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Graph partitioning method for multiple autonomous surveillance robots

Graph partitioning method for multiple autonomous surveillance robots

My presentation slides for ITC-CSCC 2021 (https://itc-cscc2021.org/2021/)

204f36383109212baaedfabb8abcfc9e?s=128

konakalab

June 28, 2021
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  1. Graph partitioning method for multiple autonomous surveillance robots Yoshimasa Takahashi,

    Eiji Konaka* Meijo University ITC-CSCC 2021
  2. Outline •Background •Proposed method • Outline • Node partitioning •

    Edge partitioning • Add redundant edges •Numerical experiments ITC-CSCC 2021 (The speech in this presentation is generated by https://ondoku3.com/ja/)
  3. Background • Autonomous surveillance robot • A part of security

    system • Scope: Multiple robots • Improve monitoring efficiency • Objective • Provide a solution for a graph partitioning problem for multiple autonomous surveillance robots. • Each subgraph is monitored by one robot. ALSOK “REBORG-Z” Mira Robotics “ugo” https://www.alsok.co.jp/corporate/robot/reborg-x/ https://www.ugo.plus/ ITC-CSCC 2021
  4. Background • Autonomous surveillance robot • A part of security

    system • Scope: Multiple robots • Improve monitoring efficiency • Objective • Provide a solution for a graph partitioning problem for multiple autonomous surveillance robots. • Each subgraph is monitored by one robot. ALSOK “REBORG-Z” Mira Robotics “ugo” https://www.alsok.co.jp/corporate/robot/reborg-x/ https://www.ugo.plus/ ITC-CSCC 2021
  5. Obstacle 0. Map to undirected graph 1. Node partitioning 3.

    Make route by solving Chinese Postman Problem (CPP) 2. Edge partitioning & add shortcuts 𝑃1 = (0, 4, 8, 4, 5, 9, 5, 1, 0) 𝑃2 = (1, 2, 6, 5, 6, 7, 3, 2, 1) 𝑃3 = (8, 12, 13, 14, 13, 9, 10, 9, 8) 𝑃4 = (6, 10, 14, 15, 11, 7, 11, 10, 6) 1 0 4 5 3 2 7 6 15 14 11 10 12 13 9 8 𝑮𝟏 𝑮𝟐 𝑮𝟑 𝑮𝟒 1 0 8 4 5 9 1 3 2 7 6 5 10 12 13 14 9 8 15 14 11 10 7 6 𝑮𝟏 𝑮𝟐 𝑮𝟑 𝑮𝟒 Floor Outline of the proposed method ITC-CSCC 2021
  6. Obstacle 0. Map to undirected graph 1. Node partitioning 3.

    Make route by solving Chinese Postman Problem (CPP) 2. Edge partitioning & add shortcuts 𝑃1 = (0, 4, 8, 4, 5, 9, 5, 1, 0) 𝑃2 = (1, 2, 6, 5, 6, 7, 3, 2, 1) 𝑃3 = (8, 12, 13, 14, 13, 9, 10, 9, 8) 𝑃4 = (6, 10, 14, 15, 11, 7, 11, 10, 6) 1 0 4 5 3 2 7 6 15 14 11 10 12 13 9 8 𝑮𝟏 𝑮𝟐 𝑮𝟑 𝑮𝟒 1 0 8 4 5 9 1 3 2 7 6 5 10 12 13 14 9 8 15 14 11 10 7 6 𝑮𝟏 𝑮𝟐 𝑮𝟑 𝑮𝟒 Floor Outline of the proposed method ITC-CSCC 2021
  7. Map to undirected graph Aisle→Edge, Crossing→Node, Aisle length→Weight of edge

    Undirected graph G Ex: Edge Node ITC-CSCC 2021
  8. Requirement on surveillance route Surveillance routes should satisfy the followings:

    • One travel must include every edge on the (sub)graph • Path:= A sequence of edges. • Travel:= A path out from and end to one node. • Start/End node of travel: charging station Chinese Postman Problem (CPP) for undirected graph CPP: A problem of determining the minimum-length run covering all edges in the graph at least once ITC-CSCC 2021
  9. Obstacle 0. Map to undirected graph 1. Node partitioning 3.

