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Approximation Algorithms for Demand-Robust and Stochastic Min-Cut Problems

Approximation Algorithms for Demand-Robust and Stochastic Min-Cut Problems. Vineet Goyal Carnegie Mellon University Based on, [Golovin, G, Ravi] (STACS’06) [G, Gupta] (Manuscript’06). Why Robust?. Uncertainty in problem data and constraints for optimization.

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Approximation Algorithms for Demand-Robust and Stochastic Min-Cut Problems

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  1. Approximation Algorithms for Demand-Robust and Stochastic Min-Cut Problems Vineet Goyal Carnegie Mellon University Based on, [Golovin, G, Ravi] (STACS’06) [G, Gupta] (Manuscript’06)

  2. Why Robust? • Uncertainty in problem data and constraints for optimization. • Optimize for the worst case realization of uncertainties. • Stochastic optimization models uncertainty but does not address worst-case future.

  3. Demand-Robust Model • Two stage model motivated from two-stage stochastic optimization (IKMM’04, RS’04, GPRS’04, SS’04). • Models uncertainty in problem constraints or demands as well as the problem data.

  4. Demand-Robust Min-Cut Problem • Given an undirected graph G, a cost function c on edges, a root vertex r and a set of k future scenarios • Scenario i specifies • a terminal ti, and • an inflation factor i, i.e. cost in scenario i, ci(e)= i¢c(e)

  5. Demand-Robust Min-Cut Problem • Goal : • first stage edges Ef and • second stage edges Esi, 8 scenario i s.t. Ef[Esi separates r from ti • Objective : min [c(Ef) + maxii¢c(Esi)]

  6. Root vertex r Four possible scenarios terminal in scenario i is ti Inflation factor = 3 for all scenarios An Example r 1 1.5 1 1.5 v1 v2 t3 t4 2 2 t1 t2

  7. Buy edges (r,v1) and (r,v2) in the first stage. First stage cost = 1.5+1.5 = 3 A Feasible Solution r 1 1.5 1 1.5 v1 v2 t3 t4 2 2 t1 t2

  8. Buy edges (r,v1) and (r,v2) in the first stage. First stage cost = 1.5+1.5 = 3 A Feasible Solution r 1 1 v1 v2 t3 t4 2 2 t1 t2

  9. Buy edges (r,v1) and (r,v2) in the first stage. First stage cost = 1.5+1.5 = 3 Maximum second stage cost is 3. Total cost = 3 + 3 = 6 A Feasible Solution r 1 1 v1 v2 t3 t4 2 2 t1 t2

  10. Buy nothing today. First stage cost = 0 Maximum second stage cost is 4.5. Total cost = 0 + 4.5 = 4.5 Another Feasible Solution r 1 1.5 1 1.5 v1 v2 t3 t4 2 2 t1 t2

  11. Previous Work • Two stage model of demand-robustness was introduced in [Dhamdhere, G, Ravi, Singh] (FOCS’05). • Prove a structural result for the first stage solution for a general covering problem. • Give approximation algorithms for min-cut, multicut, Steiner tree and facility location problems in the demand-robust model.

  12. Structural Theorem [DGRS’05] • There exists a first stage solution Ef which is a minimal solution for a subset of scenarios and it can be extended in the second stage to obtain a 2-approximate solution to the problem.

  13. Results [DGRS’05] • O(log n)-approximation for robust min-cut, • O(log n¢ loglog n)-approximation for robust multicut, • O(1)-approximation for Steiner tree and facility location problems. • Algorithms for robust min-cut (multicut) extend to stochastic min-cut (multicut) to give similar guarantees

  14. Improved Results • 2-approximation for the robust min-cut problem [Golovin, G, Ravi] (STACS’06). • 4-approximation for the stochastic min-cut problem [G, Gupta] (Manuscript’06).

  15. Robust Min-Cut: 2-approximation For simplicity, we assume same inflation factor  for all scenarios. • Basic Idea: Suppose the maximum second stage cost in some optimal solution is C • Scenarios for which the individual best solution costs at most C/ (with respect to cost c) can be ignored: “they do not need any first stage help”. • Rest of the scenarios, OPT must help in the first stage.

  16. Warm-up (Exact Algorithm for Trees) • Suppose the maximum second stage cost in some optimal solution is C. • Ignore terminals for which the min-cut costs less than C/i (we can cut in second stage). • Separate the remaining terminals from the root by a minimum cost cut in the first stage. 2

  17. Algorithm (General case) • Guess the second stage cost (say C) of some optimal solution. • Ignore terminals for which the individual min-cut costs at most 2C/ (with respect to cost c). • Separate the rest of the terminals (say R) from root in the first stage by a minimum cost cut.

  18. Analysis • Clearly, maximum second stage cost incurred by our algorithm is at most 2C. • We need to prove that the minimum cut separating root from all the terminals in R is at most twice the first stage cost in OPT.

