8a shortest paths more

7/28/2019 8a Shortest Paths More

1/22

18a-ShortestPathsMore

Shortest Paths

in a Graph (contd)Fundamental Algorithms


2/22


All-Pairs Shortest Paths

We now want to compute a table giving thelength of the shortest path between any two

vertices. (We also would like to get the shortest

paths themselves.)

We could just call Dijkstra or Bellman-Ford |V|

times, passing a different source vertex each

time.

It can be done in (V3), which seems to be asgood as you can do on dense graphs.


3/22


Doing APSP with SSSP

Dijkstra would take time(VV2) = (V3) (standard version)

(V (VlgV+ E)) = (V2lgV+ VE))

(modified, Fibonacci heaps),

but doesnt work with negative-weight edges.

Bellman-Ford would take (VVE) = (V2E).


4/22


The Floyd-Warshall AlgorithmRepresent the directed, edge-weighted graph in

adjacency-matrix form.

W= matrix of weights =

w w w

w w w

w w w

11 12 13

21 22 23

31 32 33

wijis the weight of edge (i, j), or if there is nosuch edge.

Return a matrix D, where each entry dij is d(i,j).

Could also return a predecessor matrix, P, whereeach entry pij is the predecessor of j on theshortest path from i.


5/22


Floyd-Warshall: Idea

Considerintermediate vertices of a path:

i j

Say we know the length of the shortest path

from i to j whose intermediate vertices are only

those with numbers 1, 2, ..., k-1. Call this

lengthNow how can we extend this from k-1 to k? In

other words, we want to compute . Can we

use , and if so how?

dijk( )1

dijk( )

dijk( )1


6/22


Floyd-Warshall Idea, 2

Two possibilities:

1. Going through the vertex kdoesnt help thepath through vertices 1...k-1 is still the shortest.

2. There is a shorter path consisting of twosubpaths, one from ito kand one from ktoj.Each subpath passes only through vertices

numbered 1 to k-1.

j

k

i


7/22


Floyd-Warshall Idea,3

Thus,

(since there are no intermediate vertices.)

When k= |V|, were done.

ijk

ijk

ikk

kjk

ij ij

d d d d

d w

( ) ( ) ( ) ( )

( )

min( , )

1 1 1

0Also,


8/22


Dynamic Programming

Floyd-Warshall is a dynamic programming

algorithm:

Compute and store solutions to sub-

problems. Combine those solutions to

solve larger sub-problems.

Here, the sub-problems involve finding the

shortest paths through a subset of the vertices.


9/22


Code for Floyd-WarshallFloyd-Warshall(W)

simple.)areoperationsbecauseality,proportion

ofconstant(Small(Vzippya:timeRunning

return

tofordo5to1ifordo

tofor3

verticesofnumber//

3

)(

)(

)(

).

7

),min(6

14

1

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][1

)1()1()1(

0

n

k

kj

k

ik

k

ij

k

ij

D

njn

nk

WD

Wrowsn

dddd


10/22


Example of Floyd-Warshall


11/22


Johnsons Algorithm

Makes clever use of Bellman-Ford and Dijkstrato do All-Pairs-Shortest-Paths efficiently on

sparse graphs.

Motivation:By running Dijkstra |V| times, wecould do APSP in time (V2lgV+VElgV)(Modified Dijkstra), or(V2lgV+VE) (Fibonacci

Dijkstra). This beats (V3) (Floyd-Warshall)

when the graph is sparse.

Problem: negative edge weights.


12/22


The Basic Idea

Reweight the edges so that:

1. No edge weight is negative.

2. Shortest paths are preserved. (Ashortest path in the original graph is still

one in the new, reweighted graph.)An obvious attempt: subtract the minimumweight from all the edge weights. E.g. if theminimum weight is -2:

-2 - -2 = 03 - -2 = 5

etc.


13/22


Counterexample

Subtracting the minimum weight from every

weight doesnt work.

Consider:

Paths with more edges are unfairly penalized.

-2 -1

-2

0 1

0


14/22


Johnsons Insight

Add a vertex s to the original graph G, with

edges of weight 0 to each vertex in G:

Assign new weights to each edge as follows:

(u, v) = w(u, v) +d

(s, u) -d

(s, v)

s0

0

0


15/22


Question 1

Are all the s non-negative? Yes:

Otherwise, s u v would beshorter than the shortest path

from s to v.

s

u

v

w(u, v)

0),(),(),(

),(),(),(

vsusvuw

vsvuwus

dd

dd

:Rewriting

),( vuw

),(),(),( vsvuwus dd bemust


16/22


Question 2Does the reweighting preserve shortest paths? Yes: Consider any

path kvvvp ,,, 21

),(),()(

),(),(),(

),(),(),(

),(),(),(

),()(

1

11

3232

2121

1

1

1

k

kkkk

k

i

ii

vsvspw

vsvsvvw

vsvsvvw

vsvsvvw

vvwpw

dd

dd

dd

dd

the sum

telescopes

A value that depends only on

the endpoints, not on the path.

In other words, we have adjusted the lengths of all paths by the same

amount. So this will not affect the relative ordering of the paths

shortest paths will be preserved.


17/22


Question 3

How do we compute the d(s, v)s?

Use Bellman-Ford.

This also tells us if we have a negative-weight

cycle.


18/22


Johnsons: Algorithm

1. Compute G, which consists of G augmented

with s and a zero-weight edge from s to

every vertex in G.

2. Run Bellman-Ford(G, w, s) to obtain the

d(s,v)s

3. Reweight by computing for each edge

4. Run Dijkstra on each vertex to compute

5. Undo reweighting factors to compute d

d


19/22


Johnsons: CLRS


20/22


Johnson: reweighting

(u, v) = w(u, v) + d(s, u) - d(s, v)


21/22


Johnson using Dijkstra


22/22


Johnsons: Running Time

1. Computing G: (V)

2. Bellman-Ford: (VE)

3. Reweighting: (E)

4. Running (Modified) Dijkstra: (V2lgV+VElgV)

5. Adjusting distances: (V2)

Total is dominated by Dijkstra: (V2lgV+VElgV)

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