Approximation Algorithms

Load Balancing k-center selection Pricing Method Vertex Cover Set Cover Bin Packing TSP

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Approximation Algorithms

Q. Suppose I need to solve an NP-hard problem. What should I do?A. Theory says you're unlikely to find a poly-time algorithm.

Must sacrifice one of three desired features. Solve problem to optimality. Solve problem in poly-time. Solve arbitrary instances of the problem.

-approximation algorithm. Guaranteed to run in poly-time. Guaranteed to solve arbitrary instance of the problem Guaranteed to find solution within ratio of true optimum.

Challenge. Need to prove a solution's value is close to optimum, without even knowing what optimum value is!

11.1 Load Balancing

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Load Balancing

Input. m identical machines; n jobs, job j has processing time tj. Job j must run contiguously on one machine. A machine can process at most one job at a time.

Def. Let J(i) be the subset of jobs assigned to machine i. Theload of machine i is Li = j J(i) tj.

Def. The makespan is the maximum load on any machine L = maxi Li.

Load balancing. Assign each job to a machine to minimize makespan.

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List-scheduling algorithm. Consider n jobs in some fixed order. Assign job j to machine whose load is smallest so far.

Implementation. O(n log n) using a priority queue.

Load Balancing: List Scheduling

List-Scheduling(m, n, t1,t2,…,tn) { for i = 1 to m { Li 0 J(i) }

for j = 1 to n { i = argmink Lk

J(i) J(i) {j} Li Li + tj

}}

jobs assigned to machine i

load on machine i

machine i has smallest load

assign job j to machine i

update load of machine i

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Load Balancing: List Scheduling Analysis

Theorem. [Graham, 1966] Greedy algorithm is a 2-approximation. First worst-case analysis of an approximation algorithm. Need to compare resulting solution with optimal makespan L*.

Lemma 1. The optimal makespan L* maxj tj.

Pf. Some machine must process the most time-consuming job. ▪

Lemma 2. The optimal makespan Pf.

The total processing time is j tj . One of m machines must do at least a 1/m fraction of total

work. ▪

L * 1m t jj .

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Load Balancing: List Scheduling Analysis

Theorem. Greedy algorithm is a 2-approximation.Pf. Consider load Li of bottleneck machine i.

Let j be last job scheduled on machine i. When job j assigned to machine i, i had smallest load. Its load

before assignment is Li - tj Li - tj Lk for all 1 k m.

j

0L = LiLi - tj

machine i

blue jobs scheduled before j

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Load Balancing: List Scheduling Analysis

Theorem. Greedy algorithm is a 2-approximation.Pf. Consider load Li of bottleneck machine i.

Let j be last job scheduled on machine i. When job j assigned to machine i, i had smallest load. Its load

before assignment is Li - tj Li - tj Lk for all 1 k m. Sum inequalities over all k and divide by m:

(correct the second eqn. to j) Now ▪

L i t j 1m Lkk

1m tkk

L *

Li (Li t j ) L*

t j

L*

2L *.

Lemma 2

Lemma 1

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Load Balancing: List Scheduling Analysis

Q. Is our analysis tight?A. Essentially yes.

Ex: m machines, m(m-1) jobs length 1 jobs, one job of length m

machine 2 idle

machine 3 idle

machine 4 idle

machine 5 idle

machine 6 idle

machine 7 idle

machine 8 idle

machine 9 idle

machine 10 idle

list scheduling makespan = 19

m = 10

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Load Balancing: List Scheduling Analysis

Q. Is our analysis tight?A. Essentially yes.

Ex: m machines, m(m-1) jobs length 1 jobs, one job of length m

m = 10

optimal makespan = 10

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Load Balancing: LPT Rule

Longest processing time (LPT). Sort n jobs in descending order of processing time, and then run list scheduling algorithm.

LPT-List-Scheduling(m, n, t1,t2,…,tn) {

Sort jobs so that t1 ≥ t2 ≥ … ≥ tn

for i = 1 to m { Li 0 J(i) }

for j = 1 to n { i = argmink Lk

J(i) J(i) {j} Li Li + tj

}}

jobs assigned to machine i

load on machine i

machine i has smallest load

assign job j to machine i

update load of machine i

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Load Balancing: LPT Rule

Observation. If at most m jobs, then list-scheduling is optimal.Pf. Each job put on its own machine. ▪

Lemma 3. If there are more than m jobs, L* 2 tm+1.