  19. Analysis • Fix an optimal solution, say edges Ef, Es1, Es2,…, Esk such that the maximum second stage cost is C. • We will construct a cut separating root from the terminals in R such that its cost is at most 2c(Ef). R: set of terminals whose individual min-cut cost is more than 2C/

  20. Analysis • Remove edges Ef from G and consider the Gomory-Hu tree H on GnEf .

  21. Analysis • Root H at the root vertex r.

  22. Analysis • Root H at the root vertex r. • Consider terminals which do not have any parent terminals. r t1 t3 t2

  23. Analysis r • Consider min-cuts from r corresponding to these maximal terminals t1 t3 t2

  24. Analysis Gn Ef r Es1 t1 t3 t2 ti v

  25. Analysis G r Es1 t1 t3 t2 ti v

  26. Analysis G • Blue edges are edges of Ef • We know r-t1 min-cut in G has cost more than 2C/ • Also, c(Es1) · C/ • Thus, • c(blue edges) >C/ r Es1 t1 t3 t2 ti v

  27. Analysis G • Thus, we can buy Es1 and charge against the blue edges. r Es1 t1 t3 t2 ti v

  28. Analysis G • Similarily, we can buy edges Es2 and Es3 and charge against corresponding blue edges. r Es1 t1 Es3 t3 Es2 t2 ti v

  29. Analysis G • Similarily, we can buy edges Es2 and Es3 and charge against corresponding blue edges. • Red edges and blue edges form the final first stage cut. • Note that each blue edge is charged at most twice. r Es1 t1 Es3 t3 Es2 t2 ti v

  30. Analysis G • Thus, • c(red edges) · 2 c(Ef) • Hence, we have a cut separating all terminals in R from root and has cost at most 3c(Ef) r Es1 t1 Es3 t3 Es2 t2 ti v

  31. Analysis G • In fact, we can construct a separating cut of cost at most 2c(Ef). • Thus, we get a 2-approximation for the problem. r Es1 t1 Es3 t3 Es2 t2 ti v

  32. Stochastic Min-Cut • Each scenario i has an associated probability pi and the objective is to minimize the expected cost of the solution (instead of maximum over all scenarios). • We obtain a 4-approximation for the stochastic min-cut problem.

  33. Natural IP formulation • Consider the following IP formulation, • xf(e) is the first stage variable and • xsi(e) is the second stage variable for scenario i min e2 E c(e)¢ xf(e) + i=1k pii¢e2 E c(e)¢ xsi(e) (xf+xsi)(P) ¸ 1, 8 r-ti paths P, 8 i xf, xsi 2 {0,1} 8 e, i

  34. MIP Formulation • Let i be the cost of root to ti min-cut in G with respect to cost c. Consider the following MIP: • x(e) is the first stage variable and • yi denotes whether terminal i is cut in first stage or not. min e2 E c(e)¢ x(e) + i=1k pii¢(1-yi)¢i dx(e)(r,ti) ¸ yi, 8 i=1,…,k yi2 {0,1}, 8 i=1,…,k 0 · x(e) · 1, 8 e2 E

  35. MIP Formulation min e2 E c(e)¢ x(e) + i=1k piI ¢ (1-yi) ¢ i dx(e)(r,ti) ¸ yi, 8 i=1,…,k yi2 {0,1}, 8 i=1,…,k 0 · x(e) · 1, 8 e2 E • Note that this is an approximate MIP. • We can show a feasible solution of cost at most 2OPT.

  36. MIP Feasible Solution • Consider an optimal solution, say Ef, Es1,…,Esk. Thus, OPT = c(Ef) + i=1k pii ¢ c(Esi). • Let Efi be the edges that scenario i uses from first stage. • If c(Esi) ¸ c(Efi), then let yi=0 (we can buy a min-cut for ti in the second stage for at most twice the cost that OPT pays for Esi). • For the remaining terminals, let yi=1. We show that all the terminals in R can be separated from root by a cut of cost at most 2c(Ef).

  37. Analysis • Remove edges Ef from G and consider the Gomory-Hu tree H on GnEf . • Root H at the root vertex r.

  38. Analysis Gn Ef r Es1 t1 t3 t2 ti v

  39. Analysis G r • Blue edges are edges of Ef1 • We know c(Ef1) > c(Es1) • Thus, we can buy red edges in the first charging against the blue edges. Es1 t1 t3 t2 ti v

  40. Analysis G • Red edges and blue edges form the final cut. • From the previous argument, we know that the cost of the final first stage cut · 2c(Ef). r Es1 t1 Es3 t3 Es2 t2 ti v

  41. MIP is 2-approximate min e2 E c(e)¢ x(e) + i=1k pii¢(1-yi)¢i dx(e)(r,ti) ¸ yi, 8 i=1,…,k yi2 {0,1}, 8 i=1,…,k 0 · x(e) · 1, 8 e2 E • Thus, opt(MIP) is at most 2OPT.

  42. LP Relaxation : Rounding min e2 Ec(e)¢ x(e) + i=1k pii¢(1-yi)¢i dx(e)(r,ti) ¸ yi, 8 i=1,…,k 0 · yi· 1, 8 i=1,…,k 0 · x(e) · 1, 8 e2 E • If yi < ½ , then ignore ti in the first stage. • Separate rest of the terminals from root in the first stage by a minimum cut.

  43. Rounding the LP • We can round the LP within a factor of 2. Therefore, we obtain a 4-approximation for the stochastic min-cut problem.

  44. Conclusions and Open Problems • Our technique “guess and plan” crucially exploits the structure of the demand-robust problem: every scenario in the second stage can pay up to the maximum second stage cost of OPT without worsening the solution cost. • Can we use these ideas for the multi-cut and Steiner tree problems in the demand-robust model? • Hardness of demand-robust and stochastic min-cut on undirected graphs is unknown. • Natural LP relaxation has an integrality gap. • Directed versions are NP-hard.

  45. Thank You

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