Algebraically grid-like graphs have large tree-width

By the Grid Minor Theorem of Robertson and Seymour, every graph of suﬃciently large tree-width contains a large grid as a minor. Tree-width may therefore be regarded as a measure of ’grid-likeness’ of a graph. The grid contains a long cycle on the perimeter, which is the F 2 -sum of the rectangles inside. Moreover, the grid distorts the metric of the cycle only by a factor of two. We prove that every graph that resembles the grid in this algebraic sense has large tree-width: Let k, p be integers, γ a real number and G a graph. Suppose that G contains a cycle of length at least 2 γpk which is the F 2 -sum of cycles of length at most p and whose metric is distorted by a factor of at most γ . Then G has tree-width at least k .


Introduction
For a positive integer n, the (n × n)-grid is the graph G n whose vertices are all pairs (i, j) with 1 ≤ i, j ≤ n, where two points are adjacent when they are at Euclidean distance 1. The cycle C n , which bounds the outer face in the natural drawing of G n in the plane, has length 4(n − 1) and is the F 2 -sum of the rectangles bounding the inner faces. This is by itself not a distinctive feature of graphs with large tree-width: The situation is similar for the n-wheel W n , the graph consisting of a cycle D n of length n and a vertex x / ∈ D n which is adjacent to every vertex of D n . There, D n is the F 2 -sum of all triangles xyz for yz ∈ E(D n ). Still, W n only has tree-width 3.
The key difference is the fact that in the wheel, the metric of the cycle is heavily distorted: any two vertices of D n are at distance at most two within W n , even if they are far apart within D n . In the grid, however, the distance between two vertices of C n within G n is at least half of their distance within C n .
In order to incorporate this factor of two and to allow for more flexibility, we equip the edges of our graphs with lengths. For a graph G, a length-function on G is simply a map ℓ : E(G) → R >0 . We then define the ℓ-length ℓ(H) of a subgraph H ⊆ G as the sum of the lengths of all edges of H. This naturally induces a notion of distance between two vertices of G, where we define d ℓ G as the minimum ℓ-length of a path containing both. A subgraph H ⊆ G is ℓ-geodesic if it contains a path of length d ℓ G (a, b) between any two vertices a, b ∈ V (H). When no length-function is specified, the notions of length, distance and geodecity are to be read with respect to ℓ ≡ 1 constant.
On the grid-graph G n , consider the length-function ℓ which is equal to 1 on E(C n ) and assumes the value 2 elsewhere. Then C n is ℓ-geodesic of length ℓ(C n ) = 4(n − 1) and the sum of cycles of ℓ-length at most 8. We show that any graph which shares this algebraic feature has large tree-width. Theorem 1. Let k be a positive integer and r > 0. Let G be a graph with rational-valued length-function ℓ. Suppose G contains an ℓ-geodesic cycle C with ℓ(C) ≥ 2rk, which is the F 2 -sum of cycles of ℓ-length at most r. Then the tree-width of G is at least k.
The starting point of Theorem 1 was a similar result of Matthias Hamann and the author [2]. There, it is assumed that not only the fixed cycle C, but the whole cycle space of G is generated by short cycles.
Theorem 2 ([2, Corollary 3]). Let k, p be positive integers. Let G be a graph whose cycle space is generated by cycles of length at most p. If G contains a geodesic cycle of length at least kp, then the tree-width of G is at least k.
It should be noted that Theorem 2 is not implied by Theorem 1, as the constant factors are different. In fact, the proofs are also quite different, although Lemma 5 below was inspired by a similar parity-argument in [2].
It is tempting to think that, conversely, Theorem 1 could be deduced from Theorem 2 by adequate manipulation of the graph G, but we have not been successful with such attempts.

Proof of Theorem 1
The relation to tree-width is established via a well-known separation property of graphs of bounded tree-width, due to Robertson and Seymour [3].

