1 Weak coloring number and augmentations
Let us start by showing the properties of the graphs declared in Lemma 5. Let us fix the notation as in the construction of the graph. For a vertex , let denote the set of vertices reachable from by paths in not containing vertices of , and for the unique , let . We have and .
Proof of Lemma 5.
Consider the ordering of the vertices of where is the smallest vertex, followed by vertices of in any order, vertices of in any order, and finally all other vertices of in any order. Clearly for every . If , then when . If , then when and when . For any other vertex , we have when and when . We conclude that and .
Note that any two vertices of are at distance at most from one another, as shown by a path through . If for some , then the distance between and in is , as shown by a path through . Hence, if each two vertices of are contained in a common edge, then any independent set in has at most two vertices (one in and one in ), and thus .
No vertex of is at distance at most from in . If , then no vertex of is at distance at most from in . If , or and is the unique element of , then no vertices of are at distance at most from . Hence, for each , there exists such that all vertices of at distance at most from in belong to . If is a dominating set in , it follows that . If no edges of cover all vertices of , then we conclude that .
Finally, setting , for all , and for all gives a feasible solution to the program defining , which implies that . ∎
In the rest of the paper, it is convenient to perform the arguments in terms of certain directed graphs rather than weak coloring numbers; this also makes the connection to the result of Bansal and Umboh [3] more transparent.
For a positive integer , an augmentation of a graph is an orientation of a supergraph of with such that each edge is assigned length with the following properties:

Each vertex of is incident with a loop of length .

For any nonnegative integer and vertices , the distance between and in is at most if and only if there exists a common inneighbor of and in and .
Because of (LOOP), if , then (DIST) applied with shows that the distance between and in is at most . Essentially, we augment by adding (directed) edges representing certain paths in , with being the length of the corresponding path between and . It follows that for all with , and that if and are adjacent vertices of , then at least one of the directed edges or appears in with length .
For a nonnegative integer and , let be the number of inneighbors of with , and let be the maximum of over all vertices of . Let us note a connection between weak coloring numbers and augmentations.
Observation 8.
Consider any ordering of vertices of a graph and a nonnegative integer . Let be the directed graph in which iff , and let be the minimum such that . Then is an augmentation of and is equal to the weak coloring number of the ordering for any nonnegative .
Proof.
Note that is equal to the weak coloring number of the ordering for every by the choice of , and thus it suffices to argue that is an augmentation.
Since for all , (LOOP) is satisfied by . If vertices have a common inneighbor in , then their distance in is at most the sum of distances from to and , which is at most . Conversely, suppose that the distance between and is . Let be a path of length from to , and let be the smallest vertex of in the considered ordering. Then , where and are the lengths of the subpaths of from to and , and . We conclude that (DIST) holds. ∎
Note that it is possible to obtain augmentations in other ways, e.g., using the transitive fraternal augmentation procedure of Nešetřil and Ossona de Mendez [13].
2 Domination
We are now ready to give the approximation argument for .
Theorem 9.
Let be a positive integer. If is an augmentation of a graph , then
Proof.
Let . Consider any optimal solution to the linear program defining , and let be the set of vertices such that in this solution. Let , …, be any ordering of vertices of . For , if a vertex at distance at most from belongs to , then let ; otherwise, is obtained from by adding all inneighbors of such that .
Clearly, is an dominating set of ; hence, it suffices to bound its size. We have
To bound , we perform a charge redistribution argument. Vertices start with zero charge. For , if , then we increase by the charge of each vertex such that . Let denote the total amount of charge added in this step. Observe that since , none of vertices in belongs to . In particular, no inneighbor of belongs to , and thus . Since we are considering a solution to the linear program defining , we have the following bound on the charge increase.
Consequently,
and letting be the total amount of charge created, we have
On the other hand, by (DIST), when and , then and have a common inneighbor and . Consequently, whenever the charge of is increased, some inneighbor of (distinct from ) is added to the dominating set, and thus the final charge of is at most . Furthermore, as we observed before, charge is only added to vertices not belonging to . Summing over all vertices of , we obtain
Combining these bounds, we obtain
and thus
as required. ∎
Note that , and thus if , then . Hence, . By Observation 8, Theorem 9 has the following consequence.
