Tight Chromatic Upper Bound for {3 Times K1, {2 Times K1 + (K2 UNION K1)}-free Graphs
TTight Chromatic Upper Bound for {3K , 2K +(K K )}-free Graphs Medha Dhurandhar [email protected]
Abstract
Problem of finding an optimal upper bound for the chromatic no. of 3K -free graphs is still open and pretty hard. It was proved by Choudum et al that upper bound on the chromatic no. of {3K , 2K +(K K )}-free graphs is 2ω. We improve this by proving for such graphs ≤ 8 = for ω = 5 and ≤ for ω
5, where ω is the size of a maximum clique in G. We also give examples of extremal graphs.
1. Introduction:
In [1], [2], [4], [5] chromatic bounds for graphs are considered especially in relation with and . Gyárfás [6] and Kim [7] show that the optimal -binding function for the class of graphs with independence number 2 has order ω /log(ω). Independence number 2 implies that the graph is 3K -free. If we forbid additional induced subgraphs, the order of the optimal -binding function may drop below ω /log(ω). We consider in this paper only finite, simple, connected, undirected graphs. The vertex set of G is denoted by V(G), the edge set by E(G), the maximum degree of vertices in G by Δ(G), the maximum clique size by (G) and the chromatic number by G). N(u) denotes the neighbourhood of u and )( uN = N(u)+u. In [1] it was proved that is 2ω is an upper bound on the chromatic no. of {3K , 2K +(K K )}-free graphs and the problem of finding tight chromatic upper bound for a {3K , 2K +(K K )}-free graph was stated as open. For further notation please refer to Harary [3].
2. Main Result:
Before proving the main result we prove some Lemmas.
Lemma 1:
Let G be {3K , 2K +(K K )}-free. Then V(G) = j M i where 1≤ j≤ 4 and
Let b M and c, c’ M be s.t. bc, bc’ E(G). If m M M is s.t. cm, c’m E(G) (cm, c’m E(G)), then bm E(G) (b m E(G)).
Proof: If G is complete, then the
Lemma is trivially true. Let v, w V(G) s.t. vw E(G). Let A = {x V(G)/xv, xw E(G)}, B = {x V(G)/xv E(G), xw E(G)}, and C = {x V(G)/xw E(G), xv E(G)}. Further let D A be s.t.
Let D = M , Y = M , B = M and C = M . As G is 3K -free, ,
G). Let if possible b, c have at the most |Y|-2 common adjacent vertices from Y Y’. W.l.g. let by i ’, cy i G for ’, w, y ’, c> = 2K +(K K ), a contradiction. W.l.g. l et if possible b c, c’ C s.t. bc E(G) and bc’ E(G) and by i E(G) i. As G is 3K -free, c’y i ’ E(G) i. As |Y| ≥ 4, by , cy i E(G) for at least three i, a contradiction to . If bm E(G), then
We call the pair (v, w) described in
Lemma
1, a partitioning pair.
Henceforth let X = i i x ; Y = i i y , Y’ = i i y ' with y i ’y i E(G); B = i i b and C = i i c be as defined in Lemma 1.
Lemma : If G is {3K , 2K +(K K )}-free, then (A) ω = 4 (G) ≤ 6 (B) ω = 5 (G) ≤ 8. Proof: Let if possible G be a smallest {3K , 2K +(K K )}-free graph not satisfying the result. Let deg v = ∆, (v, w) be a partitioning pair . Then every u V(G) is non-adjacent to at least |C|+1 but not more than ω vertices in G. Also as
I (A)
Let ω = 4. Then ∆ ≤ 8 and If ∆ ≤ 7, then ≤ = 6. Hence let ∆ = 8. Then by I , |X Y Y’| ≥ 5 |X| ≤ 3 and 2 ≤ |Y| = |Y’| ≤ 3. First let |Y| = 3. Then as every b i (c j ) is non-adjacent to three vertices from Y Y’ and w (v), B C> is complete. Also as ω = 4, by I , |X| = 0 and as ∆ = 8, |B| = 2 |C| ≤ 2. If |C| ≤ 1, then color the pairs (y i ’, y i ) (1≤i≤3); (v, c ); (w, b ); and b by a new color to get = 6, a contradiction. Hence let |C| = 2 y i ’ (y i ) is non-adjacent to at least two but not more than three vertices of B C. Thus by of Lemma 1 , every y i ’ (y i ) is non-adjacent to one b i and one c k W.l.g. let c y i ’ E(G) i. Then by of Lemma 1 , w.l.g. y ’b i E(G) i = 1, 2, a contradiction. Next let |Y| = 2. Then |X| = 1, |B C| = 3 and hence |B| ≥ 2. If |B| = 2, then color the pairs (y i , y i ’) (1≤i≤2); (w, b ); (v, c ) by same colors and x, b by two new colors to get ≤ 6, a contradiction. Hence |B| = 3. W.l.g. let xb E(G). Then color the pairs (x, b ); (v, w); (y i , y i ’) (1≤i≤2) by same colors and b , b by two new colors to get ≤ 6, a contradiction. (B) Let = 5. Then ≤ 13 and 9 ≤ ≤ ≤ ∆ ≤ 13. Also |C| ≤ |B| ≤
