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traveling salesman problem

traveling salesman problem. ZIP-Method deductive approach of an optimal solution of symmetrical Traveling-salesman-problems. http://www.jochen-pleines.de. new ideas. das Rundreiseproblem bisherige Lösungen New ideas Beispiel mit 6 Knoten Beispiel mit 10 Knoten

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traveling salesman problem

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  1. travelingsalesman problem ZIP-Method deductive approach of an optimal solution of symmetrical Traveling-salesman-problems http://www.jochen-pleines.de

  2. new ideas ... • das Rundreiseproblem • bisherige Lösungen • New ideas • Beispiel mit 6 Knoten • Beispiel mit 10 Knoten • Beispiel mit 26 Knoten (Ergebnisse) • Schlussfolgerungen und Ausblick

  3. new ideas ... • but first a task for you: • Please note in arbitrary order the numbers from 1 to 6. 5 3 6 2 1 4

  4. new ideas ... • If you connect the nodes, then it develops: • 1-component G • with 6 edges • with 6 nodes • each node has the grade 2 5 5 3 6 2 1 4

  5. new ideas ... we remember: if n = 6 then: n! = 720 (n-1)! = 120 (n-1)! / 2 = 60 5 5 3 6 2 1 4

  6. new ideas ... now we add the values of the nodes: as well each node accurs twice: 5 5 3 6 2 • at the beginning of an edge (1-6) • and at the endof an edge (4-1). 1 1 4 and that is evil !

  7. new ideas ... If each node accursonly once, thenoriginate from a graph: a partial-graph with all of the 6 nodes, but only with 3 edges: for example:edges: 1-6, 5-3, 2-4 5 5 3 6 2 1 4

  8. new ideas ... ... a complement partial- graph with the same structure remains like the edges: 1-4, 2-3, 5-6 5 5 3 6 2 1 4

  9. new ideas ... • So the graph gets together: • from the partial-graph with edges:1-6,5-3,2-4 • and the partial-graph with edges :1-4,2-3,5-6 5 5 3 6 2 1 4

  10. we remember: With n = 6 nodes there exist 120 graphs respectively 60 symmetric graphs How many partial-graphs actual exist ? 1. partial-graph 1 - 2 3 - 4 5 - 6 2. partial-graph 1 - 2 3 - 5 4 - 6 3. partial-graph 1 - 2 3 - 6 4 - 5 new ideas ... 4. partial-graph 1 - 3 2 - 4 5 - 6 5. partial-graph 1 - 3 2 - 5 4 - 6 6. partial-graph 1 - 3 2 - 6 4 - 5 7. partial-graph 1 - 4 2 - 3 5 - 6 8. partial-graph 1 - 4 2 - 5 3 - 6 9. partial-graph 1 - 4 2 - 6 3 - 5 10. partial-graph 1 - 5 2 - 3 4 - 6 11. partial-graph 1 - 5 2 - 4 3 - 6 12. partial-graph 1 - 5 2 - 6 3 - 4 13. partial-graph 1 - 6 2 - 3 4 - 5 14. partial-graph 1 - 6 2 - 4 3 - 5 15. partial-graph 1 - 6 2 - 5 3 - 4 not more !

  11. By using the the Symmetry-rule and the Sort-rule each of the 120 Graphscan be devorced into 2 of the 15partial-graphs. 1. partial-graph 1 - 2 3 - 4 5 - 6 2. partial-graph 1 - 2 3 - 5 4 - 6 3. partial-graph 1 - 2 3 - 6 4 - 5 4. partial-graph 1 - 3 2 - 4 5 - 6 5. partial-graph 1 - 3 2 - 5 4 - 6 6. partial-graph 1 - 3 2 - 6 4 - 5 7. partial-graph 1 - 4 2 - 3 5 - 6 8. partial-graph 1 - 4 2 - 5 3 - 6 9. partial-graph 1 - 4 2 - 6 3 - 5 10. partial-graph 1 - 5 2 - 3 4 - 6 11. partial-graph 1 - 5 2 - 4 3 - 6 12. partial-graph 1 - 5 2 - 6 3 - 4 13. partial-graph 1 - 6 2 - 3 4 - 5 14. partial-graph 1 - 6 2 - 4 3 - 5 15. partial-graph 1 - 6 2 - 5 3 - 4 new ideas ... Please try it with your own example !

