BOUNDED-DEGREE PLANAR GRAPHS DO NOT HAVE BOUNDED-DEGREE PRODUCT STRUCTURE

. Product structure theorems are a collection of recent results that have been used to resolve a number of longstanding open problems on planar graphs and related graph classes. One particularly useful version states that every planar graph G is contained in the strong product of a 3-tree H , a path P , and a 3-cycle K 3 ; written as G ⊆ H ⊠ P ⊠ K 3 . A number of researchers have asked if this theorem can be strengthened so that the maximum degree in H can be bounded by a function of the maximum degree in G . We show that no such strengthening is possible. Specifically, we describe an infinite family G of planar graphs of maximum degree 5 such that, if an n -vertex member G of G is isomorphic to a subgraph of H ⊠ P ⊠ K c where P is a path and H is a graph of maximum degree ∆ and treewidth t , then t ∆ c ⩾ 2 Ω ( √ loglog n ) .


Introduction
Recently, product structure theorems have been a key tool in resolving a number of longstanding open problems on planar graphs.Roughly, a product structure theorem for a graph family G states that every graph in G is isomorphic to a subgraph of the product of two or more "simple" graphs.As an example, there are a number of graph classes G for which there exists integers t and c such that, for each G ∈ G there is a graph H of treewidth1 t and a path P such that G is isomorphic to a subgraph of the strong product2 of H, P , and a clique K of order c.This is typically written as G ⊆ H ⊠ P ⊠ K c , where the notation G 1 ⊆ G 2 is used to mean that G 1 is isomorphic to some subgraph of G 2 .See references [2, 3, 5, 8-13, 15, 17] for examples.In some applications of product structure theorems it is helpful if, in addition to having treewidth t, the graph H has additional properties, possibly inherited from G. For example, one very useful version of the planar graph product structure theorem states that for every planar graph G there exists a planar graph H of treewidth 3 and a path P such that G ⊆ H ⊠ P ⊠ K 3 [8,Theorem 36(b)].The planarity of H in this result has been leveraged to obtain better constants and even asymptotic improvements for graph colouring and layout problems, including queue number [8], p-centered colouring [7], and ℓ-vertex ranking [1].
In this vein, the authors have been repeatedly asked if H can have bounded degree whenever G does; that is: For each ∆ ∈ N, let G ∆ be the family of planar graphs of maximum degree ∆.Do there exist functions t : N → N, d : N → N, and c : N → N such that, for each ∆ ∈ N and each G ∈ G ∆ there exists a graph H of treewidth at most t(∆) and maximum degree d(∆) and a path P such that G ⊆ H ⊠ P ⊠ K c(∆) ?
In the current paper we show that the answer to this question is no, even when ∆ = 5.
Theorem 1.For infinitely many integers n ⩾ 1, there exists an n-vertex planar graph G of maximum degree 5 such that, for every graph H of treewidth t and maximum degree ∆, every path P , and every integer c, if G ⊆ H ⊠ P ⊠ K c then t∆c ⩾ 2 Ω( √ log log n) .
The graph family G := {G h : h ∈ N} that establishes Theorem 1 consists of complete binary trees of height h augmented with edges to form, for each i ∈ {1, . . ., h}, a path D i that contains all vertices of depth i. See Figure 1.

Proof of Theorem 1
Throughout this paper, all graphs G are simple and undirected with vertex-set V (G) and edge-set E(G).For a set S, G[S] denotes the subgraph of G induced by S ∩V (G) and Inside of asymptotic notation, log n := max{1, log 2 n}.