4, |X Y Y’| ≥ 7, |X Y| ≤ 4; |Y| ≥ 3 and |X| ≤ 1. If |X| = 1, then |Y| = 3 and |B| ≥ 4 |B| = 4 b i s.t. xb i E(G) (else
Lemma 3 . Main Result:
If G is {3K , 2K +(K K )}-free and ≥ 6, then (G) ≤ . Proof: Let if possible G be a smallest such graph with (G) > . Let v V(G) be with deg v = ∆ and s = |V(G)|-∆-1. Now s ≥ 1 (else G-v) ≤ vG and G) ≤ G-v)+1 ≤ G +1 ≤ ). Again s ≥ 2 (else let vw E(G) in any ( )-coloring of G-w, v receives a unique color say . By coloring w by , we get (G) ≤ ). Hence ∆ ≤ |(V(G)|-3 and as < (G) ≤ , ≥ 2 -1 ∆+s+1 ≥ 2ω+2. Also ≥ 6, +3 ≤ < (G). I Let v, w V(G) be s.t. deg v = and vw E(G). Let (v, w) be a partitioning pair. Then |X|+|Y|+1, |B|+1, |C|+1 ≤ and as s ≥ 2, 1 ≤ |B| ≤ |C|. Also |X Y Y’| ≥ ∆-ω+1 and hence by I , |X Y Y’| ≥ |Y| > 0. II Case 1: b i c j E(G) i and j. W.l.g. let b y i E(G) i c j y k ’ E(G) j, k b j y k E(G) j, k. Also let X = {x X/ xb j E(G) j} and X = X-X . Clearly
If b mi c nj , b mi y , c nj y ’ E(G), then x p b mi E(G) iff x p c nj E(G). If x p b mi E(G) and x p c nj E(G), then
If some b (c ) is non-adjacent to two vertices of C (B ) and b y j E(G) j, then b y j , c y n ’ E(G) k, j, m, n. Let b c , b c E(G). Then c y p ’, c y p ’ E(G) p. If b s.t. say y b E(G), then y ’b E(G) and b c b c E(G) c s.t. c b E(G) c y E(G) and c b E(G). But then = 2K +(K K ), a contradiction. Claim 3:
If every b is non-adjacent to exactly one c and vice versa, then j, x j is adjacent to exactly one of b and c where b c E(G).
Let if possible x j b , x j c E(G). Let b B-b . Then as c b E(G), x j b E(G) (else
Claim 4:
If |Y| ≥ 2, then every b is non-adjacent to exactly one c and vice versa. If not, then | B | ≥ 2 or |C | ≥ 2 and by Claim 2 w.l.g. let b y j , c y j ’ E(G) i, j. As |Y| ≥ 2, by of Lemma 1, b c E(G) i, j. Also | B | ≥ 2 (else | C | ≥ 2 and as |B| ≥ |C|, |B | > 0 and b y j ’ E(G) i, j = 2K +(K K )). Let X = {x i / x i b E(G) for some j} and X = X-X . Then |B C | ≥ 2 (else ∆+s+1 = |X C Y’ w|+|X B Y v|+|B C | ≤ 2 +1). Also |B | ≥ 1 (else |C | ≥ 2, by of Lemma 1, c y j E(G) i, j and
1, c mn y j E(G) m, n, j b y j ’ E(G) i, j and every b is adjacent to at the most one c and vice versa, |Y| = 1 (else = 2K +(K K )) and by of Lemma
1, |C | ≤ 1. Also by II , |X| > 0. As s ≥ 2, for each b say x i s.t. x i b E(G). Again x i x j for i j (else x i b , x i b E(G), and if x i b E(G) then = 2K +(K K ) and if x i c E(G) then
2. Now |B | ≥ 3 (else color vertices of X by |X| new colors and pairs (y , b ); (y ’, b ); those of corresponding non-adjacent vertices of Y-y and Y’-z; (b , c ); (w, b ); (v, c) by same colors to get ≤ |X|+|Y|+1+3 ≤ +3). learly some c is non-adjacent to more than one vertex of B and by Claim 2 , |Y| = 1 and is complete. Then by of Lemma
1, c is adjacent to at the most one b . Also as s ≥ 2, x X s.t. xb E(G) x is adjacent to at the most one b (else
1, x c E(G) x b E(G) i. If x , x s.t. x c , x c E(G), then x i b E(G) for j and i = 2, 3 t > 1, x m b E(G) for some m {2, 3} (else = 2K +(K K )). W.l.g. let x b E(G). But then
1, x b E(G). Then |B | ≤ 1 (else x i b E(G) for i = 1, 2 and = 2K +(K K )). Again for j > 1, no two b s have a common non-adjacent x X (else if xc E(G), then
