traveling salesman problem

traveling salesman problem

ZIP-Methoddeductive 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– Beispiel mit 26 Knoten (Ergebnisse)– Schlussfolgerungen und Ausblick

new ideas ...

• but first

a task for you:

• Please note in

arbitrary order the

numbers from 1 to

6.

5

2

1

4

3

6

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

55

2

1

4

3

6

new ideas ...

we remember:

if n = 6 then:

n! = 720

(n-1)! = 120

(n-1)! / 2 = 60

55

2

1

4

3

6

new ideas ...

now we add the values of the nodes:

as well each node accurs twice:

55

2

1

4

3

6

– at the beginning

of an edge (1-6)

– and at the end

of an edge (4-1).

and that is evil !

1

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

55

2

1

4

3

6

new ideas ...

... a complement partial-

graph with the same

structure remains like

the

edges: 1-4, 2-3, 5-6

55

2

1

4

3

6

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

55

2

1

4

3

6

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 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

not more !

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 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

Please try it with your own example !

new ideas ...

55

2

1

4

3

6

• 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)

new ideas ...

55

2

1

4

3

6

• 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)

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, 14altogether 8 (= 2 x 4)

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 ...

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.

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 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 ...

There perhaps the lowest graph is found !

But only: perhaps!

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)

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

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


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 -

example with 6 nodes60 symmetric graphs

60 partial-graphs 60 partial-graphs

minimal graph

to divide

to sort

to compute

15 partial-graphs 15 partial-graphs


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 6610. 1 - 5 2 - 3 4 - 6 6211. 1 - 5 2 - 4 3 - 6 6912. 1 - 5 2 - 6 3 - 4 6113. 1 - 6 2 - 3 4 - 5 6914. 1 - 6 2 - 4 3 - 5 6815. 1 - 6 2 - 5 3 - 4 67

Partial-graphNr. 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 6610. 1 - 5 2 - 3 4 - 6 6211. 1 - 5 2 - 4 3 - 6 6912. 1 - 5 2 - 6 3 - 4 6113. 1 - 6 2 - 3 4 - 5 6914. 1 - 6 2 - 4 3 - 5 6815. 1 - 6 2 - 5 3 - 4 67

compl.-partial-graph

Length of the graph: 111 (49 + 62) : 2 = 55,5


– 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

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! · 24

(10 – 1)! ————————————— = (5 – 1)! · 2(5-1)


(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 = x1

example with 10 nodesnachvon

1 2 3 4 5 6 7 8 9 10

1 - 16 44 93 1 30 30 5 78 42

2 - 68 61 42 77 41 79 22 32

3 - 39 48 21 36 28 40 80

4 - 43 8 66 46 30 35

5 - 67 69 11 84 91

6 - 97 43 63 67

7 - 85 89 18

8 - 2 85

9 - 5

cur.No. length 1. edge 2.edge 3.edge 4.edge 5.edge Notice1 76 1 - 2 3 - 7 4 - 6 5 - 8 9 -102 77 1 - 5 2 - 9 3 - 8 4 - 6 7 -103 79 1 - 5 2 -10 3 - 7 4 - 6 8 - 94 83 1 - 5 2 - 7 3 - 8 4 - 6 9 -105 92 1 - 2 3 - 5 4 - 6 7 -10 8 - 96 93 1 - 2 3 - 9 4 - 6 5 - 8 7 -107 96 1 - 2 3 - 6 4 - 9 5 - 8 7 -108 96 1 - 8 2 - 5 3 - 7 4 - 6 9 -109 97 1 - 5 2 - 3 4 - 6 7 -10 8 - 9

10 100 1 - 2 3 - 6 4 - 5 7 -10 8 - 911 100 1 - 5 2 - 7 3 - 6 4 -10 8 - 9 min.p-gkomp

12 101 1 - 8 2 - 9 3 - 5 4 - 6 7 -1013 103 1 - 3 2 - 9 4 - 6 5 - 8 7 -10

....until 945

min.p-g

min. p-g + min.p-gkomp = 176; 176 / 2 = 88Sorting of the 945 partial-graphs after edge-length



10 100 1 - 2 3 - 6 4 - 5 7 -10 8 - 911 100 1 - 5 2 - 7 3 - 6 4 -10 8 - 912 101 1 - 8 2 - 9 3 - 5 4 - 6 7 -1013 103 1 - 3 2 - 9 4 - 6 5 - 8 7 -10

....until 945



10 100 1 - 2 3 - 6 4 - 5 7 -10 8 - 911 100 1 - 5 2 - 7 3 - 6 4 -10 8 - 9

min.p-gkomp

12 101 1 - 8 2 - 9 3 - 5 4 - 6 7 -1013 103 1 - 3 2 - 9 4 - 6 5 - 8 7 -10

....until 945

min. p-g

min. p-g + min.p-gkomp = 175; 175 / 2 = 87,5



10 100 1 - 2 3 - 6 4 - 5 7 -10 8 - 911 100 1 - 5 2 - 7 3 - 6 4 -10 8 - 912 101 1 - 8 2 - 9 3 - 5 4 - 6 7 -1013 103 1 - 3 2 - 9 4 - 6 5 - 8 7 -10

....until 945

opt. p-g

opt. p-gkomp



0

20

40

60

80

100

120

61-80

101-120

141-160

181-200

221-240

261-280

301-320

341-360

381-400

421-440

Qua

ntity

of

part

ial-

grap

hs

Sum of the values of edges after the 1. pass through: only 11 of 945 partial-graphs of altogether 181.440 graphs


• 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!!


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...7-10or or

7-8...7-10 8-9...8-10or

9-10.


Anfangs-Knoten

1.Kante 2.Kante 3.Kante 4.Kante 5.KanteSumme je

KnotenNr.1 945 - - - - 945Nr.2 - 840 - - - 840Nr.3 - 105 630 - - 735Nr.4 - - 270 360 - 630Nr.5 - - 45 360 120 525Nr.6 - - - 180 240 420Nr.7 - - - 45 270 315Nr.8 - - - - 210 210Nr.9 - - - - 105 105

Nr.10 - - - - 0 0Summe

derKanten

945 945 945 945 945


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)


40

90

45

50

30

0 10 20 30 40 50 60 70 80 90

1. Kante

2. Kante

3. Kante

4. Kante

5. Kante

Example: node xi with his 5 edges

Numbering-ruleMinimal-edge-rule

traveling salesman problem

Thank you for your interest

traveling salesman problem

Documents

lowest partial graph

complement partial graph

partial graphs

lowest graph

partialgraphs actual

smallest completegraph

belonging lowest comppartial

symmetric graphs