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Chromatic Number of Ordered Graphs with Forbidden Ordered Subgraphs

机译:具有禁止有序子图的有序图的色数

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It is well-known that the graphs not containing a given graph H as a subgraph have bounded chromatic number if and only if H is acyclic. Here we consider ordered graphs, i.e., graphs with a linear ordering ? on their vertex set, and the function where Forb(?)(H) denotes the set of all ordered graphs that do not contain a copy of H.If H contains a cycle, then as in the case of unordered graphs, f(?)(H)=. However, in contrast to the unordered graphs, we describe an infinite family of ordered forests H with f(?)(H) =. An ordered graph is crossing if there are two edges uv and uv with u ? u ? v ? v. For connected crossing ordered graphs H we reduce the problem of determining whether f(?)(H) to a family of so-called monotonically alternating trees. For non-crossing H we prove that f(?)(H) if and only if H is acyclic and does not contain a copy of any of the five special ordered forests on four or five vertices, which we call bonnets. For such forests H, we show that f(?)(H)2(|V(H)|) and that f(?)(H)2|V (H)|-3 if H is connected.
机译:众所周知,不包含给定图H作为子图的曲线图具有界定的色度,如果且仅当H是无循环时。在这里,我们考虑有序图,即用线性排序的图表?在它们的顶点集中,福音(?)(h)表示不包含H.if H副本的所有订购图表的集合包含一个周期,那么就像无序图一样,f(? )(h)=。然而,与无序的图形相比,我们描述了一个具有f(h)=的无限森林H.如果有两个边缘UV和UV与U具有uv?你? v? v。对于连接的交叉订购图H,我们减少了确定f(a)(h)到一个所谓的单调交替树的家庭的问题。对于非横穿H,我们证明了f(?)(h)如果h是非循环的,并且不包含四个或五个顶点的五个特殊订购森林中的任何一个副本,我们呼叫帽子。对于这种森林H,我们表明F(α)(H)2(| V(H)|)和该F(α)(H)2 | V(H)| -3如果连接。

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