    Make route by solving Chinese Postman Problem (CPP) 2. Edge partitioning & add shortcuts 𝑃1 = (0, 4, 8, 4, 5, 9, 5, 1, 0) 𝑃2 = (1, 2, 6, 5, 6, 7, 3, 2, 1) 𝑃3 = (8, 12, 13, 14, 13, 9, 10, 9, 8) 𝑃4 = (6, 10, 14, 15, 11, 7, 11, 10, 6) 1 0 4 5 3 2 7 6 15 14 11 10 12 13 9 8 𝑮𝟏 𝑮𝟐 𝑮𝟑 𝑮𝟒 1 0 8 4 5 9 1 3 2 7 6 5 10 12 13 14 9 8 15 14 11 10 7 6 𝑮𝟏 𝑮𝟐 𝑮𝟑 𝑮𝟒 Floor Outline of the proposed method ITC-CSCC 2021
  10. Node partitioning Ward’s method (hierarchical clustering method) Based on edge

    weight 𝑑 𝐼𝐽 𝐾 : distance between clusters (𝐼𝐽) and 𝐾 𝑑 𝐼𝐽 𝐾 = 𝑛𝐼 + 𝑛𝐾 𝑇 𝑑𝐼𝐾 2 + 𝑛𝐽 + 𝑛𝐾 𝑇 𝑑𝐽𝐾 2 − 𝑛𝐾 𝑇 𝑑𝐼𝐽 2 𝑇 = 𝑛𝐼 + 𝑛𝐽 + 𝑛𝐾 𝑛𝐼 , 𝑛𝐽 , 𝑛𝐾 : number of elements in each cluster ITC-CSCC 2021
  11. Node partitioning: example Undirected graph 𝐺 1 0 4 5

    3 2 7 6 12 13 9 8 15 14 11 10 𝑮𝟏 𝑮𝟐 𝑮𝟑 𝑮𝟒 Ex: partitioned into 4 subgraphs ITC-CSCC 2021
  12. Obstacle 0. Map to undirected graph 1. Node partitioning 3.

    Make route by solving Chinese Postman Problem (CPP) 2. Edge partitioning & add shortcuts 𝑃1 = (0, 4, 8, 4, 5, 9, 5, 1, 0) 𝑃2 = (1, 2, 6, 5, 6, 7, 3, 2, 1) 𝑃3 = (8, 12, 13, 14, 13, 9, 10, 9, 8) 𝑃4 = (6, 10, 14, 15, 11, 7, 11, 10, 6) 1 0 4 5 3 2 7 6 15 14 11 10 12 13 9 8 𝑮𝟏 𝑮𝟐 𝑮𝟑 𝑮𝟒 1 0 8 4 5 9 1 3 2 7 6 5 10 12 13 14 9 8 15 14 11 10 7 6 𝑮𝟏 𝑮𝟐 𝑮𝟑 𝑮𝟒 Floor Outline of the proposed method ITC-CSCC 2021
  13. Edge partitioning 1 0 4 5 12 13 9 8

    15 14 11 10 3 2 7 6 1 0 4 5 3 2 7 6 12 13 9 8 15 14 11 10 𝑮𝟏 𝑮𝟐 𝑮𝟑 𝑮𝟒 ⅰ. Add every edge whose terminal nodes are in the same subgraph 𝑮𝟏 𝑮𝟐 𝑮𝟑 𝑮𝟒 ITC-CSCC 2021
  14. Edge partitioning 1 0 4 5 12 13 9 8

    15 14 11 10 3 2 7 6 1 0 4 5 3 2 7 6 12 13 9 8 15 14 11 10 𝑮𝟏 𝑮𝟐 𝑮𝟑 𝑮𝟒 ⅰ. Add every edge whose terminal nodes are in the same subgraph 𝑮𝟏 𝑮𝟐 𝑮𝟑 𝑮𝟒 ITC-CSCC 2021
  15. Edge partitioning 6 5 1 0 8 4 5 9

    1 3 2 7 6 𝑮𝟏 𝑮𝟐 𝑮𝟏 𝑮𝟐 1 3 2 7 6 5 ⅱ. Add every remaining edge into the subgraph that has the smallest total edge weights 1 0 8 4 5 9 Weights=5 Weights=6 ITC-CSCC 2021
  16. Edge partitioning: example Ex: 𝐺 is partitioned into 4 subgraphs

