Network Flow
Analysis of Algorithms
Network Flow
•A flow network is a directed graph in whicheach directed edge (u,v) has a positivecapacity c(u,v) > 0
– Capacity is the maximum rate of flow that anyedge can carry
•Two special nodes: source, s, and target, t
Network Flow
•Flow is the rate at which material moves fromthe source to the target
–For example, liquid through pipes, electricalcurrent through a circuit, data through a network,or many other types of information
•The value of flow f is defined to be the totalflow leaving the source vertex
–Can be positive, negative, or zero
–f is also the value of the in-flow at the target
Network Flow
•A flow network is a directed graph with a sourcevertex s, a target vertex t, and every edge labeledwith a positive capacity, c
Initial network with capacities on the edges. No flow yet.
Network Flow
•A flow is then induced at the source
•At every vertex except s and t, the inflow mustbe equal to the outflow at that vertex
Network Flow
•Every flow must satisfy the following 3 properties:
–Capacity constraint: for allu,v V, f(u,v) ≤ c(u,v)
–Skew symmetry: for all u,v V,
f(u,v) = - f(v,u)
–Flow conservation: for all u V - {s,t},
vV f(u,v) = 0
•The maxflow problem: given a directed graph(network) with a capacity on each edge, find a flowof maximum value
Ford-Fulkerson Algorithm
•Ford-Fulkerson is a greedy, iterative methodfor solving the maxflow problem:
–Initially, flow = 0 for all u,v V
–At each stage, an augmenting path is found
•An augmenting path is a path from s to t for which wecan push more flow
–Repeat until no more augmenting paths can befound
Ford-Fulkerson Algorithm
•The residual network
•Consider a flow network (G,c) together with a flow f
•Define a new flow network (Gf,cf), where Gf has the samenode set as G, and cf(u,v) = c(u,v) - f(u,v)
– The capacities of Gf are the unused capacities of G
•Gf is called the residual network, cf is called the residualcapacity
•If we can find an additional flow f’(u,v) in Gf, thenwe can define f*(u,v) to be the current total flowin the network G, where f*(u,v) = f(u,v) + f’(u,v)
Ford-Fulkerson Algorithm
•The “leftover” capacity in the original networkwith flow f creates an augmenting pathavailable for more flow (the residual network)
When the capacity of anedge is used up, we canreverse the direction of theedge to allow data to flowin the opposite direction.
Ford-Fulkerson Algorithm
•The residual capacity cf = the capacity that isleft over after a flow has “run” along an edge
•The bottleneck capacity b = the minimumresidual capacity of the edges along theaugmenting path
b = min { cf(u,v): (u,v) is on path p }
Ford-Fulkerson Algorithm
•Initialize: Start with a flow of 0 on each edge
•Repeat: Push more flow along an augmentingpath until a maximum flow has been found
Ford-Fulkerson Algorithm
•An augmenting path is any undirected pathfrom s to t such that either:
–Flow can be increased along a forward edge (an edge withcapacity that is not full) or
–Flow can be decreased along a backward edge (a reversededge with a flow that is not zero)
Send 10 units offlow from s to t.
Ford-Fulkerson Algorithm
Send 10 units offlow from s to t.
Send 5 units offlow from s to t.
Ford-Fulkerson Algorithm
Send 3 units offlow from s to t.
Terminate when allpaths from s to t areblocked by either afull forward edge oran empty backwardedge
Ford-Fulkerson Algorithm
t
s
4
2
10/10
6/10
10/10
6/10
8/8
6/6
8/9
2/2
Question: In the flow network shown below, how many differentaugmenting paths are there with respect to the given flow f ?
1 0/43
Ford-Fulkerson Algorithm
When the capacities are integers, Ford-Fulkerson runs in timeproportional to the number of edges in the graph times themaximum flow in the graph.
Analysis of Ford-Fulkerson
•How do you choose an augmenting path?
•Will any sequence of paths do just as well?
