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What should be taught in approximation algorithms courses? Guy Kortsarz, Rutgers Camden. Advanced issues presented in many lecture notes and books:. Coloring a 3- colorable graph using vectors. Paper by Karger , Motwani and Sudan . Things a student needs to know:

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what should be taught in approximation algorithms courses guy kortsarz rutgers camden

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

advanced issues presented in many lecture notes and books
Advanced issues presented in many lecture notes and books:
  • Coloring a 3-colorable graph using vectors.
  • Paper by Karger, Motwani and Sudan.
  • Things a student needs to know:

Separation oracle for: A is PSD.

Getting a random vector inRn.

This is done by choosing the Normal

distributionat every entry.

Given unit vector v, v . ris normal


things a student needs to know
Things a student needs to know:
  • There is a choice of vectors vi for every i V so thatso that for every(i, j) E, vi · vj -1/2.
a student needs to know
A student needs to know:
  • S={i | r · vi }, threshold method, by now standard.
  • Sum of two normal distributions also normal.
  • Two inequalities (non trivial) about the normal distribution.
  • The above can be used to find a large independent set.
  • Combined with the greedy algorithm gives about n1/4 ratio approximation algorithm.
advanced methods are also required in the following topics often taught
Advanced methods are also required in the following topics often taught:
  • The seminal result of Jain. With the simplification of Nagarajan et. al. 2-ratio for Steiner Network.
  • The beautiful 3/2 ratio by Calinesco, Karloff and Rabani, for Multiway Cuts: geometric


  • Facharoenphol, Rao and Talwar, optimal random tree embedding. With this can get

O(log n)for undirected multicut.

how to teach sparsest cut
How to teach sparsest cut?
  • Many still teach the embedding of a metric into L1, withO(log n)distortion. By Lineal, London, Rabinovich.
  • Advantage: relatively simple.
  • The huge challenge posed by the Arora, Rao and Vazirani result. Unweighted sparsest cutsqrt{log n}
  • Teach the difficult lemma? Very advance. Very difficult.
  • A proof appears in the book of Shmoys and Williamson.
simpler topics
Simpler topics?
  • I can not complain if it is TAUGHT! Of course not. Let me give a list of basic topics that always taught
  • Ratio3/2 for TSP, the simple approximation of 2 for min cost Steiner tree.
  • Set-Cover , simple approximation ratio.
  • Knapsack,PTAS. Bin packing, constant ratio.
  • Set-Coverage. BUT:only costs 1.
knapsack set coverage
Knapsack Set-Coverage
  • The Set-Coverage problem is given a set system and a numberkselectk sets that cover as many elemnts as possible.
  • Knapsack version, not that known:
  • Each set has cost c(s) and there is a bound B on the maximum sum of costs, of sets we can choose.
  • Maximize number of elements covered.
result due to khuller moss and naor 1997 ipl
Result due to Khuller , Moss and,  Naor, 1997, IPL
  • The (1-1/e) ratio is possible.
  • In the usual algorithm & analysis (1-1/e) only follows if we can add the last set in the greedy choice. Thus, fails.
  • Because most times, adding the last set will give cost larger than B.
  • Trick: guess the 3 sets in OPTof least cost. Then apply greedy (don’t go over budget B).
why do i know this paper
Why do I know this paper?
  • I became aware of this result only several years after published. And only because I worked on Min Power Problems. No conference version!
  • This result seems absolutely basic to me. Why is it no taught?
  • Remark: Choosing one (least cost) element of OPT gives unbounded ratio. Choosing two sets of smallest cost gives ratio ½. Guessing the three sets of least cost and then greedy gives(1-1/e).
first general neglected topic
First general neglected topic
  • Important and not taught: Maximizing a submodular non-decreasing function under Matroid Constrains, ratio 1/2, Fischer, Nemhauser, Wolsey, 1977.
  • Improved in 2008(!) to best possible (1-1/e) by Vondrak in a brilliant paper.
first story a submission i refereed
First story: a submission I refereed
  • I got a paper to referee, and it was obvious that it is maximize Submodular function under Matroid constrains
  • If memory serves, the capacity 1, of the following Matroid: G(V,E), edge capacities, fix S  V. T reaches S if every vertex in Tcan send one unit of flow toS.
  • The set of all T that reach S a special Matroid called Gammoid. Everything in this paper, known!
  • Asked Chekuri (everybody must have an oracle) what is the Matroid, and Chekuri answered. Paper erased.
story 2 a worse outcome
Story 2: a worse outcome.
  • Problem. Input like Set-Cover but S= Si.
  • Required: choose at most one set of everySi 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 rounding
Related to pipage rounding
  • Due to Ageev, Sviridenko.
  • Dependent rounding, is a generalization of