Pf. Consider first m+1 jobs t1, …, tm+1. Since the ti's are in descending order, each takes at least tm+1

time. There are m+1 jobs and m machines, so by pigeonhole

principle, at least one machine gets two jobs. ▪ tj <= t(m+1) <= ½ L*

Theorem. LPT rule is a 3/2 approximation algorithm.Pf. Same basic approach as for list scheduling.

▪

L i (Li t j ) L*

t j

12L*

3

2 L *.

Lemma 3( by observation, can assume number of jobs > m )

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Load Balancing: LPT Rule

Q. Is our 3/2 analysis tight?A. No.

Theorem. [Graham, 1969] LPT rule is a 4/3-approximation.Pf. More sophisticated analysis of same algorithm.

Q. Is Graham's 4/3 analysis tight?A. Essentially yes.

Ex: m machines, n = 2m+1 jobs, 2 jobs of length m+1, m+2, …, 2m-1 and one job of length m.

11.2 Center Selection

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center

r(C)

Center Selection Problem

Input. Set of n sites s1, …, sn.

Center selection problem. Select k centers C so that maximum distance from a site to nearest center is minimized.

site

k = 4

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Center Selection Problem

Input. Set of n sites s1, …, sn.

Center selection problem. Select k centers C so that maximum distance from a site to nearest center is minimized.

Notation. dist(x, y) = distance between x and y. dist(si, C) = min c C dist(si, c) = distance from si to closest

center. r(C) = maxi dist(si, C) = smallest covering radius.

Goal. Find set of centers C that minimizes r(C), subject to |C| = k.

Distance function properties. dist(x, x) = 0 (identity) dist(x, y) = dist(y, x) (symmetry) dist(x, y) dist(x, z) + dist(z, y) (triangle inequality)

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centersite

Center Selection Example

Ex: each site is a point in the plane, a center can be any point in the plane, dist(x, y) = Euclidean distance.

Remark: search can be infinite!

r(C)

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Greedy Algorithm: A False Start

Greedy algorithm. Put the first center at the best possible location for a single center, and then keep adding centers so as to reduce the covering radius each time by as much as possible.

Remark: arbitrarily bad!

greedy center 1

k = 2 centers sitecenter

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Center Selection: Greedy Algorithm

Greedy algorithm. Repeatedly choose the next center to be the site farthest from any existing center.

Observation. Upon termination all centers in C are pairwise at least r(C) apart.Pf. By construction of algorithm.

Greedy-Center-Selection(k, n, s1,s2,…,sn) {

C = repeat k times { Select a site si with maximum dist(si, C)

Add si to C

} return C}

site farthest from any center

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Center Selection: Analysis of Greedy Algorithm

Theorem. Let C* be an optimal set of centers. Then r(C) 2r(C*).Pf. (by contradiction) Assume r(C*) < ½ r(C).

For each site ci in C, consider ball of radius ½ r(C) around it. Exactly one ci* in each ball; let ci be the site paired with ci*. Consider any site s and its closest center ci* in C*. dist(s, C) dist(s, ci) dist(s, ci*) + dist(ci*, ci) 2r(C*). Thus r(C) 2r(C*). ▪

C*sites

½ r(C)

ci

ci*s

r(C*) since ci* is closest center

½ r(C)

½ r(C)

-inequality

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Center Selection

Theorem. Let C* be an optimal set of centers. Then r(C) 2r(C*).

Theorem. Greedy algorithm is a 2-approximation for center selection problem.

Remark. Greedy algorithm always places centers at sites, but is still within a factor of 2 of best solution that is allowed to place centers anywhere.

Question. Is there hope of a 3/2-approximation? 4/3?

e.g., points in the plane

Theorem. Unless P = NP, there no -approximation for center-selectionproblem for any < 2.

11.4 The Pricing Method: Vertex Cover

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Weighted Vertex Cover

Weighted vertex cover. Given a graph G with vertex weights, find a vertex cover of minimum weight.