Corollary 10.
For any positive integer and a graph ,
On the other hand, the ratio cannot be bounded in terms of , as shown by the following example. Let
be an odd integer and let
be the hypergraph with vertex sets consisting of all subsets of of size , with edges , …, such that for , the edge consists of the sets in that contain . For any of size , the vertex is not incident with any of the edges for ; hence, Lemma 5 implies . Each vertex of is incident with edges and , and thus . Also, Lemma 5 implies .Corollary 10 implies that can be approximated in polynomial time within factor of . This bound can be improved in the special case . If has an orientation with maximum indegree at most , then giving each edge length and adding loops of length on all vertices results in a augmentation with and . Hence, we have the following.
Corollary 11.
If a graph has an orientation with maximum indegree at most , then can be approximated in polynomial time within factor of .
3 Independence
To prove a bound on the ratio , we use a result of Parekh and Pritchard [15] on generalized hypergraph matching. Let be a hypergraph and let be a positive integer. A matching in is a set of edges of such that each vertex of is incident with at most edges of . Let denote the maximum size of a matching in . Let be the fractional relaxation of this parameter, defined as
subject to  
for all  
for all 
Clearly, . Conversely, we have the following.
Theorem 12 (Parekh and Pritchard [15]).
If is a hypergraph with all edges of size at most and is a positive integer, then
Furthermore, a matching of size at least can be found in polynomial time.
Corollary 13.
If is a hypergraph with all edges of size at most , then then .
Proof.
Consider an optimal solution to the linear program defining . Let and . Clearly, is a matching in , and thus . If , the desired bound on follows, and thus assume that .
Let for all such that and for all other . This gives a feasible solution to the program defining , and thus
By Theorem 12, we have
as required. ∎
We use this result to find sets intersecting outneighborhoods in augmentations only in a bounded number of vertices.
Lemma 14.
Let be a graph, let be an augmentation of , and let . There exists a set such that each vertex of has at most outneighbors in and
Proof.
For a vertex , let be the set of inneighbors of in . Let be the hypergraph with vertex set and edge set ; each edge of has size at most . Note that is a matching in if and only if each vertex of has at most outneighbors in . Hence, it suffices to prove that .
Consider an optimal solution to the linear program defining , and for every , let . For each , we have
and thus this gives a feasible solution to the program defining . We conclude that , and thus the claim follows from Corollary 13. ∎
We are now ready to show existence of large independent sets.
Theorem 15.
Let be a graph, let be an augmentation of , and let . We have
Proof.
Let , and let be the set obtained by applying Lemma 14, such that every vertex of has at most outneighbors belonging to . For any and , (DIST) implies that either is an inneighbor of in and , or and have a common inneighbor with . Hence, we have
and thus is a independent set in . Consequently,
as required. ∎
The following lemma clarifies the relationship with Theorem 4.
Lemma 16.
If is a augmentation of , then for every positive integer ,
Proof.
Let be a independent set in of size . Let be the graph with vertex set and distinct vertices adjacent iff their distance in is at most . Orient the edges of as follows: if is an inneighbor of in and , then direct the edge from to . Since is independent, we have for each inneighbor of , and the inequality is strict when . Hence, the maximum indegree of is less than . Furthermore, all edges of are directed in at least one direction by (DIST). Consequently, each subgraph of has less than edges, and thus is degenerate. Consequently, , and thus contains an independent set of size at least . Observe that is a independent in , which gives the required lower bound on . ∎
Composing Theorems 9 and 15 with Lemma 16, and using Observation 8, we obtain the following inequalities, implying Theorem 7.
Corollary 17.
Let be a graph, let be an augmentation of and let be a augmentation of . Let . Then
In particular,
and if has an orientation with maximum indegree at most , then
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