Pipage rounding by Gandhi, Khuller, Parthasarathy, Srinivasan.

  • Say that we have an LP and a constraint

xi=k. RR can not derive exact equality.

  • Pipage Rounding : instead of going to a larger set of solutions like IP to LP, we replace the objective function.
the principals of pipage rounding
The principals of pipage rounding
  • We start with LP maximization with function L(X).
  • Define a non linear function F.
  • Show that the maximum of F is integral.
  • Show that integral points of F belong to the Polyhedra of L. Namely feasible for L as long as it is integral, and feasible for F.
the principals of pipage rounding1
The principals of pipage rounding
  • Then, show that F(Xint ) ≥L(X* )/, for

> 1.

  • HereXint is the (integral) optimumof F and

X*the optimum fractional solution for L

  • Because Xint is known to be feasible for L(x) due to its integrality, it is feasible for L and thus  approximation.
example max coverage
Example: Max Coverage
  • Max j wi zi


 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 f
The function F
  • F(x)=j wi (1-element j belongs to set i(1-xi) )
  • Definea function on a cycle.
  • As a function of .
  • The idea is to make plus  and then minus  over the cycle.
  • Make one entryonthe cycle smaller by  and another larger by .
the function f1
The function F
  • 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 entryonthe cycle smaller by  and another larger by .
  • The  appears as 2 in this term.
the function f2
The function F
  • 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 two by two
Changing the  two by two
  • Putting plus and minus  alternating along a cycle make at least one entry integral.
  • Moreover, we can decompose a cycle into two matching and there are two ways to increase and decrease by .
  • One direction of the two makes the function not smaller.
  • This implies that the optimum of F is integral.
thus the optimum of f is integral
Thus the optimum of F is integral
  • Its not hard to see that on integral vectors F and L have the same value.
  • Another inequality that is quite hard to prove is that: 1-i=1 to k (1-xi)≥(1-(1-1/k)k)L(X)
  • This gives a slightly better than 1-1/e ratio if k is small.
submodularity related to very basic technique
Submodularity: related to very basic technique.
  • f is submodular if f(A)+f(B)f(AB)+f(AB)
  • Makes a lot of difference if non-decreasing or not. If not, in my opinion represent concave.
  • If non-decreasing, brings us to the next lost simple subject: Submodular cover problems.
  • Input: U and submodular non-decreasing function f and cost c(u) per item u.
  • Required: a set S of minimum cost so that f(S)=f(U).
wolsey 1982 did much better
Wolsey , 1982, did much better
  • Each iteration pick item u so that helpu(S)/c(u) is maximum.
  • The ratio is max{uU}ln f(u)+1.
  • Example: For Set-Coverln|s|+1,s largest set.
  • Example: Same for Set-Cover with hard capacities. A paper in 1991, and one in 2002, did this result again (second was 20 years after Wolsey). Special case after 20 years! But its worse, yet.
  • Wolsey did better than that. Natural LP unbounded ratio even for Set-Cover with hard capacities.
  • Wolsey found a fabulous LP of gap max{uU}ln f(u)+1.
more general density
More general: density
  • Not taught at all but just cited. Why?
  • Here is a formal way:
  • Universe U and a function f: 2UR+
  • Each element in U has a cost c(u).
  • The function f not decreasing.
  • We want to find a minimum cost W so that f(W)=f(U).
  • We usually say, S U, c(S)=uS c(u)
  • But it works for an subadditive cost function
the density claim
The density claim
  • Say that we already created a set S via a greedy algorithm.
  • Now say that at any iteration we are able to find some Z so that:


  • 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 δ=1.
  • Say that the current set S has no intersection with the optimum.
  • Then if we add all of OPT to S we certainly get a feasible solution.
  • Then clearlyf(S+OPT)=f(U)
  • 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 samedensity as adding OPT.
proof continued
Proof continued
  • f(U) - j≤ i-1 f(Sj)≥ 1.
  • We may assume that the cost of every set added is at most

opt, therefore c(Sj ) ≤opt

  • Therefore it remains to bound:

 j≤ i-1 c(Zi)

Let us concentrate on what happens before

Si is added.

by the previous claims
By 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 mine
A paper of mine
  • Min c x subject to ABx b, with A positive entries and B flow matrix. Ratio logarithmic.
  • We got much more general results. The above I was sure then and sure now, KNOWN and presented as known.
  • Referees: Cite, or prove submodularity! We had to prove (referees did not agree its known!).
  • Example: gives log n for directed Source Location. Maybe first time stated but I considered it known.
  • This log n was proved at least 4 times since then.
  • The bad thing about these 4 papers is not that did not know our paper (to be expected) but that they would think such a simple result is NOTKNOWN.
  • It is good to know the result of Wolsey: for example, used it recently (Hajiaghayi ,Khandekar,K , Nutov) to give a lower bound of about log 2 n for a problem in fashion: Capacitated NetworkDesign (Steiner network with capacities). First lower bound for hard capacities.
dual fitting and a mistake we all make
Dual fitting and a mistake we all make
  • 1992. GK to Noga Alon:
  • This (spanner) result bares similarities to the proof done by Lovats for set-cover.
  • Noga Alon (seems very unhappy, maybe angry): Give me a break! That is folklore. Lovats told me he wrote it so he would have something to cite.......
  • Everybody cites Lovats here. Its simply not true.
  • We don’t know the basics. Result known many years before 1975.
  • Should we cite folklore? Yes!
how to teach dual fitting for set cover unweighted
HOW to teach dual fitting for set cover, unweighted?
  • Let S be the collection of sets and Tthe elements.
  • The dual, costs 1: MaximizetT yt
  • Subject to: ts yt  c(s)=1
  • We define a dual: if the greedy chose a star of length i, each element in the set gets 1/i






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





















primal dual of gw
Primal Dual of GW
  • Goemans and Williamson gave a rather well known Primal-Dual algorithm. Always taught, and should be.
  • A question I asked quite several researchers and I don’t remember a correct response: Why reverse delete?
  • Why not Michael Jackson?
  • GW primal dual imitates recursion.
  • In LR reverse delete follows from recusrsion.
local ratio for covering problems
Local Ratio for covering problems
  • Give weights to items so that every minimal. solution is a  approximation. Reduce items costs by weights chosen.
  • Elements of cost 0 enter the solution.
  • Make minimal.
  • Recurse.
  • No need for reverse delete. Recursion implies it.
  • Simpler for Steiner Network in my opinion.
local ratio
Local Ratio
  • Without it I don’t think we could find a ratio 2 for Vertex feedback set.
  • A recent result of K, Langberg, Nutov. Minor result (main results are different) but solves an open problem of a very smart person: Krivelevich.
  • Covering triangles, gap2for LP (polynomial).
  • Open problem: tight?
  • Not only we showed tight family but showed as hard as approximatingVC.Used LR in proof.
group steiner problem on trees
Group Steiner problem on 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 konjevod and ravi
The theorem of Garg Konjevod 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 lp
The 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 xeare 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 gkr
The rounding method of GKR
  • Consider xe and say that its parent edge is (par(v),v)
  • Independently for every e, add it to the solution with probability xe/xpar(v)v
  • We show that the expected cost is bounded by the LP cost.
  • The probability that an edge gets to the root

is a telescopic multiplication.

