1. 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. 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. 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. 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 t i, and  an inflation factor  i, i.e. cost in scenario i, c i (e)=  i ¢c(e)

5. 5 Demand-Robust Min-Cut Problem Goal :  first stage edges E f and  second stage edges E s i, 8 scenario i s.t. E f [E s i separates r from t i Objective : min [c(E f ) + max i  i ¢c(E s i )]

6. 6 An Example Root vertex r Four possible scenarios  terminal in scenario i is t i Inflation factor = 3 for all scenarios 1.5 1 r t1t1 t2t2 t3t3 t4t4 1 22 v1v1 v2v2

7. 7 A Feasible Solution Buy edges (r,v 1 ) and (r,v 2 ) in the first stage. First stage cost = 1.5+1.5 = 3 1.5 1 r t1t1 t2t2 t3t3 t4t4 1 22 v1v1 v2v2

8. 8 A Feasible Solution Buy edges (r,v 1 ) and (r,v 2 ) in the first stage. First stage cost = 1.5+1.5 = 3 1 r t1t1 t2t2 t3t3 t4t4 1 22 v1v1 v2v2

9. 9 A Feasible Solution Buy edges (r,v 1 ) and (r,v 2 ) in the first stage. First stage cost = 1.5+1.5 = 3 Maximum second stage cost is 3. Total cost = 3 + 3 = 6 1 r t1t1 t2t2 t3t3 t4t4 1 22 v1v1 v2v2

10. 10 Another Feasible Solution Buy nothing today. First stage cost = 0 Maximum second stage cost is 4.5. Total cost = 0 + 4.5 = 4.5 1.5 1 r t1t1 t2t2 t3t3 t4t4 1 22 v1v1 v2v2

11. 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. 12 Structural Theorem [DGRS’05] There exists a first stage solution E f 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. 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. 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. 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. 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. 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. 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. 19 Analysis Fix an optimal solution, say edges E f, E s 1, E s 2,…, E s k 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(E f ). R: set of terminals whose individual min-cut cost is more than 2C/ 

20. 20 Analysis Remove edges E f from G and consider the Gomory-Hu tree H on GnE f.

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

22. 22 Analysis Root H at the root vertex r. Consider terminals which do not have any parent terminals. t1t1 t3t3 t2t2 r

23. 23 Analysis Consider min-cuts from r corresponding to these maximal terminals t1t1 t2t2 r t3t3

24. 24 Analysis t1t1 t2t2 r t3t3 Es1Es1 titi v G n E f

25. 25 Analysis t1t1 t2t2 r t3t3 Es1Es1 titi v G

26. 26 Analysis t1t1 t2t2 r t3t3 Es1Es1 titi v G Blue edges are edges of E f We know r-t 1 min-cut in G has cost more than 2C/  Also, c(E s 1 ) · C/  Thus, c(blue edges) >C/ 

27. 27 Analysis t1t1 t2t2 r t3t3 Es1Es1 titi v G Thus, we can buy E s 1 and charge against the blue edges.

28. 28 Analysis t1t1 t2t2 r t3t3 Es1Es1 titi v G Similarily, we can buy edges E s 2 and E s 3 and charge against corresponding blue edges. Es2Es2 Es3Es3

29. 29 Analysis t1t1 t2t2 r t3t3 Es1Es1 titi v G Similarily, we can buy edges E s 2 and E s 3 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. Es2Es2 Es3Es3

30. 30 Analysis t1t1 t2t2 r t3t3 Es1Es1 titi v G Thus, c(red edges) · 2 c(E f ) Hence, we have a cut separating all terminals in R from root and has cost at most 3c(E f ) Es2Es2 Es3Es3

31. 31 Analysis t1t1 t2t2 r t3t3 Es1Es1 titi v G In fact, we can construct a separating cut of cost at most 2c(E f ). Thus, we get a 2- approximation for the problem. Es2Es2 Es3Es3

