Vertex Cover, Dominating set, Clique, Independent set Set problems Vertex Cover, Dominating set, Clique, Independent set

Basic problems Several basic graph problems: Relations Finding subsets of vertices or edges with simple property – as small or as large as possible Coloring (next time) Relations Approximation Algorithms Set problems

Vertex cover Set of vertices W Í V with for all {x,y} Î E: x Î W or y Î W. Vertex Cover problem: Given G, find vertex cover of minimum size Set problems

Dominating Set Set of vertices W such that for all v Î V: v Î W or v has a neighbor w with w ÎW. Dominating Set problem: find dominating set of minimum size E.g.: facility location problems Set problems

Clique and Independent Set Clique: Set of vertices W with for all v, w Î W: {v,w} Î E. Independent set: Set of vertices W with for all v, w Î W: {v,w} Ï E. Clique problem: find largest clique. Independent set problem: find largest independent set Set problems

Relations The following are equivalent G has an independent set with at least k vertices The complement of G has a clique with at least k vertices G has a vertex cover with at most n-k vertices Set problems

Approximation Clique, IS are very hard to approximate Ratio 2 approximation algorithm for Vertex Cover H = G, S = Æ repeat until H is empty Select an edge {v,w} Î E. Put v and w in S, and remove v, w, and all adjacent edges from H Does not translate to approx. algorithm for Clique or Independent Set E.g., take a graph with 1000 vertices and a vertex cover of size 498. Set problems

Vertex cover in bipartite graphs In a bipartite graph: the minimum size of a vertex cover equals the maximum size of a matching If we have a matching: each edge must have an endpoint in the vertex cover In the other direction: Let X be a vertex cover. Take flow model. Look at the edges from s to vertices in X, and from vertices in X to t. These form a cut. Now use the min-cut max-flow theorem. Set problems

Bipartite graphs We saw: min vertex cover in bipartite graph equals maximum matching Corollary: Independent Set in bipartite graphs is polynomial time computable Set problems

Special cases for Independent Set For many types of graphs (e.g., planar) IS stays NP-complete when restricted to these graphs. Some special cases are polynomial Trees, bounded treewidth Bipartite graphs Interval graphs, circular arc graphs Set problems

Interval graphs Each vertex represents an interval Edge between two vertices when their intervals have non-empty intersection Independent set on interval graphs models room-booking problem Set problems

Greedy algorithms on interval graphs Greedy algorithm for IS on interval graphs is optimal: Repeat: Select vertex v with leftmost right-endpoint of interval Remove v and its neighbors Clique is trivial on interval graphs Set problems

Circular arc graphs Like interval graphs, but now `intervals on a circle’. If we remove a vertex and its neighbors from an circular arc graph, then we get an interval graph Algorithm, e.g.: for all vertices v do Compute the maximum independent set of G minus v and its neighbors If the largest i.s. is found for v*, output that set with v* added. Set problems

Perfect graphs A graph G is perfect if for each induced subgraph H of G: the size of the maximum clique in H equals the chromatic number of H (nb. of colors needed to `vertex color H’). Strong perfect graph theorem (conjectured Berge 1966, proven Chudnowsky et al., 2002). G is perfect, if and only if G does not contain an induced subgraph that is a cycle without additional edges of odd length at least four, and the complement of G also does not contain such a cycle. (`No odd hole or antihole’) Set problems

Coloring and independent set in perfect graphs Polynomial time algorithm for finding the maximum independent set and the coloring number of a perfect graph Grötschel, Lovasz, Schrijver (1981) Uses Linear Programming methods Set problems

Unit disk graphs Vertices are disks on plane with radius 1 Edge when disks intersect Model radio transmitters and inferences or connections Set problems

More on unit disk graphs Finding the intersection model is hard (NP-hard; not yet known to be in NP). Approximation algorithms for Independent Set, Dominating Set, … Greedy algorithm is a 5-approximation Simple modification gives 3 approximation (next) Set problems

3-approximation for Independent set on Unit Disk graphs W = V S = Æ repeat until W = Æ Choose a vertex v from W whose center is as much as possible to the left S = S È {v} W = W – N(v) – {v} Return S Leftmost Greedy Algorithm Set problems

Ratio 3 The maximum independent set is never larger than 3 times the set, given by the leftmost greedy algorithm. If we choose v, then N(v) has an independent set of size at most 3 Between successive nonintersecting neighbors, there is an angle that is larger than p/3 Set problems

Polynomial Time Approximation A PTAS For every e>0, there is a polynomial time algorithm for Independent set that gives an independent set of size at least (1-e)*OPT. Uses geometric separation technique PTAS = Polynomial Time Approximation Scheme Set problems

Bounded regions Suppose all disks are in a region of size r by r. Then the maximum size of an independent set in this graph is at most 4r2/p2. Each disk takes space p2/4. The disks in the solution do not overlap. Then, the maximum independent set problem can be solved in O(nO(r*r)) time. Check each subset of size at most 4r2/p2. Take the largest independent one of these. Set problems

Removing strips and getting small area subgraphs Remove, at steps of r+1, for a 1-wide horizontal strip all disks intersecting the strip. Remove, at steps of r+1, for a 1-wide vertical strip all disks intersecting the strip. On the remaining graph, each connected component fits in an r by r area Set problems

