1. © B.Raghavachari & J.Veerasamy, UTD 1 Euler tours, postman tours and mixed graphs Jeyakesavan Veerasamy* * Joint work with Balaji Raghavachari Samsung Telecommunications America, Inc. and The University of Texas at Dallas veerasam@utdallas.edu

2. © B.Raghavachari & J.Veerasamy, UTD 2 Approximation Algorithms An algorithm for solving a minimization problem that –runs in polynomial time and outputs a feasible solution S –performance ratio = Cost(S) / Cost(Opt) In practice, if problem is NP-hard (intractable), –we can apply exact algorithms (if time permits), or –apply heuristic algorithms and approximation algorithms, and select best solution obtained. Critical part of approximation algorithm analysis is lower- bound analysis of optimal solution Such lower bounds can be used in practice to speed up exact algorithms such as branch and bound, and evaluate performance of heuristic algorithms.

3. © B.Raghavachari & J.Veerasamy, UTD 3 Euler Tour Problem Find a tour that traverses all edges exactly once. Even degree node A InputOutput A graph is Eulerian if and only if it is connected and the degree of each vertex is even (Euler, 1736!)

4. © B.Raghavachari & J.Veerasamy, UTD 4 Non-Eulerian graph A B Even degree node Odd degree node Walk starting at A got stuck at B

5. © B.Raghavachari & J.Veerasamy, UTD 5 Directed Euler Tour Problem A Balanced node Find a tour in a directed graph that traverses all edges exactly once. A directed graph is Eulerian if and only if it is strongly connected and indegree = outdegree at each vertex

6. © B.Raghavachari & J.Veerasamy, UTD 6 Non-Eulerian Directed Graph A B Balanced node Deficit node Surplus node Walk starting at A got stuck at B

7. © B.Raghavachari & J.Veerasamy, UTD 7 Chinese Postman Problem (CPP) Find shortest tour visiting all edges at least once. Even-degree node Odd-degree node A InputAugmentedOutput Add a minimum-weight T-join which has odd degree at odd-degree nodes and even-degree at even-degree nodes

8. © B.Raghavachari & J.Veerasamy, UTD 8 Directed CPP Balanced node Deficit node Surplus node InputOutput Add a minimum-weight subgraph which has outdegree = indegree + surplus at surplus nodes indegree = outdegree + deficit at deficit nodes

9. © B.Raghavachari & J.Veerasamy, UTD 9 Directed CPP: Another example Balanced node Deficit node Surplus node Output Input

10. © B.Raghavachari & J.Veerasamy, UTD 10 Mixed Postman Problem (MPP) Balanced node Unbalanced node Solution 1Solution 2 Includes both undirected and directed edges Input NP-hard!

11. © B.Raghavachari & J.Veerasamy, UTD 11 Applications: City Management Elm Commerce Main Jackson Houston Street cleaning, Mail delivery, Snow removal, Courier delivery route

12. © B.Raghavachari & J.Veerasamy, UTD 12 Previous work Edmonds, Johnson, 1973 Frederickson, 1979: –Designed two approximation algorithms –Select the best of the two solutions computed –Worst-case performance ratio = 5/3 Christofides, 1984 Ralphs, 1993: Linear Programming formulation Nobert, Picard, 1996: Applied LP-based solution to schedule snow removal in Montreal Raghavachari, Veerasamy, 1998: –Improved Frederickson’s algorithm for MPP –Performance ratio = 3/2

13. © B.Raghavachari & J.Veerasamy, UTD 13 Properties of Eulerian Graphs Even-degree condition: degree of each node is even Balanced set condition: For any cut (S, V \ S), the difference between number of directed edges crossing cut in each direction is at most the number of undirected edges crossing cut Eulerian mixed graphs can be detected in polynomial time SV \ S Given a graph satisfying even-degree condition, Frederickson showed how to find minimum-cost augmentation to make it Eulerian.

