An Efficient Solution to Travelling Salesman Problem using ...
Floyd’s Algorithm can be used to help solve Travelling Salesman
48
page 1 M B L A X 48 20 10 19 18 7 10 23 Floyd’s Algorithm can be used to help solve Travelling Salesman and other shortest routes problems. Floyd’s uses a matrix form.
-
Upload
mary-gilmore -
Category
Documents
-
view
22 -
download
1
description
M. 48. B. 19. 10. 23. 20. L. 10. 7. 18. A. X. Floyd’s Algorithm can be used to help solve Travelling Salesman and other shortest routes problems. Floyd’s uses a matrix form. To look back at the network click here:. D( 0 ). R( 0 ). This is the route matrix. - PowerPoint PPT Presentation
Transcript of Floyd’s Algorithm can be used to help solve Travelling Salesman
page 1
MB
L
AX
48
20
10 19
18
710
23
Floyd’s Algorithm can be used to help solve Travelling Salesman and other shortest routes problems.
Floyd’s uses a matrix form.
page 2
D( 0 ) R( 0 )
This is the distance matrix. It initially shows the direct distance between one vertex and the others. The infinity sign denotes no direct route.
This is the route matrix.It will eventually show the next vertex that needs to be taken on route to finding theshortest route to another vertex.
•To view the process step by step click here
•To view the completed matrices click here
A M L B X
A 23 10 18
M 23 10 48
L 10 10 19 7
B 48 19 20
X 18 7 20
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
To look back at the network click here:
D( 0 ) R( 0 )
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
Click here to see step:
A M L B X
A 23 10 18
M 23 10 48
L 10 10 19 7
B 48 19 20
X 18 7 20
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
D( 0 ) R( 0 )
Click here to see step:
23 + 23 = 46, less than so change.
A M L B X
A 23 10 18
M 23 10 48
L 10 10 19 7
B 48 19 20
X 18 7 20
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
23 + 23 = 46, less than so change.
A M L B X
A 23 10 18
M 23 46 10 48
L 10 10 19 7
B 48 19 20
X 18 7 20
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
10 + 23 = 33, so we leave this item
A M L B X
A 23 10 18
M 23 46 10 48
L 10 10 19 7
B 48 19 20
X 18 7 20
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
When is involved we leave the item.
A M L B X
A 23 10 18
M 23 46 10 48
L 10 10 19 7
B 48 19 20
X 18 7 20
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
18 + 23 = 41, so we replace the item.
A M L B X
A 23 10 18
M 23 46 10 48
L 10 10 19 7
B 48 19 20
X 18 7 20
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
18 + 23 = 41, so we replace the item.
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 19 7
B 48 19 20
X 18 7 20
C A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
23 + 10 = 33, so we leave this item.
48 41
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 19 7
B 48 19 20
X 18 7 20
C A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
10 + 10 = 20, so we replace this item.
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 7 20
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
Infinity is involved so we leave this item.
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 7 20
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
18 + 10 = 28 which is more than 7 so we leave this item.
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 7 20
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
Infinity is involved so we leave this item.
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 7 20
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
Infinity is involved so we leave this item.
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 7 20
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
Infinity is involved so we leave this item.
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 7 20
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
Infinity is involved so we leave this item.
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 7 20
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
We replace this item with23 + 18 = 41
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 7 20
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
We replace this item with23 + 18 = 41
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 41 7 20
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
10 + 18 = 28, 7 is less than thisso leave it.
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 41 7 20
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
Infinity is involved so leave this item.
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 41 7 20
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
18 + 18 = 36, so replace this item.
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 41 7 20
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
18 + 18 = 36, so replace this item.
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 41 7 20 36
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
We highlight the first column and row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column. If the sum is less than the item then it should be replaced with the sum.
