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Approximation Algorithms

EPIT 2007. Approximation Algorithms. EPIT 2007. Independent Set. Instance: G=(V,E), k  N Question: Is there an independent set of size k in G (i.e. a subset V’  V such that no two vertices in V’ are joined by an edge) ? This problem is NP-Complete Instance: G=(V,E)

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Approximation Algorithms

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  1. EPIT 2007 Approximation Algorithms

  2. EPIT 2007 Independent Set Instance: G=(V,E), k N Question: Is there an independent set of size k in G (i.e. a subset V’ V such that no two vertices in V’ are joined by an edge) ? This problem is NP-Complete Instance: G=(V,E) Question: Find an independent set of maximal cardinality NP-hard

  3. EPIT 2007 Definition of Approximation Algorithms Find a «good solution» in polynomial-time What is a good solution? Let S be a solution given by a heuristic Let S* be an optimal solution The aim: S/S*≈1

  4. EPIT 2007 Another problem: vertex cover Instance: G=(V,E), k N Question: Find a set cover of edges with minimal size (i.e. a subset V’  V such that, for each edge {u,v}  E, at least one of u and v belongs to V’).

  5. EPIT 2007 These two problems seem to be equivalent independent set = vertex cover and vertex cover = independent set Max independent set ≈ Min vertex cover An existence of an independent set of size k <=> an existence of the vertex cover of size n-k.

  6. EPIT 2007 In fact these two problems are very different in term of approximability: there exists an algorithm for the vertex cover which gives a solution S to a factor two of a optimal solution S* at most => 2  S/S* No constant ratio exists for the independent set

  7. EPIT 2007 Definition An algorithm A is (n)-approximated If Max (S/S*,S*/S)  (n) For all instances of size n of the problem where S is the solution given by A and S* an optimal solution.

  8. EPIT 2007 FPTAS and PTAS Q  PTAS: >0, Asuch that A is (1+)-approximated. Moreover, if the complexity of A is polynomial in 1/, then Q  FPTAS.

  9. EPIT 2007 Some problems of PTAS Cover of points by minimum number of circles of radius r Knapsack problem

  10. EPIT 2007 APX and no-APX Q  APX:   R, A such that A is a -approximation algorithm Q  no-APX :   R, there exists no A such that A is -approximation algorithm

  11. EPIT 2007 Some problems APX and no-APX Vertex cover of a graph is APX Travelling salesman problem is APX Independent set of maximum cardinality is no-APX Vertices coloring is no-APX

  12. EPIT 2007 The world if N  NP No-APX APX PTAS FPTAS P

  13. EPIT 2007 An absolute approximation Algorithm The edge coloring of a graph: It exists a 4/3-approximated algorithm Vizing’s theorem : It is always possible to color the edges with (G)+1 colors

  14. EPIT 2007 An absolute approximation Algorithm This result is the best possible Indeed : the problem to know if a graph with (G)=3 can be colored with 3 colors is NP-complete. It is easy to see that an algorithm -approximated with <4/3 permits to solve this problem.

  15. EPIT 2007 An absolute approximation Algorithm Indeed if the solution is c<4 then c* ≤3, and the graph can be colored with 3 colors. If the solution is c4, since c/c*<4/3 we have c*>3 and the graph cannot be colored with 3 colors.

  16. EPIT 2007 The problem of the k-center Instance: Let V be a set of n sites, D be a matrix where dij is the distance between site i and site j, and let kN. Problem : Find S V, |S|=k which minimizes maxiV(dist(i,S)) with dist(i,S)=minjS(dij)

  17. EPIT 2007 Remark There exists no -approximated algorithm with  <2. Proof : If such an algorithm existed, it could be used to solve the problem of the existence of a dominating set of size k.

  18. EPIT 2007 Some definitions We call Gd=(V,Ed) the graph where {i,j} Ed if and only if d  dij . Let G=(V,E), we call G2=(V,E’) the graph such that {i,j} E’ if and only if either {i,j} E or k such that {i,k} E and {k,j} E.

  19. EPIT 2007 Here is a 2-approximated algorithm: Sort the dij : d1 < d2 < … < dm , i :=0 Repeat i := i+1 ; d := di Construct Gd2 Find a maximal stable S of Gd2 until k  |S| Take S as solution

  20. EPIT 2007 • Remark : • Let i0 be the value of the last i in the loop. Then all the sites are at distance at most 2di0 from S. • The optimal solution is at least di0. • Indeed Gd2 with d=di0-1, has a stable of cardinality at least k+1.

  21. EPIT 2007 Conclusion Thus the algorithm is a 2-approximated. That is the best possible result.

  22. EPIT 2007 Most of the scheduling problems are NP-Hard Take the following simple problem: Instance: Independent tasks (no precedence constraint), each task ti has a processing time pi Question: Scheduling these tasks on two processors while minimizing the length of the schedule

  23. EPIT 2007 This problem is NP-Hard (partition is a particular case). But there exists a FPTAS

  24. EPIT 2007 A intermediate problem Let be the sum problem: a list E of n integers, an integer t. The goal is to find a sublist E’  E which the sum of the integers in E’ is maximum but less than t (Find E’ E, such that max(∑ iE’ i) t)

  25. EPIT 2007 A pseudo-polynomial-time algorithm L0 = {0} L1 = {0, e1 } Li = Li-1 {x |y Li-1, x=y+ei, x t} The solution is expressed as y*= max (Ln) The complexity order of this algorithm is in O (n * min(t,2n))

  26. EPIT 2007 How can we reduce the size of the lists? The idea is to filter the list Example. Suppose that the two integers 45 and 47 are in a list Li. The sum which we would able to reach from 45 will be closed to their from 47. Thus 45 will represent 47

