Single Source Shortest-Path:
The General Case (with negative edges)
Bellman-Ford algorithm.
Iteratively relax all edges |V|-1 times
If no negative cycles, then: 1) d(v)= (v)
2) triangle inequality should hold
COSC 3101N J. Elder
All-Pairs Shortest Paths
Given a directed graph G = (V, E), weight function w : E R, |V| = n.→ Assume no negative weight cycles.
Goal: create an n × n matrix of shortest-path distances δ(u, v). Could run BELLMAN-FORD once from each vertex:
O(V2E)—which is O(V4) if the graph is dense (E = (V2)).
If no negative-weight edges, could run Dijkstra’s algorithm once from
each vertex:
O(V E lg V) with binary heap—O(V3 lg V) if dense,
All Pairs Shortest Path – Floyd-Warshall Algorithm
Dynamic programming approach.
Use optimal substructure of shortest paths: Any subpath of
a shortest path is a shortest path.
Create a 3-dimensional table:
Let dij(k) –shortest path weight of any path from i to j where all
intermediate vertices are from the set {1,2, …, k}.
Computing d
ij(k) Base condition: dij(0) = ?
dij(0) = wij.
For k>0:
Let p=<vi, . . . , vj> be a shortest path from vertex i to vertex j
with all intermediate vertices in {1,2, …, k}.
If k is not an intermediate vertex, then all intermediate
vertices are in {1,2, …, k-1}.
If k is an intermediate vertex, then p is composed of 2
Algorithm
Running time = ?
O(n3).
Memory required = ?
All-Pairs Shortest Path: Johnson’s Algorithm
Idea: If the graph is sparse (|E|<<|V|2), it pays to run
Dijkstra’s algorithm once from each vertex.
O(VE log V) using binary heap, O(V2 log V + V E) using
Fibonacci heap.
But Dijkstra’s algorithm does not handle negative edges.
Johnson’s Algorithm: reweight edges to form equivalent
graph with non-negative edges.
Floyd-Warshall still has advantages:
very simple implementation
no fancy data structures
The Problem
Use a graph to model material that flows through conduits.
Each edge represents one conduit, and has a capacity, which
is an upper bound on the flow rate = units/time.
Can think of edges as pipes of different sizes. But flows
don’t have to be of liquids.
Want to compute max rate that we can ship material from a
What is a Flow Network?
Each edge (u,v) has a nonnegative capacity c(u,v). If (u,v) is not in E, assume c(u,v)=0.
We have a source s, and a sink t.
Assume that every vertex v in V is on some path from s to t.
What is a Flow in a Network?
For each edge (u,v), the flow f(u,v) is a real-valued function
that must satisfy 3 conditions:
Note that skew symmetry condition implies that f(u,u)=0.
Capacity constraint: u v V f u v, , ( , ) c u v( , )
Skew symmetry: u v V f u v, , ( , ) f v u( , )
Flow conservation: { , }, ( , ) 0 v V
Example of a Flow:
f(v2, v1) = 1, c(v2, v1) = 4.
f(v1, v2) = -1, c(v1, v2) = 10.
f(v3, s) + f(v3, v1) + f(v3, v2) + f(v3, v4) + f(v3, t) =
0 + (-12) + 4 + (-7) + 15 = 0
flow
The Value of a flow
The value of a flow is given by
V
v
V
v
t
v
f
v
s
f
f
|
(
,
)
(
,
)
|
Example:
|f| = f(s, v1) + f(s, v2) + f(s, v3) + f(s, v4) + f(s, t) = 11 + 8 + 0 + 0 + 0 = 19
|f|= f(s, t) + f(v1, t) + f(v2, t) + f(v3, t) + f(v4, t) =
A flow in a network
We assume that there is only flow in one direction at a time.
Sending 7 trucks from Edmonton to Calgary and 3 trucks from
Multiple Sources Network
We have several sources and several targets.
The Ford-Fulkerson Method
Try to improve the flow, until we reach the maximum.
The residual capacity of the network with a flow f is given by:
)
,
(
)
,
(
)
,
(
u
v
c
u
v
f
u
v
c
f
Example of residual capacities
Network:
Residual Network:
The residual network
The edges of the residual network are the edges on which
Augmenting Paths
An augmenting path p is a simple path from s to t on the
residual network.
We can put more flow from s to t through p.
We call the maximum capacity by which we can increase the
flow on p the residual capacity of p.
}
on
is
)
,
(
:
)
,
(
min{
)
(
p
c
u
v
u
v
p
Augmenting Paths - example
The residual capacity of p = 4.
Example
Flow(1)
Residual(1)
Flow(2)
Residual(2)
Augmenting Paths – example
The maximum possible flow through the cut = 12 + 7 + 4 = 23
The network has a capacity of at most 23.
This is called a
minimum cut
.
Cuts of Flow Networks
A cut in a network is a partition of V into S and T=V-S so
The net flow through a cut (S,T)
f(S,T) = 12 – 4 + 11 = 19
The capacity of (S,T)
c(S,T)= 12+ 0 + 14 = 26
Lemma 26.5
Corollary 26.6
The value of any flow f in a flow network G is bounded from
Theorem 26.7 (Max-flow min-cut theorem)
If f is a flow in a flow network G=(V,E), with source s and
sink t, then the following conditions are equivalent:
1. f is a maximum flow in G.
2. The residual network Gf contains no augmented paths.
Example
Resulting
Flow =
Original Network
augmenting path
Example
Resulting
Flow = 4
Residual Network
Example
Residual Network
Example
Resulting
Flow = 11
Residual Network
Example
Residual Network
Example
Resulting
Flow = 19
Residual Network
Example
Residual Network
Residual Network
Example
Resulting
Flow = 23
Analysis
O(E)
Analysis
If capacities are all integer, then each augmenting path
raises |f| by ≥ 1.
If max flow is f*, then need ≤ |f*| iterations time is O(E|⇒
f*|).
Note that this running time is not polynomial in input size. It
depends on |f*|, which is not a function of |V| or |E|.
If capacities are rational, can scale them to integers.
The basic Ford-Fulkerson Algorithm
With time O ( E |f*|), the algorithm is not polynomial.
This problem is real: Ford-Fulkerson may perform very badly
if we are unlucky:
Run Ford-Fulkerson on this example
Augmenting Path
Run Ford-Fulkerson on this example
Augmenting Path
Run Ford-Fulkerson on this example
The Edmonds-Karp Algorithm
A small fix to the Ford-Fulkerson algorithm makes it work in
polynomial time.
The Edmonds-Karp Algorithm
Compute the path in line 4 using breadth-first search on
residual network.
The augmenting path p is the shortest path from s to t in
the residual network (treating all edge weights as 1).
The Edmonds-Karp Algorithm - example
Further Improvements
Push-relabel algorithm ([CLRS, 26.4]) – O(V2 E).
