Shortest Paths
A
8
0
4
2
B
8
2
Shortest Paths
7
5
E
C
3
2
1
9
D
8
F
3
5
1
Outline and Reading
Weighted graphs (§7.1)


Shortest path problem
Shortest path properties
Dijkstra’s algorithm (§7.1.1)


Algorithm
Edge relaxation
The Bellman-Ford algorithm (§7.1.2)
Shortest paths in dags (§7.1.3)
All-pairs shortest paths (§7.2.1)
Shortest Paths
2
Weighted Graphs
In a weighted graph, each edge has an associated numerical
value, called the weight of the edge
Edge weights may represent, distances, costs, etc.
Example:

In a flight route graph, the weight of an edge represents the
distance in miles between the endpoint airports
SFO
PVD
ORD
LGA
HNL
LAX
DFW
Shortest Paths
MIA
3
Shortest Path Problem
Given a weighted graph and two vertices u and v, we want to
find a path of minimum total weight between u and v.

Length of a path is the sum of the weights of its edges.
Example:

Shortest path between Providence and Honolulu
Applications



Internet packet routing
Flight reservations
Driving directions
SFO
PVD
ORD
LGA
HNL
LAX
DFW
Shortest Paths
MIA
4
Shortest Path Properties
Property 1:
A subpath of a shortest path is itself a shortest path
Property 2:
There is a tree of shortest paths from a start vertex to all the other
vertices
Example:
Tree of shortest paths from Providence
SFO
PVD
ORD
LGA
HNL
LAX
DFW
Shortest Paths
MIA
5
Dijkstra’s Algorithm
The distance of a vertex
v from a vertex s is the
length of a shortest path
between s and v
Dijkstra’s algorithm
computes the distances
of all the vertices from a
given start vertex s
Assumptions:



the graph is connected
the edges are
undirected
the edge weights are
nonnegative
We grow a “cloud” of vertices,
beginning with s and eventually
covering all the vertices
We store with each vertex v a
label d(v) representing the
distance of v from s in the
subgraph consisting of the cloud
and its adjacent vertices
At each step


Shortest Paths
We add to the cloud the vertex
u outside the cloud with the
smallest distance label, d(u)
We update the labels of the
vertices adjacent to u
6
Edge Relaxation
Consider an edge e = (u,z)
such that


d(u) = 50
u is the vertex most recently
added to the cloud
z is not in the cloud
s
The relaxation of edge e
updates distance d(z) as
follows:
u
e
z
d(u) = 50
d(z)  min{d(z),d(u) + weight(e)}
s
Shortest Paths
u
d(z) = 75
d(z) = 60
e
z
7
Example
A
8
0
4
A
8
2
B
8
7
2
C
2
1
D
9
E

F
A
8
4
5
B
8
7
5
E
2
C
3
B
2
7
5
E
D
8
A
8
3
5
0
4
2
C
3
1
9
0
4
2
F
2
8
4
2
3

0
2
1
9
D
11
F
3
5
B
2
Shortest Paths
7
7
5
E
C
3
2
1
9
D
8
F
5
8
3
Example (cont.)
A
8
0
4
2
B
2
7
7
C
3
5
E
2
1
9
D
8
F
3
5
A
8
0
4
2
B
2
Shortest Paths
7
7
C
3
5
E
2
1
9
D
8
F
3
5
9
Dijkstra’s Algorithm
A priority queue stores
the vertices outside the
cloud


Key: distance
Element: vertex
Locator-based methods


insert(k,e) returns a
locator
replaceKey(l,k) changes
the key of an item
We store two labels
with each vertex:


Distance (d(v) label)
locator in priority
queue
Algorithm DijkstraDistances(G, s)
Q  new heap-based priority queue
for all v  G.vertices()
if v = s
setDistance(v, 0)
else
setDistance(v, )
l  Q.insert(getDistance(v), v)
setLocator(v,l)
while Q.isEmpty()
u  Q.removeMin()
for all e  G.incidentEdges(u)
{ relax edge e }
z  G.opposite(u,e)
r  getDistance(u) + weight(e)
if r < getDistance(z)
setDistance(z,r)
Q.replaceKey(getLocator(z),r)
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Analysis
Graph operations

Method incidentEdges is called once for each vertex
Label operations


We set/get the distance and locator labels of vertex z O(deg(z)) times
Setting/getting a label takes O(1) time
Priority queue operations


Each vertex is inserted once into and removed once from the priority
queue, where each insertion or removal takes O(log n) time
The key of a vertex in the priority queue is modified at most deg(w)
times, where each key change takes O(log n) time
Dijkstra’s algorithm runs in O((n + m) log n) time provided the
graph is represented by the adjacency list structure

Recall that
Sv deg(v) = 2m
The running time can also be expressed as O(m log n) since the
graph is connected
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11
Extension
Using the template
method pattern, we
can extend Dijkstra’s
algorithm to return a
tree of shortest paths
from the start vertex
to all other vertices
We store with each
vertex a third label:

parent edge in the
shortest path tree
In the edge relaxation
step, we update the
parent label
Algorithm DijkstraShortestPathsTree(G, s)
…
for all v  G.vertices()
…
setParent(v, )
…
for all e  G.incidentEdges(u)
{ relax edge e }
z  G.opposite(u,e)
r  getDistance(u) + weight(e)
if r < getDistance(z)
setDistance(z,r)
setParent(z,e)
Q.replaceKey(getLocator(z),r)
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Why Dijkstra’s Algorithm
Works
Dijkstra’s algorithm is based on the greedy
method. It adds vertices by increasing distance.