    ITC-CSCC 2021
  17. Edge partitioning ⅲ. Add shortcut edge 40 42 41 22

    21 40 42 41 22 21 ITC-CSCC 2021
  18. Edge partitioning ⅲ. Add shortcut edge 40 42 41 22

    21 40 42 41 22 21 40 42 41 22 21 40 42 41 22 21 ITC-CSCC 2021
  19. Edge partitioning ⅲ. Add shortcut edge 40 42 41 22

    21 40 42 41 22 21 40 42 41 22 21 40 42 41 22 21 Solve Chinese Postman Problem (CPP) ITC-CSCC 2021
  20. Obstacle 0. Map to undirected graph 1. Node partitioning 3.

    Make route by solving Chinese Postman Problem (CPP) 2. Edge partitioning & add shortcuts 𝑃1 = (0, 4, 8, 4, 5, 9, 5, 1, 0) 𝑃2 = (1, 2, 6, 5, 6, 7, 3, 2, 1) 𝑃3 = (8, 12, 13, 14, 13, 9, 10, 9, 8) 𝑃4 = (6, 10, 14, 15, 11, 7, 11, 10, 6) 1 0 4 5 3 2 7 6 15 14 11 10 12 13 9 8 𝑮𝟏 𝑮𝟐 𝑮𝟑 𝑮𝟒 1 0 8 4 5 9 1 3 2 7 6 5 10 12 13 14 9 8 15 14 11 10 7 6 𝑮𝟏 𝑮𝟐 𝑮𝟑 𝑮𝟒 Floor Outline of the proposed method ITC-CSCC 2021
  21. Make surveillance route: solve CPP • CPP: A problem of

    determining the minimum-length run covering all edges in the graph at least once • Final output: Solution of CPP for each subgraph 𝑃1 = (0, 4, 8, 4, 5, 9, 5, 1, 0) Ex: 1 0 8 4 5 9 𝑮𝟏 1 0 8 4 5 9 𝑷𝟏 ITC-CSCC 2021
  22. Numerical experiments • Two cases (maps) • Case 1 :

    lattice • Case 2 : made from a drugstore • Performance indices • Total sum of lengths of the surveillance routes • Maximum duration between visits • Number of robots • For all indices, smaller values are preferable ITC-CSCC 2021
  23. Case 1 (lattice) • Number of robots: 2, 4, and

    8 ITC-CSCC 2021
  24. Performance indices in Case 1 (lattice) Visit interval: gets smaller

    as N Total distance: minimum in N=2, 4 ITC-CSCC 2021
  25. Partition result of Case 1 Graph partitioning result (𝑁 =

    4) • Symmetric • No redundant edges ITC-CSCC 2021
  26. Case 2 (drugstore) • Graph modeled a real drugstore •

    Number of robots: 𝑁 = 3, 4, 5, ⋯ , 14 ITC-CSCC 2021
  27. Performance indices in Case 2 (drugstore) Visit interval: gets smaller

    as N Total distance: minimum in N=6 • Not monotonic as N • Larger N -> many overlaps ITC-CSCC 2021
  28. Partition result of Case 2 Graph partitioning result (𝑁 =

    6) Total distance: minimum in N=6 • Not monotonic as N • Redundant short edges are added • Difficult design for human designers ITC-CSCC 2021
  29. Summary and future work • Summary - Provide a solution

    for a graph partitioning problem for multiple autonomous surveillance robots. - The proposed method can design surveillance routes for multiple robots for a real drugstore map. • Future work -Scalability: Applied to larger map, e.g., airport. ITC-CSCC 2021
  30. Thank you for attention! ITC-CSCC 2021

  31. Appendix ITC-CSCC 2021

  32. Assumptions on robot surveillance • Floor: No moving obstacles. No

    doors to be opened by the robot • Robot has the map on the floor. It can run autonomously between designated two points. • Sensing range of the robot is wider than the width of the aisles. ITC-CSCC 2021
  33. Clustering result (Case 1) ITC-CSCC 2021

  34. Clustering result (Case 2) ITC-CSCC 2021