What should be taught What should be taught in approximation in approximation algorithms courses?algorithms courses?Guy Kortsarz, Rutgers Guy Kortsarz, Rutgers CamdenCamden

Advanced issues presented Advanced issues presented in many lecture notes and in many lecture notes and

books:books:• Coloring a Coloring a 3-3-colorable graph using vectors.colorable graph using vectors.• Paper by Paper by KargerKarger, , Motwani Motwani andand Sudan Sudan..• Things a student needs to know:Things a student needs to know: Separation oracle for: Separation oracle for: AA is is PSDPSD.. Getting a random vector inGetting a random vector in RRnn.. This is done by choosing the This is done by choosing the Normal Normal distributiondistribution at every entry.at every entry. Given unit vector Given unit vector vv, , v . rv . r is normalis normal distribution. distribution.

Things a student needs to Things a student needs to know:know:

• There is a choice of vectors There is a choice of vectors vvi i for for everyevery i i V V so thatso that so that for everyso that for every ((i, j)i, j) E E, , vvi i · v· vjj -1/2 -1/2..

A student needs to know:A student needs to know:• S={i | r S={i | r ·· v vi i }}, , threshold method, by now threshold method, by now

standard.standard.• Sum of two normal distributions also Sum of two normal distributions also

normal.normal.• Two inequalities (non trivial) about the Two inequalities (non trivial) about the

normal distribution.normal distribution.• The above can be used to find a large The above can be used to find a large

independent set.independent set.• Combined with the greedy algorithm gives Combined with the greedy algorithm gives

about about nn1/4 1/4 ratio approximation algorithm.ratio approximation algorithm.

Advanced methods are also Advanced methods are also required in the following required in the following

topics often taught:topics often taught:• The seminal result of The seminal result of JainJain. With the . With the

simplification of simplification of NagarajanNagarajan et. al. et. al. 2-ratio2-ratio for for Steiner NetworkSteiner Network..

• The beautiful The beautiful 3/23/2 ratio by ratio by CalinescoCalinesco, , Karloff Karloff and and Rabani, Rabani, for for Multiway CutsMultiway Cuts: : geometric geometric

embeddings. embeddings. • FacharoenpholFacharoenphol,, Rao and Talwar, Rao and Talwar, optimal optimal

random tree embedding. With this can getrandom tree embedding. With this can get O(log n)O(log n) for undirected multicut.for undirected multicut.

How to teach sparsest How to teach sparsest cut?cut?

• Many still teach the embedding of a metric into Many still teach the embedding of a metric into LL11 , with, with O(log n)O(log n) distortion. By distortion. By LinealLineal,, London London,, RabinovichRabinovich..

• Advantage: relatively simple.Advantage: relatively simple.• The huge challenge posed by the The huge challenge posed by the Arora, RaoArora, Rao and and

VaziraniVazirani result. result. Unweighted sparsest cutUnweighted sparsest cut sqrt{log sqrt{log n}n}

• Teach the difficult lemma? Very advance. Very Teach the difficult lemma? Very advance. Very difficult.difficult.

• A proof appears in the book of A proof appears in the book of ShmoysShmoys and and Williamson.Williamson.

Simpler topics?Simpler topics?• I can not complain if it is I can not complain if it is TAUGHT!TAUGHT! Of Of

course not. Let me give a list of basic course not. Let me give a list of basic topics that always taughttopics that always taught

• RatioRatio3/23/2 for for TSPTSP, the simple approximation , the simple approximation of of 22 for min cost for min cost SteinerSteiner tree. tree.

• Set-CoverSet-Cover , simple approximation ratio. , simple approximation ratio.• KnapsackKnapsack,, PTAS.PTAS. Bin packing Bin packing, constant , constant

ratio.ratio.• Set-Coverage. Set-Coverage. BUT:BUT: only costsonly costs 1 1..

Knapsack Set-CoverageKnapsack Set-Coverage• The The Set-Coverage problem Set-Coverage problem is given ais given a set set

system system and a numberand a number kk selectselect k k sets that sets that cover as many elemnts as possible. cover as many elemnts as possible.

• Knapsack version, not that known: Knapsack version, not that known: • Each set has cost Each set has cost c(s)c(s) and there is a and there is a

bound bound BB on the maximum sum of costs, of on the maximum sum of costs, of sets we can choose. sets we can choose.

• Maximize number of elements covered.Maximize number of elements covered.

Result due to Result due to KhullerKhuller , , MossMoss and,  and,  Naor Naor, 1997, , 1997, IPLIPL

• The The (1-1/e)(1-1/e) ratio is possible. ratio is possible.• In the usual algorithm & analysis In the usual algorithm & analysis (1-1/e)(1-1/e)

only follows if we can add the last set in only follows if we can add the last set in the greedy choice. Thus, fails.the greedy choice. Thus, fails.

