© B.Raghavachari & J.Veerasamy, UTD 1 Approximation Algorithms for Postman Problems Jeyakesavan Veerasamy* * Joint work withBalaji Raghavachari Samsung Telecommunications America, Inc. and The University of Texas atDallas

© B.Raghavachari & J.Veerasamy, UTD 2 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!)

© B.Raghavachari & J.Veerasamy, UTD 3 Non-Eulerian graph A B Even degree node Odd degree node Walk starting at A got stuck at B

© B.Raghavachari & J.Veerasamy, UTD 4 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

© B.Raghavachari & J.Veerasamy, UTD 5 Non-Eulerian Directed Graph A B Balanced node Deficit node Surplus node Walk starting at A got stuck at B

© B.Raghavachari & J.Veerasamy, UTD 6 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

© B.Raghavachari & J.Veerasamy, UTD 7 Directed CPP Balanced node Deficit node Surplus node Output Input Add a minimum-weight subgraph which has outdegree = indegree + surplus at surplus nodes indegree = outdegree + deficit at deficit nodes

© B.Raghavachari & J.Veerasamy, UTD 8 Mixed Postman Problem (MPP) Balanced node Unbalanced node Solution 1Solution 2 Includes both undirected and directed edges Input NP-hard!

© B.Raghavachari & J.Veerasamy, UTD 9 Windy Postman Problem (WPP) Even degree node Odd degree node Undirected edges only, but traversal costs vary based on direction. NP-hard!

© B.Raghavachari & J.Veerasamy, UTD 10 Applications: City Management Elm Commerce Main Jackson Houston Street cleaning, Mail delivery, Snow removal, Courier delivery route

© B.Raghavachari & J.Veerasamy, UTD 11 Previous work Edmonds and Johnson, 1973: –Algorithms for undirected CPP and directed CPP –Optimal algorithm for even degree mixed graphs –Suggested a 2-approximation algorithm for MPP MPP: Frederickson, 1979: –Designed two approximation algorithms –Select the best of the two solutions computed –Worst-case performance ratio = 5/3 WPP: Win, 1989: –Optimal algorithm for Eulerian graphs –2-approximation algorithm for general graphs

© B.Raghavachari & J.Veerasamy, UTD 12 Our results 3/2 approximation algorithm for Mixed Postman Problem 3/2 approximation algorithm for Windy Postman Problem Optimal algorithm for Mixed Postman Problem with duplication restricted to a subset of edges and arcs for even degree input graphs. 3/2 approximation algorithm for Mixed Postman Problem with duplication is restricted to edges. Euler Tour Problem with restriction on turns is NP- complete. log n approximation algorithms for Directed Rural Postman Problem, General Routing Problem, and Postman Problem with restricted turns.

© 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.

© 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

© 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

© 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

© B.Raghavachari & J.Veerasamy, UTD 17 Remarks Input G Flow IN=OUT Directed edges D Undirected edges U Cost(Opt)  C(D) + C(U) = C FRED 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)

© 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

© 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 FRED + C(X)

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

© B.Raghavachari & J.Veerasamy, UTD 21 Alternating cycles - Properties Symmetric difference of two INOUT solutions: arcs can be decomposed into alternating cyles. Each alternating cycle (AC) contains components (R and B) from both graphs. R and B satisfy the same deficit/surplus of nodes in the solution graphs. Let G R and G B be the solution graphs. Then, G R - R + B and G B - B + R are also INOUT solution graphs. If C(G R ) < C(G B ), then C(R) < C(B) for at least one alternating cycle.

© B.Raghavachari & J.Veerasamy, UTD 22 Lower bound proof (continued) G IO H G -B+R B R C(R) < C(B)  C(G IO -B+R) < C(G IO ) Since INOUT is an optimal algorithm, this is false. So, adding any edge in U to G does not decrease INOUT augmentation cost.

© B.Raghavachari & J.Veerasamy, UTD 23 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

© B.Raghavachari & J.Veerasamy, UTD 24 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

© B.Raghavachari & J.Veerasamy, UTD 25 Performance of Algorithms C(D) Cost of solution C(Opt) 2C(Opt) C(new S 1 ) C(S 2 ) C(S 1 ) C(Opt)

© B.Raghavachari & J.Veerasamy, UTD 26 Windy Postman Problem (WPP) Even degree node Odd degree node Undirected edges only, but traversal costs vary based on direction. NP-hard!

