On The Achromatic Number
Problem
Guy Kortsarz
Rutgers University, Camden
On Independent Sets And
Cliques
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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
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INPUT: An undirected graph G(V,W)
Goal: A partition of G(V,W) into as
many subsets V=U Di as possible
such that the sets Di are
independent sets, pairwise disjoint, and
there is an edge between every pair of
sets Di Dj ij.
Example
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In the following graph, the optimum is
4.
A Partition To Independent
Sets Is Called A Coloring
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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
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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 nc2 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
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A Complete Bipartite Graph:
About the Achromatic Number
Problem
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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
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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?
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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
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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 2n operations for large n.
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For n=300 this is more than the number of atoms in
the known universe.
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Some rule of thumb usually works great, in the sense
that is %5 from optimum (how do we know that?).
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We want a provable multiplicable distance from the
optimum. This is called Approximation ratio.
Approximation Ratio For
Maximization Problems
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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
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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
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In a paper with Krauthgamer,
min{sqrt{} ,n1/3} approximation
algorithm for graph with girth at least 5.
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!
Bipartite graphs
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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 ) 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
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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.
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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?
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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.
Graph Theory
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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
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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
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For a leaf  and a star S,  is adjacent
to at most one leaf in S.
From Partial Coloring To A
Complete Coloring
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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?
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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
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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
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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
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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
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We show A,B and xV2 such that x is
adjacent to (say) B but is not to A.
A
B
x
C
The Set A,B And The Vertex x
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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 V1 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 V2 (so independent, and
independent with respect to A and B).
Finding Some Adjacent Sets
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The vertex x has about sqrt{}
equivalent copies.
Put (a copy of) x in any equivalence set
that does not contain x.
This partitions V1 to equivalence classes
S that contain x and T that do not
contain x .
How To Continue
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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?
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The sets in S were not separated by x.
The sets in T were not separated by x.
Therefore V2 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
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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?