  27. EPIT 2007 How do we reduce the size of the lists ? Filtering (L, ) consists in suppressing all the integers y in L such that there exists an integer z in L with y  z (1- )y Example: L = {0, 1, 2, 4, 5, 11, 12, 20, 21, 23, 24, 25, 26} and =0.2

  28. EPIT 2007 How do we reduce the size of the lists ? Filtering (L, ) consists in suppressing all the integers y in L such that there exists an integer z in L with y  z (1- )y Example: L = {0, 1, 2, 4, 5, 11, 12, 20, 21, 23, 24, 25, 26} and =0.2. Then Filtering(L,)= {0,1,2, 4, 11, 20, 26}

  29. EPIT 2007 How do we reducethe size of the lists ? We are given the pseudo-polynomial algorithm, and in each step we filter with =/n. Let L’i be the intermediate lists. Then the approximated solution is y = max(L’n)

  30. EPIT 2007 The sizes of the lists are small L’i = {0, z1 , z2 , … zm+1} we have by construction : zi/ zi-1 > 1/(1-) and t  zm+1 andz1 1 Thus t  zm+1/z1 = i=2..m+1zi/ zi-1  (1/(1- ) )m

  31. EPIT 2007 The sizes of the lists are small Thus t  (1/(1- ) )m log (t)  - m * log (1- ) Finally m < n log(t)/  (because - log(1-x)>x)

  32. EPIT 2007 The sizes of the lists are small 1. m < n log(t)/  2. the size of the lists is in O(m) The algorithm becomes polynomial.

  33. EPIT 2007 The error is small We can prove by induction that : yLi , zL’i such that y z (1-)i y

  34. EPIT 2007 The error is small We can prove by induction that : yLi , zL’i such that y z (1-)i y This result is true for i=n and y=y*

  35. EPIT 2007 The error is small We can prove by induction that : yLi , zL’i such that y z (1-)i y This result is true for i=n and y=y*  zL’n z  (1-)n y*

  36. EPIT 2007 The error is small  zL’n z  (1-)n y* and y  z Conclusion: y / y*  (1-/n)n  1-

  37. EPIT 2007 Back to our problem Let B be the sum of the processing time of all the tasks. P1 P2  B/2 The solution S is B/2+  If pi becomes ei and t=B/2 : the solution S’ is B/2- 

  38. EPIT 2007 Back to our problem  B/2 The solution S is B/2+ , S* is B/2+ with  If pi becomes ei and t=B/2 : the solution S’ is B/2-  and the best solution S’* is B/2- It suffices to note that S/S*<S’*/S’ to show that our problem is FPTAS

  39. EPIT 2007 A non-APX problem Instance: Some tasks with the same processing times and some incompatibility constraints. Question: Find a optimal schedule satisfing the incompatibility constraints. Let G be the conflict graph where vertices represent tasks a scheduling ≈ vertex coloring. This problem is not APX

  40. EPIT 2007 An APX problem without PTAS Scheduling on a given number of processors Instance: Let G=(V,A) be a d.a.g. with the chronological constraints and let m be the number of processors Question: Minimize the length of the schedule

  41. EPIT 2007 No -approximated algorithm if <4/3 We are going to show that : Even if the graph is bipartite and the duration of all the tasks is equal to 1, knowing whether the graph can be scheduled in time 3 is an NP-complete problem The proof of this result suffices to conclude

  42. EPIT 2007 The proof The problem of well-balanced independent set in bipartite graphs is NP-complete Instance : G=(V1V2,E) a bipartite graph Question : does there exist an independent set S in G such that |S|=|V1| and |V1|=|V2| and |V1S|= |V2S|?

  43. EPIT 2007 W with |W|=n/2 X with |X|=n Y with |Y|=n T with |T|=n/2 The edges of G Complete Graph

  44. EPIT 2007 X with |X|=n W with |W|=n/2 Y with |Y|=n T with |T|=n/2 Scheduling this DAG in three steps on n processors is equivalent to showing the existence of a well-balanced independent set in the graph G.

  45. EPIT 2007 X with |X|=n W with |W|=n/2 m=n Y with |Y|=n T with |T|=n/2 Scheduling in 3 steps : no idle time

  46. EPIT 2007 X with |X|=n W with |W|=n/2 m=n Y with |Y|=n T with |T|=n/2 Scheduling in 3 steps : no idle time Step 1 : W and X1 ( X1 X, |X1|=n/2) must be executed.

  47. EPIT 2007 X with |X|=n W with |W|=n/2 m=n Y with |Y|=n T with |T|=n/2 Scheduling in 3 steps : no idle time Step 1 : W and X1 ( X1 X, |X1|=n/2) must be executed. Step 3 : T and Y1 ( Y1 Y, |Y1|=n/2) must be executed.

  48. EPIT 2007 X with |X|=n W with |W|=n/2 m=n Y with |Y|=n T with |T|=n/2 Scheduling in 3 steps : no idle time Step 1 : W and X1 ( X1 X, |X1|=n/2) must be executed. Step 2 : X2 and Y2 ( |X2|=|Y2|=n/2 ) must be executed. Step 3 : T and Y1 ( Y1 Y, |Y1|=n/2) must be executed.

  49. EPIT 2007 X with |X|=n W with |W|=n/2 m=n Y with |Y|=n T with |T|=n/2 Scheduling in 3 steps : no idle time Step 2 : X2 and Y2 ( |X2|=|Y2|=n/2 ) must be executed. X2 Y2 is a well balanced independent set in the graph G

  50. EPIT 2007 2-approximated list scheduling Graham’s Theorem : all lists scheduling is a 2-approximation for this problem. Proof: We can construct by induction a path in the d.a.g. such that at all the times, either a task in the path is executing or all the processors are working.

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