1.5 The Crossing-Lemma Averaging Technique Given a graph G = ( V, E - - PDF document

1.5 The Crossing-Lemma Averaging Technique

The Crossing Lemma. Given a graph G = (V, E), let cr(G) denote the crossing number

• f G, which is the minimum of the number of crossings of any embedding of G in R2. By

definition, planar graphs have crossing number 0. The crossing lemma states the following: Theorem 11 (Crossing Lemma). For any graph G with m edges and n vertices, cr(G) = Ω(m3/n2). It is easy to see that, for any value of m and n, this bound is tight. Consider for example, a graph G which is a union of n2/m cliques, each with m/n vertices. G has n vertices, and m edges; furthermore, an arbitrary embedding of each clique gives at most O((m/n)4) crossings, and so cr(G) = O((m/n)4 · n2/m) = O(m3/n2). The crossing lemma can be proved by a double-counting argument. To get the main idea, consider first the easy case of understanding the crossing number of the complete graph on n vertices, Kn: every subset of five vertices induce a K5, and so will have a crossing. Therefore, there are n

5

• f crossings, except that each crossing can be double-counted several (at most

n − 4) times, so dividing gives us the lower-bound of Ω(n4). Formally, fix the embedding of Kn minimizing the crossing number. We will double-count the following quantity: the sum of the crossing numbers of graphs induced by every subset

• f five vertices of Kn. For Kn, the subgraph induced by every five vertices is K5, which is

non-planar. Therefore, we have:

• G′⊆Kn,

|G′|=5

cr(G′) = n 5

• · cr(K5) =

n 5

• On the other hand, for each G′, consider its embedding exactly as laid out in the fixed

embedding of Kn. By double-counting, the sum of crossings for all G′ realized by this fixed embedding is exactly equal to the number of subsets G′ in which each crossing of Kn appears. A fixed crossing ⊠ of Kn requires four vertices (forming the two crossing edges) and so:

• G′⊆Kn,

|G′|=5

cr(G′) ≤

• G′⊆Kn,

|G′|=5

Crossings of G′ in the fixed embedding of Kn =

• Crossing ⊠

Number of G′ in which crossing ⊠ appears =

• Crossing ⊠

n − 4 1

• = cr(Kn) · (n − 4)

Putting the upper- and lower- bounds together, we get the required lower-bound for cr(Kn). 18

The same idea works for an arbitrary graph G. As before, we’d like to consider subsets of G, and show that each subset has at least some number of crossings. Then summing this up, and eliminating the crossings which are double-counted will give a lower-bound for the crossing number of G. However, unlike the Kn case, each subgraph G′ of size t can have very different number of edges, and so consequently two subgraphs on t vertices can have widely different crossing

• numbers. However, as a function of number of edges of G′, one can still get a lower-bound

for cr(G′): Claim: For any graph G with n vertices and m edges, cr(G) ≥ m − 3n. Take the embedding of G realizing cr(G). For each crossing, delete one of the edges. The remaining graph has at least m−cr(G) edges. Since it is planar, we have m−cr(G) ≤ 3n−6, and we get the desired bound. Now, exactly as before, it remains to count the crossing numbers of all subgraphs of G induced by t vertices, the parameter t to be fixed optimally later on. Consider all subsets V ′ ⊆ V of size t. Denote the i-th subset as Vi, with induced edges Ei, and the induced subgraph Gi. By the above claim, cr(Gi) ≥ |Ei| − 3 · |Vi| ∀ i Summing up over all i, we get a lower-bound on the sum of crossing numbers:

