Edge Covering problems with budget constrains

By R. Gandhi and G. Kortsarz

Presented by: Alantha Newman

First problem, unweighted caseInput: A graph G(V,E) and a

number kRequired: A subset U of V of size

k that is touched by the minimum number of edges.

Touching edges of U: An edge (u,v) so that either uU or uU

We denote this number by t(U)

What is the properties of a good solution?

Let us check the case of a d regular graphs. The

question is if the edges are internal or external

S

What is the properties of a good solution

In this example most of the edges in S stay inside. Which means that t(S) is close to kd/2

U

What is the properties of a good solution

But S can behave badly. Namely most edges go

to V-UIn this case t(S) close to dk.

U

A trivial ratio of 2

•Let OPT, |OPT|=k, be the best solution •Let U be the k least degrees vertices , thus deg(OPT)≥ deg(U)• Clearly t(OPT)≥deg(OPT)/2• Therefore: t(U)≤deg(U)≤deg(OPT) ≤2t(OPT)

We tried to improve the 2 but there is a problem

The following we became aware of only recently.

Let G be a d-regular graph. Consider only “small” S and:

= e(S,V-S)/deg(S)The without restriction: the

Sparsest Cut problem and admits a sqrt{log n} by Arora Rao and Vazirani

The case of small S What happens if we bound the

size of S?The question of if there is a small

set S with bad expansion may be harder.

Let 0≤≤1/2And allow only sets with at most n vertices. What is the worst expansion?

Let G be a d-regular graph. Let ≤0.5.

Let =Min |S|≤ n e(S,V-S)/deg(S)The SSEC: It is hard to tell between the following two cases, for small : ≥ 1- and ≤

Due to Raghavendra and Steurer.

The Small Set Expansion Conjecture

For a given , k= n.

If =e(S,V-S)/deg(S) is roughly one most edges are outside S.

For our problem the value of the solution about kd

Breaking the ratio of 2 is equivalent to disproving SSEC

The good case is S with small expansion

If =e(S,V-S)/deg(S) close to 0 almost all edges are internal to S.

This means the value for our problem may be very close to kd/2. Good for SSEC good for us.

The other case

The approximation of 2 is optimal(?)An approximation better than 2

means that the Small Set Expansion Conjecture fails!

This problem reduces to the unique game conjecture by Khot.

Seems hugely hard but easier than the famed Unique Game conjecture.

In the words of the movie “Marathon Man:” Is it safe? (To assume its hard?)

Is the Small Graph Expasion conjecture set reliable?

Opinions vary. I think: VERY RELIABLE. We tried to disprove the SSGE and

failed The first problem with ratio 2 so that

breaking the ratio 2 disproves the conjecture is the At least size k Densest subgraph. I think unpublished (U. Feige).

I suspect that in order to give good ratio for the SSEC you need a good algorithm for the Dense k-subgraph

Our second main problem Say that we are given a graph G(V,E) and

a number M. Find the maximum number of vertices that are touched by at most M edges.

Again for this problem a ratio 2 is simple.To the best of our knowledge this problem

was first studied for approximation by Goldshmidt and Hochbaum

This problem is motivated by application in loading of semi-conductor components to be assembled into products

The weighted caseVertices have weights. Largest weight under M touching

edges: Find a set U of maximum weight so that the number of edges touching U is at most M.

Minimum edges under cost at least k: Find a set U of cost at least k and minimize the number of edges touching U.

Our results

Minimum edges under cost at least k admits a polynomial time 2 ratio. Improves 2(1+) by Carnes and Shmoys.

Maximum weight under at most M edges: admits a polynomial time algorithm with ratio 2. Improves ratio 3 by Goldschmidt and Hochbaum.

Lower boundsGoldschmidt and Hochbaum

show that these two problems are NPC.

Under the SSEC we show that our approximation is optimal for both problems.

We show: Both problems even in the unweighted case admits no 2- approximation for any constant >0 unless the SSEC fails.

Our results

The natural LP for Weight at least k and at most M edges in the unweighted case, has integrality gap 2. Not a surprise given the SSEC.

However showing integrality gap 2 is quite non trivial. Uses the probabilistic method.

Comparing the two problemsWe show that a ratio for Minimum edges cost at least k implies a ratio of for Maximum weight at most M edges. It seems that the reverse does not hold.

Further results We show that the density version of Minimum edges for cost at least k can be solved by flow (only LP solution was known).

Given a selected already set S the goal is to add a set U and minimize

(e(U)+e(U,S))/deg(S)

Some ideas of how to give ratio 2 for Maximum cost at most M edges

We use Dynamic Programing.We guess the number P of edges between the optimum set OPT and V-OPT.

We guess the sum of degree of OPT whom may be 2M. A serious technical problem: we are only able to compute A[n, P, M].

The reason for thatThis is the only way, it seems, to

assure feasibility.Indeed if deg(U)≤M then t(U)≤M.The question is do we loose a lot

by bounding the sum of degrees by M while the sum may be 2M?

One more detail: we need to guess the highest cost vertex in OPT and add it the our solution.

If deg(U)≤M, how much cost we loose?

Let OPT= A {x} B so that deg(A+x) is the first to be above M

Thus A is a feasible solution for M.Clearly B too is a feasible solution for

M because deg(A+x)>M and the total at most 2M

One of A or B has ½ the weight. The fact that we guess the highest cost vertex in OPT compensate for x.

Thus ratio 2.

Thank you for your attentionAny Questions?