Lecture 8 Shortest Path
•
Shortest-path problems
•
Single-source shortest path
•
All-pair shortest path
Overview
• Shortest-path problems
• Single-source shortest path algorithms
‣
Bellman-Ford algorithm
‣
Dijkstra algorithm
• All-Pair shortest path algorithms
‣
Floyd-Warshall algorithm
Weighted graph
Beijing
Qingdao
Lhasa
•Single-source shortest path
•All-pair shortest path
Shanghai
Guangzhou
Shortest Path
Shortest Path
Optimal substructure:
Subpaths of shortest paths are shortest paths.
Cycles. Can a shortest path contains cycles?
Negative weights.
Where are we?
• Shortest-path problems
• Single-source shortest path algorithms
‣
Bellman-Ford algorithm
‣
Dijkstra algorithm
• All-Pair shortest path algorithms
‣
Floyd-Warshall algorithm
Relaxing
The process of relaxing an edge (u,v) consists
of testing whether we can improve the shortest
path to v found so far by going through u.
Shortest-path properties
Notation: We fix the source to be s.
λ[v]: the length of path computed by our algorithms
from s to v.
δ[v]: the length of the shortest path from s to v.
Triangle property
δ[v] <= δ[u] + w(u,v) for any edge (u,v).
Upper-bound property
δ[v] <= λ[v] for any vertex v.
Convergence property
If s⇒u→v is a shortest path, and if λ[u] = δ[u], then
after relax(u,v), we have λ[v] = δ[v].
Bellman-Ford algorithm
E = {(t,x),(t,y),(t,z),(x,t),(y,x),(y,z),(z,x),(z,s),(s,t),(s,y)}
Negative cycle
Correctness
Correctness of Bellman-Ford
If G contains no negative-weight cycles reachable
from s, then the algorithm returns TRUE, and for all
v, λ[v] = δ[v], otherwise the algorithm returns FALSE.
Proof.
No negative cycle. By the fact that the length of simple paths
is bounded by |V| - 1.
After i-th iteration of relax, λ[vi] = δ[vi].
s
v1
Vj
vi
Negative cycle. After i-th iteration of relax,
λ[vi] + w(vi,vj) < λ[vj].
The length of negative cycle is at most |V|.
v
Observation 1
If there is no cycle (DAG), ...
Relax in topological order.
s
v1
vj
vi
v
Weighted DAG application
seam
Observation 2
If there is no negative edge, ...
vj
vi
s
v1
If (s, vi) is the lightest edge sourcing from s, then λ[vi] = δ[vi]
Dijkstra algorithm
Edsger Wybe Dijkstra in 2002
Dijkstra algorithm
Correctness
Assume u is chosen in Step 7, and
1. λ[u] > δ[u]
2. s ⇒ x → y ⇒ u is the shortest path
δ[u] = δ[x] + w(x,y) + w(p2)
= λ[x] + w(x,y) + w(p2)
≥ λ[x] + w(x,y)
≥ λ[y] ≥ λ[u]
Time complexity
What is the time complexity?
How about use d-heap?
Comparisons
Dijkstra
Array
Binary heap
d-ary heap
O(n2)
O(mlogn)
O(dnlogdn + mlogdn)
Dense graph
• dense: m = n1+ε, ε is not too small.
• d-heap, d = m/n
• complexity: O(dnlogdn + mlogdn) = O(m)
Where are we?
• Shortest-path problems
• Single-source shortest path algorithms
‣
Bellman-Ford algorithm
‣
Dijkstra algorithm
• All-Pair shortest path algorithms
‣
Floyd-Warshall algorithm
All-pairs shortest path
Floyd-Warshall algorithm
Floyd-Warshall algorithm
Conclusion
Dijkstra’s algorithm.
• Nearly linear-time when weights are nonnegative.
Acyclic edge-weighted digraphs.
• Faster than Dijkstra’s algorithm.
• Negative weights are no problem.
Negative weights and negative cycles.
• If no negative cycles, can find shortest paths via BellmanFord.
• If negative cycles, can find one via Bellman-Ford.
All-pair shortest path.
• can be solved via Floyd-Warshall
• Floyd-Warshall can also compute the transitive closure of
directed graph.
Which of the following statements are true for shortest path?
♠. If you run Dijkstra's algorithm on an edge-weighted DAG
with positive weights, the order in which the vertices are
relaxed is a topological order.
♥. Let P be a shortest path from some s to t in an edgeweighted digraph G. If the weight of each edge in G is
increased by one, then P will still be a shortest path from s to t
in the modified digraph G'.
♣. Let G be a digraph with positive edge weights. Suppose that
you increase the length of an edge by x. Then, the length of
the shortest path from s to t can increase by more than x.
♦. Bellman-Ford finds the shortest simple path from s to every
other vertex, even if the edge weights are positive or negative
integers, provided there are no negative cycles.