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Investigating the maximum difference between the chromatic number and maximum clique that can be reached by graphs with at most n vertices. Analyzing the duality gap, perfect, critical, and extremal graphs, as well as the Ramsey numbers.
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The chromatic gap and its extremes András Sebő, CNRS, G-SCOP,Grenoble with AndrásGyárfás MTA SZTAKI, Budapest, Nicolas Trotignon, CNRS, LIAFA, Paris
THE PROBLEM What is the maximum of the differencechromatic number – max cliquethat can be reached by graphs on at most n vertices ?
(G) := min cover by cliques = chromatic number of complement (G) := max stable (independent) set of vertices (no induced edge) gap(G):= (G) - (G) 0 This is the(integer linear programming) duality gap of a related linear program
Perfect, critical, extremal Perfect :gap = 0 induced subgraph Gap is not necessarily monotonous: gap-critical: G, gap(H)<gap(G) induced H Minimal imperfect : gap-critical with gap = 1. t-extremal graph: of min order with gap t t-extremal gap-critical factorcritical? = 3 = 3 gap=0 = 2 gap=1
Critique, pas critique ? G facteur-critique ? C9 ? C93 ? G
gap(n)=max {gap(G) : G has n vertices} s(t) =smallest order of a graph of gap t gap-extremal ? gap(i)=0 i 4, ... gap-critical ! gap(5)=1, gap(10)=2, … ??? s(1)=5, s(2)=10, … , s(t)=5t?
R13= (3,5)-Ramsey … s(3)=15 ? Counterexample: n=13, =2, =4, =7 gap = 3 s(3)= 13
Ramsey: number, graph Ramsey number:R(,):=the smallest integer s.t.: every graph on at least R(,) vertices has either an -clique or an -stable set. (Ramsey theorem: it is finite.) Ramsey graph : Graph on R(,) – 1 vertices having neither an -clique nor an -stable set. Example : R(3,3) =6; (3, 3) Ramsey graph R(3,2), R(3,3), … : 2, 6, 9, 14, 18, 23, 28, 36, ? R(4,4), R(4,5) : 18, 25, ?
EASY FACTS Fact 1 : If C1 , … , Ck the components of G, gap(G) = gap(C1) + … + gap (Ck) G is gap-critical all of its components are. Fact 2 : If G is a graph and Q a clique : (G) (G-Q) (G)-1, (G) (G-Q) (G)-1 gap(G) + 1 gap(G-Q) gap(G) - 1 Fact 3 : If a t-extremal graph has a k-clique, s(t) s(t-1) + k
How big should be for a big gap ? gap= - n / large ~ small ~ large (Ramsey) How does gap = - change if varies ? Conjecture:t-extremal graphs are -free (=2). gap2 (n) : max gap of a triangle-free on n vertices s2 (t) : min order of a triangle-free of gap t. Theorem : s(t) = 2t + (sqrt (t log t ) ) = s2 (t) gap(n)=gap2 (n), s(t)=s2 (t) for all n, t
Inverse Ramsey (n) := min {(G): G triangle-free of order n} If R(3, ) n < R(3, + 1 ) ,then (n) = For instance : (5) = 2 , (6)= 3 Fact: If n = n1 + n2 , then (n) (n1) + (n2) If ‘’=‘’ n is Ramsey-perfect; (10)=(5) + (5) Conjecture: This is the only nontriv example
A version of the main result Theorem: n / 2 - (n)gap(n)n / 2 - (n)+3 and with the exception of Ramsey numbers + 1, … , 14,there is equality with the lower bound The extremal graphs are triangle-free.
The proof relies on two lucky facts : (1) gap2 is relatively easy to determine (2) Whenever gap > gap2 the growth of gap slows down while gap2 grows constantly except at Ramsey If s(t+1) s2 (t+1), then s(t+1) s(t)+3 s2 (t+1)=s2 (t) + 2, unless s2 (t)+1 or +2 is Ramsey
The gap of triangle-free graphs n / 2 The more you want, the less you get : For gap-critical graphs,(if they are connected): = n / 2 , so gap =n / 2 - (n) If not connected, apply to components : the components are connected, gap-critical
If G is gap-critical, (G) > (G-v) for all v. If G is triangle-free, connected and (G) > (G-v) for all v, then (G) = n / 2 Gallai : For triangle-free : + =n Gallai : If (G-v) = (G) for every vertex v, then G-v has a perfect matching –’’--. In particular, n is odd
Triangle-free, connected cont’d : THEN GAP-CRITICAL FACTOR-CRITICAL Proof of Gallai’s Lemma : … Equivalent to Tutte’s theorem ( Tutte-set)
The gap of triangle free graphs Theorem : gap2 (n) =n / 2 - (n) , or +1 the latter n is even Ramsey-perfect. Proof: At most 2 comp: from inequalities about Ramsey: if n =n1 + n2 + n3 , then (n) < (n1) + (n2) + (n3) 2 comp : if of only if n is even, Ramsey- perfect, and the components have odd size n1, n2 , (n) = (n1) + (n2) For the connected components by Gallai: = n / 2
The proof relies on two lucky facts : (1) gap2 is relatively easy to determine (2) Whenever gap > gap2 the growth of gap slows down while gap2 grows constantly except at Ramsey If s(t+1) < s2 (t+1), then s(t+1) s(t)+3 n / 2 - (n)
Theorem: 0 gap(n) - gap2 (n) 2and they are equal everywhere but on small constant size intervals after Ramsey numbers. Corollary: All subgraphs of (3, ) Ramsey-graphs of order at least R(3, -1) are perfectly matchable. T 1 2 3 4 5 9 10 s(t) 5 10 13 17 20 or 21, 32 or 33 35 R 6, 9, 14, 18, 23,28, 36
0 s2(t) - s (t)10n / 2-(n)gap(n)n / 2 - (n)+3 NOT YET THE END The conjectures ? Dramatic corollaries for Ramsey ? s(5) = 20 or 21 (=s2(5) ) ? = almost always