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

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

- 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

distribution.

- There is a choice of vectors vi for every i V so thatso that for every(i, j) E, vi · vj -1/2.

- 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.

- 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
embeddings.

- Facharoenphol, Rao and Talwar, optimal random tree embedding. With this can get
O(log n)for undirected multicut.

- 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.

- 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.

- 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.

- 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).

- 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).

- 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.

- 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.

- 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.

- 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.

- 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.

- 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.

- 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.

- 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 .

- 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.

- 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.

- 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.

- 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.

- f is submodular if f(A)+f(B)f(AB)+f(AB)
- 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).

- Each iteration pick item u so that helpu(S)/c(u) is maximum.
- The ratio is max{uU}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{uU}ln f(u)+1.

- Not taught at all but just cited. Why?
- Here is a formal way:
- Universe U and a function f: 2UR+
- 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)=uS c(u)
- But it works for an subadditive cost function

- 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:
(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

- 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.

- 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.

- 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.

- Min c x subject to ABx 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.

- 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!

- Let S be the collection of sets and Tthe elements.
- The dual, costs 1: MaximizetT yt
- Subject to: ts 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

.2

.2

.2

.2

.2

1/2

1/2

1

1/7

1/6

1/7

1/5

1/7

1/4

1/4

1/3

1/12

1/12

1/11

1/12

1/10

1/9

1/12

1/8

1/12

- 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.

- 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.

- 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.
- 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.

- 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.

- 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.

- 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.

- 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?

- 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 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.

- (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)

Therefore:

P(v)≤Π(1-xe·fgv’ /(xpar(v)v(h-i)·xe)=

Π (1-fgv’ /(xpar(v)v(h-i))

- 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)))

- 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

- 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

- 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

- 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.

- 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 hopt/z.
- Can make h O(1/) with ratio penalty n1/ (Zelikovsky). Time: larger but in the ball park of nh = nO(1/).

- 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.

- 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.

- 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.

- In case >opt algorithm returnsa solution of cost at most opt.
- Binary search: fails for /2but succeeds with . As /2<opt,and return a solution of cost at most opt, the ratio is 2
- Referees of my papers failed to understand that, many many times. Convention does not seem to be known to all. Should be!

- 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.

- 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.

- 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.

- Cover NS 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.

(k,n)

(k,n1)

(k,n2)

(k,np)

z

p

q

p

z

q

opt

opt

opt

n

s’

sqrt{k}

sqrt{k}

t

W

k

t’

sqrt{k}

1

1

- 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