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traveling salesman problem ZIP-Method deductive approach of an optimal solution of symmetrical Traveling-salesman-problems

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

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new ideas... but first a task for you: Please note in arbitrary order the numbers from 1 to

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

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

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

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new ideas... If each node accurs only once, then originate from a graph: a partial-graph with all of the 6 nodes, but only with 3 edges: for example: edges: 1-6, 5-3,

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

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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,

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new ideas... we remember: With n = 6 nodes there exist 120 graphs respectively 60 symmetric graphs How many partial-graphs actual exist ? 1. partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph not more !

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new ideas... By using the the Symmetry-rule and the Sort-rule each of the 120 Graphs can be devorced into 2 of the 15 partial-graphs. 1. partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph Please try it with your own example !

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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)

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

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new ideas... 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 No.5, 6, 8, 9, 10, 11, 13, 14 altogether 8 (= 2 x 4) 1. partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph

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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.

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new ideas... How to find out the lowest graph ? 1. step: on find out the lowest partial - graph 2. step: on find out the belonging lowest Comp-partial-graph. 1. partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph partial-graph

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

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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) a and/or b < (c/2) and a a < (c/2)

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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

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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

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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 from from from from from 6 -

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

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example with 6 nodes Nr.1.K. 2.K. 3.K. K.-Länge Partial-graph Nr.1.K. 2.K. 3.K. K.-Länge compl.-partial-graph Length of the graph: 111 ( ) : 2 = 55,5

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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

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

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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 x 2 x 2 x 2 x 2 (Symmetry) Begin-edge = x 1

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example with 10 nodes

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cur.No.length1. edge2.edge3.edge4.edge5.edgeNotice min.p-g komp until 945 min.p-g min. p-g + min.p-g komp = 176; 176 / 2 = 88 Sorting of the 945 partial-graphs after edge-length example with 10 nodes

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cur.No.length1. edge2.edge3.edge4.edge5.edgeNotice until 945 example with 10 nodes

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cur.No.length1. edge2.edge3.edge4.edge5.edgeNotice until 945 example with 10 nodes

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cur.No.length1. edge2.edge3.edge4.edge5.edgeNotice min.p-g komp until 945 min. p-g min. p-g + min.p-g komp = 175; 175 / 2 = 87,5 example with 10 nodes

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cur.No.length1. edge2.edge3.edge4.edge5.edgeNotice until 945 opt. p-g opt. p-g komp example with 10 nodes

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Quantity of partial-graphs Sum of the values of edges after the 1. pass through: only 11 of 945 partial-graphs of altogether graphs

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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 earliest stop if possible!!

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example with 10 nodes Relation between begin-node and place of the edge: 1. Edge 2. Edge 3. Edge 4. Edge 5. Edge orororor ororor oror or 9-10.

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example with 10 nodes

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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)

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example with 6 nodes Example: node x i with his 5 edges Numbering-ruleMinimal-edge-rule

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traveling salesman problem Thank you for your interest

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