Edge Covering problems with budget constrains By R
Edge Covering problems with budget constrains By R. Gandhi and G. Kortsarz Presented by: Alantha Newman
First problem, unweighted case �Input: A graph G(V, E) and a number k �Required: 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 u U or u U �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 k d/2 U
What is the properties of a good solution But S can behave badly. Namely most edges go to V-U In 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(OPT) ≤ 2 t(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≤ ≤ 1/2 �And allow only sets with at most n vertices. What is the worst expansion?
The Small Set Expansion Conjecture �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.
Breaking the ratio of 2 is equivalent to disproving SSEC a given , k= n. If =e(S, V-S)/deg(S) is roughly � For one most edges are outside S. For our problem the value of the solution about k d
The other case �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 k d/2. Good for SSEC good for us.
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
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 case �Vertices 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 bounds �Goldschmidt 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 problems 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. �We
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 2 M. A serious technical problem: we are only able to compute A[n, P, M].
The reason for that �This 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 2 M? �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? 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 2 M �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. �Let
Thank you for your attention �Any Questions?
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