  12. new ideas ... • Symmetry-Rule: • The begin-node of an edge has the lower number then the end-node: i < j: ->f(1-6) + f(5-3) + f(2-4) = f(1-6) + f(3-5) + f(2-4) 5 5 3 6 2 1 4

  13. new ideas ... • Sort-Rule: • die edges will be sorted by the begin-node of the edges. • 1. edge2. edge3. edge->f(1-6)+ f(3-5) + f(2-4) = f(1-6)+ f(2-4) + f(3-5) 5 5 3 6 2 1 4

  14. How many partial graphs „matching“ to a partial graph, i.o. do all of the graphs form one together again ? Example: 1-2, 3-4, 5-6 1. partial-graph 1 - 2 3 - 4 5 - 6 2. partial-graph 1 - 2 3 - 5 4 - 6 3. partial-graph 1 - 2 3 - 6 4 - 5 4. partial-graph 1 - 3 2 - 4 5 - 6 5. partial-graph 1 - 3 2 - 5 4 - 6 6. partial-graph 1 - 3 2 - 6 4 - 5 7. partial-graph 1 - 4 2 - 3 5 - 6 8. partial-graph 1 - 4 2 - 5 3 - 6 9. partial-graph 1 - 4 2 - 6 3 - 5 10. partial-graph 1 - 5 2 - 3 4 - 6 11. partial-graph 1 - 5 2 - 4 3 - 6 12. partial-graph 1 - 5 2 - 6 3 - 4 13. partial-graph 1 - 6 2 - 3 4 - 5 14. partial-graph 1 - 6 2 - 4 3 - 5 15. partial-graph 1 - 6 2 - 5 3 - 4 new ideas ... altogether 8 (= 2 x 4) No.5, 6, 8, 9, 10, 11, 13, 14

  15. new ideas ... Further considerations: The smallest (complete-)graph gets together: either: from the two found partial-graphs(smallest partial-graph with accompanying smallest comp-partial-graph) or: from two partial-graphs lying between this.

  16. How to find out the lowest graph ? 1. step:on find out the lowest partial - graph 2. step:on find out the belonginglowest Comp-partial-graph. new ideas ... 1. partial-graph 1 - 2 3 - 4 5 - 6 2. partial-graph 1 - 2 3 - 5 4 - 6 3. partial-graph 1 - 2 3 - 6 4 - 5 4. partial-graph 1 - 3 2 - 4 5 - 6 5. partial-graph 1 - 3 2 - 5 4 - 6 6. partial-graph 1 - 3 2 - 6 4 - 5 7. partial-graph 1 - 4 2 - 3 5 - 6 8. partial-graph 1 - 4 2 - 5 3 - 6 9. partial-graph 1 - 4 2 - 6 3 - 5 10. partial-graph 1 - 5 2 - 3 4 - 6 11. partial-graph 1 - 5 2 - 4 3 - 6 12. partial-graph 1 - 5 2 - 6 3 - 4 13. partial-graph 1 - 6 2 - 3 4 - 5 14. partial-graph 1 - 6 2 - 4 3 - 5 15. partial-graph 1 - 6 2 - 5 3 - 4

  17. new ideas ... There perhaps the lowest graph is found !But only: perhaps!

