On The Achromatic Number Problem Guy Kortsarz Rutgers University, Camden

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

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

Example In the following graph, the optimum is 4.

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).

An Example Of A Difference Between The Two Functions Say that   (G) is a universal constant c. Then we are able to solve the problem in polynomial time by exhaustive search. Search in time n c2 for 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.

How Do We Find A Feasible Solution?

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

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.

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.

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.

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. Computers can not run an algorithm that performs 2 n operations for 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.

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 polynomial time and if for every instance I: Val(OPT(I))/Val(A(I))   It is conjecture by Chadhary and Vishwanathan that the Achromatic number problem admits a sqrt{   } approximation.

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.

Some Known Approximation Algorithms, Continued In a paper with Krauthgamer, min{sqrt{   },n 1/3 } approximation algorithm for graph with girth at least 5. This improves a min{sqrt{   },n 3/8 } ratio approximation algorithms by Krysta and Lorys. And they even looked at an easier problem: Girth at least 6!

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(n 4/5 ) ratio approximating 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.

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.

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.

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 n 1- , then P=NP. Thus a lower bound of n 1-  by Magnus. M. Halldorsson.

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/n with 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.

Proof That    m/n In A Graph Of Girth At Least 5 Remove all vertices with degree strictly less then 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.

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

From Partial Coloring To A Complete Coloring Say that we have a partial achoromatic coloring with p colors 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.

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.

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{   }.

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.

Small Size Classes Don’t Count Recompute n. Classes of size at most n/(2sqrt{   }) can contain at most n/2 vertices. Therefore equivalence classes of size at least n/(2sqrt{   }) vertices contain 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.

The Algorithm We show A,B and x  V 2 such that x is adjacent to (say) B but is not to A. A B C x

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 V 1 into sets that are adjacent to x and to set that are not adjacent to x. A PROPER partition. As we shall see vertices are added to A and B but only from V 2 (so independent, and independent with respect to A and B).

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 V 1 to equivalence classes S that contain x and T that do not contain x.

How To Continue Note: every equivalence set in S is adjacent to every equivalence set in T. This is because of the edges from x to T 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?).

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 V 2 contains, 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.

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?