• Because most times, adding the last set Because most times, adding the last set will give cost larger than will give cost larger than BB..

• Trick: guess the Trick: guess the 33 sets in sets in OPTOPT of least of least costcost. Then apply greedy (don’t go over . Then apply greedy (don’t go over budget budget BB).).

Why do I know this Why do I know this paper?paper?

• I became aware of this result only several I became aware of this result only several years after published. And only because I years after published. And only because I worked on worked on Min Power ProblemsMin Power Problems. No . No conference version!conference version!

• This result seems absolutely basic to me. Why This result seems absolutely basic to me. Why is it no taught?is it no taught?

• Remark: Choosing one (least cost) element of Remark: Choosing one (least cost) element of OPT OPT gives gives unbounded ratiounbounded ratio. Choosing two sets . Choosing two sets of smallest cost gives ratio of smallest cost gives ratio ½½. Guessing the . Guessing the three sets of least cost and then greedy givesthree sets of least cost and then greedy gives (1-1/e)(1-1/e)..

First First generalgeneral neglected neglected topictopic

• Important and not taught: Important and not taught: Maximizing a submodular non-Maximizing a submodular non-decreasing function under Matroid decreasing function under Matroid ConstrainsConstrains, ratio , ratio 1/21/2, , FischerFischer,, NemhauserNemhauser,, Wolsey Wolsey, 1977., 1977.

• Improved in 2008(!) to best possible Improved in 2008(!) to best possible (1-1/e)(1-1/e) by by VondrakVondrak in a brilliant in a brilliant paper.paper.

First story: a submission I First story: a submission I refereedrefereed

• I got a paper to referee, and it was obvious that it I got a paper to referee, and it was obvious that it is maximize Submodular function under Matroid is maximize Submodular function under Matroid constrainsconstrains

• If memory serves, the capacity If memory serves, the capacity 11, of the following , of the following Matroid: Matroid: G(V,E)G(V,E),, edge capacities, fix edge capacities, fix S S V V. . TT reaches reaches S S if every vertex in if every vertex in TT can send one unit of can send one unit of flow toflow to SS..

• The set of all The set of all TT that reach that reach S S a special Matroid a special Matroid called called GammoidGammoid. Everything in this paper, . Everything in this paper, known!known!

• Asked Asked ChekuriChekuri (everybody must have an oracle) (everybody must have an oracle) what is the Matroid, and what is the Matroid, and ChekuriChekuri answered. answered. Paper erased.Paper erased.

Story 2: a worse Story 2: a worse outcome.outcome.

• Problem. Input like Problem. Input like Set-CoverSet-Cover but but S=S= S Sii..• Required: choose at most one set of everyRequired: choose at most one set of every SSi i

and maximize the number of elements covered.

• Paper gave ratio ½. This is maximizing submodular cover subject to partition Matroid. PLEASE!!! Do not try to check who the authors are. Not ethical. Unfair to authors, as well.

• Nice applications, but was accepted and ratio not new.

Related to pipage roundingRelated to pipage rounding• Due to Due to AgeevAgeev, , SviridenkoSviridenko..• Dependent roundingDependent rounding, is a generalization of, is a generalization of Pipage rounding byPipage rounding by Gandhi Gandhi, , KhullerKhuller, ,

ParthasarathyParthasarathy, , SrinivasanSrinivasan..• Say that we have an Say that we have an LPLP and a constraint and a constraint xxii=k=k. . RRRR can not can not derive exact equality. derive exact equality.• Pipage RoundingPipage Rounding : instead of going to a : instead of going to a

larger set of solutions like IP to LP, we larger set of solutions like IP to LP, we replace the objective function.replace the objective function.

The principals of pipage The principals of pipage roundingrounding

• We start with LP maximization with We start with LP maximization with function function L(X)L(X)..

• Define a non linear function Define a non linear function FF..• Show that the maximum of Show that the maximum of FF is is

integral.integral.• Show that integral points of Show that integral points of FF belong to belong to

the Polyhedra of the Polyhedra of LL. Namely feasible for . Namely feasible for LL as long as it is integral, and feasible as long as it is integral, and feasible for for FF..

The principals of pipage The principals of pipage roundingrounding

• Then, show that Then, show that F(XF(Xint int ) ≥L(X) ≥L(X* * )/)/,, for for > 1> 1..• HereHere XXint int is the (integral) optimumis the (integral) optimum ofof F F

and and XX** the optimum fractional solution forthe optimum fractional solution for L L• Because Because XXint int is known to be feasible for is known to be feasible for

L(x)L(x) due to its integrality, it is feasible due to its integrality, it is feasible for for LL and thus and thus approximation.approximation.

