On The Achromatic Number Problem

1. On The Achromatic Number Problem Guy Kortsarz Rutgers University, Camden

2. On Independent Sets And Cliques • Independent set: a set W so that no two vertices in W share an edge. In aCLIQUEevery two vertices are adjacent

3. Problem Definition • INPUT: An undirected graph G(V,W) • Goal: A partition of G(V,W) into as many subsetsV=U Di as possible such that the setsDi are independent sets, pairwise disjoint, and there is an edge between every pair of sets Di Dj ij.

4. Example • In the following graph, the optimum is4.

5. A Partition To Independent Sets Is Called A Coloring • The Chromatic number problem is to color the graph with as few colors as possible, so that each color set is an independent set. Namely, to decompose G to as few independent sets as possible. • It is denoted by (G). • The maximum achromatic number is a coloring as well and is denoted by (G). • Clearly (G) (G).

6. An Example Of A Difference Between The Two Functions • Say that (G) is a universal constantc. • Then we are able to solve the problem in polynomial time by exhaustive search. • Search in time nc2for the edges that cross between sets. This gives a partial achromatic coloring. Later we shall see how to complete the partial achromatic coloring into a full one. • For Chromatic (namely minimum size) coloring the problem is NP-Hard even if we are promised that (G)=3.

7. How Do We Find A Feasible Solution?

8. The Achromatic Number Is ‘Sensitive’ To Few Edges • A Complete Bipartite Graph:

9. About the Achromatic Number Problem • The problem is very extensively studied in the Branch of math called Graph Theory. • Some times in Graph Theory the authors are more concerned with proving a statement then finding a structure in polynomial time. • As the problem is central in Graph Theory, there are two surveys on the problem, one by MacGillivary and one by Eduards.

10. The complexity Of The Problem • The decision version of the problem is NPC even if the underlying graph is a tree! (Cairnie and Eduards). • For trees we can find -1 size partition! • We do not know (yet?) how to find the optimum for NP-Hard questions in polynomial time.

11. About P=NP? • It seems that we gave up on trying to show that an NPC problems, say, the Traveling Sales Person, can be solved in polynomial time. • I would imagine most people in theory believe that PNP. • Still if we classified a problem to a certain complexity class, it does not go away.

12. Approximation Algorithms • But some times (only on hard inputs!) the running time indeed does not allow us to solve NP-Hard problems even in practice.Computerscan not runan algorithm that performs 2n operationsfor large n. • For n=300 this is more than the number of atoms in the known universe. • Some rule of thumb usually works great, in the sense that is %5 from optimum (how do we know that?). • We want a provable multiplicable distance from the optimum. This is called Approximation ratio.

13. Approximation Ratio For Maximization Problems • Consider some abstract problem P with infinitely many inputs • An approximation algorithm has an approximation ratio  if it runs in polynomialtime and if for every instanceI: Val(OPT(I))/Val(A(I))  • It is conjecture byChadhary and Vishwanathan that the Achromatic number problem admits a sqrt{} approximation.

14. Some Known Approximation Algorithms • For general graphs, very bad situation. • An algorithm with ratio n/sqrt{log n} was given by Chaudhary and Vishanathan. • Krauthgamer and I, gave a roughly n/log n ratio approximation algorithm, improving the above result.

15. Some Known Approximation Algorithms, Continued • In a paper with Krauthgamer, min{sqrt{} ,n1/3}approximation algorithm for graph with girth at least5. • This improves a min{sqrt{} ,n3/8} ratio approximation algorithms by Krysta and Lorys. And they even looked at an easier problem: Girth at least 6!

16. Bipartite graphs • For bipartite graphs (as for graphs of girth at most 5) we were able to get a truly sub-linear ratio approximation. • We give an O(n4/5 ) ratioapproximating algorithm. • This is joint work with Shende. • The algorithm is truly complex and uses a previous result that we are going to see in this talk. The result is from a previous paper.

17. A Lower Bound • What is a lower bound on the approximability? • If can be approximated by  then can be solved in polynomial time. Thus P=NP. • We show (a paper with Radhakrishnan and Sivasubramanian) that if you can approximate the achromatic number problem within sqrt{log n},you can solve the problem. • Called sqrt{log n} lower bound.

