Flyod's algorithm for finding shortest path
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Prof. Madhumita Tamhane Floyd's ALGORITHM
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Transcript of Flyod's algorithm for finding shortest path
Prof. Madhumita Ta
mhane
Floyd's ALGORITHM
Prof. Madhumita Ta
mhaneMB
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. • Above problem is explained to find shortest route.
Floyd's ALGORITHM
Prof. Madhumita Ta
mhaneD( 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 the shortest route to another vertex.
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
∞
∞∞
∞∞ ∞
∞
∞∞
Prof. Madhumita Ta
mhaneD( 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.
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
∞ ∞∞ ∞
∞∞ ∞
∞ ∞
Prof. Madhumita Ta
mhaneD( 0 ) R( 0 )
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.
∞ ∞∞ ∞
∞∞ ∞
∞ ∞
Prof. Madhumita Ta
mhaneD( 0 ) R( 0 )
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.
∞ ∞∞
∞∞ ∞
∞ ∞
Prof. Madhumita Ta
mhaneD( 0 ) R( 0 )
18 + 18 = 36, so replace this item. !Hence finally we have changed all desired cells in first iteration where sum is less than the cell data. !Now whichever cell is changed, we put 1 (for node A) in corresponding cells in matrix R(0). Finally at the end of first iteration…..
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.
∞ ∞
∞ ∞
Prof. Madhumita Ta
mhaneR( 0 )
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
∞ ∞
∞ ∞
Prof. Madhumita Ta
mhaneWe have now completed one iteration. We rename the new matrices:
Subsequent iterations are now shown completed:
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
∞ ∞
∞ ∞
Prof. Madhumita Ta
mhane
In second iteration, second row and column are highlighted. Similarly other cells are compared with Sum of corresponding row and column and changed as before. Corresponding cell at R(0) will have 2 (for node M). Result is …
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
∞
∞ ∞
∞
Prof. Madhumita Ta
mhaneD( 1 ) R( 1 )
The items have been altered accordingly:
Subsequent iterations are now shown completed:
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
Prof. Madhumita Ta
mhaneD( 2 ) R( 2 )
We can now rename the matrices:
Similarly do 3rd, 4th and 5th iteration by highlighting 3rd, 4th and 5th row and column respectively. Change cells of R(0) to 3, 4 or 5 accordingly. Final matrix is …
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
Prof. Madhumita Ta
mhaneD( 2 ) R( 2 )
Next iteration:
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
Prof. Madhumita Ta
mhaneD( 2 ) R( 2 )
Next iteration, the items are altered appropriately :
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
Prof. Madhumita Ta
mhaneD( 3 ) R( 3 )
The matrices are renamed :
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
Prof. Madhumita Ta
mhaneThe next iteration :
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
Prof. Madhumita Ta
mhaneIn this iteration, you do not need to change any of the items, so we go onto the next iteration :
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
Prof. Madhumita Ta
mhaneA 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 )
Prof. Madhumita Ta
mhaneThere 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
Prof. Madhumita Ta
mhaneThese 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:
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
Prof. Madhumita Ta
mhane
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.
!The route matrix gives us an idea about the next vertex to visit on route - 1 represents A, 2 - M, etc.
Prof. Madhumita Ta
mhane
1
5
4
3
2
75 35
32
15
40
30
70
Q1.
Prof. Madhumita Ta
mhane1 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?
Prof. Madhumita Ta
mhane
1
5
4
3
2
20 15 12
35
50
50
10
Q2.
Prof. Madhumita Ta
mhaneD( 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