32. 32 Stochastic Min-Cut Each scenario i has an associated probability p i 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. 33 Natural IP formulation Consider the following IP formulation,  x f (e) is the first stage variable and  x s i (e) is the second stage variable for scenario i min  e2 E c(e)¢ x f (e) +  i=1 k p i  i ¢  e2 E c(e)¢ x s i (e) (x f +x s i )(P) ¸ 1, 8 r-t i paths P, 8 i x f, x s i 2 {0,1} 8 e, i

34. 34 MIP Formulation Let  i be the cost of root to t i min-cut in G with respect to cost c. Consider the following MIP:  x(e) is the first stage variable and  y i denotes whether terminal i is cut in first stage or not. min  e2 E c(e)¢ x(e) +  i=1 k p i  i ¢ ( 1- y i )¢  i d x(e) (r,t i ) ¸ y i, 8 i=1,…,k y i 2 {0,1}, 8 i=1,…,k 0 · x(e) · 1, 8 e2 E

35. 35 MIP Formulation min  e2 E c(e)¢ x(e) +  i=1 k p i  I ¢ (1-y i ) ¢  i d x(e) (r,t i ) ¸ y i, 8 i=1,…,k y i 2 {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. 36 MIP Feasible Solution Consider an optimal solution, say E f, E s 1,…,E s k. Thus, OPT = c(E f ) +  i=1 k p i  i ¢ c(E s i ). Let E f i be the edges that scenario i uses from first stage. If c(E s i ) ¸ c(E f i ), then let y i =0 (we can buy a min-cut for t i in the second stage for at most twice the cost that OPT pays for E s i ). For the remaining terminals, let y i =1. We show that all the terminals in R can be separated from root by a cut of cost at most 2c(E f ).

37. 37 Analysis Remove edges E f from G and consider the Gomory- Hu tree H on GnE f. Root H at the root vertex r.

38. 38 Analysis t1t1 t2t2 r t3t3 Es1Es1 titi v G n E f

39. 39 Analysis t1t1 t2t2 r t3t3 Es1Es1 titi v G Blue edges are edges of E f 1 We know c(E f 1 ) > c(E s 1 ) Thus, we can buy red edges in the first charging against the blue edges.

40. 40 Analysis t1t1 t2t2 r t3t3 Es1Es1 titi v 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(E f ). Es2Es2 Es3Es3

41. 41 MIP is 2-approximate min  e2 E c(e)¢ x(e) +  i=1 k p i  i ¢(1-y i )¢  i d x(e) (r,t i ) ¸ y i, 8 i=1,…,k y i 2 {0,1}, 8 i=1,…,k 0 · x(e) · 1, 8 e2 E Thus, opt(MIP) is at most 2OPT.

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

43. 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. 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. 45 Thank You

46. 46 Warm-up (Exact Algorithm for Trees) Let  =2. Here, C=3. r t1t1 t2t2 t3t3 t4t4 2 3 6 4 6 5 8 1 2 4 2 1.5 v

47. 47 Warm-up (Exact Algorithm for Trees) Let  =2. Here, optimal C=3. We can ignore t 3 and t 4. r t1t1 t2t2 t3t3 t4t4 2 3 6 4 6 5 8 1 2 4 2 1.5 v

48. 48 Warm-up (Exact Algorithm for Trees) Let  =2. Here, optimal C=3. We can ignore t 3 and t 4. r t1t1 t2t2 t3t3 t4t4 2 3 6 4 6 5 8 1 2 4 2 1.5 v

49. 49 Warm-up (Exact Algorithm for Trees) Let  =2. Here, optimal C=3. We can ignore t 3 and t 4. Terminals t 1 and t 2 need first stage help ) t 1 and t 2 need to be cut in the first stage. Cost = c(r,v) + 3 = 9 (optimal). r t1t1 t2t2 t3t3 t4t4 2 3 6 4 6 5 8 1 2 4 2 1.5 v