Example: Max CoverageExample: Max Coverage• Max j wi zi

S.T element j belongs to set i xi≥zj

set i xi=p xi and zj are integral

In Set Coverage we bound the number of sets.

The function The function FF• F(x)=F(x)= j wi (1-element j belongs to set i(1-xi) )• Define a function on a cycle.• As a function of . • The idea is to make plus and then

minus over the cycle.• Make one entry on the cycle smaller

by and another larger by .

The function The function FF• F(x)=F(x)= j wi (1-element j belongs to set i(1-xj) )• The idea is to make plus and then

minus all over the cycle.• But to show convexity we make just

one entry on the cycle smaller by and another larger by .

• The appears as 2 in this term.

The function The function FF• As appears as 2 in this term, the

second derivative is positive.• Thus F is convex.• Which means that the maximum is in

the borders. • For example for x2 between -4 and

3.• The maximum is in the border -4.

Changing the Changing the two by two by twotwo

• Putting plus and minus Putting plus and minus alternating along alternating along a cycle make at least one entry integral.a cycle make at least one entry integral.

• Moreover, we can decompose a cycle into Moreover, we can decompose a cycle into two matching and there are two ways to two matching and there are two ways to increase and decrease by increase and decrease by ..

• One direction of the two makes the One direction of the two makes the function not smaller.function not smaller.

• This implies that the optimum of This implies that the optimum of FF is is integral.integral.

Thus the optimum of Thus the optimum of FF is is integralintegral

• Its not hard to see that on integral Its not hard to see that on integral vectors F and L have the same value. vectors F and L have the same value.

• Another inequality that is quite hard Another inequality that is quite hard to prove is that: to prove is that: 1-1-i=1 to k i=1 to k (1-x(1-xii)≥(1-)≥(1-(1-1/k)(1-1/k)kk)L(X) )L(X)

• This gives a slightly better than This gives a slightly better than 1-1/e1-1/e ratio if ratio if kk is small. is small.

Submodularity: related to Submodularity: related to very basic technique.very basic technique.

• f f is is submodularsubmodular if if f(A)+f(B)f(A)+f(B)f(Af(AB)+f(AB)+f(AB)B)• Makes a lot of difference if non-decreasing Makes a lot of difference if non-decreasing

or not. If not, in my opinion represent or not. If not, in my opinion represent concaveconcave..

• If non-decreasing, brings us to the next lost If non-decreasing, brings us to the next lost simple subject: simple subject: Submodular coverSubmodular cover problems. problems.

• Input: Input: U U and submodular non-decreasing and submodular non-decreasing function function ff and cost and cost c(u)c(u) per item per item uu..

• Required: a set Required: a set SS of minimum cost so that of minimum cost so that f(S)=f(U)f(S)=f(U)..

WolseyWolsey , 1982, did much , 1982, did much betterbetter

• Each iteration pick item Each iteration pick item uu so that so that helphelpuu(S)/c(u)(S)/c(u) is is maximum.maximum.

• The ratio is The ratio is maxmax{u{u U}U}ln f(u)+1ln f(u)+1.. • Example: For Example: For Set-CoverSet-Cover ln|s|+1ln|s|+1,, ss largest set. largest set.• Example: Same for Example: Same for Set-Cover with hard capacitiesSet-Cover with hard capacities. .

A paper in 1991, and one in 2002, did this result A paper in 1991, and one in 2002, did this result again (second was 20 years after again (second was 20 years after WolseyWolsey). Special ). Special case after 20 years! But its worse, yet.case after 20 years! But its worse, yet.

• WolseyWolsey did better than that. Natural LP did better than that. Natural LP unbounded ratio even for Set-Cover with hard unbounded ratio even for Set-Cover with hard capacities.capacities.

• WolseyWolsey found a fabulous LP of gap found a fabulous LP of gap maxmax{u{u U}U}ln f(u)ln f(u) +1+1..

