T HE INTERESTING BEHAVIOR OF THE SOURCE LOCATION PROBLEM G. Kortsarz, Rutgers Camden

D EFINITION G(V,E), capacities c(e) demands d(v) a cost w(v) for every v. Find Min Cost S so that (S,v)≥d(v). v d(v)= S

I F ALL COSTS ARE 1 THE PROBLEM IS POLYNOMIAL The Algorithm: Start with S=V. The greedy move: remove a vertex v of minimum demand, if S-{v} is still feasible, and iterate. This simple algorithm is optimal. We now present a run of the algorithm

I F ALL COSTS ARE 1 THE PROBLEM IS POLYNOMIAL Say that these are the demands and S=V at start

I F ALL COSTS ARE 1 THE PROBLEM IS POLYNOMIAL Removing the vertex of demand

I F ALL COSTS ARE 1 THE PROBLEM IS POLYNOMIAL Removing the vertex of demand

I F ALL COSTS ARE 1 THE PROBLEM IS POLYNOMIAL Removing two of the 4 vertices, still valid. Vertex 3 gets flow via

I F ALL COSTS ARE 1 THE PROBLEM IS POLYNOMIAL The 5 can not be removed, as in this case |S|<

S OME HINTS FOR WHY THE ALGORITHM IS OPTIMAL Deficient sets of vertices: For a vertex set W let d(W) be the maximum demand in W Let (W) be the number of edges leaving W W is deficient if: d(W)> (W) S  W  all deficient W.

U NCROSSING : A G LIMPSE If S is feasible then it must be that S  W  for all deficient W. Say that W 1 and W 2 are deficient sets. We say that two deficient sets W 1 and W 2 cross if 1) W 1  W 2  2) No set contains the other.

A CRUCIAL PROPERTY IN CONNECTIVITY We wish that minimal deficient sets will not cross In this case the deficient sets form a laminar family (tree). Indeed here, inclusion minimal deficient sets can not cross. Let v 1 is the maximum demand vertex in W 1 and v 2 for W 2

G ETTING A CONTRADICTION By definition of deficient sets: d(v 1 )+d(v 2 )> (W 1 )  + (W 2 ) For the following inequality is known: d(v 1 )+d(v 2 )> (W 1 -W 2 )  + (W 2 -W 1 ) This means that W 1 -W 2 or W 2 -W 1 are deficient. Contradiction

I DEA OF PROOF Theorem (easy): For every v j that cant be removed there is a deficient W so that S  W= v j Match W and critical v so that W are pairwise vertex disjoint Thus all the v in S required. But S is also enough.

I F ALL DEMANDS ARE THE SAME ALSO POLYNOMIAL This time the algorithm quite complex. What about general demands and general costs? I was hoping constant ratio possible because undirected graphs. Circa M. Sakashita, K. Makino, and S. Fujishige.  (log n) hard

T HE CACTUS REPRESENTATION OF CUTS A tree of cycles. Every 2 vertex connected components: edges or cycle. All cycles intersect on at most one vertex. Global minimum cuts are edges in the tree and pairs of edges in the cycle. Due to Dinits et al Implies O(n 2 ) minimum global cuts. Today easier proof by Karger.

A N INTERESTING CASE K, Nutov: general costs and demands but maximum degree at most 3. Using the cactus theory reduce to tree of cycles and then: Change to

A TWO STEPS REDUCTION OF THE GRAPH TO A TREE. The solution of SL GD GC on trees very complex dynamic programming. Weakly Polynomial. In our case all numbers are small. Complex: 6 entries in the array. We do not know a PTAS for SL on trees because the demands can be large numbers.

T HE VAST NUMBER OF PROBLEM IN SL Could require capacity on vertices. For example VERTEX DISJOINT PATHS. Could be directed and undirected Could be the case that vertices chosen to S still need flow.

T HE INTERESTING WORK OF F UKUNAGA Fukunaga studied an interesting version in which if v  S it does not get full flow. There is a number p(v) of free flow that a vertex v  S gets. The motivation was p=1 The rest of the flow namely d(v)- p(v) has to come from S-v. Fukunaga’s idea: give a ratio when max demand k is small.

T HE RESULTS OF F UKUNAGA Fukunaga gave an O(k  log k) ratio for arbitrary p in the vertex disjoint model. In connectivity problem the vertex case many times is harder than the edge case. Not in SL as the problems are of unrelated difficulty. No f(k) was known for the edge case.

M Y MOST RECENT PAPER K,N UTOV All the problems I described are a special case of submodular-SL  (S,v)  d v so that  (S,v) submodular. All the function we saw are submodular. Our ratio proof is half a page (some of the proof of submodularity are very long)..

G ENERALIZING F UKUNAGA Besides p, we are able to handle capacities q v and assumed unit capacity edges. O(k  log k) ratio. For arbitrary p,q. We get and k ratio for the edge disjoint set, the first f(k) ratio (easy. Known?)

O NE OF OUR MAIN RESULT : SMALL P Ratio, O(p  log 2 k) with p the maximum free flow. The case p=1 studied by Fukunaga. We improve his result from O(k  log) to O(log 2 k).

F OLKLORE O( LOG N ) RATIO A FUNCTION f IS MONOTE SUBMODULAR IF: For every S,T, so that S  T,  x  T-S (f(S+x)-f(S))  f(T) In words: the sum of increases of individual elements to T-S is at least as large the increase of S when added to T-S. The DK-Subgraph is an example when this does not happen. A set S can have edges inside S.