D( 0 ) R( 0 )
Click here to see step:
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 41 7 20 36
A M L B X
A 1 2 3 4 5
M 1 2 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
R( 0 )
Click here to see step:
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 41 7 20 36
A M L B X
A 1 2 3 4 5
M 1 1 3 4 5
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
D( 0 )
R( 0 )
Click here to see step:
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 41 7 20 36
D( 0 )
A M L B X
A 1 2 3 4 5
M 1 1 3 4 1
L 1 2 3 4 5
B 1 2 3 4 5
X 1 2 3 4 5
R( 0 )
Click here to see step:
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 41 7 20 36
D( 0 )
A M L B X
A 1 2 3 4 5
M 1 1 3 4 1
L 1 2 1 4 5
B 1 2 3 4 5
X 1 2 3 4 5
R( 0 )
Click here to see step:
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 41 7 20 36
D( 0 )
A M L B X
A 1 2 3 4 5
M 1 1 3 4 1
L 1 2 1 4 5
B 1 2 3 4 5
X 1 1 3 4 5
R( 0 )
Click here to see step:
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 41 7 20 36
D( 0 )
A M L B X
A 1 2 3 4 5
M 1 1 3 4 1
L 1 2 1 4 5
B 1 2 3 4 5
X 1 1 3 4 1
We have now completed one iteration. We rename the new matrices:
Click here to see step:
Subsequent iterations are now shown completed:
Click here to see final matrices:
R( 1 )D( 1 )
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 41 7 20 36
A M L B X
A 1 2 3 4 5
M 1 1 3 4 1
L 1 2 1 4 5
B 1 2 3 4 5
X 1 1 3 4 1
Click here to see step:
Subsequent iterations are now shown completed:
Click here to see final matrices:
R( 1 )D( 1 )
A M L B X
A 23 10 18
M 23 46 10 48 41
L 10 10 20 19 7
B 48 19 20
X 18 41 7 20 36
A M L B X
A 1 2 3 4 5
M 1 1 3 4 1
L 1 2 1 4 5
B 1 2 3 4 5
X 1 1 3 4 1
D( 1 ) R( 1 )
The items have been altered accordingly:
Click here to see step:
Subsequent iterations are now shown completed:
Click here to see final matrices:
A M L B X
A 46 23 10 71 18
M 23 46 10 48 41
L 10 10 20 19 7
B 71 48 19 96 20
X 18 41 7 20 36
A M L B X
A 1 2 3 4 5
M 1 1 3 4 1
L 1 2 1 4 5
B 1 2 3 4 5
X 1 1 3 4 1
D( 2 ) R( 2 )
We can now rename the matrices:
Click here to see step:
Click here to see final matrices:
A M L B X
A 46 23 10 71 18
M 23 46 10 48 41
L 10 10 20 19 7
B 71 48 19 96 20
X 18 41 7 20 36
A M L B X
A 2 2 3 2 5
M 1 1 3 4 1
L 1 2 1 4 5
B 2 2 3 2 5
X 1 1 3 4 1
D( 2 ) R( 2 )
Next iteration:
Click here to see step:
Click here to see final matrices:
A M L B X
A 46 23 10 71 18
M 23 46 10 48 41
L 10 10 20 19 7
B 71 48 19 96 20
X 18 41 7 20 36
A M L B X
A 2 2 3 2 5
M 1 1 3 4 1
L 1 2 1 4 5
B 2 2 3 2 5
X 1 1 3 4 1
D( 2 ) R( 2 )
Next iteration, the items are altered appropriately :
Click here to see step:
Click here to see final matrices:
A M L B X
A 20 20 10 29 17
M 20 20 10 29 17
L 10 10 20 19 7
B 29 29 19 38 20
X 17 17 7 20 14
A M L B X
A 3 3 3 3 3
M 3 3 3 3 3
L 1 2 1 4 5
B 3 3 3 3 5
X 3 3 3 4 3
D( 3 ) R( 3 )
The matrices are renamed :
Click here to see step:
Click here to see final matrices:
A M L B X
A 20 20 10 29 17
M 20 20 10 29 17
L 10 10 20 19 7
B 29 29 19 38 20
X 17 17 7 20 14