© B.Raghavachari & J.Veerasamy, UTD 27 Additional complexity of WPP Unlike MPP, undirected cycles cannot be oriented to get optimal Eulerian subtours   Intermediate graph Optimal solution Solution with  cost

© B.Raghavachari & J.Veerasamy, UTD 28 LP formulation for WPP  Goal is to minimize total traversal cost (c ij x ij + c ji x ji ) (i,j)  E satisfying following constraints: x ij + x ji  1 (  (i,j)  E) Usage of each edge Flow preservation at each node (x ij - x ji ) j : (i, j)  E  = Non-negative flow x ij, x ji  0 (  (i,j)  E)

© B.Raghavachari & J.Veerasamy, UTD 29 LP output C(Input) = C(LP Output) = 34.5 Input G LP Directed edges D, Undirected edges U (x ij = x ji = 0.5) Cost(Opt)  C(G LP ) = C(D) + C(U) D U G LP LP augmentation cost C LP (G) = C(G LP ) - C(G)

© B.Raghavachari & J.Veerasamy, UTD 30 Algorithm 1 C(LP output) = 34.5 D U Duplicate edges C(S 1 ) = 57 C(S 1 ) = C(D) + 2 C(U) Even degree node Odd degree node

© B.Raghavachari & J.Veerasamy, UTD 31 Algorithm 2 D U min-cost X T-join within D Even degree node Odd degree node Components of D have to be Even degree to make the graph Eulerian.

© B.Raghavachari & J.Veerasamy, UTD 32 Algorithm 2 (continued) Hint: U and T-join within D form cycles! 2D U Duplicate D T-join within D D U U Question: Can we form a solution with cost  C(2D + U + X)?

© B.Raghavachari & J.Veerasamy, UTD 33 Algorithm 2 (continued) 2D U Directed walks C(S 2A ) = 50 C(S 2B ) = 54 C(2D+U+X) = 52 S 2 = solution with min. cost, C(S 2 )  2 C(D) + C(U) + C(X) Even degree node Odd degree node

© B.Raghavachari & J.Veerasamy, UTD 34 If C(Opt)  C(D) + C(U) + C(X), then C(S 2 )  C(Opt) + C(D) and C(S)  1.5 C(Opt) C(S 1 ) = C(D) + 2C(U)  C(Opt) - C(D) Algorithm outputs min. cost solution between S 1 and S 2. C(S 2 )  2C(D) + C(U) + C(X) X is min-cost T-join between odd-degree strongly connected Can we prove: C(Opt)  C(D) + C(U) + C(X) ? components (SCC) of G LP Remarks

© B.Raghavachari & J.Veerasamy, UTD 35 Improved lower bound Shrink each SCC to a single node, and compute T-join between odd degree SCCs.

© B.Raghavachari & J.Veerasamy, UTD 36 Improved lower bound (continued) 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 LP augmentation cost, then C(Opt)  C(D) + C(U) + C(X)

© B.Raghavachari & J.Veerasamy, UTD 37 Lower bound proof - sketch Input GNew input H LP output Can C LP (H) < C LP (G)? No. In fact, adding any edge in U does not decrease LP augmentation cost. Alternating cycle x

© B.Raghavachari & J.Veerasamy, UTD 38 Lower bound proof (continued) G IO H IO G IO -B+R B R C(R) < C(B)  C(G IO -B+R) < C(G IO ) Since INOUT is an optimal algorithm, this is false. So, adding any edge in U to G does not decrease INOUT augmentation cost.

© B.Raghavachari & J.Veerasamy, UTD 39 Future work Approximation algorithms for MPP / WPP with approximation ratio better than 1.5? Particularly, for planar graphs? Approximation algorithm for MPP with duplication restricted to a subset of edges and arcs? Approximation algorithm for Undirected CPP with restricted turns with approximation ratio better than log n? Approximation algorithm for Directed Rural Postman Problem with approximation ratio better than log n?