Single pair shortest path problem

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Single pair shortest path

problemMD Tanvir Anwoar


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Introduction:In graph theory, the shortest path problem is the problem of finding a path betweentwo vertices (or nodes) in a graph such that the sum of the weights of its constituentedges is minimized.This is analogous to the problem of finding the shortest path between two intersectionson a road map: the graph's vertices correspond to intersections and the edges

correspond to road segments, each weighted by the length of its road segment.


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Application:

* The traffic and length of the highways are path weights.* Vehicle routing problem solving* Solving network design problem* In video games, these algorithms are frequently used to find the shortest

path between two points on a map

Purpose of the problem solving:

* Improve Quality of the service, Reducing time and cost


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Problem solving methods :

Dijkstra Algorithm :

a. All edge weights should be non-negativeb. Each iteration of Dijkstra takes O(n^2) for array-based or O(m log n)

for heap-basedc. Total complexity is either O(n^3) or O(mn log n)

d. This is a case where just repeatedly using a solution to a simplerproblem works out fine.


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Bellman-Ford Algorithm:

a. If some edge weights are negativeb. It has complexity O(nm) for a single sourcec. Total sources solution is O(n^2 m), which is O(n4) for dense graphs


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Floyd-Warshall Algorithm :

a. If some edge weights are negativeb. Dynamic programming solution to compute all sources shortest pathsc. Works with negative weights (or without) we assume no negative cyclesd. Complexity O(n^3)


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Floyd-Warshall Algorithm

1. Graph should be directed

2. It may contain negative edges but no negative cycle3. Find the shortest path between all the pairs of nodes

A B


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1. Graph should be directed2. It may contain negative edges but no negative cycle

3. Find the shortest path between all the pairs of nodes

A B

- 1

- 2


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Example :

1 2

34


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Example :

1 2

34

D1

1,4 =


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Example :

1 2

34

D1

1,4 = 1 ,2,4


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Example :

1 2

34

D1

1,4 = 1 ,2,4 ; 1,4


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matrix formation :

D(k)

( I , j ) = { min (D(k-1)

( I , j ) ) ,D(k-1)

( I , k) + D(k-1)

( k , j )}

if k>= 1


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matrix formation :

D(0)0 3 8 X -4

X 0 X 1 7X 4 0 X -42 X -5 1 7X X X 0 0

1 2

35

4

3

-4

-56

8

1

47

2


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matrix formation :

1 2

35

4

D(0)0 3 8 X -4

X 0 X 1 7X 4 0 X -42 X -5 1 7X X X 0 0

3

-4

-56

8

1

47

D(0)

( 2 , 4 ) = 1

1

2


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matrix formation :

D(1)0 3 8 X -4

X 0 X 1 7X 4 0 X X2 5 -5 0 -2X X X 6 0

Adjacency Matrix

D(1)

( 4 , 5 ) = -2

1 2

35

4

3

-4

-56

8

1

47

2

-2


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matrix formation :

D(5)0 1 -3 2 -4

3 0 -4 1 -17 4 0 5 32 -1 -5 0 -28 5 1 6 0

Adjacency Matrix

D(5)

( 2 , 3 ) = -4

1 2

35

4

3

-4

-56

8

1

47

2

-4


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Pseudocode:

1. For i=1 to |V| do2. For j=1 to |V| do3. S[i,j,0] = w(i,j)

4. For k=1 to |V| do5. For i=1 to |V| do6. For j=1 to |V| do7. S[i,j,k] = min {8. S[i,j,k-1],9. S[i,k,k-1]+S[k,j,k-1]

}

10. Return S[:,:,n]# return 2d array with n = |V|


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Single pair shortest path problem

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