Lemma 3 ([3]
). Let k be a positive integer, G a graph and A ⊆ V (G). If the tree-width of G is less than k, then there exists It is not hard to see that Theorem 1 can be reduced to the case where ℓ ≡ 1. This case is treated in the next theorem.
Theorem 4. Let k, p be positive integers. Let G be a graph containing a geodesic cycle C of length at least 4⌊p/2⌋k, which is the F 2 -sum of cycles of length at most p. Then for every X ⊆ V (G) of order at most k, some component of G − X contains at least half the vertices of C.
Proof of Theorem 1, assuming Theorem 4. Let D be a set of cycles of length at most r with C = D. Since ℓ is rational-valued, we may assume that r ∈ Q, as the premise also holds for r ′ the maximum ℓ-length of a cycle in D. Take an integer M so that rM and ℓ ′ (e) := M ℓ(e) are natural numbers for every e ∈ E(G).
Obtain the subdivision G ′ of G by replacing every e ∈ E(G) by a path of length ℓ ′ (e). Denote by C ′ , D ′ the subdivisions of C and D ∈ D, respectively.
By Lemma 3, G ′ has tree-width at least k. Since tree-width is invariant under subdivision, the tree-width of G is also at least k.
Our goal is now to prove Theorem 4. The proof consists of two separate lemmas. The first lemma involves separators and F 2 -sums of cycles.
Lemma 5. Let G be a graph, C ⊆ G a cycle and D a set of cycles in G such that C = D. Let R be a set of disjoint vertex-sets of G such that for every Proof. Suppose that no D ∈ D meets two distinct R, R ′ ∈ R. Then C has no edges between the sets in R: Any such edge would have to lie in at least one D ∈ D. Let Y := R and let Q be the set of components of G − Y .
Let Q ∈ Q, R ∈ R and D ∈ D arbitrary. If D has an edge between Q and R, then D cannot meet Y \ R. Therefore, all edges of D between Q and V (G) \ Q must join Q to R. As D is a cycle, it has an even number of edges between Q and V (G) \ Q and thus between Q and R. As C = D, we find For every R ∈ R which intersects C, there are precisely two edges of C between R and V (C) \ R, because R ∩ C is connected. As mentioned above, C contains no edges between R and Y \R, so both edges join R to V (G)\Y . But C has an even number of edges between R and each component of V (G) \ Y , so it follows that both edges join R to the same Q(R) ∈ Q.
Since every component of C −(C ∩Y ) is contained in a component of G−Y , it follows that there is a Q ∈ Q containing all vertices of C not contained in Y .
To deduce Theorem 4, we want to apply Lemma 5 to a suitable family R with R ⊇ X to deduce that some component of G − X contains many vertices of C. Here, D consists of cycles of length at most ℓ, so if the sets in R are at pairwise distance > ⌊ℓ/2⌋, then no D ∈ D can pass through two of them. The next lemma ensures that we can find such a family R with a bound on | R|, when the cycle C is geodesic.
Lemma 6. Let d be a positive integer, G a graph, X ⊆ V (G) and C ⊆ G a geodesic cycle. Then there exists a family R of disjoint sets of vertices of G with X ⊆ R ⊆ X ∪ V (C) and | R ∩ V (C)| ≤ 2d|X| such that for each R ∈ R, the set R ∩ V (C) induces a (possibly empty) connected subgraph of C and the distance between any two sets in R is greater than d.
Proof. Let Y ⊆ V (G) and y ∈ Y . For j ≥ 0, let B j Y (y) be the set of all z ∈ Y at distance at most jd from y. Since |B 0 Y (y)| = 1, there is a maximum number j for which |B j Y (y)| ≥ 1 + j, and we call this j = j Y (y) the range of y in Y . Observe that every z ∈ Y \ B jY (y) has distance greater than (j Y (y) + 1)d from y.
Starting with X 1 := X, repeat the following procedure for k ≥ 1. If X k ∩ V (C) is empty, terminate the process. Otherwise, pick an x k ∈ X k ∩ V (C) of maximum range in X k . Let j k := j X k (x k ) and B k := B j k X k (x k ). Let X k+1 := X k \ B k and repeat.
Since the size of X k decreases in each step, there is a smallest integer m for which X m+1 ∩ V (C) is empty, at which point the process terminates. By construction, the distance between B k and X k+1 is greater than d for each k ≤ m. For each 1 ≤ k ≤ m, there are two edge-disjoint paths P 1 k , P 2 k ⊆ C, starting at x k , each of length at most j k d, so that B k ∩ V (C) ⊆ S k := P 1 k ∪ P 2 k . Choose these paths minimal, so that the endvertices of S k lie in B k . Note that every vertex of S k has distance at most j k d from x k . Therefore, the distance between R k := B k ∪ S k and X k+1 is greater than d.
We claim that the distance between R k and R k ′ is greater than d for any k < k ′ . Since B k ′ ⊆ X k+1 , it is clear that every vertex of B k ′ has distance greater than d from R k . Take a vertex q ∈ S k ′ \ R k ′ and assume for a contradiction that its distance to R k was at most d. Then the distance between x k and q is at most (j k + 1)d. Let a, b ∈ B k ′ be the endvertices of S k ′ . If x k / ∈ S k ′ , then one of a and b lies on the shortest path from x k to q within C and therefore has distance at most (j k + 1)d from x k . But then, since j k is the range of x k in X k , that vertex would already lie in B k , a contradiction. Suppose now that x k ∈ S k ′ . Then x k lies on the path in S k ′ from x k to one of a or b, so the distance between x k and x k ′ is at most j k ′ d. Since x k ′ ∈ X k ∩ V (C), it follows from our choice of x k that where the second inequality follows from the fact that X k ′ ⊆ X k and j Y (y) ≥ j Y ′ (y) whenever Y ⊇ Y ′ . But then x k ′ ∈ B k , a contradiction. This finishes the proof of the claim.
Finally, let R : The distance between any two sets in R is greater than d. For k ≤ m, R k ∩ V (C) = S k is a connected subgraph of C, while X m+1 ∩ V (C) is empty. Moreover, Proof of Theorem 4. Let X ⊆ V (G) of order at most k and let d := ⌊p/2⌋. By Lemma 6, there exists a family R of disjoint sets of vertices of G with X ⊆ R ⊆ X ∪ V (C) and | R ∩ V (C)| ≤ 2dk so that for each R ∈ R, the set R ∩ V (C) induces a (possibly empty) connected subgraph of C and the distance between any two sets in R is greater than d.
Let D be a set of cycles of length at most p with C = D. Then no D ∈ D can meet two distinct R, R ′ ∈ R, since the diameter of D is at most d. By Lemma 5, there is a component Q of G − R which contains every vertex of C \ R. This component is connected in G − X and therefore contained in some component Q ′ of G − X, which then satisfies Since |C| ≥ 4dk, the claim follows.

