(les09)-shortestpath

7/25/2019 (les09)-shortestpath

1/26

Algoritmen & Datastructuren2014 2015

Shortest Path

Philip Dutr

Dept. of Computer Science, K.U.Leuven


2/26

Overview

Copyright Ph.Dutr, Spring 2015

Shortest path problem Shortest Path Tree

Edge Relexation

Generic SPT Algorithm

Edsger Dijkstra

Dijkstras Algorithm

Acyclic Graphs

Negative Weights and cycles Currency Exchange


3/26

Leuven Eindhoven?


Make a graph! Every intersection is a vertex, every road is an edge

Edges are weighted (e.g. by length) and directed(one-way roads)


4/26


5/26

Shortest Path: variants


Which vertices? Source-sink: from one vertex to another

Single source: from one vertex to every other

All pairs: between all pairs of vertices.

Restrictions on edge weights? Non-negative weights Arbitrary weights

Euclidean weights

Cycles?

No cycles No "negative cycles"

Simplifying assumption:There exists a shortest path from sto each vertex v


6/26

Shortest Path


Goal: Find the shortest path froms

to every other vertex Observation: A shortest path tree (SPT) solution exists

Consequence. We can represent the SPT with twovertex-indexed arrays:

distTo[v] is length of shortest path from sto v edgeTo[v] is last edge on shortest path from sto v


7/26

Edge relaxation


Basic operation Relax edge e= vw

distTo[v]is length of shortestknown path from sto v

distTo[w]is length of shortestknown path from sto w

edgeTo[w]is last edge on shortestknown path from sto w

If e= vwgives shorter path to wthrough v,update distTo[w]and edgeTo[w]


8/26

(Very) Generic SPT Algorithm


1.

Initialize distTo[s]= 0 and distTo[v]= for all v

!=s

2. Repeat until optimality conditions (*) are satisfied: Relax any edge

Proof:

Throughout algorithm, distTo[v] is the length of a simple path from s to v andedgeTo[v] is last edge on path

Each successful relaxation decreases distTo[v] for some v

The entry distTo[v] can decrease at most a finite number of times

Problem: How to choose which edge to relax?different algorithms

(*) the shortest path for each v from s has been computed


9/26


10/26

Edsger W. Dijkstra


In their capacity as a tool, computers will be but a ripple on

the surface of our culture. In their capacity as intellectualchallenge, they are without precedent in the cultural historyof mankind.

The use of COBOL cripples the mind; its teaching should,therefore, be regarded as a criminal offence.

It is practically impossible to teach good programming tostudents that have had a prior exposure to BASIC: aspotential programmers they are mentally mutilated beyondhope of regeneration.

Asking whether computers can think is like asking whethersubmarines can swim.

Computer science is no more about computers thanastronomy is about telescopes.


11/26



12/26

Dijkstras algorithm


Consider vertices in increasingorder of distance froms(non-SPT vertex with the lowest distTo[ ] value)

Add vertex to SPT and relax all edges starting from that

vertex

(Example)


13/26

Proof of Dijkstra


Each edge e= vwis relaxed exactly once(when v is added to SPT),leaving distTo[w] distTo[v]+ e.weight()

Inequality holds until algorithm terminates because: distTo[w] cannot increase

distTo[v] will not change(because we have non-negative weights & v is not relaxedagain)

Thus, upon termination, shortest-paths optimalityconditions hold


14/26

Analysis


Almost identical to Prim! Also depends on type of priority queue

Priority queue of Vvertices

Every edge update requires a decrease key operationin the queueproportional to E.log2V)


15/26

Acyclic graphs


Better solution possible: Use topological order


16/26

Acyclic graphs


Topological sort algorithm Consider vertices in topologically order

Relax all edges starting at source vertex

Topological sort algorithm computes SPT in any edge-weighted Directed Acyclic Graph in time proportional E + V

Proof: Each edge e = vwis relaxed exactly once (when v is added),

leaving distTo[w] distTo[v] + e.weight().

Inequality holds until algorithm terminates because: distTo[w]cannot increase

distTo[v]will not change(no edge pointing to v wil be relaxed anymore)

Thus, upon termination, shortest-paths optimality conditions hold.


17/26

Acyclic graphs: longest path


Negate all weights, find shortest (negative weights dont work for Dijkstra, but they do for

Directed Acyclic Graphs)

Application: job scheduling


18/26

Critical path



19/26

Negative weights



20/26

Negative cycles


SPT exists iff nonegative cycles


21/26

New algorithm


Shortest path with negative weights Bellman-Ford algorithm

In each pass, relax all edges

Make V passes

Idea: After pass i, found shortest path containing at most i

edges

Time ~ E.V

If distTo[v]does not change during pass i, no need torelax any edge starting from v in pass i +1

Maintain queue of vertices whose distTo[] has changed

Reuses values from previous iteration


22/26



23/26



24/26

Detect negative cycles


After V-1 passes

If there are still vertices on the queue

Negative cycle is detected

Check optimality condition for all edges


25/26

Currency exchange


How to change gold in $?


26/26

Currency exchange


Find path that maximizes productof weights

Maximum ProductMinimal Sum?

Weights- ln (weight)

(les09)-shortestpath

Documents