S UBMODULAR COVER Given a universe U and a submodular monotone f, and costs over the elements, find a minimum cost S so that f(S)=f(U). Wolsey, 1982 ln (f(v)) ratio. Set cover with hard capacities is an example. Was subbmited in 1992, 10 years after Wolsey and in years after Wolsey.

D ENSITY : UNIT COTS The following inequality holds:  x  OPT-S (f(S+x)-f(S))  f(OPT) Let |OPT|=opt Averaging: there exists an x so that f(S+x)-f(S)  f(OPT)/opt

T HE APPROXIMATION How many sets will be picked until we get to f(OPT)=f(U)? At every iteration f(OPT)- f(S)  f(OPT)(1-1/opt) After opt  ln f(OPT) iterations f(OPT)-f(S)  1 For unit costs this means f(S)=f(OPT)=f(U). ln(f(OPT)) ratio.

S ET C OVER WITH HARD CAPACITIES Input as set cover S sets and E elements and each set has a hard capacity c(S). After S is chosen assign elements to sets so that no S is assigned more than S(C) elements. Find minimum cost feasible S

G IVEN S CHECKING FEASIBILITY IS POLYNOMIAL : FLOW t S c(S) s

T HE ALGORITHM 1) Start with the empty set S  2) While flow(S)<|E| do 2.1 Find s  S so that (flow(S+s)-flow(S)) /c(s) is maximum 2.2 S  S+s 2.2 Recompute flow(S) 3) Return S

D ETAILS flow(S) is monotone submodular. Thus the ratio is ln(  )+1 The same argument works for SL. ln(  )+1 ratio. Vertex capacities, compute flow with vertex capacities Works for undirected and directed. If it has free flow f(v) needs dummy edges into t.

A CLOSELY RELATED PROBLEM Steiner Network G(V,E) with costs c(e) over edges and demands D={d us }, over V  V Required: a minimum cost subgraph G(V,E’) do that for every u and v there are at least d us vertex disjoint paths between u and v. The edge disjoint case has ratio 2. Jain The vertex disjoint path Labelcover hard K,Krautghamer and Lee

E XAMPLE a b c d ab =2, d ab =2, d bc =2.

E LEMENT CONNECTIVITY Let T be the set of vertices with positive demand to at least one other vertex. Element connectivity: be disjoint of edges and vertices of V-T. Do not need to be vertex disjoint on T. This problem has ratio 2. Fleischer et al.

W HY IS IT CALLED ELEMENT CONNECTIVITY ? A vertex that is not a terminal is called an element An edge is called an element Thus the solution has to be element disjoint When I saw this problem I said: Why is this question interesting? Boy. I was so badly mistaken.

B RILLIANT IDEA C HUZHOY K HANNA Let k be the maximum demand The cost for the Element Connectivity solution is no larger than the Steiner Network with vertex disjoint paths. Randomly create many G i But treat them as element connectivity instance. But will the solution be feasible?

T HE DIFFICULTY X separates s and t and solution should be disjoint on X, and will not be since X  T i X TiTi

T HE DIFFICULTY It is even a big problem if X just intersects T i X TiTi

T HE DIFFICULTY It is not a problem if the two do not intersect. Since s,t have to be disjoint on X. X TiTi

L ET P =1/128 K 3  LOG (|T|) Take p copies of G But every vertex v  T randomly belongs to G i with probability 1/64 k 2 log(|T|). Let  (s) the indices s chose. Consider a pair s,t and an instance G i. A set is s,t-safe if X  T i = .

T HE IDEA OF THE PROOF The idea is that w.h.p  (s)   (t)   (X) The instances that have both s,t as a pair have at least one index that no vertices of X were chosen to as terminals. Thus s,t are X-Safe. Thus there is an instance in which both s,t are terminals and all path have to be disjoint on X. We use |X|<k.

S OME MORE OF THE CALCULATIONS Let  (X) be all indices chosen by X, |X|< k. E(  (X)   (t))<q/2 Thus with very high probability is at most 3q/4 (Chernoff) Recall s chose q so (  (s)-  (X))≥q/4 Easy to see that Pr(  (s)   (t)   (X)) very small. Use union bound over all X.

A GOOD QUESTION : CAN WE GET F ( K ) RATIO ? We are thinking of k as a constant which is reasonable. An f(k) approximation means O(1) approximation. A result by K, Nutov: if all edges that you can add are a star, ratio O(log 2 k) (undirected graphs) Interesting in its own right: the first f(k),albeit a restrictive case suggested by Fukunaga Used in our O(p  log 2 k) ratio.

DIFFICULTY We want to do o(p)  polylog(k) A set is most violated if its residual demand is largest. d S is the deficiency of S. If p v =10 and v  S then d S goes down by 10.

D IFFICULTY We only cover most deficient sets Say that the maximum deficient value is 20. And we have vertices v so that p v =10. If v  S, d S then d S becomes 20-10=10 But we only cover most deficient sets There may be a deficient set with deficiency d S =19.

O PEN PROBLEMS Is there is a Steiner Network algorithm whose ratio depends only on k? Does the SL, with general p and q, problem admit a polylog(k) ratio?