More general: More general: densitydensity• Not taught at all but just cited. Why?Not taught at all but just cited. Why?• Here is a formal way:Here is a formal way:• Universe Universe UU and a function and a function f: 2f: 2UUR+R+• Each element in Each element in UU has a cost has a cost c(u) c(u)..• The function The function ff not decreasing. not decreasing.• We want to find a minimum cost We want to find a minimum cost W W so thatso that

f(W)=f(U).f(W)=f(U).• We usually say, We usually say, SS U, c(S)= U, c(S)=uuSS c(u) c(u) • But it works for an subadditive cost functionBut it works for an subadditive cost function

The The densitydensity claim claim• Say that we already created a set Say that we already created a set S S via a via a

greedy algorithm.greedy algorithm.• Now say that at any iteration we are able Now say that at any iteration we are able

to find some to find some ZZ so that: so that: (f(Z+S)-f(S))/c(Z)≥(f(U)-f(S))/(δ·opt)

• Then the final set S has cost bounded by

(δ ln(U)+1) opt

What does it mean?What does it mean?• Think for the moment of Think for the moment of δδ=1=1..• Say that the current set Say that the current set S S has no has no

intersection with the optimum.intersection with the optimum.• Then if we add all of Then if we add all of OPTOPT to to SS we we

certainly get a feasible solutioncertainly get a feasible solution..• Then clearlyThen clearly f(S+OPT)=f(U)f(S+OPT)=f(U)• And And • (f(S+OPT)-f(S))/c(Z)≥ (f(OPT)-f(S))/c(OPT)• =(f(U)-f(S))/f(OPT)• It means that we found a solution to add

that has the same density as adding OPT.

Proof continuedProof continued• f(U) -f(U) - j≤ i-1 f(Sj)≥ 1.• We may assume that the cost of every We may assume that the cost of every

set added is at most set added is at most optopt, therefore , therefore c(Sc(Sjj ) ) ≤ opt• Therefore it remains to bound: Therefore it remains to bound: j≤ i-1 c(Zi) Let us concentrate on what happens Let us concentrate on what happens

before before Si is added.

By the previous claimsBy the previous claims• 1 ≤ f(U)-f(Z1+Z2 +……Zi-1)≤ Πj≤ i-1(1-c(Zi)/δ·opt)· f(U)• 1/f(U) ≤ Πj≤ i-1(1-c(Zi)/δ·opt)·

• Take ln and use ln(1+x) ≤ x: -ln( f(U))≤ i≤ j -c(Zi) )/δ·opt i≤ j c(Zi) ≤ opt δ ln( f(U)) and so the ratio of (δ ln( f(U))+1)

follows.

A paper of mineA paper of mine• Min c Min c xx subject to subject to ABABxx b b, with , with A A positive positive

entries and entries and BB flow matrix. Ratio logarithmic. flow matrix. Ratio logarithmic.• We got much more general results. The above I We got much more general results. The above I

was sure then and sure now, was sure then and sure now, KNOWN KNOWN and and presented as known.presented as known.

• Referees: Cite, or prove submodularity! We had Referees: Cite, or prove submodularity! We had to prove (referees did not agree its known!).to prove (referees did not agree its known!).

• Example: gives Example: gives log nlog n for directed Source for directed Source Location. Maybe first time stated but I Location. Maybe first time stated but I considered it considered it knownknown..

• This This log nlog n was proved at least was proved at least 44 times since then. times since then.

RemarksRemarks• The bad thing about these The bad thing about these 44 papers is not that papers is not that

did not know our paper (to be expected) but did not know our paper (to be expected) but that they would think such a simple result is that they would think such a simple result is NOTNOT KNOWNKNOWN..

• It is good to know the result of It is good to know the result of WolseyWolsey: for : for example, used it recently example, used it recently ((HajiaghayiHajiaghayi , ,KhandekarKhandekar,,KK , , NutovNutov) to give a ) to give a lower bound of about lower bound of about log log 2 2 nn for a problem in for a problem in fashion: fashion: Capacitated NetworkCapacitated Network Design Design (Steiner (Steiner network with capacities). First lower bound network with capacities). First lower bound for hard capacities.for hard capacities.

Dual fitting and a mistake Dual fitting and a mistake we all makewe all make

• 1992. 1992. GKGK to to Noga AlonNoga Alon::• This (spanner) result bares similarities to the This (spanner) result bares similarities to the

proof done by proof done by LovatsLovats for set-cover. for set-cover.• Noga AlonNoga Alon (seems very unhappy, maybe angry): (seems very unhappy, maybe angry):

Give me a break! That is folklore. Give me a break! That is folklore. LovatsLovats told me told me he wrote it so he would have something to he wrote it so he would have something to cite.......cite.......

• Everybody cites Everybody cites LovatsLovats here. Its simply not true. here. Its simply not true.• We don’t know the basics. Result known many We don’t know the basics. Result known many

years before 1975.years before 1975.• Should we cite folklore? Yes!Should we cite folklore? Yes!

HOWHOW to teach dual fitting to teach dual fitting for set cover, unweighted?for set cover, unweighted?

