Graphs with Large Girth and Large Chromatic Number

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Graphs with Large Girth and Large Chromatic Number GRAPHS WITH LARGE GIRTH AND LARGE CHROMATIC NUMBER CHEUK TO TSUI Abstract. This paper investigates graphs that have large girth and large chromatic number. We first give a construction of a family of graphs that do not contain cycles of length 3, but have unbounded chromatic number. Next, we prove the existence of graphs that have large girth and large chro- matic number. With existence proven, we give an explicit construction of such graphs. Contents 1. Introduction 1 2. Triangle-Free Graphs with Large Chromatic Number 2 3. Existence of Graphs with Large Girth and Large Chromatic Number 3 4. Construction of Graphs with Large Girth and Large Chromatic Number 5 4.1. Lower Bound on Girth of Xp 8 4.2. Lower Bound on Chromatic Number of Xp,q 11 Acknowledgments 13 References 14 1. Introduction Finding a lower bound for the chromatic number of a given graph is, in general, difficult to do. There are few techniques that show that a certain number of colors are not enough to color a graph. Because of this, it is interesting to consider how we may construct graphs with large chromatic number. One way to do so is to include many cycles; since every even cycle requires at least 2 colors and every odd cycle requires at least 3, it is easy to construct graphs with large chromatic number as long as cycles are permitted (take the complete graphs, for instance). But the matter becomes more complicated when we do not permit cycles below a given length, and want to construct graphs with large girth and large chromatic number. In the remainder of Section 1, we define the terminology that allows us to begin addressing this problem. In section 2, we address the simplified problem of showing that there are triangle-free graphs with arbitrarily large chromatic number. In section 3, we show more generally that there exist graphs that have girth l and chromatic number k for any k and l. In section 4, we explicitly construct≥ graphs with large girth and≥ chromatic number, show the basic method of finding a lower bound for their girth, and find a lower bound for their chromatic number. 1 2CHEUKTOTSUI Definition 1.1. A graph G =(V,E) consists of a vertex set V and an edge set E, where elements of the edge set E are unordered pairs vi,vj ,withvi = vj and v ,v V . { } 6 i j 2 Definition 1.2. Two vertices vi and vj are adjacent if there is an edge between them, that is, v ,v E. We can also write v v . { i j} 2 i ⇠ j Definition 1.3. A cycle is a sequence of vertices that begins and ends at the same vertex, such that successive vertices are adjacent and no vertex except the first repeats. The number of distinct vertices in the sequence is called the length of the cycle. A cycle of length k is denoted Ck. Definition 1.4. The girth of a graph is the length of the shortest cycle contained in it. If a graph contains no cycles, its girth is defined to be . 1 Definition 1.5. A graph G is triangle-free if it does not contain a cycle of length 3. Definition 1.6. A set of vertices S is independent if no two vertices in S are adjacent. Definition 1.7. The independence number ↵(G) of a graph G is the maximum size of an independent vertex set. Definition 1.8. A (proper) k-coloring of a graph is a function f : V 1, 2,...,k such that if v and v are adjacent, then f(v ) = f(v ). ! { } i j i 6 j Definition 1.9. The chromatic number χ(G) of a graph G is the minimum value of k such that a k-coloring exists. 