Partitions
Let G and H be graphs.An H-partition whose parts are indexed by the vertices of H with the property that, if vw is an edge of G with v ∈ B x and w ∈ B y then x = y or xy ∈ E(H).The width of H is the size of its largest part; that is, max{|B x | : x ∈ V (H)}.If H is in a class G of graphs then we may call H a G-partition of G. Specifically, if H is a tree, then H is a tree-partition of G and if H is a path, then H is a path-partition of G.A path-partition P := {P x : x ∈ V (P )} of G is also referred to as a layering of G and the parts of P are referred to as layers.A set of layers {P x 1 , . . ., P x q } ⊆ P is As in previous works, we make use of the following relationship between H-partitions and strong products, which follows immediately from the preceding definitions.
Observation 2. For every integer c ⩾ 1, and all graphs G, H, and The following important result of Ding and Oporowski [4] (also see [6,16]) allows us to focus on the case where the first factor in our product is a tree.
Theorem 3 (Ding and Oporowski [4]).If H is a graph with maximum degree ∆ and treewidth t, then H has a tree-partition of width at most 24∆(t + 1).
where H has treewidth t and maximum degree ∆ then there exists a tree T such that G ⊆ T ⊠ P ⊠ K 24c∆(t+1) .
Proof.By Theorem 3, H has a tree-partition T := {B x : x ∈ V (T )} of width at most 24∆(t+1).By Observation 2, The length of a path is the number of edges in it.Given two vertices v, w ∈ V (G), dist G (v, w) denotes the minimum length of a path in G that contains v and w, or dist Observation 5. Let G be a graph, let R ⊆ V (G), and let L be a layering of G. Then there exists a layer Proof.By the definition of layering, the vertices in R are contained in a set of at most diam G (R) + 1 consecutive layers of L. The result then follows from the Pigeonhole Principle.
We also make use of the following basic fact about tree-partitions: Observation 6.Let G be a graph, let T := {B x : x ∈ V (T )} be a tree-partition of G, let x ∈ V (T ), and let v, w ∈ N G (B x ) be in the same component of G − B x .Then T contains an edge xy with v, w ∈ B y .
Proof.Suppose that v ∈ B y and w ∈ B z for some y, z ∈ V (T ).Since v, w ∈ N G (B x ), T contains the edges xy and xz.All that remains is to show that y = z.For the purpose of contradiction, assume y z.Since v and w are in the same component of G − B x , G contains a path from v to w that avoids all vertices in B x , which implies that T contains a path P yz from y to z that does not include x.This is a contradiction since then P yz and the edges xy and yz form a cycle in T , but T is a tree.

Percolation in Binary Trees
The depth of a vertex v in a rooted tree T is the length of the path A T (v) from v to the root of T .Each vertex a ∈ V (A T (v)) is an ancestor of v, and v is a descendant of each vertex in V (A T (v)).We say that a set B ⊆ V (T ) is unrelated if no vertex of B is an ancestor of any other vertex in B.
For each h ∈ N, let T h denote the complete binary tree of height h; that is, the rooted ordered tree with 2 h leaves, each having depth h and in which each non-leaf vertex has exactly two children, one left child and one right child.Note that the ordering of T h induces an ordering on every unrelated set B ⊆ V (T h ), which we refer to as the left-to-right ordering.Specifically, v ∈ B appears before w ∈ B in the left-to-right ordering of B if and only if there exists a common ancestor a of both v and w such that the path from a to v contains the left child of a and the path from a to w contains the right child of a.
We use the following two percolation-type results for T h .

Lemma 7. Let h ⩾ 1, let r be the root of T h , and let
(ii) v r and the parent of v is in S ∪ {r}; and (iii) T h − S contains a path from v to a leaf of T h .