• i

cr(Gi) ≥

• i

|Ei| −

• i

3|Vi| where

• i

|Vi| =

• vertices v

# of subsets v’s in =

• v

n − 1 t − 1

• = n ·

n − 1 t − 1

• i

|Ei| = m · n − 2 t − 2

• The upper-bound follows exactly as for Kn: consider the embedding of each Gi exactly as

laid out in the fixed embedding of G realizing cr(G). Then each crossing ⊠ of G appears as a crossing in exactly n−4

t−4

• subgraphs Gi, and so
• i

cr(Gi) ≤

• i

Crossings of Gi in the fixed embedding of G = cr(G) n − 4 t − 4

• Putting the upper- and lower- bounds together:

cr(G) n − 4 t − 4

• ≥

m · n − 2 t − 2

• − 3n ·

n − 1 t − 1

• cr(G)

≥ m n−2

t−2

• n−4

t−4

− 3n n−1

t−1

• n−4

t−4

= c1 · mn2 t2 − c2 · n4 t3 Setting t = c3 · n2

m for a large-enough constant c3, the above gives cr(G) = Ω(m3/n2).

19

Two other flavors. The averaging idea in the above argument can be ‘implemented’ in two

• ther ways: probabilistic, and inductive. The argument above can be seen as an ‘unfolding’
• f the inductive argument, which averages out one vertex at a time via induction.

The probabilistic one is, in my opinion, much nicer, and it captures the abstract averaging idea very well: pick a subgraph induced by a random subset of vertices. The number of edges and crossings in this subgraph will be the average of values computed above, and so just applying the bound from above claim on this subgraph will give the required lower-bound

• n cr(G).

Or seen another way: if cr(G) is small (say just m − 3n) for a graph G, then there will exist an induced subgraph, say G′, of G with (relatively) lots of edges, but much fewer crossings. This gives a contradiction, since even in the induced subgraph, the weak crossing bound holds (i.e., cr(G′) ≥ |E′| − 3|V ′|). Therefore, cr(G) will have to be quite large. Crossing triangles in R3. Here’s another application of the crossing lemma-type tech-

• nique. Given a set P of n points in R3, let T be a set of t triangles spanned by these points.

Two triangles ∆1 and ∆2 cross if and only if they intersect and don’t share a vertex. Let cr(T) be the number of crossing triangles of T. Theorem 12 (Triangle-Crossing Lemma). Given a set P of n points in R3, cr(T) ≥ t4/n6. The proof is exactly the same as the Crossing-Lemma proof. We first need the following weaker claim. Claim: Given a set P of n points in R3 and any set T of t triangles, cr(T) ≥ t − 3n2. First observe that there can be only 3n2 triangles if no pair of triangles cross: imagine a very small ball B around each point p ∈ P. Then each triangle with one vertex as p will intersect B in an edge, and so the triangles with one vertex as p map to edges over n − 1 vertices

• n this ball. As no two triangles cross, this graph is planar, and has at most 3n edges, and

so there are 3n triangles with p as one of their vertices. Summing up over all n points, we get 3n2 triangles. But each triangle is counted three times, so in fact there are at most n2 triangles in T ′ 4. Now, given any set T, for each crossing pair, remove one triangle. There are at least t − cr(T) triangles remaining, and since now there are no crossing triangles, t − cr(T) ≤ 3n2. Given P, add each point to our sample R with probability p. The expected number of induced triangles in R are t · p3, and the expected number of crossings are cr(T) · p6. By Claim above and Linearity of Expectation, we have cr(T) · p6 ≥ t · p3 − 3(np)2

4Actually, this is not completely accurate. If no two triangles cross, then still the graph on B need not

be planar, as the two intersecting triangles could share the same vertex p, and so would not be considered

• crossing. But then, one edge, say edge e, of one of the two intersecting triangles must intersect the other

triangle, say ∆2. And there can be at most 3 triangles using the edge e (with the third vertex from ∆2), as

• therwise any other triangle using e would properly cross ∆2. Remove these three triangles incident to e;
• verall, we lose only 3 ·

n

2

• ≤ 2n2 triangles.

20

Setting p = c1 · n2/t, for a large-enough constant c1, we get the required bound. Final Remarks. The probabilistic proof of the crossing lemma is well-known, and can be found in several books (e.g., in Proofs from the Book or in [M]). The inductive proof can be found in Pach and Agarwals book. 21