the probability that an edge is chosen
The probability that an edge is chosen
  • All terms cancel but the first and the last. The First is xe. The last is the flow from ‘ The parent of r to r ’ which we may assume is 1.
  • Since this is the case, xe contributes

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

fvg /((h-i+1)· xpar(v)v)

let p v be this probability that the group is not covered
Let P(v) be this probability that the group is not covered
  • Let P(v) be the probability that there is no path from v to a leaf in group g. In the next inequalities a vertex v’ is always a child of v and the corresponding edge is e=(v,v’).
  • P(v)=Πv’ (1-(xe· (1-p(v’))/xpar(v)v )
  • Explanation: The probability for a group to get connected tov’ for some child v’ of v is (1-P(v’)). Given that, the probability that the edge (v,v’) gets selected is xe·/xpar(v)v . The multiplication is because the events are independent for different children.
proof continued1
Proof continued
  • (1-P(v’)) is the probability that v’ can reach a leaf of g by a path after the randomized process.
  • By the induction assumption:

(1-P(v’)) ≥fgv’ /((h-i+1)· xpar(v)v)


P(v)≤Π(1-xe·fgv’ /(xpar(v)v(h-i)·xe)=

Π (1-fgv’ /(xpar(v)v(h-i))

proof continued2
Proof continued
  • We use the inequality 1-x≤exp(-x) to get the inequality:

P(v) ≤ exp(- fgv’ /(xpar(v)v(h-i))

  • From the constrains of the LP we get:
  • P(v) ≤exp( -fgv/(xpar(v)v(h-i)))
ending the proof
Ending the proof
  • Use the inequality exp(1/(1-x))≤1-1/x to get:
  • P(v)≤ 1- fvg/((h-i-1) · xpar(v)v)
  • This ends the proof.
  • We now only have to consider v=r
proof continued3
Proof continued
  • For the root we may think ofxpar(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 probability 1/p we can take every uncovered group and join it by a shortest path to r. A shortest path from any group member to r is at most opt.
  • Thus the expected cost of this final stage is:

1/p· p · opt=opt

  • Thus the expected cost is (h+1)ln p· opt+opt
making the h log n
Making the h=log n
  • Question: If the input for Group Steiner is a very tall tree to begin with. How do we get O(log2 n) ratio?
  • Use FRT? Looses a log n and complicated.
  • Basic but probably not widely known: Chekuri Even and Kortsarz show how to reduce the height of any tree to log n with a penalty 8 on the cost. Combinatorial!
  • In summary, we get an elementary analysis of

O(log n· log p) approximation ratio for the Group

Steiner on trees.