4

9

2

2

4

9

2

2

weight = 2 + 2 + 4 weight = 9

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Weighted Vertex Cover

Pricing method. Each edge must be covered by some vertex i. Edge e pays price pe 0 to use vertex i.

Fairness. Edges incident to vertex i should pay wi in total.

Claim. For any vertex cover S and any fair prices pe: e pe

w(S).

Proof. ▪

4

9

2

2

sum fairness inequalitiesfor each node in S

each edge e covered byat least one node in S

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Pricing Method

Pricing method. Set prices and find vertex cover simultaneously.

Weighted-Vertex-Cover-Approx(G, w) { foreach e in E pe = 0

while ( edge i-j such that neither i nor j are tight) select such an edge e increase pe without violating fairness }

S set of all tight nodes return S}

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Pricing Method

vertex weight

Figure 11.8

price of edge a-b

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Pricing Method: Analysis

Theorem. Pricing method is a 2-approximation.Pf.

Algorithm terminates since at least one new node becomes tight after each iteration of while loop.

Let S = set of all tight nodes upon termination of algorithm. S is a vertex cover: if some edge i-j is uncovered, then neither i nor j is tight. But then while loop would not terminate.

Let S* be optimal vertex cover. We show w(S) 2w(S*).

all nodes in S are tight S V,prices 0

fairness lemmaeach edge counted twice

Extra Slides

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Machine 2

Machine 1a d f

b c e g

yes

Load Balancing on 2 Machines

Claim. Load balancing is hard even if only 2 machines.Pf. NUMBER-PARTITIONING P LOAD-BALANCE.

a d

f

b c

ge

length of job f

Time L0

machine 1

machine 2

NP-complete by Exercise 8.26

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Center Selection: Hardness of Approximation

Theorem. Unless P = NP, there is no -approximation algorithm formetric k-center problem for any < 2.

Pf. We show how we could use a (2 - ) approximation algorithm for k-center to solve DOMINATING-SET in poly-time.

Let G = (V, E), k be an instance of DOMINATING-SET. Construct instance G' of k-center with sites V and distances

– d(u, v) = 2 if (u, v) E– d(u, v) = 1 if (u, v) E

Note that G' satisfies the triangle inequality. Claim: G has dominating set of size k iff there exists k centers

C* with r(C*) = 1. Thus, if G has a dominating set of size k, a (2 - )-

approximation algorithm on G' must find a solution C* with r(C*) = 1 since it cannot use any edge of distance 2.

see Exercise 8.29

Vertex Cover Approximation

A vertex cover is a subset of vertices such that every edge in the graph is incident to at least one of these vertices.

The vertex cover optimization problem is to find a vertex cover of minimum size.

For a good strategy, a heuristic is needed

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Vertex Cover

Consider an arbitrary edge (u, v) in the graph. One of its two vertices must be in the cover, but we do not know which one.

The idea of this heuristic is to simply put both vertices into the vertex cover.

Then we remove all edges that are incident to u and v (since they are now all covered), and recurse on the remaining edges.

For every one vertex that must be in the cover, we put two into our cover, so it is easy to see that the cover we generate is at most twice the size of the optimum cover.

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Proof of aprroximation ratio

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Claim : approx VC yields a factor-2 approximationProof: Consider the set C output by ApproxVC. Let C* be the optimum VC. Let A be the set of edges selected by the line marked with “(*)” in the figure. Observe that the size of C is exactly 2|A|because we add two vertices for each such edge. However note that in the optimum VC one of these two vertices must have been added to the VC, and thus the size of C* is at least |A|. Thus we have:

|C| ---- = |A| <= |C*| 2Therefore :

|C|---- <= 2|C*|

Example

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Approximate VC Algorithm : Naive Approach

ApproxVC {

C = empty-set while (E is nonempty) do { (*) let (u,v) be any edge of E add both u and v to C remove from E all edges incident to either u or v

} return C; }

Can we improve on it ?

Why not consider vertices with higher degrees first (Greedy Strategy)

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Greedy VC

Greedy Approximation for VC GreedyVC(G=(V,E)) {

C = empty-set; while (E is nonempty) do {

let u be the vertex of maximum degree in G; add u to C; remove from E all edges incident to u;

}

return C; }

For the example, it yields the optimum solution

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Greedy VC Example

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Can we prove Greedy VC outperforms the other one ? NO !