• Let Let SS be the collection of sets and be the collection of sets and TT the elements.the elements.

• The dual, costs The dual, costs 11: : MaximizeMaximize ttTT y ytt

• Subject to: Subject to: ttss y yt t c(s)=1 c(s)=1• We define a dual: if the greedy chose We define a dual: if the greedy chose

a star of length a star of length ii, each element in , each element in the set gets the set gets 1/i1/i

.2

.2.2.2.2

The bound on the sum of The bound on the sum of elements of a given setelements of a given set

1/121/12

1/12 1/12

1/121/11

1/10

1/9 1/8 1/7 1/71/71/6

1/51/41/41/3 1/2 1/21

Primal Dual of GWPrimal Dual of GW• Goemans and WilliamsonGoemans and Williamson gave a rather gave a rather

well known well known Primal-Dual algorithm. Primal-Dual algorithm. Always taught, and should be.Always taught, and should be.

• A question I asked quite several A question I asked quite several researchers and I don’t remember a researchers and I don’t remember a correct response: correct response: Why reverse delete? Why reverse delete?

• Why not Michael Jackson?Why not Michael Jackson?• GW primal dual imitates recursion.GW primal dual imitates recursion.• In LR reverse delete In LR reverse delete follows follows from from

recusrsion.recusrsion.

Local Ratio for covering Local Ratio for covering problemsproblems

• Give weights to items so that every minimal. Give weights to items so that every minimal. solution is a solution is a approximation. Reduce items approximation. Reduce items costs by weights chosen.costs by weights chosen.

• Elements of costElements of cost 0 0 enter the solution. enter the solution.• Make minimal.Make minimal.• Recurse. Recurse. • No needNo need for reverse delete. Recursion for reverse delete. Recursion

implies it.implies it.• Simpler for Simpler for Steiner NetworkSteiner Network in my opinion. in my opinion.

Local RatioLocal Ratio• Without it I don’t think we could find a ratioWithout it I don’t think we could find a ratio 2 2

for for Vertex feedback setVertex feedback set..• A recent result of A recent result of K, Langberg, NutovK, Langberg, Nutov. Minor . Minor

result (main results are different) but solves an result (main results are different) but solves an open problem of a very smart person: open problem of a very smart person: KrivelevichKrivelevich..

• Covering triangles, gapCovering triangles, gap 22 for LP (polynomial).for LP (polynomial).• Open problem: tight?Open problem: tight?• Not only we showed tight family but showed as Not only we showed tight family but showed as

hard as approximatinghard as approximating VCVC.. Used Used LRLR in proof. in proof.

Group Steiner problem Group Steiner problem on treeson trees

• Group Steiner problem on trees.• Input : An undirected weighted rooted by r tree T = (V; E) and subsets S1,……,Sp V.• Goal: Find a tree in G that connects at least one

vertex from each Si to r. • The Garg, Konjevod and Ravi proof while quite

simple can be much much further simplified. In both proofs:

O(log n· log p) ratio.• The easier (unpublished) proof is by Khandekar

and Garg.

The theorem of Garg The theorem of Garg Konjevod and RaviKonjevod and Ravi

• There is an O(h log p)-approximation algorithm for Group Steiner on trees.

T= (V; E) rooted at r has depth h.• Simple observation: we may assume

that the groups only contain leaves by adding zero cost edges.

• The GKR result uses an LP methods.

The fractional LPThe fractional LP• Minimize e cost(e)· xe

frg=1 For every g. feg≤ xe

fvg ≤ v’ child of v fvv’(g) fvg = fpar(v) v(g)

The xe are capacities. Under that, the sum of flows from r to the leaves that belong to g is 1. If we set xe =1 for the edges of the optimum we get an optimum solution.

Thus the above (fractional) LP is a relaxation.

The rounding method of The rounding method of GKRGKR

• Consider Consider xxee and say that its parent and say that its parent edge is edge is (par(v),v)(par(v),v)

• Independently for every Independently for every ee, add it to the , add it to the solution with probabilitysolution with probability x xee//xpar(v)v

• We show that the expected cost is We show that the expected cost is bounded by the LP cost.bounded by the LP cost.

• The probability that an edge gets to the root

is a telescopic multiplication.

The probability that an The probability that an edge is chosenedge is chosen

• All terms cancel but the first and the last. All terms cancel but the first and the last. The First is The First is xxe. The last is the flow from ‘ . The last is the flow from ‘ The parent of The parent of rr to to r r ’ which we may ’ which we may assume is assume is 11..

• Since this is the case, Since this is the case, xxe contributes xxe· cost(e) to the expected cost.• Therefore, the expected cost is the LP

value which is at most the integral optimum. However: what is the probability that a group is covered?