Remarks
We have described the content of Theorem 1 as an algebraic criterion for a graph to have large tree-width. The reader might object that the cycle C being ℓ-geodesic is a metric property and not an algebraic one. Karl Heuer has pointed out to us, however, that geodecity of a cycle can be expressed as an algebraic property after all. This is a consequence of a more general lemma of Gollin and Heuer [1], which allowed them to introduce a meaningful notion of geodecity for cuts.
Proposition 7 ([1]). Let G be a graph with length-function ℓ and C ⊆ G a cycle. Then C is ℓ-geodesic if and only if there do not exist cycles Finally, we'd like to point out that Theorem 1 does not only offer a 'one-way criterion' for large tree-width, but that it has a qualitative converse. First, we recall the Grid Minor Theorem of Robertson and Seymour [4], phrased in terms of walls . For a positive integer t, an elementary t-wall is the graph obtained from the 2t × t-grid as follows. Delete all edges with endpoints (i, j), (i, j + 1) when i and j have the same parity. Delete the two resulting vertices of degree one. A t-wall is any subdivision of an elementary t-wall. Note that the (2t× 2t)grid has a subgraph isomorphic to a t-wall.
Theorem 8 (Grid Minor Theorem [4]). For every t there exists a k such that every graph of tree-width at least k contains a t-wall.
Here, then, is our qualitative converse to Theorem 1, showing that the algebraic condition in the premise of Theorem 1 in fact captures tree-width.
Corollary 9. For every L there exists a k such that for every graph G the following holds. If G has tree-width at least k, then there exists a rational length-function on G so that G contains a ℓ-geodesic cycle C with ℓ(C) ≥ L which is the F 2 -sum of cycles of ℓ-length at most 1.
Proof. Let s := 3L. By the Grid Minor Theorem, there exists an integer k so that every graph of tree-width at least k contains an s-wall. Suppose G is a graph of tree-width at least k. Let W be an elementary s-wall so that G contains some subdivision W ′ of W , where e ∈ E(W ) has been replaced by some path P e ⊆ G of length m(e).
The outer cycle C of W satisfies d C (u, v) ≤ 3d W (u, v) for all u, v ∈ V (C). Moreover, C is the F 2 -sum of cycles of length at most six.
Define a length-function ℓ on G as follows. Let e ∈ E(G). If e ∈ P f for f ∈ E(C), let ℓ(e) := 1/m(f ). Then ℓ(P f ) = 1 for every f ∈ E(C). If e ∈ P f for f ∈ E(W ) \ E(C), let ℓ(e) := 3/m(f ). Then ℓ(P f ) = 3 for every f ∈ E(W ) \ E(C). If e / ∈ E(W ′ ), let ℓ(e) := 10s 3 , so that ℓ(e) > ℓ(W ′ ). It is easy to see that the subdivision C ′ ⊆ G of C is ℓ-geodesic in G. It has length ℓ(C ′ ) = |C| ≥ 6s and is the F 2 -sum of the subdivisions of 6-cycles of W . Each of these satisfies ℓ(D) ≤ 18. Rescaling all lengths by a factor of 1/18 yields the desired result.