18. Complte partitions of graphs. • It is defined just as Achromatic number but the sets do not have to be independent set • The problem is fully understood up to a constant with respect to approximation • O({sqrt{log n}) upper bound • ( {sqrt{log n} ) lower bound • Such a ratio is called TIGHT.

19. Why Not Approximate Via Small Maximal Independent Sets? • The algorithm that we saw finds a maximal independent set, gives it color 1, and then recurses. • Obviously, there is a maximal independent set of size n/ or less. • Big difficulty: If you can approximate the minimum maximal independent set by a ratio of n1-,then P=NP. Thus a lower bound of n1-by Magnus. M. Halldorsson.

20. Graph Theory • We show that if the girth of the graph is 5 or more then there is always an achromatic number of size m/nwith n=|V| and m=|E|. • As it happens a lot in Discrete Math, the proof is in fact an algorithm. • Note, this result is tight. The example is a complete bipartite graph without a perfect matching.

21. Proof That m/nIn A GraphOf Girth At Least5 • Remove all vertices with degree strictlylessthen m/n. • The graph does not turn empty. • The minimum degree is at least m/n. • Consider the 2 BFS layers of an arbitrary vertex v.

22. How Do We Color The Graph:Example • For a leaf  and a star S, is adjacent to at most one leaf in S.

23. From Partial Coloring To A Complete Coloring • Say that we have a partial achoromatic coloring with pcolors of a partial set. • This means that only part of the vertices are inside independent sets. • Also the independent sets form a partial achromatic coloring. They admit the property that every two independent sets share an edge.

24. How to complete the coloring? • Consider v that does not belong to any partial independent set. • If there is some (partial) independent set with no neighbors of v, put v in this set. Clearly the partial coloring is still legal. • If a neighbor of v appears in every independent set, color v with a new color. • We again have a legal partial coloring which is larger by one (contains one more vertex). • Recurse.

25. An Equivalence Relation And Bipartite Graphs • We say that v  u if N(v)=N(u). • Note that v and u can not be neighbors. • We are again (not be talking about approximation but rather) show a result in graph theory on bipartite graphs. • Say that q=sqrt{}.In this case the conjecture holds; We find a partition of size q=sqrt{}. But in general its n/q for q sqrt{}.

26. No Huge Equivalence Classes • We may assume that the size of every equivalence class is at most sqrt{}. • If a class has sqrt{}+1 copies of a vertex x and we use them all, each achromatic set may contain a copy of x. • But there will be a class with two copies of x. This is not needed.

27. Small Size Classes Don’t Count • Recompute n. • Classes of size at most n/(2sqrt{}) cancontain at mostn/2vertices. • Therefore equivalence classes of size at leastn/(2sqrt{}) verticescontain at least n/2 vertices. Recall their size is also at most sqrt{}. • Thus there are at least n/(2sqrt{}) equivalence sets with size at least n/(2sqrt{}) sqrt{}/2.

28. The Algorithm • We show A,B and xV2 such that x is adjacent to (say) B but is not to A. A B x C

29. The Set A,B And The Vertex x • The sets A and B from previous slide are not equivalent. • Therefore there must be some x that that is independent of A but not of B. • We are going to partition V1into sets that are adjacent to xand to set that are not adjacent to x. APROPER partition. • As we shall see vertices are added toAand B but only from V2 (so independent, and independent with respect to A and B).

30. Finding Some Adjacent Sets • The vertex x has about sqrt{}equivalent copies. • Put (a copy of) x in any equivalence set that does not contain x. • This partitions V1to equivalence classes S that contain x and T that do not contain x .

31. How To Continue • Note: every equivalence set in S is adjacent to every equivalence set in T. • This is because of the edges from xtoT • The sets in T itself do not share an edge nor do the sets that belong to S . • Thus recurse on S and T (Quicksort?).

32. Why Can We Continue The Recursion? • The sets in S were not separated by x. • The sets in T were not separated by x. • Therefore V2contains, a vertex y that is adjacent to some A’ in S but not to some B’ in S. Again put a copy of y in all sets that are independent of y. • Same for T .

33. Open Problems • For general graphs the upper bound of n/log n approximation ratio, is hugely far from the lower bound of sqrt{log n}. • If this is very hard to narrow, then perhaps one can narrow the gap for bipartite graphs?