The probability a group The probability a group is coveredis covered

•Let v be a vertex at level I in Let v be a vertex at level I in the tree, then the probability the tree, then the probability that after rounding there is a that after rounding there is a path from v to a vertex in g is path from v to a vertex in g is at least:at least:

fvg /((h-i+1)· xpar(v)v)

Let Let P(v)P(v) be this probability be this probability that the group is not that the group is not

coveredcovered• Let Let P(v)P(v) be the probability that there is no path be the probability that there is no path fromfrom v v to a leaf in group to a leaf in group gg. In the next . In the next inequalities a vertex inequalities a vertex v’v’ is always a child of is always a child of vv and and the corresponding edge is the corresponding edge is e=(v,v’)e=(v,v’). .

• P(v)=P(v)=ΠΠv’ v’ (1-(x(1-(xee· (1-p(v’))/x· (1-p(v’))/xpar(v)v par(v)v ))

• Explanation: Explanation: The probability for a group to get The probability for a group to get connected toconnected to v’v’ for some child for some child v’v’ of of v v is is (1-P(v’))(1-P(v’)). . Given that, the probability that the edge Given that, the probability that the edge (v,v’)(v,v’) gets selected is gets selected is xxee·/x·/xpar(v)v par(v)v . The multiplication is . The multiplication is because the events are independent for different because the events are independent for different children.children.

Proof continuedProof continued• (1-P(v’))(1-P(v’)) is the probability that is the probability that v’v’ can can

reach a leaf of reach a leaf of g g by a path after the by a path after the randomized process.randomized process.

• By the induction assumption:By the induction assumption: (1-P(v’)) (1-P(v’)) ≥≥ fgv’ /((h-i+1)· xpar(v)v) Therefore: Therefore:

P(v)≤P(v)≤ΠΠ(1-x(1-xee·· fgv’ /(xpar(v)v(h-i)·xe)= ΠΠ (1- (1-fgv’ /(xpar(v)v(h-i))

Proof continuedProof continued• We use the inequality We use the inequality 1-x≤exp(-x) 1-x≤exp(-x) to to

get the inequality:get the inequality: P(v) ≤ exp(-P(v) ≤ exp(- fgv’ /(xpar(v)v(h-i))• From the constrains of the LP From the constrains of the LP

we get:we get:• P(v)P(v) ≤exp( - ≤exp( -fgv/(xpar(v)v(h-i)))

Ending the proofEnding the proof

•Use the inequality exp(1/(1-x))≤1-1/x to get:

• P(v) ≤ 1- ffvgvg/((h-i-1) · /((h-i-1) · xpar(v)v)•This ends the proof.•We now only have to consider v=r

Proof continuedProof continued• For the root we may think ofFor the root we may think of

xpar(r)r=1•For the root frg=1 and thus the

probability that a group is covered is at least 1/(h+1). The probability that a group is not covered in (h+1)· ln p iterations is at most

•(1-1/(h+1))(h+1)·ln p exp(-ln p)=1/p

End of proof.End of proof.• Since a group is not covered with Since a group is not covered with

probability probability 1/p 1/p we can take every we can take every uncovered group and join it by a shortest uncovered group and join it by a shortest path to path to rr. A shortest path from any group . A shortest path from any group member to r is at most member to r is at most optopt..

• Thus the expected cost of this final stage is:Thus the expected cost of this final stage is: 1/p· p · opt=opt1/p· p · opt=opt• Thus the expected cost is Thus the expected cost is (h+1)ln p(h+1)ln p· ·

opt+optopt+opt

Making the Making the h=log nh=log n• QuestionQuestion: If the input for Group Steiner is a very : If the input for Group Steiner is a very

tall tree to begin with. How do we get tall tree to begin with. How do we get O(logO(log2 2 n)n) ratio?ratio?

• Use Use FRTFRT? Looses a ? Looses a log nlog n and complicated. and complicated.• Basic but probably not widely known: Basic but probably not widely known: Chekuri Chekuri

EvenEven and and KortsarzKortsarz show how to reduce the height show how to reduce the height of any tree to of any tree to log nlog n with a penalty with a penalty 88 on the cost. on the cost. Combinatorial!Combinatorial!

• In summary, we get an elementary analysis of In summary, we get an elementary analysis of O(log n· log p)O(log n· log p) approximation ratio for the Group approximation ratio for the Group Steiner on trees.Steiner on trees.

