cells that intersect . Every good cell in must belong to exactly one of the sets , and . Observe that every cell in is at -distance at most from , and every cell in is at -distance at least from . Therefore, if and , then and must be at -distance at least . In particular, no cell in is adjacent to any cell in with respect to the graph of cells .

Claim 1.

Every box contains a good cell.

Since is a topologically connected union of boxes and pairs of cells contained in adjacent boxes are adjacent in the graph of cells, Claim 1 implies that must induce a connected subgraph of . Recall that the graph induced by the set is also connected by the definition of good cells. Hence, since there are no edges between and in the graph of cells, we deduce that also induces a connected graph. Let be the set of good cells at -distance at most from cell . Observe that . Moreover, and every cell must intersect a box which is adjacent to a box , so is adjacent in the graph of cells to some cell in . Hence, induces a connected graph as well.

We proceed to bound the diameter (as a graph) of . Consider any two cells and a path of cells in of minimal length. If two cells in the path intersect the same box, then they are adjacent in the graph of cells, so they must be consecutive in the path by our minimal length assumption. Similarly, we cannot have more than two consecutive cells in the path intersecting one same box. Therefore, we deduce that the number of cells in the path is at most twice the number of boxes that may be potentially intersected by cells in , which is at most

 2((6r0+s)/t+2)d≤2(20d)d.

This gives the desired upper bound on the diameter of the graph.

It only remains to prove Claim 1. In order to do so, suppose our ugly cell is at distance at most from exactly facets of the cube , for some . By (10) in the proof of Lemma 10, we can find a set of at least sparse cells within -distance at most from , where is a positive constant.

Pick any box . Every cell contained in is at -distance (and thus at -distance) at least from , so it cannot belong to . Suppose that all cells contained in are sparse. Then, by (11), we have at least

 (2d−i−1/d)ε−1(1−2αε1/d)+ε−1/(θ(2d)d+1)≥(2d−i−1/d)ε−1(1+ε)≥d−id(1+ε)ε−1

sparse cells at -distance at most from . Thus, these sparse cells are within -distance from each other and at distance at most from facets of . This contradicts the a.a.s. conclusion of Lemma 7 (with parameters and ). Therefore, every box in must contain at least one dense cell, which must be also good in view of Lemmas 9 and 11. This proves Claim 1, and completes the proof of the lemma. ∎

We say that a collection of paths in a graph covers a vertex if the vertex belongs to some path in . The next lemma provides us with an appropriate collection of paths that covers the ugly vertices and will be a crucial ingredient in the construction of a rainbow Hamilton cycle.

Lemma 13.

Let denote the set of vertices in that belong to ugly cells. Given any constant , a.a.s. there is a collection of vertex-disjoint paths in such that:

1. covers all vertices of ;

2. covers at most two vertices inside of each non-ugly cell;

3. every vertex in that is covered by is at graph-distance at most from some vertex in with respect to the graph ;

4. for each path , there is a good cell such that the two endvertices of lie in cells that are adjacent (in the graph of cells) to ;

5. every two different paths in are at -distance at least