Each connected component fits in one white area All disks intersecting a green strip are removed. We end up with the graph, formed by the disks fitting in the white area For fixed r, we can solve the IS problem on the remaining graph in polynomial time Each connected component fits in one white area Set problems

Set problems

Graphs Gi,j Remove horizontal strips at spaces i, i+(r+1), i + 2*(r+1), … (I.e., start at i.) Remove vertical strips at spaces j, j+(r+1), j + 2*(r+1), … (I.e., start at j.) Let Gi,j = (Vi,j,Ei,j) be the resulting graph Algorithm: try all (r+1)2 choices for (i,j), for each solve IS optimally for Gi,j and output the largest IS from these. Set problems

Quality of heuristic (1) Let X be an independent set of maximum size. Let v Î X. There are at most 4(r+1) graphs Gi,j that do not contain v. v can be removed by a horizontal strip (upside, downside intersects) or a vertical strip (left side, right side intersects) . Set problems

Quality of heuristic (2) Total size over all m.i.s.’s over all Gi,j is at least the sum over all |Vi,j Ç X| ³ (r+1)2 *|X| - 4(r+1)*|X| There must be at least one Gi,j that contains an independent set of size at least ((r+1)2 – 4)/(r+1)2 * |X| Maximum is at least average Set problems

Setting r Let r be the smallest integer such that For fixed e, this is a constant – gives a polynomial time algorithm: PTAS Set problems

Variants Other type of intersecting objects Other type of surfaces Other problems Example: Dominating set. Instead of removing strips, we duplicate vertices in strips giving them to the left, and right area Leads also to PTAS Set problems

Approximating Dominating Set on general graphs Can be done with ratio O( log n) Here: algorithm for generalization: set covering problem Given: set X, collection C of subsets of X Find: smallest number of sets from C such that its union is X. From DS to Set Covering: For each vertex v, take a set Sv in the collection, with Sv = {v} È N(v). Set problems

Greedy approximation of Set Covering problem U = X A = Æ while U ¹ Æ do Select an S from C with maximum | S Ç U| A = A È {S} U = U – S Output A Set problems

Ratio H(n) = Si=1…n 1/i Theorem. Greedy algorithm gives a number of sets that is at most H( max{|S| | S Î C}) times the optimal number of sets. Corollary: Greedy algorithm has ratio O(log n). Set problems

Proof (1) Assign a cost 1 to each selected set. Spread this cost evenly over the newly covered elements, cx. Say A* is optimal solution, A solution by algorithm. |A| = Sx Î X cx £ SS Î A* Sx Î S cx We’ll proof: Sx Î S cx £ H(|S|) That proves the theorem. Set problems

Sx Î S cx £ H(|S|) Consider some S in C. Write ui = number of elements of S that still are in U after i sets have been taken = |S – (S1 È … È Si)|. u0 = |S|. Suppose uk = 0; uk-1 >0. ui-1 – ui > 0 elements from S were covered for the first time by Si. Sx Î S cx = Si=1…k (ui-1 – ui)* 1/(|Si –((S1 È … È Si-1)| £ Si=1…k (ui-1 – ui)/ ui-1 Set problems

Proof continues Sx Î S cx £ Si=1…k (ui-1 – ui)/ ui-1 £ Si=1…k (H(ui-1) – H(ui)) = H(u0) – H(uk) = H(|S|) – 0. Use H(b) – H(a) ³ (b – a)/b Now we showed: Sx Î S cx £ H(|S|) Set problems

Comments (I)LP methods work also: log n ratio O(log n) is the best one can hope for unless P=NP Set problems

Facility location In a town, we want to place a number of facilities (e.g., fire stations, hospitals, …), such that the maximum distance to a facility is small Models as a graph problem Set problems

Facility location problem Given: complete graph G, distances d between vertices fulfilling triangle inequality, integer k Question: find a set S of k vertices, such that the maximum over all vertices v of minw ÎS d(v,w) is as small as possible Set problems

Restating the problem Find the smallest r and a set S such that S is a dominating set of size at most k for the graph Gr formed by the edges of length at most r. OPT is smallest r. G2 is square of a graph: edge in G2 when there is a path of length 2 Set problems

A Lemma Let I be an independent set in H2. Then |I| is at most the minimum size of a dominating set in H. Proof: Let D be a minimum dominating set of H. For each v Î D, N(v)È {v} contains at most one vertex from I. Set problems

Approximation algorithm for facility location Sort the values of the lengths of the edges in G, say d1 £ … £ dm for i = 1 to m do Compute Gx with x = di Compute Gx2 Find a maximal independent set I in Gx2 If it has size at most k, then return I; stop. Set problems

Analysis The optimal solution has cost at least x=di For y<x: the minimum size of a dominating set in Gy is k+1 or more (use lemma) Ratio of algorithm is 2 Each maximal independent set is a domination set, so I is a dominating set in Gx2. Each vertex has a path of length at most 2 to a vertex in I in Gx: so, I is a k-center with `cost’ at most 2x. Set problems

Graphs that are not complete Given: graph G, distance for each edge d(e), integer k. Question: find a set S of k vertices, such that the maximum over all vertices of the length of the shortest path to a vertex in S is as small as possible. Can be translated to previous problem: make complete graph with d(v,w) the length of the shortest path from v to w Set problems

Conclusions Several techniques for approximation problems like vertex cover, dominating set, independent set Work often for other problems too (with modifications) Set problems