THE INTERESTING BEHAVIOR OF THE SOURCE LOCATION PROBLEM

G. Kortsarz, Rutgers Camden

DEFINITION 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)= 29

7 3

S

46

3

4 3

IF 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

IF ALL COSTS ARE 1 THE PROBLEM IS POLYNOMIAL

Say that these are the demands and S=V at start.

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24

IF ALL COSTS ARE 1 THE PROBLEM IS POLYNOMIAL

Removing the vertex of demand 2.

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5

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IF ALL COSTS ARE 1 THE PROBLEM IS POLYNOMIAL

Removing the vertex of demand 3.

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IF ALL COSTS ARE 1 THE PROBLEM IS POLYNOMIAL

Removing two of the 4 vertices, still valid.

Vertex 3 gets flow via 43

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5

24

IF ALL COSTS ARE 1 THE PROBLEM IS POLYNOMIAL

The 5 can not be removed, as in this case |S|<5.

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SOME 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) SW all deficient W.

UNCROSSING: A GLIMPSE If S is feasible then it must be that SW for all deficient W.

Say that W1 and W2 are deficient

sets. We say that two deficient sets W1 and W2 cross if

1) W1 W2 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 v1 is the maximum demand vertex in W1 and v2 for W2

GETTING A CONTRADICTIONBy definition of deficient sets: d(v1)+d(v2 )> (W1)+(W2 )For the following inequality is known:d(v1)+d(v2 )> (W1 -W2)+ (W2 -W1 )This means that W1 -W2 or

W2 -W1 are deficient.

Contradiction

IDEA OF PROOF

Theorem (easy): For every vj

that cant be removed there is a deficient W so that SW= vj

Match W and critical v so that W are pairwise vertex disjoint

Thus all the v in S required.

But S is also enough.

IF 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 2004.

M. Sakashita, K. Makino, and S. Fujishige. (log n) hard. 2006.

THE 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(n2) minimum global cuts.Today easier proof by Karger.

AN 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.

THE 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.

THE INTERESTING WORK OF FUKUNAGA

Fukunaga studied an interesting version in which if vS 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.

THE RESULTS OF FUKUNAGA

Fukunaga gave an O(klog 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.

MY MOST RECENT PAPER K,NUTOV

All the problems I described are a special case of submodular-SL (S,v) dv 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).

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GENERALIZING FUKUNAGA

Besides p, we are able to handle capacities qv and assumed unit capacity edges. O(klog k) ratio. For arbitrary p,q.

We get and k ratio for the edge disjoint set, the first f(k) ratio (easy. Known?)

ONE OF OUR MAIN RESULT: SMALL P

Ratio, O(plog2 k) with p the maximum free flow.

The case p=1 studied by Fukunaga.

We improve his result from O(klog) to O(log2 k)

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FOLKLORE O(LOG N) RATIO

A FUNCTION f IS MONOTE SUBMODULAR IF:

For every S,T, so that ST, xT-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.

SUBMODULAR 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 2002 20 years after Wolsey.

DENSITY: UNIT COTS The following inequality holds:

xOPT-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

THE APPROXIMATIONHow 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 optln f(OPT) iterations f(OPT)-f(S) 1 For unit costs this means f(S)=f(OPT)=f(U). ln(f(OPT)) ratio.

SET COVER 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

GIVEN S CHECKING FEASIBILITY IS POLYNOMIAL: FLOW

t

S

c(S)

1 1 1 1 11 1

s

1 1 11 1

THE ALGORITHM 1) Start with the empty set S2) While flow(S)<|E| do 2.1 Find sS so that (flow(S+s)-flow(S)) /c(s) is maximum 2.2 SS+s 2.2 Recompute flow(S)3) Return S

DETAILS 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={dus}, over VV

Required: a minimum cost subgraph G(V,E’) do that for every u and v there are at least dus vertex disjoint paths between u and v.

The edge disjoint case has ratio 2. JainThe vertex disjoint path Labelcover

hard K,Krautghamer and Lee

EXAMPLE

a

b

c

dab=2, dab=2, dbc=2.

ELEMENT 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

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WHY IS IT CALLED ELEMENT CONNECTIVITY?A vertex that is not a terminal is called an element

An edge is called an elementThus 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.

BRILLIANT IDEA CHUZHOY KHANNA

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 Gi

But treat them as element connectivity instance.

But will the solution be feasible?

THE DIFFICULTY X separates s and t and solution should be

disjoint on X, and will not be since XTi

X

Ti

THE DIFFICULTY It is even a big problem if X just

intersects Ti

X

Ti

THE DIFFICULTY It is not a problem if the two do

not intersect. Since s,t have to be disjoint on X.

X

Ti

LET P=1/128 K3 LOG(|T|)

Take p copies of GBut every vertex vT randomly belongs to Gi with probability 1/64 k2 log(|T|).

Let (s) the indices s chose. Consider a pair s,t and an instance Gi .

A set is s,t-safe if X Ti =.

THE IDEA OF THE PROOFThe 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.

SOME MORE OF THE CALCULATIONS

Let (X) be all indices chosen by X, |X|< k.

E((X) (t))<q/2Thus with very high probability is

at most 3q/4 (Chernoff)Recall s chose q so ((s)-(X))≥q/4Easy 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(log2k) (undirected graphs)

Interesting in its own right: the first f(k),albeit a restrictive case suggested by Fukunaga

Used in our O(p log2k) ratio.

DIFFICULTY

We want to do o(p)polylog(k)

A set is most violated if its residual demand is largest.

dS is the deficiency of S. If pv=10 and v S then dS

goes down by 10.

DIFFICULTYWe only cover most deficient setsSay that the maximum deficient

value is 20.And we have vertices v so that

pv=10. If vS, dS then dS becomes 20-10=10

But we only cover most deficient sets

There may be a deficient set with deficiency dS=19.

OPEN PROBLEMSIs 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?