recursive greedy
Recursive greedy
  • Never taught. Directed Steiner, basic problem.
  • A gem by Charikar et al. Say that the number of terminals to be covered is z. There is a child u inT* whose density is at most opt/z.
  • Let z’be the number of terminals in T*u
  • The analysis stops once we cover at leastz’/(h-1)terminals.Details omitted but gives telescopic multiplication that means density returned at most hopt/z.
  • Can make h O(1/) with ratio penalty n1/ (Zelikovsky). Time: larger but in the ball park of nh = nO(1/).
alternative approximation algorithm for directed steiner
Alternative approximation algorithm for Directed Steiner
  • This was known (Chekuri told me) apparently in more complex form, since 1999.
  • The simpler way (as far as I know) Mendel and Nutov.
  • Create a graph H in which each path from the root r to some terminal u of length at most 1/,is a node.
  • There is a directed edge between p’ andp if pextendsp’ by one edge.
  • By the theorem of Zelikovsky, a solution of cost at mostO(n 1/ )opt isembedded inH.
a non recursive greedy approximation for directed steiner
A non recursive greedy approximation for Directed Steiner
  • For every terminal t, make a group Ht of all paths of length at most 1/ that start at r and end at t.
  • This reduces the problem to Group Steineron trees:Connect at least one terminal of Htby a path fromr , for every t . Our analysis works and it’s a page and a half.
  • This gives a non Recursive Greedy algorithm of two pages for Directed Steiner with same ratio: n. Only black box is the (very complex) height reduction CEK and the Zelikovsky theorem.
certificate of failure
Certificate of failure
  • Many papers say that: 'The value opt of OPT is KNOWN'.
  • Knowing opt?? How can we know opt? Absurd. Means P=NP.
  • I first saw this in a paper of Hochbaum and Shmoys from J.ACM 1984. The paper is called: Powers of graphs: A powerful approximation technique for bottleneck problems.
  • Certificate of failure. Take. If  <opt the algorithm may return a set of size  opt.
  • Alternatively, may return failure. In this case  <opt. and then this hold true (this is why its certificate).
  • If mu\geq opt returns alpha\cdot opt cost solution.
  • Binary search. Get to mu and mu/2 with failure.
  • For mu, alpha\cdot opt cost solution.
certificate of failure1
Certificate of failure
  • In case  >opt algorithm returnsa solution of cost at most  opt.
  • Binary search: fails for /2but succeeds with  . As /2
  • Referees of my papers failed to understand that, many many times. Convention does not seem to be known to all. Should be!
density lp useful and basic
Density LP: useful and basic
  • Say that you have an LP for a covering problem that has some good ratio.
  • But now you only want to cover k of the elements. For every element x, there will be a variable yx that says how much x is taken.
  • We write yx=k but then divide the sum by k which means that the objective value is also divided by k. Thus we try to solve a densityLP.
density lp
Density LP
  • You can get the original ratio with penalty in the ratio of O(log 2 n)
  • Number of items inside the solution may be much more than k therefore if we can get exactly k may depend on the possibility of density decomposition.
  • I first was shown this (by Chekuri) about 6 years ago. What do I not know about LP now? I fear that a lot.
application of the basics example 1
Application of the basics, example 1
  • Broadcast problem, directed graph, Steiner set S.
  • A vertex r knows a message and the goal is to transmit it to all of S. Let K be the set that know the message and N those who don’t. At every round a directed matching from K to N.
  • The endpoint in N of the matching join K.
  • Minimize number of rounds.
  • Let k=|S|. Remark: Result obtained with Elkin.
  • Find u that has at least sqrt{k} terminals at distance at most opt from u.
  • Remove Tu with exactly sqrt{k} terminals from G and height at mostopt.Let N remaining vertices.
  • Iterate untill no such u.
  • Let K’ be the union of trees, R be the roots. Clearly number of roots at most sqrt{k}.
  • Can not employ recursion but can inform allK’ in 2sqrt{k}+2opt rounds.
to finish enough to inform distance opt dominating set d n
To finish enough to inform distance opt dominating set DN
  • Cover NS by trees rooted at D.No vertex in those trees has more than sqrt{k} terminals at distance opt. So informing the rest of N given D K,requires opt+sqrt {k} rounds.
  • How do we inform a distance opt dominating set?
  • Reduce to the minimization version of maximazing a non-decreasing submodular function under partition Matroid.
define a new graph
Define a new graph















finding k s arborescence from r with minimum maximum outdegree
Finding k<|S|Arborescence from r with minimum maximum outdegree











  • Solution obtained with Khandekar and Nutov.
  • Edges that carry flow and an arborescence from r to W. Flow(W) non-decreasing submodular
  • We prove there exists a size sqrt{k/} feasible W. Non-trivial proof, omitted.
  • The capacity of vertices and edges, divided by is alsosqrt{k/}.
  • By the Wolsey theorem about sqrt{k/} ratio approximation.The LP gap is sqrt{k}!
  • It goes without saying that my opinions bound me only.
  • My intention is not to change courses for real. Will be presumptuous.
  • Will I follow my own advice? Yes.
  • Can not only use the wonderful existing slides.
  • The little man always had to struggle in very difficult circumstances.
  • Thank you