lecture-37 all pairs shortest path problem

8/3/2019 Lecture-37 All Pairs Shortest Path Problem

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All Pairs Shortest

Path Problem


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All pairs shortest path The problem: find the shortest path between

every pair ofvertices ofa graph

The graph: may contain negative edges butno negative cycles

Arepresentation

:a weight matrix whereW(i,j)=0 if i=j.

W(i,j)=g if there is no edge between i and j.W(i,j)=weight ofedge


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The weight matrix and the

graph1 2 3 4 5

1 0 1 1 5

2 9 0 3 2

3 0 4

4 2 0 3

5 3 0

g

gg g g

g g

g g g

v1

v2

v3

v4

v5

32

2

4

1

3

1

93

5


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Solution Dijkstras Algorithm can be used.

How?

Set each vertex as source and call

Dijkstras algo

But we may solve the problem using direct

approach (Algorithm By Floyd)


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Floyds AlgoAssume vertices are numbered as

1,2,3n for simplicity

Uses nn matrix D (n is the number of

vertices in the graph G)

Shortest paths are computed in matrix D

After running the algorithm D[i,j] contains

the shortest distance (cost) between

vertices i and j


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Floyds Algo

Initially we set D[i,j]=W[i,j] Remember

W[i,j]=0 if i==j

W(i,j)= if there is no edge between i andj

W(i,j)=weight of edge

We make n iteration over matrix D

After kth iteration D[i,j] will store the valueof minimum weight path from vertex i tovertex j that does not pass through avertex numbered higher than k


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Floyds Algo In kth iteration we use following formula tocompute D

!

],[],[

],[min],[

11

1

jkDkiD

jiDjiD

kk

k

k

Subscript k denotes the value of matrix D after

the kth iteration (It should not be assumed thereare n different matrices of size nn)


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ViVj

Vk

Shortest Path using intermediate vertices { V1, . . . Vk-1 }

Shortest path using intermediate vertices

{V1, . . . Vk}


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Floyeds AlgorithmFloyd1. DnW // initialize D array to W[ ]

2.

3. forkn 1 to n

4. do forin 1 to n

5. do forjn 1 to n

6. if (D[ i,j] > D[ i, k ] + D[ k,j] ) then7. D[ i,j]n D[ i, k ] + D[ k,j]


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Recovering Paths Use another matrix P

P[i,j] hold the vertex k that led Floyd to find

the smallest value of D[i,j]

If P[i,j]=0 there is no intermediate edge

involved, shortest path is direct edge

between i and j

So modified version of Floyd is given


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Recovering PathsFloyd1. DnW // initialize D array to W[ ]

2. Pn 0

3. forkn 1 to n4. do forin 1 to n5. do forjn 1 to n6

. if

(D

[ i,j] >D

[ i,k

] +D

[k,j] ) then7. D[ i,j]n D[ i, k ] + D[ k,j]

8. P[ i, j] n k


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Example

W = D0 =40 5

2 0 w

w -3 0

1 2 3

1

2

3

00 0

0 0 0

0 0 0

1 2 3

1

2

3

P =

1

2

3

5

-3

24


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D1=40 5

2 0 7

w -3 0

1 2 3

1

2

3

00 0

0 0 1

0 0 0

1 2 3

1

2

3

P =

1

2

3

5

-32

4

k = 1

Vertex 1 can be

intermediate node

D1[2,3] = min( D0[2,3], D0[2,1]+D0[1,3] )

= min (w

, 7)= 7

D1

[3,2] = min( D0

[3,2], D0

[3,1]+D0

[1,2] )= min (-3,w)

= -3

40 5

2 0 w

w -3 0

1 2 3

1

2

3

D0 =


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D2=40 5

2 0 7

-1 -3 0

1 2 3

1

2

3

00 0

0 0 1

2 0 0

1 2 3

1

2

3

P =

D2[1,3] = min( D1[1,3], D1[1,2]+D1[2,3] )

= min (5, 4+7)= 5

D2[3,1] = min( D1[3,1], D1[3,2]+D1[2,1] )

= min (w, -3+2)

= -1

1

2

3

5

-32

4

D1=40 5

2 0 7

w -3 0

1 2 3

1

2

3

k = 2

Vertices 1, 2can be

intermediate


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D3=20 5

2 0 7

-1 -3 0

1 2 3

1

2

3

30 0

0 0 1

2 0 0

1 2 3

1

2

3

P =

D3[1,2] = min(D2[1,2], D2[1,3]+D2[3,2] )

= min (4, 5+(-3))

= 2

D3[2,1] = min(D2[2,1], D2[2,3]+D2[3,1] )

= min (2, 7+ (-1))

= 2

D2=

40 5

2 0 7

-1 -3 0

1 2 3

1

2

3

1

2

3

5

-32

4

k = 3

Vertices 1, 2, 3can be

intermediate


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Printing intermediate nodes on

shortest path from i to jPath (i, j)

k= P[ i, j ];

if (k== 0)return;

Path (i , k);

print( k);Path (k, j);

30 0

0 0 1

2 0 0

1 2 3

1

2

3

P =

1

2

3

5

-3

24

lecture-37 all pairs shortest path problem

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