It can even perform poorly than it.

However, it should also be pointed out that the vertex cover constructed by the greedy heuristic is (for typical graphs) smaller than that one computed by the 2-for-1 heuristic, so it would probably be wise to run both algorithms and take the better of the two.

Third Attempt: Use Matching

A matching is a subset of edges that have no vertices in common

A matching is maximal if no more edges can be added to it.

Maximal matchings will help us find good vertex covers, and moreover, they are easy to generate: repeatedly pick edges that are disjoint from the ones chosen already, until this is no longer possible.

Any vertex cover of a graph G must be at least as large as the number of edges in any matching in G; that is, any matching provides a lower bound on OPT. This is simply because each edge of the matching must be covered by one of its endpoints in any vertex cover!

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Example

Figure below shows how to convert from Maximal Matching to Vertex Cover

a) A matching b) Completion to MaxMatch c) Its VC

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Vertex Cover from Matching

let S be a set that contains both endpoints of each edge in a maximal matching M.

Then S must be a vertex cover—if it isn’t, that is, if it

doesn’t touch some edge e ∪ E, then M could not possibly be maximal since we could still add e to it. But our cover S has 2|M| vertices

We know that any vertex cover must have size at least |M|.

Algorithm :

Find a maximal matching M ∞ E Return S = {all endpoints of edges in M}

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Vertex cover from Matching

This simple procedure always returns a vertex cover whose size is at most twice optimal!

In summary, even though we have no way of finding the best vertex cover, we can easily find another structure, a maximal matching, with two key properties:

1. Its size gives us a lower bound on the optimal vertex

cover. 2. It can be used to build a vertex cover, whose size can

be related to that of the optimal cover using property 1.

Alpha <= 2

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Set Cover Problem Revisited

Given a pair (X,F) where X ={x1,x2,...,xm} is a finite set (a domain of elements) and F ={S1,S2,...,Sn} is a family of subsets of X, such that every element of X belongs to at least one set of F.

For C ∈ F. (This is a collection of sets over X.) We say

that C covers the domain if every element of X is in some set of C

The problem is to find the minimum-sized subset C of F that covers X.

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Set Cover

Vertex Cover is a type of set cover problem. The domain to be covered are the edges, and each vertex covers the subset of incident edges.

Decision-problem formulation of set cover (“does there exist a set cover of size at most k?”) is NP-complete

There is a factor-2 approximation for the vertex cover

problem, but it cannot be applied to generate a factor 2 approximation for set cover.

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Set Cover

It is known that there is no constant factor approximation to the set cover problem

There is however the greedy heuristic, which achieves an approximation bound of ln m, where m = |X|, the size of the underlying domain, we will leave the proof.

A simple greedy approach to set cover works by at each

stage selecting the set that covers the greatest number of “uncovered” elements

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Set Cover : The Approx. Algorithm

Greedy-Set-Cover(X, F)

{ U = X // U are the items to be covered C = empty // C will be the sets in the cover

while (U is nonempty) { // there is someone left to cover

select S in F that covers the most elements of U addS to C U=U-S

} return C

}

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Set Cover : Bad Example

The optimal set cover consists of sets S5 and S6, each of size 16. Initially all three sets S1, S5, and S6 have 16 elements. If ties are broken in the worst possible way, the greedy algorithm will first select set S1. We remove all the covered elements. Now S2, S5 and S6 all cover 8 of the remaining elements. Again, if we choose poorly, S2 is chosen. The pattern repeats, choosing S3 (size 4), S4 (size 2) and finally S5 and S6 (each of size 1).

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Bin Packing

Bin packing is another well-known NP-complete problem, which is a variant of the knapsack problem

Given a set of n objects, where si denotes the size of the ith object (0 < si < 1. for simplification) , put objects into bins

Size of a bin is 1 at max.

Use fewest bins as possible

Ex : Fit object sto a truck etc.

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Bin Packing : Example

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Bin Packing : Approximation Factor

Theorem: The first-fit heuristic achieves a ratio bound of 2.