Proof. The proof is by induction on h. When
For h ⩾ 2, let ℓ be the maximum integer such that S ∪ {r} contains all 2 ℓ vertices of depth ℓ.Observe that 2 ℓ ⩽ |S|, so ℓ ⩽ log 2 |S| < h.Let L be the set of 2 ℓ+1 depth-(ℓ + 1) vertices in T h .By the Pigeonhole Principle some vertex r ′ ∈ L is the root of a complete binary tree If V (T ′ ) ∩ S = ∅ then choosing v := r ′ satisfies the requirements of the lemma.Otherwise, by applying induction on T ′ and S ′ := S ∩ V (T ′ ) we obtain a vertex v ′ of depth at most ℓ + 1 + log 2 (|S ′ |) + 1 ⩽ log 2 |S| + 1 whose parent is in S ∪ {r ′ }, and such that T h − S contains a path from v ′ to a leaf of T h .Thus v ′ satisfies requirements (i) and (iii).If the parent of v ′ is in S then v ′ also satisfies requirement (ii) and the lemma is proven, with v := v ′ .Otherwise, the parent of v ′ is r ′ , in which case r ′ satisfies requirements (i)-(iii) and we are done, with v := r ′ .Lemma 8. Let h ⩾ 1, let r be the root of T h , and and let S ⊆ V (T h ) with 1 ⩽ |S| < 2 h−1 .Then there exist two unrelated vertices v 1 and v 2 of T h such that, for each i ∈ {1, 2}: Proof.Let T 1 and T 2 be the two maximal subtrees of T h rooted at the children r 1 and r 2 , respectively of r. (Each of T 1 and T 2 is a complete binary tree of height h − 1.)For each i ∈ {1, 2}, let i is not in S, then the parent of v ′ i is r i S and r i satisfies (i)-(iii), so we set v i := r i .Finally, since v 1 ∈ V (T 1 ) and v 2 ∈ V (T 2 ), v 1 and v 2 are unrelated.

A Connectivity Lemma
The x × y grid G x×y is the graph with vertex-set V (G x×y ) := {1, . . ., x} × {1, . . ., y} and that contains an edge with endpoints (x 1 , y 1 ) and (x 2 , y 2 ) if and only if Proof.For each i ∈ {1, . . ., x − 1}, in order to separate column i from column i + 1, S must contain at least y subdivision vertices on the horizontal edges between columns i and i + 1.Since |S| < py, this implies that there are at most p − 1 values of i ∈ {1, . . ., x − 1} for which columns i and i + 1 are in different components of G − S.These at most p − 1 values of i partition {1, . . ., x} into at most p intervals, at least one of which contains at least x/p consecutive columns that are contained in a single component of G − S.