A M L B X
A 3 3 3 3 3
M 3 3 3 3 3
L 1 2 1 4 5
B 3 3 3 3 5
X 3 3 3 4 3
The next iteration :
Click here to see step:
Click here to see final matrices:
D( 3 ) R( 3 )
A M L B X
A 20 20 10 29 17
M 20 20 10 29 17
L 10 10 20 19 7
B 29 29 19 38 20
X 17 17 7 20 14
A M L B X
A 3 3 3 3 3
M 3 3 3 3 3
L 1 2 1 4 5
B 3 3 3 3 5
X 3 3 3 4 3
In this iteration, you do not need to change any of the items, so we go onto the next iteration :
Click here to see step:
Click here to see final matrices:
D( 3 ) R( 3 )
A M L B X
A 20 20 10 29 17
M 20 20 10 29 17
L 10 10 20 19 7
B 29 29 19 38 20
X 17 17 7 20 14
A M L B X
A 3 3 3 3 3
M 3 3 3 3 3
L 1 2 1 4 5
B 3 3 3 3 5
X 3 3 3 4 3
Click here to see step:
Click here to see final matrices:
A M L B X
A 20 20 10 29 17
M 20 20 10 29 17
L 10 10 20 19 7
B 29 29 19 38 20
X 17 17 7 20 14
A M L B X
A 3 3 3 3 3
M 3 3 3 3 3
L 1 2 1 4 5
B 3 3 3 3 5
X 3 3 3 4 3
D( 4 ) R( 4 )
Click here to see step:
Click here to see final matrices:
There is only one item that we need to change in this case:
D( 4 ) R( 4 )
A M L B X
A 20 20 10 29 17
M 20 20 10 29 17
L 10 10 14 19 7
B 29 29 19 38 20
X 17 17 7 20 14
A M L B X
A 3 3 3 3 3
M 3 3 3 3 3
L 1 2 1 4 5
B 3 3 3 3 5
X 3 3 3 4 3
These are the final matrices, remember that you can use them to redraw the original network. You can then use this to help us solve travelling salesman problems:
Click here to see new network:
D( 5 ) R( 5 )
A M L B X
A 20 20 10 29 17
M 20 20 10 29 17
L 10 10 14 19 7
B 29 29 19 38 20
X 17 17 7 20 14
A M L B X
A 3 3 3 3 3
M 3 3 3 3 3
L 1 2 5 4 5
B 3 3 3 3 5
X 3 3 3 4 3
page 1
MB
L
AX
29
20
10 19
17
710
20
29
17
This network now gives you a better idea of the quickest routes.
Click below to try a question:
The route matrix gives us an idea about the next vertex to visit on route - 1 represents A, 2 - M, etc.
1
5
4
3
2
75 35
32
15
40
30
70
Try this one! Click below when you have completed it to check the answers:
1 2 3 4 5
1 60 30 40 45 77
2 30 30 50 15 47
3 40 50 70 35 67
4 45 15 35 30 32
5 77 47 67 32 64
1 2 3 4 5
1 2 2 3 2 2
2 1 4 4 4 4
3 1 4 4 4 4
4 2 2 3 2 5
5 4 4 4 4 4
D( 5 ) R( 5 )
These are the completed matrices. Are yours correct?
1
5
4
3
2
20 15 12
35
50
50
10
Qu2.
Answers…………….
D( 5 ) R( 5 )
These are the completed matrices. Are yours correct?
1 2 3 4 5
1 100 50 50 65 60
2 50 40 20 35 30
3 50 20 20 15 10
4 65 35 15 24 12
5 60 30 10 12 20
1 2 3 4 5
1 2 2 3 3 3
2 1 3 3 4 3
3 1 2 5 4 5
4 3 2 3 5 5
5 3 3 3 4 3
1
4
3
2
8
3 5
4
3
2
Qu2.
Answers…………….
D( 4 ) R( 4 )
These are the completed matrices. Are yours correct?
1 2 3 4
1 6 3 6 4
2 3 6 5 3
3 6 5 4 2
4 4 3 2 4
1 2 3 4
1 2 2 4 4
2 1 1 3 4
3 4 2 4 4
4 1 2 3 3
![Travelling Salesman Problemweb.tecnico.ulisboa.pt/.../legacy/PRINT/NotebookSC_5TSP_06pp.pdf · Travelling Salesman Problem [:6] 3 This is, however, not a solution to the TSP, because](https://static.fdocuments.net/doc/165x107/5fab2a9ca11f0c17d66c3999/travelling-salesman-travelling-salesman-problem-6-3-this-is-however-not-a-solution.jpg)