Recursive greedyRecursive greedy• Never taught. Never taught. Directed SteinerDirected Steiner, basic problem., basic problem.• A gem by A gem by CharikarCharikar et al. Say that the number of et al. Say that the number of

terminals to be covered is terminals to be covered is zz. There is a child . There is a child u u inin TT* * whose density is at most whose density is at most opt/zopt/z..

• Let Let z’z’ be the number of terminals in be the number of terminals in TT**uu

• The analysis stops once we cover at leastThe analysis stops once we cover at least z’/(h-1)z’/(h-1) terminals.terminals. Details omitted but gives telescopic Details omitted but gives telescopic multiplication that means density returned at multiplication that means density returned at most most hhopt/zopt/z..

• Can makeCan make h h O(1/O(1/) ) with ratio penaltywith ratio penalty n n1/1/

((ZelikovskyZelikovsky). Time: larger but in the ball park of ). Time: larger but in the ball park of nnhh = = nnO(1/O(1/))..

Alternative approximation Alternative approximation algorithm for Directed algorithm for Directed

SteinerSteiner• This was known (This was known (ChekuriChekuri told me) apparently in told me) apparently in

more complex form, since 1999.more complex form, since 1999.• The simpler way (as far as I know) The simpler way (as far as I know) Mendel and Mendel and

NutovNutov..• Create a graph Create a graph HH in which each path from the in which each path from the

root root rr to some terminal to some terminal uu of length at most of length at most 1/1/ ,, is a node.is a node.

• There is a directed edge betweenThere is a directed edge between p’ p’ andand p p if if pp extendsextends p’ p’ by one edgeby one edge..

• By the theorem of By the theorem of ZelikovskyZelikovsky, a solution of cost at , a solution of cost at mostmost O(n O(n 1/1/ )opt )opt isis embedded inembedded in HH..

A A non recursive greedynon recursive greedy approximation for Directed approximation for Directed

SteinerSteiner• For every terminal For every terminal tt, make a group , make a group HHtt of all of all

paths of length at most paths of length at most 1/1/ that start at that start at rr and and end at end at tt..

• This reduces the problem to This reduces the problem to Group SteinerGroup Steiner on on treestrees:: Connect at least one terminal of Connect at least one terminal of HHtt by a by a path frompath from r r , , for everyfor every t t . Our analysis works . Our analysis works and it’s a page and a half.and it’s a page and a half.

• This gives a non This gives a non Recursive GreedyRecursive Greedy algorithm algorithm of two pages for of two pages for Directed SteinerDirected Steiner with same with same ratio: ratio: nn. Only black box is the (very complex) . Only black box is the (very complex) height reduction height reduction CEK CEK and theand the Zelikovsky Zelikovsky theorem.theorem.

Certificate of failureCertificate of failure• Many papers say that: 'The value Many papers say that: 'The value optopt of of OPTOPT is KNOWN'. is KNOWN'.• Knowing Knowing optopt?? How can we know ?? How can we know optopt? Absurd. Means ? Absurd. Means P=NPP=NP..• I first saw this in a paper of I first saw this in a paper of HochbaumHochbaum and and ShmoysShmoys from from

J.ACM 1984. The paper is called: J.ACM 1984. The paper is called: Powers of graphs: A Powers of graphs: A powerful approximation technique for bottleneck problemspowerful approximation technique for bottleneck problems..

• Certificate of failure. TakeCertificate of failure. Take . If . If < <optopt the algorithm the algorithm maymay return a set of size return a set of size opt opt..

• Alternatively, may return Alternatively, may return failurefailure. In this case . In this case < <optopt. and . and then this hold true (this is why its certificate).then this hold true (this is why its certificate).

• If mu\geq opt returns alpha\cdot opt cost solution.If mu\geq opt returns alpha\cdot opt cost solution.

• Binary search. Get to mu and mu/2 with failure.Binary search. Get to mu and mu/2 with failure.• For mu, alpha\cdot opt cost solution.For mu, alpha\cdot opt cost solution.

Certificate of failureCertificate of failure• In case In case > >opt opt algorithm returnsalgorithm returns a a

solution of cost at most solution of cost at most opt opt.. • Binary search: fails for Binary search: fails for /2/2 but succeeds but succeeds

with with . . AsAs /2<opt/2<opt,, and return a solution of and return a solution of cost at most cost at most opt, opt, the ratio isthe ratio is 2 2

• Referees of my papers failed to Referees of my papers failed to understand that, many many times. understand that, many many times. Convention does not seem to be known Convention does not seem to be known to all. to all. Should be!Should be!

Density LP: useful and Density LP: useful and basicbasic

• Say that you have an Say that you have an LPLP for a covering for a covering problem that has some good ratio.problem that has some good ratio.