The Proof
Recall that, for each h ∈ N, G h is the planar supergraph of the complete binary tree T h of height h obtained by adding the edges of a path D i that contains all vertices of depth i, in left-to-right order, for each i ∈ {1, . . ., h}.Since T h is a spanning subgraph of G h , the depth of a vertex v in G h refers to the depth of v in T h .The height of a depth-d vertex of T h is h − d.We are now ready to prove the following result that, combined with Corollary 4 is sufficient to prove Theorem 1: Theorem 10.For every h ∈ N, every tree T , and every path It is worth noting that, unlike Theorem 1, there is no restriction on the maximum degree of the tree T .
Before diving into technical details, we first sketch our strategy for proving Theorem 10.We may assume that c ⩽ 2 √ log 2 h since, otherwise there is nothing to prove.
Recall Observation 2, which states that if G h ⊆ T ⊠ P ⊠ K c then G h has a tree-partition T := {B x : x ∈ V (T )} and a path-partition (that is, layering) P := {P y : y ∈ V (P )} with . This upper bound on |B x | is used to establish all of the results described in the following paragraph.
Refer to Figure 2. We will construct a sequence of sets R 1 , . . ., R t+1 and a sequence of nodes x 1 , . . ., x t+1 of T , where each R i is a family of unrelated sets in T h such that ∪R i ⊆ B x i .The first family R 1 consists of q 1 ⩾ h/(25c) singleton sets whose union is an unrelated set in T h .For each i ∈ {2, . . ., t + 1}, R i has size q i ⩾ q 1 /(10c) i−1 − 3.For each R i := {R i,1 , . . ., R i,q i }, each R i,j ⊆ V (T ) is an unrelated set in T h of size 2 i−1 that has a common ancestor a i,j of height at least h/5 that is at distance at most (i − 1)(log 2 (ch) + 2) from every element in R i,j .Furthermore, {a i,1 , . . ., a i,q i } is an unrelated set.These properties imply that ∪R i is also an unrelated set.
We do this for some appropriately chosen integer t ∈ Θ( log h) in order to ensure that q t+1 ⩾ 1, so R t+1 contains at least one part R of size 2 t .By Observation 5, there exists some y ∈ V (P ) such that √ log h therefore leads to the conclusion that c ⩾ 2 Ω( √ log h) , which establishes Theorem 1.
We now proceed with the details of the proof outlined above.The next two lemmas will be used to obtain the set R 1 that allows us to start the argument.Informally, the first lemma says that every balanced separator S of G h must contain a vertex of depth i for each i ∈ {i 0 , . . ., h}, where i 0 ∈ O(log |S|).
Proof.Let C be the vertex set of a component of G h − S that maximizes C ∩ V (D h ).For each i ∈ {0, . . ., h}, let C i := C ∩V (D i ) and let S i := S ∩V (D i ).We will show that, for each i ⩾ i 0 , C i is non-empty but does not contain all 2 i vertices in For each i ∈ {0, . . ., h − 1}, the vertices in C i+1 are adjacent to at least for i ⩾ 2 log 2 |S| + 2. Since i 0 ⩾ 2 log 2 |S| + 2, this establishes that C i is non-empty for each i ∈ {i 0 , . . ., h}.
The following lemma shows that every tree-partition of G h must have a part with a large unrelated set that is far from the leaves of T h and will be used to obtain our first set R 1 .
Lemma 12.For every α ∈ (0, 1/4), there exists h 0 such that the following is true, for all integers Let T Y be the minimal (connected) subtree of T h that spans Y , and let L be the set of leaves of T Y (excluding the root of T Y if this happens to be contained in Y ).Observe that L ⊆ Y is an unrelated set.Therefore, L satisfies (i) and, by definition, each vertex in L has height at least h/4 > αh, so L satisfies (iii).If |L| ⩾ αh ⩾ α 2 h/c then L also satisfies (ii).In this case, we can take R := L and we are done.We now assume that |L| < αh.
where the second equality is a standard fact about rooted trees.Rewriting this, we get Combining these two formulas, we obtain Refer to Figure 3.For each r ∈ Z, Lemma 7 applied to the subtree of T h rooted at r with S = B x implies that r has a descendant v such that (a) the parent of v is in Form the set Z ′ using the following rule for each r ∈ Z: If the vertex v described in the preceding paragraph is a child of r then place r in Z ′ , otherwise place v in Z ′ .Since each r ∈ Z is a child of some vertex in Y ⊆ B x , this ensures that the parent of v is in B x for each v ∈ Z ′ .Since Z is an unrelated set and Z ′ is obtained by replacing each vertex in Z with one of its descendants, Z ′ is an unrelated set.Since α < 1/4, for sufficiently large h, |Z ′ | ⩾ h/4 − O(log(ch + 1)) ⩾ αh and each vertex in Z ′ has height at least h/4 − O(log(ch + 1)) ⩾ αh.Now observe that the union of the paths Q v for each v ∈ Z ′ and the paths D h−⌈αh⌉+1 , . . ., D h contains a subgraph G ′ isomorphic to a graph that can be obtained from the grid G ⌈αh⌉×⌈αh⌉ by subdividing horizontal edges.Since B x does not contain any vertex of Q v for each v ∈ Z ′ , B x ∩ V (G ′ ) contains only vertices corresponding to subdivision vertices.Therefore, by Lemma 9, some component of G − B x contains a subset R ⊆ Z ′ of size at least α 2 h/c.Each element in R has a parent in B x .By Observation 6 some neighbour y of x in T has a bag B y that contains all of R.This completes the proof.
