Dijkstra algorithm ds 57612334t4t44.ppt

Dijkstra’s Algorithm
Slide Courtesy: Uwash, UT
1

Single-Source Shortest Path Problem
Single-Source Shortest Path Problem - The
problem of finding shortest paths from a source
vertex v to all other vertices in the graph.

Applications
- Maps (Map Quest, Google Maps)
- Routing Systems

Dijkstra's algorithm
Dijkstra's algorithm - is a solution to the single-source
shortest path problem in graph theory.
Works on both directed and undirected graphs. However,
all edges must have nonnegative weights.
Input: Weighted graph G={E,V} and source vertex v∈V,
such that all edge weights are nonnegative
Output: Lengths of shortest paths (or the shortest paths
themselves) from a given source vertex v∈V to all other
vertices

Approach
• The algorithm computes for each vertex u the distance to u from
the start vertex v, that is, the weight of a shortest path between
v and u.
• the algorithm keeps track of the set of vertices for which the
distance has been computed, called the cloud C
• Every vertex has a label D associated with it. For any vertex u,
D[u] stores an approximation of the distance between v and u.
The algorithm will update a D[u] value when it finds a shorter
path from v to u.
• When a vertex u is added to the cloud, its label D[u] is equal to
the actual (final) distance between the starting vertex v and
vertex u.
5

Dijkstra pseudocode
Dijkstra(v1, v2):
for each vertex v: // Initialization
v's distance := infinity.
v's previous := none.
v1's distance := 0.
List := {all vertices}.
while List is not empty:
v := remove List vertex with minimum distance.
mark v as known.
for each unknown neighbor n of v:
dist := v's distance + edge (v, n)'s weight.
if dist is smaller than n's distance:
n's distance := dist.
n's previous := v.
reconstruct path from v2 back to v1,
following previous pointers.
6

7
Example: Initialization
A
G
F
B
E
C D
4 1
2
10
3
6
4
2
2
8
5
1
0 ∞
∞ ∞
∞
Pick vertex in List with minimum distance.
∞ ∞
Distance(source) = 0 Distance (all vertices
but source) = ∞

8
Example: Update neighbors'
distance
A
G
F
B
E
C D
4 1
2
10
3
6
4
2
2
8
5
1
0 2
∞ ∞
1
∞ ∞
Distance(B) = 2
Distance(D) = 1

9
Example: Remove vertex with
minimum distance
Pick vertex in List with minimum distance, i.e., D
A
G
F
B
E
C D
4 1
2
10
3
6
4
2
2
8
5
1
0 2
∞ ∞
1
∞ ∞

10
Example: Update neighbors
A
G
F
B
E
C D
4 1
2
10
3
6
4
2
2
8
5
1
0 2
3 3
1
9 5
Distance(C) = 1 + 2 = 3
Distance(E) = 1 + 2 = 3
Distance(F) = 1 + 8 = 9
Distance(G) = 1 + 4 = 5

11
Example: Continued...
A
G
F
B
E
C D
4 1
2
10
3
6
4
2
2
8
5
1
0 2
3 3
1
Pick vertex in List with minimum distance (B) and update neighbors
9 5
Note : distance(D) not
updated since D is
already known and
distance(E) not updated
since it is larger than
previously computed

12
A
G
F
B
E
C D
4 1
2
10
3
6
4
2
2
8
5
1
0 2
3 3
1
9 5
No updating
Pick vertex List with minimum distance (E) and update neighbors

13
A
G
F
B
E
C D
4 1
2
10
3
6
4
2
2
8
5
1
0 2
3 3
1
8 5
Pick vertex List with minimum distance (C) and update neighbors
Distance(F) = 3 + 5 = 8

14
A
G
F
B
E
C D
4 1
2
10
3
6
4
2
2
8
5
1
0 2
3 3
1
6 5
Distance(F) = min (8, 5+1) = 6
Previous distance
Pick vertex List with minimum distance (G) and update neighbors

15
Example (end)
A
G
F
B
E
C D
4 1
2
10
3
6
4
2
2
8
5
1
0 2
3 3
1
Pick vertex not in S with lowest cost (F) and update neighbors
6 5

THE KNOWN
CLOUD
v
P: Next shortest path from
inside the known cloud
v'
Correctness :“Cloudy” Proof
• If the path to v is the next shortest path, the path to v' must be at
least as long. Therefore, any path through v' to v cannot be shorter!
Source
Least cost node
competitor
When a vertex is added to the cloud, it has shortest
distance to source.

26
Dijkstra’s Correctness
• We will prove that whenever u is added to S, d[u] =
(s,u), i.e., that d[u] is minimum, and that equality is
maintained thereafter
• Proof
– Note that v, d[v] ≥ (s,v)
– Let u be the first vertex picked such that there is a shorter
path than d[u], i.e., that d[u]  (s,u)
– We will show that this assumption leads to a contradiction

27
Dijkstra Correctness (2)
• Let y be the first vertex V – S on the actual
shortest path from s to u, then it must be that d[y] =
(s,y) because
– d[x] is set correctly for y's predecessor x S on the
shortest path (by choice of u as the first vertex for which d
is set incorrectly)
– when the algorithm inserted x into S, it relaxed the edge
(x,y), assigning d[y] the correct value

28
• But if d[u] > d[y], the algorithm would have chosen y
(from the Q) to process next, not u  Contradiction
• Thus d[u] = (s,u) at time of insertion of u into S, and
Dijkstra's algorithm is correct
Dijkstra Correctness (3)
[ ] ( , ) (initial assumption)
( , ) ( , ) (optimal substructure)
[ ] ( , ) (correctness of [ ])
[ ] (no negative weights)
d u s u
s y y u
d y y u d y
d y
 
  
 


29
Dijkstra’s Pseudo Code
• Graph G, weight function w, root s
relaxing
edges

Time Complexity: Using List
The simplest implementation of the Dijkstra's algorithm stores
vertices in an ordinary linked list or array
– Good for dense graphs (many edges)
• |V| vertices and |E| edges
• Initialization O(|V|)
• While loop O(|V|)
– Find and remove min distance vertices O(|V|)
• Potentially |E| updates
• Update costs O(1)
Total time O(|V2
| + |E|) = O(|V2
| )

31
Time Complexity: Priority Queue
For sparse graphs, (i.e. graphs with much less than |V2| edges)
Dijkstra's implemented more efficiently by priority queue
• Initialization O(|V|) using O(|V|) buildHeap
• While loop O(|V|)
• Find and remove min distance vertices O(log |V|) using O(log |V|)
deleteMin
• Potentially |E| updates
• Update costs O(log |V|) using decreaseKey
Total time O(|V|log|V| + |E|log|V|) = O(|E|log|V|)
• |V| = O(|E|) assuming a connected graph

Dijkstra algorithm ds 57612334t4t44.ppt

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