  18. new ideas ... example: • lowest partial-graph has a length of edge: 20 • der lowest belonging Compl.-partial-graph: 40 • result a graph: 60 • that means, • (a+b) < c -> a and/or b < (c/2) • and a < b or a = b -> a < (c/2)

  19. new ideas ... Further iteration steps:(up to half the edge length of the so far smallest found complet graph) • starting out from the smallest partial-graph it this one is respectively greater partial-graph with his complement partial-graph checked most nearly, wether from this a smaler complete-graph can be put together. • if yes: the new complete-graph is initial value for further iteration steps. • if no: the smallest graph is already found

  20. example with 6 nodes • das Rundreiseproblem • bisherige Lösungen • neue Überlegungen • example with 6 nodes • Beispiel mit 10 Knoten • Beispiel mit 26 Knoten (Ergebnisse) • Schlussfolgerungen und Ausblick

  21. example with 6 nodes Given 6 nodes with the values of the edges: • to 1 to 2 to 3 to 4 to 5 to 6 from 1 - 12 25 30 28 22 from 2 - 16 20 22 10 from 3 - 23 26 21 from 4 - 31 18 from 5 - 14 from 6 -

  22. to divide 60 partial-graphs 60 partial-graphs to sort 15 partial-graphs 15 partial-graphs to compute example with 6 nodes 60 symmetric graphs minimal graph

  23. compl.-partial-graph Nr. 1.K.2.K.3.K.K.-Länge 1. 1 - 2 3 - 4 5 - 6 49 2. 1 - 2 3 - 5 4 - 6 56 3. 1 - 2 3 - 6 4 - 5 64 4. 1 - 3 2 - 4 5 - 6 59 5. 1 - 3 2 - 5 4 - 6 65 6. 1 - 3 2 - 6 4 - 5 66 7. 1 - 4 2 - 3 5 - 6 60 8. 1 - 4 2 - 5 3 - 6 73 9. 1 - 4 2 - 6 3 - 5 66 10. 1 - 5 2 - 3 4 - 6 62 11. 1 - 5 2 - 4 3 - 6 69 12. 1 - 5 2 - 6 3 - 4 61 13. 1 - 6 2 - 3 4 - 5 69 14. 1 - 6 2 - 4 3 - 5 68 15. 1 - 6 2 - 5 3 - 4 67 example with 6 nodes Partial-graph Nr. 1.K.2.K.3.K.K.-Länge 1. 1 - 2 3 - 4 5 - 6 49 2. 1 - 2 3 - 5 4 - 6 56 3. 1 - 2 3 - 6 4 - 5 64 4. 1 - 3 2 - 4 5 - 6 59 5. 1 - 3 2 - 5 4 - 6 65 6. 1 - 3 2 - 6 4 - 5 66 7. 1 - 4 2 - 3 5 - 6 60 8. 1 - 4 2 - 5 3 - 6 73 9. 1 - 4 2 - 6 3 - 5 66 10. 1 - 5 2 - 3 4 - 6 62 11. 1 - 5 2 - 4 3 - 6 69 12. 1 - 5 2 - 6 3 - 4 61 13. 1 - 6 2 - 3 4 - 5 69 14. 1 - 6 2 - 4 3 - 5 68 15. 1 - 6 2 - 5 3 - 4 67 (49 + 62) : 2 = 55,5 Length of the graph: 111

  24. example with 10 nodes • das Rundreiseproblem - Fragestellung • Problem und bisherige Lösungen • neue Überlegungen • Beispiel mit 6 Knoten • Example with 10 nodes • Beispiel mit 26 Knoten (Ergebnisse) • Schlussfolgerungen und Ausblick

  25. example with 10 nodes development of the ZIP-term with n = 10: 1 · 3 · 5 · 7 · 9 = 1 ·2· 3 · 4· 5 · 6 · 7 · 8 · 9 ————————————— = 2 · 4 · 6 · 8 9! ————————————— = (1 · 2) · (2 · 2) · (3 · 2) · (4 · 2) 9! ————————————— = ( 1 · 2 · 3 · 4 ) · ( 2·2·2·2) 9! ————————————— = 4! · 24 (10 – 1)! ————————————— = (5 – 1)! · 2(5-1)

  26. example with 10 nodes (n – 1)! ————————— ( n/2 – 1 ) ! · 2 n/2 - 1 by that: { ( n/2 – 1 ) ! }(Sort-rule) Partial-graph: —— — — — (number of edges =n/2)1 x2 x 2 x 2 x 2 (Symmetry) Begin-edge = x1