Proof: Consider an instance {s1,...,sn} of the bin packing problem. Let S =Σ i si denote the sum of all the

object sizes. Let b* denote the optimal number of bins, and bff denote the number of bins used by first-fit.

b* >= S : since no bin can hold a total capacity of more than 1 unit, and even if we were to fill each bin exactly to its capacity, we would need at least S bins

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Bin Packing : Analysis

We claim that bff <= 2S. To see this, let ti denote the total size of the objects that

first-fit puts into bin i. Consider bins i and i + 1 filled by first-fit. Assume that

indexing is cyclical, so if i is the last index (i = bff ) then

i+1 = 1. We claim that ti + ti +1 > = 1. If not, then the contents of

bins i and i + 1 could both be put into the same bin, and hence first-fit would never have started to fill the second bin, preferring to keep everything in the first bin. Thus we have:

bffΣi=1 ( ti + ti+1 ) > = bff

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Bin Packing : Analysis

But this sum adds up all the elements twice, so it has a total value of 2S. Thus we have 2S >= bff .

Combining this with the fact that b* >= S we have :

bff <= 2S <= 2b* showing bff /b* <= 2 as required

best fit attempts put the object into the bin in which it fits most closely with the available space (approx. ratio 17/10)

first fit decreasing, in which the objects are first sorted in decreasing order of size (approx. ratio 11/9)

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Traveling Salesman Problem (TSP)

In the TSP, given a complete undirected graph with nonnegative edge weights,

Find a cycle that visits all vertices and is of minimumcost. (NP-Complete)

Distances should satisfy the triangle inequality

for all u, v, w → c(u, w) <= c(u, v)+c(v, w) (c(u,v) = cost on edge uv or cost of shortest path)

There is an approx. Algorithm forTSP with a ratio of 2 (the tour that it produces cannot be worse than twice the cost of the optimal tour)

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TSP Observations

A TSP with one edge removed is a spanning tree (not necessarily a minimum spanning tree)

Therefore, the cost of the minimum TSP tour is at least as large as the cost of the MST.

MST can be computed efficiently, using, for example, either Kruskal’s or Prim’s algorithm

If we can find some way to convert the MST into a TSP tour while increasing its cost by at most a constant factor, then we will have an approximation for TSP.

We will see that if the edge weights satisfy the triangle inequality, then this is possible.

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TSP MST

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Given any free tree there is a tour of the tree called a twice around tour that traverses the edges of the tree twice, once in each direction. The figure below shows an example.

MST twice round tour Short Cut Optimal TSP

TSP

This path is not simple because it revisits vertices, but we can make it simple by short-cutting, that is, we skip over previously visited vertices

the final order in which vertices are visited using the short-cuts is exactly the same as a preorder traversal of the MST

The triangle inequality assures us that the path length will not increase when we take short-cuts.

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Approximate Algorithm for TSP

ApproxTSP(G=(V,E)) {

T = minimum spanning tree for G r = any vertex L = list of vertices visited by a preorder walk of T

starting with r

return L }

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Approx.TSP Analysis

Claim: Approx-TSP has a ratio bound of 2. Proof: Let H denote the tour produced by this algorithm

and let H* be the optimum tour. Let T be the minimum spanning tree.

We can remove any edge of H* resulting in a spanning tree, and since T is the minimum cost spanning tree we have

c(T) <= c(H*). Twice around tour of T has cost 2c(T), since every edge

in T is hit twice. By the triangle inequality, when we short-cut an edge of T to form H we do not increase the cost of the tour, and so we have

– c(H) <= 2c(T). Combining these we have c(H) /2 <= c(T) <= c(H*) , therefore c(H) / c(H*) <= 2.

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Graph Partitioning

Input: An undirected graph G = (V;E) with nonnegative edge weights; a real number α ϵ (0, 1/2].

Output: A partition of the vertices into two groups A and B, each of size at least α *|V|

Goal: Minimize the capacity of the cut (A,B).

Applications : from circuit layout to program analysis to image segmentation.

Graph Partitioning is NP –Hard

Removing the restriction on the sizes of A and B would give the MINIMUM CUT problem, which we know to be efficiently solvable using flow techniques.

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Acknowledgements

The last few algorithms are dependent on David Mount’c 451 Course, University of Waterloo

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