2. Triangle-Free Graphs with Large Chromatic Number Before addressing the general question of constructing graphs with large girth and chromatic number, we first consider the simpler problem of constructing triangle- free graphs with large chromatic number. Theorem 2.1. For any positive integer k, there exists a triangle-free graph with chromatic number k. Proof. For k = 1, a single point will work. For k = 2, two adjacent vertices will work, and for k = 3, C5 will work. For chromatic numbers larger than 3, we give a construction that, when applied to a triangle-free graph, produces a new triangle-free graph that increases the chromatic number by 1. Construction 2.2 (Mycielski. See [1]). Given a graph G, label its vertices v ,i i 2 1, 2,...,n . Create n new vertices ui and an additional vertex w, and draw an edge between{ each} u and w.Then,if v ,v is an edge, draw the two edges u ,v i { i j} { i j} and v ,u . Call this new graph G0. { i j} First, we show that if G is triangle-free, then G0 is triangle-free. Suppose there exists a C3 in G0.Vertexw cannot be a vertex of a C3, because it is adjacent only to all the ui, which are not pairwise adjacent. There are also no C3’s containing two of the ui,sincenotwoui are adjacent. Therefore, the C3 must contain two adjacent vi and one of the ui. Let the vertices be vj, vk and ul. They must be pairwise adjacent, meaning that vj,ul and vk,ul are edges of G0. By the construction, this would be the case only{ if }v ,v {and }v ,v were edges of G. But this would { j l} { k l} GRAPHS WITH LARGE GIRTH AND LARGE CHROMATIC NUMBER 3 mean that vj,vk,vl,vj would be a C3 in G, which contradicts the fact that G is triangle-free. Thus there exists no C3 in G0. Now, we show that χ(G0)=χ(G) + 1. Let χ(G)=k. This means that the vi must be colored by at least k colors. We claim that in any coloring of G0,theui must also be colored by at least k colors. Suppose that the ui can be colored by at most k 1 colors. so at least one of the v are colored by color k, while none − i of the ui are. Observe that for all i, vi and ui are adjacent to exactly the same vj’s in G. Take the vi that are colored by color k, and change them to the color of the corresponding ui. This new coloring of the vi would still be valid, because each vi has the same neighbours as the corresponding ui, meaning that none of the vj’s adjacent to each vi have the same color as the corresponding ui.Butthis new coloring (of the v ) uses at most k 1 colors, contradicting the fact that G i − must be colored by at least k colors. Thus, the ui must be colored by at least k colors as well. Since w is adjacent to all the ui, it must be colored by a (k + 1)st color. Therefore, χ(G0) χ(G) + 1. Furthermore, G0 can be colored by χ(G)+1 ≥ colors: First color the vi with k colors, then color each ui the same color as the corresponding vi. Then color w with a (k + 1)st color. So χ(G0)=χ(G) + 1. Therefore, to obtain a triangle-free graph with chromatic number k>3, we iteratively apply the construction to C k 3 times. 5 − ⇤ 3. Existence of Graphs with Large Girth and Large Chromatic Number Theorem 2.1 shows that the chromatic number of a graph can be made arbitrarily large, even if it does not contain cycles of length 3. Now we consider the more general question of whether it is possible to avoid cycles below a given length, yet still have an arbitrarily large chromatic number. As preparation, we present one probabilistic result and define some more notation. Lemma 3.1 (Markov’s Inequality). If X is a non-negative discrete random variable and a>0, then E[X] P [X a] . ≥ a Proof. We have E[X]= xP (x) x 0 X≥ = xP (x)+ xP (x) x a x<a X≥ X xP (x) ≥ x a X≥ a P (x) ≥ x a X≥ = aP [X a]. ≥ ⇤ 4CHEUKTOTSUI Definition 3.2 ([2, p. 38]). We write G G(n, p)ifG is a random graph on n ver- tices chosen by picking each pair of vertices⇠ as an edge randomly and independently with probability p. Definition 3.3 (little-oh notation). We write an = o(bn)if a lim n =0. n !1 bn Remark 3.4. an = o(1) means that limn an =0. !1 Theorem 3.5 (Erd¨os1959. See [2, pp. 38-39]). For all k, l there exists a graph G with girth(G) >land χ(G) >k. ✓ 1 Proof. Let ✓ < 1/l, and let G G(n, p)withp = n − . Let X be the number of ⇠ cycles of length at most l in G, and let Xi be the number of cycles of size exactly i. By linearity of expectation, we have l E[X]= E[Xi]. i=3 X To determine E[X], we count the possible number of cycles of length i. There are n (i 1)! i ways to choose the vertices of the cycle, and −2 ways to order them (we can start with any vertex). We must divide by 2, because the reverse of any ordering gives the same cycle. So the total number of possible cycles is n (i 1)! (n) − = i , i 2 2i ✓ ◆ where (n)i =(n)(n 1)...(n i + 1) (falling factorial).
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