• But now you only want to cover But now you only want to cover kk of the of the elements. For every element elements. For every element xx, there will be , there will be a variable a variable yyxx that says how much that says how much xx is taken. is taken.

• We write We write yyxx=k =k but then divide the sum bybut then divide the sum by k k which means that the objective value is which means that the objective value is also divided byalso divided by k. k. Thus we try to solve a Thus we try to solve a densitydensity LPLP..

Density LPDensity LP• You can get the original ratio with You can get the original ratio with

penalty in the ratio of penalty in the ratio of O(log O(log 2 2 n) n) • Number of items inside the solution Number of items inside the solution

may be much more thanmay be much more than k k therefore if therefore if we can get exactlywe can get exactly k k may depend on may depend on the possibility of density the possibility of density decomposition.decomposition.

• I first was shown this (by I first was shown this (by ChekuriChekuri) ) about about 66 years ago. What do I not know years ago. What do I not know about about LPLP now? I fear that now? I fear that a lota lot..

Application of the basics, Application of the basics, example 1example 1

• Broadcast problem, directed graph, Steiner Broadcast problem, directed graph, Steiner set set SS..

• A vertex A vertex rr knows a message and the goal is to knows a message and the goal is to transmit it to all of transmit it to all of SS. Let . Let KK be the set that be the set that know the message andknow the message and N N those who don’t. At those who don’t. At every round a directed matching from every round a directed matching from KK to to NN..

• The endpoint in The endpoint in N N of the matching join of the matching join KK..• Minimize number of rounds.Minimize number of rounds.• Let Let k=|S|k=|S|. . Remark: Result obtained with Remark: Result obtained with ElkinElkin..

AlgorithmAlgorithm• Find Find uu that has at least that has at least sqrt{k} sqrt{k} terminals at terminals at

distance at most distance at most optopt from from uu..• Remove Remove TTu u with exactlywith exactly sqrt{k} sqrt{k} terminals terminals

fromfrom G G and height at mostand height at most optopt.. Let Let NN remaining vertices.remaining vertices.

• Iterate untill no such Iterate untill no such uu..• Let Let K’K’ be the union of trees, be the union of trees, RR be the roots. be the roots.

Clearly number of roots at most Clearly number of roots at most sqrt{k}sqrt{k}..• Can not employ recursion but can inform allCan not employ recursion but can inform all

K’K’ in in 2sqrt{k}+2opt 2sqrt{k}+2opt rounds.rounds.

To finish enough to inform To finish enough to inform distance distance optopt dominating dominating

set set DDNN• Cover Cover NNS S by trees rooted at by trees rooted at DD.. No No vertex in those trees has more thanvertex in those trees has more than sqrt{k} sqrt{k} terminals at distanceterminals at distance opt. opt. So So informing the rest ofinforming the rest of N N givengiven D D K K,, requires requires opt+sqrt {k} opt+sqrt {k} rounds.rounds.

• How do we inform a distance How do we inform a distance optopt dominating set?dominating set?

• Reduce to the minimization version of Reduce to the minimization version of maximazing a non-decreasing submodular maximazing a non-decreasing submodular function under partition Matroid.function under partition Matroid.

Define a new graphDefine a new graph

(k,n)

(k,n1)(k,n2) (k,np)

z p q z p q

n

opt

opt

opt

Finding Finding k<|S|k<|S|Arborescence Arborescence from from r r with minimum with minimum maximum outdegreemaximum outdegree

sqrt{k}

sqrt{k}

s’

t’

ktW

11

sqrt{k}

SolutionSolution• Solution obtained with Solution obtained with KhandekarKhandekar and and

NutovNutov..• Edges that carry flow and an arborescence Edges that carry flow and an arborescence

from from rr to to WW. . Flow(W)Flow(W) non-decreasing non-decreasing submodularsubmodular

• We prove there exists a size We prove there exists a size sqrt{k/sqrt{k/} feasible W. Non-trivial proof, omitted.

• The capacity of vertices and edges, divided by is also sqrt{k/sqrt{k/}.

• By the Wolsey theorem about sqrt{k/sqrt{k/} ratio approximation. The LP gap is sqrt{k}!

SummarySummary• It goes without saying that my opinions It goes without saying that my opinions

bound me only.bound me only.• My intention is not to change courses for My intention is not to change courses for

real. Will be presumptuous.real. Will be presumptuous. • Will I follow my own advice? Yes.Will I follow my own advice? Yes.• Can not only use the wonderful existing Can not only use the wonderful existing

slides.slides.• The little man always had to struggle in very The little man always had to struggle in very

difficult circumstances. difficult circumstances. • Thank you Thank you