  27. example with 10 nodes

  28. min.p-g min.p-gkomp Sorting of the 945 partial-graphs after edge-length example with 10 nodes cur.No. length 1. edge 2.edge 3.edge 4.edge 5.edge Notice 1 76 1 - 2 3 - 7 4 - 6 5 - 8 9 -10 2 77 1 - 5 2 - 9 3 - 8 4 - 6 7 -10 3 79 1 - 5 2 -10 3 - 7 4 - 6 8 - 9 4 83 1 - 5 2 - 7 3 - 8 4 - 6 9 -10 92 1 - 2 3 - 5 4 - 6 7 -10 8 - 9 5 6 93 1 - 2 3 - 9 4 - 6 5 - 8 7 -10 7 96 1 - 2 3 - 6 4 - 9 5 - 8 7 -10 8 96 1 - 8 2 - 5 3 - 7 4 - 6 9 -10 9 97 1 - 5 2 - 3 4 - 6 7 -10 8 - 9 10 100 1 - 2 3 - 6 4 - 5 7 -10 8 - 9 11 100 1 - 5 2 - 7 3 - 6 4 -10 8 - 9 12 101 1 - 8 2 - 9 3 - 5 4 - 6 7 -10 13 103 1 - 3 2 - 9 4 - 6 5 - 8 7 -10 .... min. p-g + min.p-gkomp = 176; 176 / 2 = 88 until 945

  29. example with 10 nodes cur.No. length 1. edge 2.edge 3.edge 4.edge 5.edge Notice 1 76 1 - 2 3 - 7 4 - 6 5 - 8 9 -10 2 77 1 - 5 2 - 9 3 - 8 4 - 6 7 -10 3 79 1 - 5 2 -10 3 - 7 4 - 6 8 - 9 4 83 1 - 5 2 - 7 3 - 8 4 - 6 9 -10 92 1 - 2 3 - 5 4 - 6 7 -10 8 - 9 5 6 93 1 - 2 3 - 9 4 - 6 5 - 8 7 -10 7 96 1 - 2 3 - 6 4 - 9 5 - 8 7 -10 8 96 1 - 8 2 - 5 3 - 7 4 - 6 9 -10 9 97 1 - 5 2 - 3 4 - 6 7 -10 8 - 9 10 100 1 - 2 3 - 6 4 - 5 7 -10 8 - 9 11 100 1 - 5 2 - 7 3 - 6 4 -10 8 - 9 12 101 1 - 8 2 - 9 3 - 5 4 - 6 7 -10 13 103 1 - 3 2 - 9 4 - 6 5 - 8 7 -10 .... until 945

  30. example with 10 nodes cur.No. length 1. edge 2.edge 3.edge 4.edge 5.edge Notice 1 76 1 - 2 3 - 7 4 - 6 5 - 8 9 -10 2 77 1 - 5 2 - 9 3 - 8 4 - 6 7 -10 3 79 1 - 5 2 -10 3 - 7 4 - 6 8 - 9 4 83 1 - 5 2 - 7 3 - 8 4 - 6 9 -10 92 1 - 2 3 - 5 4 - 6 7 -10 8 - 9 5 6 93 1 - 2 3 - 9 4 - 6 5 - 8 7 -10 7 96 1 - 2 3 - 6 4 - 9 5 - 8 7 -10 8 96 1 - 8 2 - 5 3 - 7 4 - 6 9 -10 9 97 1 - 5 2 - 3 4 - 6 7 -10 8 - 9 10 100 1 - 2 3 - 6 4 - 5 7 -10 8 - 9 11 100 1 - 5 2 - 7 3 - 6 4 -10 8 - 9 12 101 1 - 8 2 - 9 3 - 5 4 - 6 7 -10 13 103 1 - 3 2 - 9 4 - 6 5 - 8 7 -10 .... until 945