For each i ∈ {1, . . ., q} there exists a common ancestor a i of R i such that dist T h (v, a i ) ⩽ ℓ for each v ∈ R i .3. a 1 , . . ., a q are unrelated and each has height at least m.This definition has the following implications: (i) ∪R is an unrelated set; and (ii) If a i precedes a j in the left-to-right ordering of {a 1 , . . ., a q } then every element of R i precedes every element of R j in the left-to-right order of ∪R.We say that a vertex v of T h is compatible Lemma 13.Let R := {R 1 , . . ., R q } be a (k, ℓ, m)-compact set, and let S ⊇ ∪R have size 1 ⩽ |S| < 2 m−ℓ−2 .Then, there exists a (2k, ℓ Proof.For each i ∈ {1, . . ., q} and each r ∈ R i , replace r with the descendants v 1 and v 2 of r described in Lemma 8 and call the resulting set The next lemma is the last ingredient in the proof of Theorem 1. Lemma 14.Let T := {B x : x ∈ V (T )} be a tree-partition of G h and let P := {P y : y ∈ V (P )} be a path-partition of G h .Then there exists (x, y) ∈ V (T ) × V (P ) such that Proof.Let x ∈ V (T ) be a node that maximizes then there is nothing more to prove, so we may assume that |B x | < ch where c := 2 By Lemma 12, with α := 1/5, T contains a node x 1 such that B x 1 contains an unrelated set R of size q 1 := |R| ⩾ h/(25c) where each vertex in R has height at least m := m 1 := h/5.Let R 1 := {{v} : v ∈ R}.By definition R 1 is a (1, 0, m)-compact set.R 1 will be the first in a sequence of sets R 1 , . . ., R t+1 , where t will be fixed below.For each i ∈ {1, . . ., t + 1}, R i will satisfy the following properties: Note that, by a simple inductive argument, one can show that Indeed, the base case i = 1 holds trivially, and for the inductive case (i ⩾ 2) we have q i > q It is straightforward to verify that R 1 satisfies (a)-(c).Let t := min{t 1 , t 2 } where t 1 := ⌊log 10c (q 1 /3)⌋ and t 2 := ⌊h/(10( log 2 h + log 2 h) + 2)⌋.Observe that, since c = 2 log 10c (h/75c) ∈ Ω(log c h) ⊆ Ω( log h) and that t 2 ⩾ h/(10( log 2 h+log 2 h)+2)−1 ∈ Ω(h/ log h).Therefore t ∈ Ω( log h).These specific values of t 1 and t 2 are chosen for the following reasons: (i) Since t ⩽ t 1 , q t+1 > q 1 /(10c) t 1 − 3 ⩾ 0, so q t+1 ⩾ 1.
We now describe how to obtain R i+1 from R i for each i ∈ {1, . . ., t}.By Lemma 13 (applied to R := R i and S := B x i ), T h contains a (2 i , i log 2 (ch)+2, m)-compact set R + i+1 of size q i that is compatible with B x i .For each v ∈ ∪R + i+1 , v has height at least m i+1 := m i − (⌊log 2 (ch)⌋ + 2) ⩾ h/5 − i(⌊log 2 (ch)⌋ + 2).Therefore R + i+1 satisfies (a), but does not necessarily satisfy (c).Next we show how to extract R i+1 ⊆ R + i+1 that also satisfies (b) and (c).For each v ∈ ∪R + i+1 , T h − B x i contains a path Q v from v to a leaf of T h .The union of the paths in D h−m i+1 , . . ., D h and the paths in to a graph that can be obtained from G 2 i q i ×m i+1 by subdividing horizontal edges.By Lemma 9 applied to G := G ′ with S := B x i and p := ch/m i+1 , some component Since ∪R i is unrelated, it has a left to-right-order.This order defines a total order ≺ on the paths in C i , in which Q v ≺ Q w if and only if v precedes w in left-to-right order.The resulting total order (≺, C i ) corresponds to the order of the columns in G ′ and each part in R + i+1 corresponds to 2 i consecutive columns of G ′ .There are at most two parts We define R i+1 ⊆ R + i+1 as the set of parts in R + i+1 that are completely contained in The preceding calculation shows that R i+1 satisfies (b).Since ∪R i+1 is contained in a single component X of G x − B x i and each vertex in ∪R i+1 has a neighbour (its parent in T ) in B x i , Observation 6 implies that T contains an edge x i x i+1 with ∪R i+1 ⊆ B x i+1 .Therefore R i+1 satisfies (c).This completes the definition of R 1 , . . ., R t+1 .Properties (a)-(c) imply that, R t+1 is a (2 t , t(log 2 (ch) + 2), m)-compact set of size q t+1 ⩾ 1 and ∪R t+1 ⊆ B x t+1 for some x t+1 ∈ V (T ).Let R be one of the sets in R t+1 .Since R t+1 is (2 t , t(log 2 (ch) + 2), m)-compact, |R| ⩾ 2 t and all vertices in R have a common ancestor a whose distance to each element of R is at most t(log 2 (ch) + 2).Therefore, diam G h (R) ⩽ diam T h (R) ⩽ 2t(log 2 (ch) + 2).By Observation 5, there exists some P y ∈ P with

