Densest/Heaviest k-subgraph on Interval Graphs, Chordal Graphs and Planar Graphs

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1 / 25 Densest/Heaviest k-subgraph on Interval Graphs, Chordal Graphs and Planar Graphs Presented by Jian Li, Fudan University Mar 2007, HKUST May 8, 2006

2 / 25 Problem Definition: Densest k-subgraph Problem(DS-k): Input:G(V, E), k > 0. Output: an induced subgraph D s.t. V (D) = k. Goal:Maximize E(D). Heaviest k-subgraph Problem(HS-k): Input:G(V, E),w : E R +,k > 0. Output: a induced subgraph D s.t. V (D) = k. Goal:Maximize e E(D) w(e).

3 / 25 Previous Results NP-hard even on chordal graphs(corneil,perl.1984) and planar graphs(keil,brecht.1991). n δ -approximation for some δ < 1/3 (Feige,Kortsarz,Peleg.2001). n/k-approximation(srivastav,wolf.1998;goemans,1999). No PTAS in general(khot.2004). PTAS for dense graph(arora, Karger, Karpinski.1995).

4 / 25 Previous Results Some better approximation for special k. HS-k is in P on trees(maffioli.1991), co-graphs(corneil,perl.1984) and chordal graph if its clique graph is a path(liazi,milis,zissimopoulos.2004). PTAS on chordal graph if its clique graph is a star(liazi,milis,pascual,zissimopoulos.2006). OPEN: complexity on interval graphs(even for proper interval graphs).

5 / 25 Our Results Proper interval graphs(unknown): PTAS. Chordal graphs(np-hard): Constant approximation. Planar graphs(np-hard): PTAS.

6 / 25 Proper Interval Graphs-A simple 3-approximation Densest disjoint clique k-subgraph(ddcs-k) problem: Find a (not necessarily induced) subgraph G (V, E ) such that V = k; G is composed with several vertex disjoint cliques; E is maximized.

7 / 25 Proper Interval Graphs-A simple 3-approximation DDCS-k can be solved by Dynamic Programming: Let DS(i, l) be the optimal solution of DDCS-l problem on G(V 1...i ). ( ) l x DS(i, l) = max {DS(j, x) + } (j,x) A 2 where A is the feasible integer solution set of the following constraints system: 1 j < i, 0 x l, l x i j, l x i q G (i) + 1.

8 / 25 Proper Interval Graphs-A simple 3-approximation An optimal DDCS-k solution is a 3-approximation of DS-k problem. We construct a DDCS-k solution OP T DDCS from an optimal solution OP T DS of the DS-k problem such that OP T DDCS 1/3 OP T D S.

9 / 25 Proper Interval Graphs-A simple 3-approximation Construction(Greedy): Repeatedly remove the vertices and all adjacent edges of a maximum clique from OP T DS. Take OP T DDCS as the union of these maximum cliques.

10 / 25 Proper Interval Graphs-A simple 3-approximation B A C

11 / 25 Proper Interval Graphs-PTAS Def: overlap number κ G (v) as the number of maximal cliques in G containing v. The h-overlap clique subgraph H is a subgraph of G such that κ H (v) h for all v V (H). For example, a disjoint clique subgraph is a 1-overlap clique subgraph.

12 / 25 Proper Interval Graphs-PTAS densest h-overlap clique k-subgraph(docs-(h, k)) problem: Find a (not necessarily induced) subgraph G (V, E ) such that V = k; G is a h-overlap clique subgraph of G; E is maximized. DOCS-(h, k) can be also solved by dynamic problem if h is a constant.

13 / 25 Proper Interval Graphs-PTAS Similarly, we construct a DOCS-(h, k) solution OP T DOCS from an optimal solution OP T DS of the DS-k problem such that OP T DOCS (1 4 h/2 1 ) OP T DS. So, in order to get a 1 ɛ approximation, it is enough to set h = 2 + 8/ɛ.

14 / 25 Proper Interval Graphs-PTAS C 1 C 2 j C 3 C 3 C 4 C 5 C 6 C 3,1 C 3,2 C 3,3 C 4 p GO (j) (a) (b)

15 / 25 Chordal Graph A graph is chordal if it does not contain an induced cycle of length k for k 4. A perfect elimination order of a graph is an ordering of the vertices such that P red(v) forms a clique for every vertex v, where P red(v) is the set of vertices adjacent to v and preceding v in the order. Thm: A graph is chordal if and only if it has a perfect elimination order.