  31. min.p-gkomp example with 10 nodes cur.No. length 1. edge 2.edge 3.edge 4.edge 5.edge Notice 1 76 1 - 2 3 - 7 4 - 6 5 - 8 9 -10 2 77 1 - 5 2 - 9 3 - 8 4 - 6 7 -10 min. p-g 3 79 1 - 5 2 -10 3 - 7 4 - 6 8 - 9 4 83 1 - 5 2 - 7 3 - 8 4 - 6 9 -10 92 1 - 2 3 - 5 4 - 6 7 -10 8 - 9 5 6 93 1 - 2 3 - 9 4 - 6 5 - 8 7 -10 7 96 1 - 2 3 - 6 4 - 9 5 - 8 7 -10 8 96 1 - 8 2 - 5 3 - 7 4 - 6 9 -10 9 97 1 - 5 2 - 3 4 - 6 7 -10 8 - 9 10 100 1 - 2 3 - 6 4 - 5 7 -10 8 - 9 11 100 1 - 5 2 - 7 3 - 6 4 -10 8 - 9 12 101 1 - 8 2 - 9 3 - 5 4 - 6 7 -10 13 103 1 - 3 2 - 9 4 - 6 5 - 8 7 -10 .... min. p-g + min.p-gkomp = 175; 175 / 2 = 87,5 until 945

  32. opt. p-g opt. p-gkomp example with 10 nodes cur.No. length 1. edge 2.edge 3.edge 4.edge 5.edge Notice 1 76 1 - 2 3 - 7 4 - 6 5 - 8 9 -10 2 77 1 - 5 2 - 9 3 - 8 4 - 6 7 -10 3 79 1 - 5 2 -10 3 - 7 4 - 6 8 - 9 4 83 1 - 5 2 - 7 3 - 8 4 - 6 9 -10 92 1 - 2 3 - 5 4 - 6 8 - 9 5 7 -10 6 93 1 - 2 3 - 9 4 - 6 5 - 8 7 -10 7 96 1 - 2 3 - 6 4 - 9 5 - 8 7 -10 8 96 1 - 8 2 - 5 3 - 7 4 - 6 9 -10 9 97 1 - 5 2 - 3 4 - 6 7 -10 8 - 9 10 100 1 - 2 3 - 6 4 - 5 7 -10 8 - 9 11 100 1 - 5 2 - 7 3 - 6 4 -10 8 - 9 12 101 1 - 8 2 - 9 3 - 5 4 - 6 7 -10 13 103 1 - 3 2 - 9 4 - 6 5 - 8 7 -10 .... until 945

  33. example with 10 nodes Quantity of partial-graphs after the 1. pass through: only 11 of 945 partial-graphs Sum of the values of edges of altogether 181.440 graphs

  34. example with 10 nodes • From altogether 945 partial-graphs will eliminate with back tracking of the limited Enumeration: • Stop after the 5th edge: 0 edge • Stop after the 4th edge: 1 edge • Stop after the 3th edge: 3 edges • Stop after the 2th edge: 15 edges • Stop after the first edge: 105 edges Þ earliest stop if possible!!

  35. example with 10 nodes Relation between begin-node and place of the edge: 1. Edge 2. Edge 3. Edge 4. Edge 5. Edge 1-2...1-10 2-3...2-10 3-4...3-10 4-5...4-10 5-6...5-10or or or or 3-4...3-10 4-5...4-10 5-6...5-10 6-7...6-10or or or 5-6...5-10 6-7...6-10 7-8...7-10 or or 7-8...7-10 8-9...8-10or 9-10.

  36. example with 10 nodes

  37. example with 6 nodes Some more ideas: • Numbering-rule (the greatest difference first) • Minimal-edge-rule (calculation of the smallest edge still outstanding for every single edge-place; no more for everyone – see page before: relation between begin-node and place of the edge)

  38. example with 6 nodes Example: node xi with his 5 edges Minimal-edge-rule Numbering-rule

  39. traveling salesman problem Thank you for your interest

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