Open Problems
We know that every planar graph G is contained in a product of the form H ⊠P ⊠K 3 where tw(H) ⩽ 3 [8].Theorem 10 states that, for every c, there exists a planar graph of maximum degree 5 that is not contained in any product of the form T ⊠ P ⊠ K c where T is a tree and P is a path.This leaves the following open problem: Is every planar graph G of maximum degree ∆ contained in a product of the form H ⊠P ⊠K c where the treewidth of H is 2, P is a path, and c is some function of ∆?
Figure 4 and Observation 2 show that G h is a subgraph of H ⊠P where H has treewidth 2 (and is even outerplanar) and P is a path.Our proof breaks down in this case because, unlike tree-partitions, outerplanar-partitions do not satisfy Observation 6.Indeed, the outerplanar-partition illustrated in Figure 4

Figure 1 :
Figure 1: The graph G 5 from the graph family {G h : h ∈ N} that establishes Theorem 1.

Lemma 9 .
Let x, y, p ⩾ 1 be integers, let G be a graph obtained by subdividing horizontal edges of G x×y , and let S ⊆ V (G) \ V (G x×y ) be a set of subdivision vertices of size |S| < py.Then some component of G − S contains at least x/p consecutive columns of G x×y .
h/c; and (iii) Each vertex in R has height at least αh.Proof.It is well-known and easy to show that there exists a node x of T such that G − B x has no component with more than |V (G h )|/2 vertices [14, (2.6)].Let Y be the set of vertices in B x that have height at least h/4.By Lemma 11, |Y | ⩾ 3h/4 − O(log(ch + 1)).

Figure 3 :
Figure 3: A step in the proof of Lemma 12.

Figure 4 :
Figure 4: An outerplanar-partition H and a path-partition P of G 5 for which |B ∩ P | ⩽ 1 for each B ∈ H and P ∈ P .
Figure4and Observation 2 show that G h is a subgraph of H ⊠P where H has treewidth 2 (and is even outerplanar) and P is a path.Our proof breaks down in this case because, unlike tree-partitions, outerplanar-partitions do not satisfy Observation 6.Indeed, the outerplanar-partition illustrated in Figure4contains a part B x and a component X of G h − B x with |N G h (B x )∩V (X)| = h.In a tree-partition this would imply that |N G h (B x )∩V (X)∩B y | = h for some other part B y of the partition.In contrast, for the outerplanar-partition shown in Figure4, |N G h (B x ) ∩ V (X) ∩ B y | ⩽ 1 for each y ∈ V (H).