Suppose it didn’t find all shortest
distances. Let F be the first wrong
vertex the algorithm processed.
When the previous node, D, on the
true shortest path was considered,
its distance was correct.
But the edge (D,F) was relaxed at
that time!
Thus, so long as d(F)>d(D), F’s
distance cannot be wrong. That is,
there is no wrong vertex.
Shortest Paths
A
8
0
4
2
B
2
7
7
C
3
5
E
2
1
9
D
8
F
5
13
3
Why It Doesn’t Work for
Negative-Weight Edges
Dijkstra’s algorithm is based on the greedy
method. It adds vertices by increasing distance.

If a node with a negative
incident edge were to be added
late to the cloud, it could mess
up distances for vertices already
in the cloud.
A
8
0
4
6
B
2
7
7
C
0
5
E
5
1
-8
D
9
F
5
C’s true distance is 1, but
it is already in the cloud
with d(C)=5!
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14
4
Bellman-Ford Algorithm
Works even with negative- Algorithm BellmanFord(G, s)
weight edges
for all v  G.vertices()
if v = s
Must assume directed
setDistance(v, 0)
edges (for otherwise we
else
would have negativesetDistance(v, )
weight cycles)
for i  1 to n-1 do
Iteration i finds all shortest
for each e  G.edges()
paths that use i edges.
{ relax edge e }
Running time: O(nm).
u  G.origin(e)
z  G.opposite(u,e)
Can be extended to detect
r  getDistance(u) + weight(e)
a negative-weight cycle if it
if r < getDistance(z)
exists

setDistance(z,r)
How?
Shortest Paths
15
Bellman-Ford Example
Nodes are labeled with their d(v) values
0
8
4
0
8
-2
7

-2
1

3
-2
8

7

9

0
5
-2
4
7
-2
6 1



5
4
0
8
4
-2
1
-2
3
1
9

-2
5 8
-2

3

8
4
9
9

4
-1
5
7
3
5
-2
Shortest Paths
1
1
-2
9
-1
4
9
5
16
DAG-based Algorithm
Works even with
negative-weight edges
Uses topological order
Doesn’t use any fancy
data structures
Is much faster than
Dijkstra’s algorithm
Running time: O(n+m).
Algorithm DagDistances(G, s)
for all v  G.vertices()
if v = s
setDistance(v, 0)
else
setDistance(v, )
Perform a topological sort of the vertices
for u  1 to n do {in topological order}
for each e  G.outEdges(u)
{ relax edge e }
z  G.opposite(u,e)
r  getDistance(u) + weight(e)
if r < getDistance(z)
setDistance(z,r)
Shortest Paths
17
DAG Example
Nodes are labeled with their d(v) values
and sorted in topological order
1
1
0
8
4
0
8
-2
3
2
7

-2

4
1

3
-5
3


8
2
7
9

6
5
-5

5
0
8
-5
2
1

3
6

4
9
5

6
4
5
0
8
-2
7
4
1
1
8
5

-2
3
1
3
4
4
-2
4
1
-2
4
-1
3
5
2
7
9
7

5
-5
5
Shortest Paths
0
1
3
6
4
1
-2
9
4
7
(two steps)
-1
5
5
18
All-Pairs Shortest Paths
Algorithm AllPair(G) {assumes vertices 1,…,n}
Find the distance
for all vertex pairs (i,j)
between every pair of
if i = j
vertices in a weighted
D0[i,i]  0
directed graph G.
else if (i,j) is an edge in G
We can make n calls to
D0[i,j]  weight of edge (i,j)
Dijkstra’s algorithm (if no
else
negative edges), which
D0[i,j]  + 
takes O(nmlog n) time.
for k  1 to n do
Likewise, n calls to
for i  1 to n do
Bellman-Ford would take
for j  1 to n do
Dk[i,j]  min{Dk-1[i,j], Dk-1[i,k]+Dk-1[k,j]}
O(n2m) time.
return Dn
We can achieve O(n3)
time using dynamic
Uses only vertices numbered 1,…,k
i
(compute weight of this edge)
programming (similar to
j
the Floyd-Warshall
Uses only vertices
Uses only vertices
algorithm).
numbered 1,…,k-1
k
Shortest Paths
numbered 1,…,k-1
19