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: Separation oracle for: A is PSD. Getting a random vector in Rn. This is done by choosing the Normal distribution at every entry. Given unit vector v, v. r is normal distribution.
Things a student needs to know: • There is a choice of vectors vi for every i V so that for every (i, j) E, vi · vj -1/2.
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 n 1/4 ratio approximation algorithm.
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 embeddings. • Facharoenphol, Rao and Talwar, optimal random tree embedding. With this can get O(log n) for undirected multicut.
How to teach sparsest cut? • Many still teach the embedding of a metric into L 1 , with O(log n) distortion. By Lineal, London, Rabinovich. • Advantage: relatively simple. • The huge challenge posed by the Arora, Rao and Vazirani result. Unweighted sparsest cut sqrt{log n} • Teach the difficult lemma? Very advance. Very difficult. • A proof appears in the book of Shmoys and Williamson.
Simpler topics? • I can not complain if it is TAUGHT! Of course not. Let me give a list of basic topics that always taught • Ratio 3/2 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 • The Set-Coverage problem is given a set system and a number k select k 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 • 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 OPT of least cost. Then apply greedy (don’t go over budget B).
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 • 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 • 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 T can send one unit of flow to S. • 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. • Problem. Input like Set-Cover but S= Si. • Required: choose at most one set of every Si 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 • 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 • 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 rounding • Then, show that F(Xint ) ≥L(X* )/ , for > 1. • Here Xint is the (integral) optimum of 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 • 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 F 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 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 entry on the cycle smaller by and another larger by . • The appears as 2 in this term.
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 x 2 between -4 and 3. • The maximum is in the border -4.
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 • 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. • 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 • 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-Cover ln|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 • • Not taught at all but just cited. Why? Here is a formal way: Universe U and a function f: 2 U 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 • 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
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 clearly f(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 same density as adding OPT.
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 • 1 ≤ f(U)-f(Z 1+Z 2 +……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 • 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.
Remarks • 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 NOT KNOWN. • 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 Network Design (Steiner network with capacities). First lower bound for hard capacities.
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? • Let S be the collection of sets and T the 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. 2. 2. 2 . 2
The bound on the sum of elements of a given set 1/7 1/5 1/3 1/12 1/8 1/7 1/6 1/7 1/4 1/11 1/10 1/9 1/12 1/2 1
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 • 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 • 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, gap 2 for LP (polynomial). • Open problem: tight? • Not only we showed tight family but showed as hard as approximating VC. Used LR in proof.
Group Steiner problem on trees • Group Steiner problem on trees. • Input : An undirected weighted rooted by r tree T = (V; E) and subsets S 1, ……, 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 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 • 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 • 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 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 • 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 • 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 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 to v’ 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 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) Therefore: P(v)≤Π(1 -xe· fgv’ /(xpar(v)v(h-i)·xe)= Π (1 -fgv’ /(xpar(v)v(h-i)) (1 -
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 • 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 continued For 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. • 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 • Question: If the input for Group Steiner is a very tall tree to begin with. How do we get O(log 2 n) ratio? • Use FRT? Looses a log n and complicated. • Basic but probably not widely known: Chekuri Even and Kortsarz show 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 • 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 in T* whose density is at most opt/z. • Let z’ be the number of terminals in T*u • The analysis stops once we cover at least z’/(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 n 1/ (Zelikovsky). Time: larger but in the ball park of nh = n. O(1/ ).
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’ and p if p extends p’ by one edge. • By theorem of Zelikovsky, a solution of cost at most O(n 1/ )opt is embedded in H.
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 Steiner on trees: Connect at least one terminal of Ht by a path from r , 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 • 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).
Certificate of failure • In case > opt algorithm returns a solution of cost at most opt. • Binary search: fails for /2 but 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 times. Convention does not seem to be known to all. Should be!
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 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 • 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.
Algorithm • 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 most opt. 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 all K’ in 2 sqrt{k}+2 opt rounds.
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 (k, n) (k, n 1) (k, n 2) z p q (k, np) z p q opt opt n
Finding k<|S|Arborescence from r with minimum maximum outdegree s’ sqrt{k } W sqrt{k } t’ k 1 1 t
Solution • 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 also sqrt{k/ }. • By the Wolsey theorem about sqrt{k/ } ratio approximation. The LP gap is sqrt{k}!
Summary • • • 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
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