14. © B.Raghavachari & J.Veerasamy, UTD 14 Even-degree mixed graphs Solve flow problem and obtain IN = OUT at every node Consider undirected edges and duplicated edges only Traverse cycles formed by these edges and correct parity Balanced node Unbalanced node

15. © B.Raghavachari & J.Veerasamy, UTD 15 Input G Frederickson’s Algorithm (Part 1) Match Flow H Even parity Solution S 1 C(H)  C(Opt) C(S 1 )  C(Opt) + 2C(D) Balanced node Unbalanced node

16. © B.Raghavachari & J.Veerasamy, UTD 16 Frederickson’s Algorithm (Part 2) Input G FlowDouble C(Opt)  C(D) + C(U) C(S 2 )  C(D) + 2C(U)  2C(Opt) - C(D) Solution S 2 Solution S = min(S 1, S 2 ) C(S)  5/3 C(Opt) D U

17. © B.Raghavachari & J.Veerasamy, UTD 17 Remarks Input G Flow IN=OUT Directed edges D Undirected edges U Cost(Opt)  C(D) + C(U) D may contain any number of copies of edges in G. H is obtained from G by adding a T-join of odd-degree nodes, ignoring the directions. If H-G contains only edges of D: C(S 1 )  2C(D) + C(U) C(S)  (3/2) C(Opt) Flow augmentation cost C IO (G) = C(D) + C(U) - C(G)

18. © B.Raghavachari & J.Veerasamy, UTD 18 Idea for Improvement Modify computation of S 1 as follows: –When H is computed, make cost of edges in D to be 0. –T-join prefers edges of D to U. Problem: We may be forced to duplicate some edges of U. Components of D may have odd-number of odd-degree nodes

19. © B.Raghavachari & J.Veerasamy, UTD 19 Improved lower bound Shrink each directed component to a single node, and compute T-join between odd degree components. X, subset of U, forms the T-join. Optimal solution also has to pay C(X) to satisfy even degree condition. If we can prove that adding X to G does not decrease Flow augmentation cost, then C(Opt)  C(D) + C(U) + C(X)

20. © B.Raghavachari & J.Veerasamy, UTD 20 Lower bound proof Input GNew input H INOUT solution Can C IO (H) < C IO (G)? No. In fact, adding any edge in U does not decrease INOUT augmentation cost. Alternating cycle x

21. © B.Raghavachari & J.Veerasamy, UTD 21 Our Mixed algorithm (part 2) Input G FlowDouble C(Opt)  C(D) + C(U) C(S 2 )  C(D) + 2C(U)  2C(Opt) - C(D) Solution S 2 D U

22. © B.Raghavachari & J.Veerasamy, UTD 22 Input G Our Mixed Algorithm (part 1) Match Flow H Even parity Solution S 1 C(S 1 )  C(U) + C(X) + 2C(D)  C(Opt) + C(D) Balanced node Unbalanced node Solution S = min(S 1, S 2 ) C(S)  3/2 C(Opt) X

23. © B.Raghavachari & J.Veerasamy, UTD 23 Performance of Algorithms C(D) Cost of solution C(Opt) 2C(Opt) C(new S 1 ) C(S 2 ) C(S 1 ) C(Opt)

24. © B.Raghavachari & J.Veerasamy, UTD 24 Concluding Remarks Worst case performance ratio for our algorithm is 1.5 (Analysis is tight). Windy Postman Problem: Asymmetric version of MPP LP relaxation of WPP is half-integral - gives 2 approximation algorithm. Recently we have improved ratio of WPP to 1.5 Open: Improvements? Planar Graphs? U V C(U,V)  C(V,U)

25. © B.Raghavachari & J.Veerasamy, UTD 25 Further Research Explore MPP with additional constraints: –duplicate only undirected edges –duplicate only directed edges Explore shortest path problem in mixed graphs when negative costs are allowed. Implement a few heuristic algorithms and analyze the performance.