Every b is non-adjacent to a unique c and vice versa. Then |B| ≤ 4 (else w.l.g. y b i E(G) for 1≤i≤3 and by
Claim 5 say b i = b for 1≤i≤2. Then c y E(G) and
21 1 i i b and y c E(G) c C-
21 1 i i c |C | = 0 (else as |B| ≥ |C| and |B | = |C |, |B | > 0 and = 2K +(K K )). Now |C | i is non-adjacent to at the most 4 vertices and deg x i > deg v = ), |B | > 0 (else color vertices of X by |X| new colors and pairs (v, w); (b , c ); (y i , y i ’) by same colors to get ≤ |X|+|Y|+1+|B | ≤ +3) and x i b E(G) (else deg x i > ). Then color vertices of X-x i by |X|-1 new colors and pairs (x i , b ); (v, w); (b , c ); (y i , y i ’) by same colors to get ≤ |X|+|Y|+1+3 ≤ +3, a contradiction. Case 2.2:
Some vertex B or C be non-adjacent to two vertices of C or B . Then by Claim 4 , |Y| = 1 and by
Claim 2 , b y j , c y n ’ E(G) for i, j, m, n. Then i, b c E(G) for at the most one j and vice versa. Also by II , |X| > 0. Let X = {x X/x b E(G) where b c E(G) for some k, m} and X = X-X Then |X | > 0 (else let X’ = {x X/x b E(G) where b c E(G) m} and X” = X-X’. Then ∆+s+1 = |X C y ’ w|+|X” B y v|+|B C | ≤ 2ω+2) |X | = 0. Thus every x k is non-adjacent to some b where b c E(G) for some j |B |, |C | ≥ 2 as b c , b c E(G) for some k, t. Now no two x’, x” X are non-adjacent to the same b where b c E(G) for some j. Let if possible x’b , x”b E(G). Then xb E(G) x X (else let xb E(G) and xb E(G)
Case 3.1: |C| > 1. If c, c’ s.t. zc, zc’ E(G), then by of Lemma 1 ,
As s ≥ 2, every y i ’ is non-adjacent to some vertex of B’ but by of Lemma 1 , y i ’ is not non-adjacent to more than one vertex of B’. Thus every vertex of B’ is non-adjacent to one y i ’ and vice versa |B’| = |Y’|. Also deg y i ’ = i. (b) xc E(G) x X (else as s ≥ 2, xb i E(G) for say i = 1, 2 and
1, xb’ E(G) b’ B’ (d) By of Lemma
1, every vertex of X is adjacent to at the most one vertex of B” (e)
Every vertex of B” is adjacent to at the most one vertex of X (else = 2K +(K K )). (f) If t = min{|B”|+1, |X|}, then (x i , u i ) where x i X, u i {B” c} s.t. x i u i E(G). This can be proved by induction on t using (d) and (e). (g)
If y ’b ’ E(G), then roles of X and B” are interchanged when instead of (v, w) (y ’, b ’) is considered as a partitioning pair . (h)
G ~ r1i 5 C , =3r. 2. If ω=2r+1, then let
G ~ r1i 5 C +W , =3r+1 3. If ω=5, then let G be as follows: V(G) = {v, w, i i y , ' i i y , i i b , i i c }. E(G) is defined as below: Vertex Non-adjacent to vertices Vertex Non-adjacent to vertices v w; c i i w, c , y ’, y ’, y ’ c v, b , y , y , y b w, c , y , y , y ’ c v, b , y ’, y ’, y b w, c , y , y ’, y c v, b , y ’, y , y ’ b w, c , y ’, y , y c v, b , y , y ’, y ’ y y ’, c , b , b , c y ’ y , b , c , c , b y y ’, c , b , c , b y ’ y , b , c , b , c y y ’, c , c , b , b y ’ y , b , b , c , c Edge Set E(G)
Then G is {3K , 2K +(K K )}-free, 10-regular and (G) = 8. References: [1] “Linear Chromatic Bounds for a Subfamily of 3K1-free Graphs”,S. A. Choudum, T. Karthick, M. A. Shalu,Graphs and Combinatorics 24:413–428, 2008 [2] “On the divisibility of graphs”, Chinh T. Hoang, Colin McDiarmid, Discrete Mathematics 242, 145–156, 2002 [3] Harary Frank,
Graph Theory (1969), Addison–Wesley, Reading, MA. [4] “ , , and ", B.A. Reed, J. Graph Theory 27, pp. 177-212, 1998 [5] “Some results on Reed's Conjecture about and with respect to ”, Anja Kohl, Ingo Schiermeyer, Discrete Mathematics 310, pp. 1429-1438 [6] “Problems from the world surrounding perfect graphs”, A. Gyárfás. Tanulmányok_MTA Számitástech. Automat. Kutató Int. Budapest, (177): 53, 1985. [7] “The Ramsey number R(3; t) has order of magnitude t2