16 / 25 Chordal Graph Maximum Density Subgraph Problem(MDSP) Input G(V, E), vertex weight w : v R +, Output: an induced subgraph G (V, E ). Goal: maximize the density ( v V w(v)+ E ) V. This problem can be solved optimally in polynomial time by reducing to a parametric flow problem [Gallo,Grigoriadis,Tarjan.1989]. Important Fact: w(v) + d G (v) ρ.

17 / 25 Chordal Graph The high level idea : We run the above MSDP algorithm on our given graph with w(v) = 0 for all v V we get a subgraph G of size k, we have exactly an optimal solution for the DS-k problem. If we get a smaller subgraph, we repeat the MSDP algorithm in the remaining graph and add the solution in. If we we get a larger subgraph, we need to pick some vertices in this subgraph to satisfy the cardinality constraint without losing much density.

18 / 25 Chordal Graph Densest-k-Subgraph-Chordal(G(V,E)) 1: V 0 = ; i = 0; 2: i = i + 1;run MSDP in the remaining graph G(V V i 1, E(V V i 1 ), w i (v) = d(v, V i 1 ). let the optimal subgraph(subset of vertices) be V i and the density be ρ i. 3: if V i 1 + V i < k/2 then 4: V i = V i 1 V i and go back to step 2. 5: else if k/2 V i 1 + V i k then 6: V i = V i 1 V. 7: else if V i 1 + V i > k then 8: V = P ick(v i, w i) and V i = V i 1 V ; 9: end if 10: Arbitrary take k V i remaining vertices into V i.

19 / 25 Chordal Graph Pick(V t,w) 1: Compute a perfect elimination order for V, say {v 1, v 2,..., v m }, m > k/2. V =. 2: for i=m to 1 do 3: if P red V t (v i ) ρ t /2 then 4: if V + P red V t (v i ) + 1 k/4 then 5: V = V {v i } P red V t (v i ). 6: else if k/4 < V + P red V t (v i ) + 1 k/2 then 7: V = V {v i } P red V t (v i ); return V. 8: else if V + P red V t (v i ) + 1 > k/2 then 9: Add into V v i and arbitrary its k/2 V 1 predecessors; return V. 10: end if 11: else if w(v i ) + Succ V t (v i ) > ρ t /2 then 12: V = V {v i }; If V > k/4 return V ; 13: end if 14: end for

20 / 25 Chordal Graph Suppose OP T = G (V, E ). If E(SOL V ) E /2, then the algorithm is a 1/2-approximation. If not...

21 / 25 Chordal Graph Analysis Sketch: Let I i = V i V and R i = V \ I i. Since we get the the optimal solution V i on MDSP instance G(V V i 1, E(V V i 1 ), w i (v) = d(v, V i 1 ) at step i, we have ρ i = E(V i ) +d(v i 1,V i ) V i E(R i 1) +d(v i 1,R i 1 ) R i 1 E(R i 1) +d(i i 1,R i 1 ) R i 1 E(R i 1) +d(i i 1,R i 1 ) k = E E(I i 1 ) k for all i t. E E(I t) k E 2k = ρ 2.

22 / 25 Chordal Graph we can prove ρ i ρ i+1 for all i. We can also prove if ρ i > k/4 then i = 1. If ρ t > k/4, and recall d V1 (v 1 ) ρ 1 > k/4, So, a clique of size at least k/4, a 16-approximation. If not,...

23 / 25 Chordal Graph In Pick: we can see w(v) + d V (v) = w(v) + P red V (v) + Succ V (v) ρ t /2. So ρ t = E(V )+d(v,v i 1 ) V = 1/2 v V d V (v)+ v V d(v,v i 1) V v V d V (v)+ v V w i(v) 2 V ρt 4.

24 / 25 Planar Graph Sketch: Decompose the planar graph into a series of K-outerplanar graphs. Solve the problem in each outerplanar graph. Recombine the solution. Standard Baker s technique, but some more details...omit here.

25 / 25 Thank You! thanks to Jian XIA and Yan ZHANG for discussions on proper interval graphs.