•Consider the following flow network:
v
1,000,000
s
1,000,000
t
1,000,000
1,000,000
1
s
v
uu
t
999,999
999,999
999,999
v
u
t
999,999
s
1,000,000
999,999
1,000,000
1
11
11
999,999
1
1
1
Two augmenting paths with bottleneck capacities of 1, and resultingresidual network. Running time is proportional to E f, where f is thevalue of the maximum flow; for this example, f = 2,000,000.
Improvements to Ford-Fulkerson
•Edmunds and Karp suggested two heuristicsfor improving performance by selecting betteraugmenting paths
–Always augment by a path of maximumbottleneck capacity (the fattest path) or
–Always augment by a path with the fewestnumber of edges (the shortest path)
Is Ford-Fulkerson Optimal?
•Since Ford-Fulkerson is a greedy method, howdo we know if we have found the optimal(maximum) flow?
•Consider:
–A flow is maximum if and only if its residual networkcontains no augmenting paths
–There will be no more augmenting paths when allthe bottleneck capacities have been used up
–The bottleneck capacities are found from theminimum cut of the network
Network Cuts
•Definition: A cut is a partition of the verticesinto two disjoint sets, A and B
•An st-cut is a cut where the source s is in A
and the sink t is in B
A
B
s
t
Network Cuts
•The capacity of a cut is the sum of thecapacities of the edges from A to B
•In the example below, A = {s} and B = {all othervertices}
Network Cuts
•In this example, A = {the dark-colored vertices}and B = {the light-colored vertices}
– Only include the edges from A to B whendetermining the capacity across the cut; do notinclude the edges from B to A or from A to A
Network Cuts
Question: In the flow network shown here,what is the capacity of the cut with
A = {s,2,4} and B = {1,3,t}?
Mincut problem: For all possible ways to cut a graph with s A
and t B, find the cut of minimum capacity.
1
t
s
4
2
3
10
10
10
10
8
6
9
2
4
Flows and Cuts
•The net flow across a cut (A, B) is the sum of the flows on itsedges from A (gray vertices) to B (white vertices) minus thesum of the flows on its edges from B to A
•Flow-value lemma: Let f be any flow and let (A, B) be any cut.Then, the net flow across (A, B) equals the value of f.
•In other words, every cut must have the same net flow!
Maxflow-Mincut Theorem
•According to the flow-value lemma, the netflow is the same across all s-t cuts, includingthe minimum s-t cut
– If you send as much flow as you possibly can (themaximum flow), then the capacities of theminimum cut will be saturated and you cannotsend more flow
•Therefore, the maximum flow = minimum cut
Maxflow-Mincut Theorem
•Proof:
–Suppose (A, B) is a cut such that vertex u A
–We can move u to B without altering the flow because ofthe conservation of flow (flow in = flow out), thus f(A, B) =f(A - {u}, B + {u})
–Therefore, every flow (even the maximum flow) is thesame regardless of where you cut (even the minimum cut)
=
A
B
s
w
4
u
w3
w2
w1
t
w
5
B+{u}
A-{u}
s
w
4
u
w1
w3
w2
t
w
5
Flow Equivalence Conditions
•The following three conditions are equivalentfor any flow f:
1.There exists a cut whose capacity equals the valueof the flow f
2.Flow f is a maxflow
3.There is no augmenting path with respect to f
Flow Equivalence Conditions
1.There exists a cut whose capacity equals the value of theflow f
2.Flow f is a maxflow
1 implies 2:
–Suppose that (A, B) is a cut with capacity equal to thevalue of f
–Then, the value of any other flow f ' ≤ capacity of
cut(A, B) = value of f (by assumption on 1 above)
–Thus, f is a maxflow
Flow Equivalence Conditions
2.Flow f is a maxflow
3.There is no augmenting path with respect to f
2 implies 3 using the contrapositive (3 implies 2):
–Suppose that there is another augmenting path withrespect to the current flow f
–We could send more flow along this path
–Thus, the current flow f is not a maxflow
Flow Equivalence Conditions
3.There is no augmenting path with respect to f
1.There exists a cut whose capacity equals the value ofthe flow f
3 implies 1:
–If there is no augmenting path with respect to f(i.e., if there is no path from s to t where we couldsend more flow), then this creates a cut
–The capacity of this cut defines the bottleneck,and has flow with value f (from the flow-valuelemma)
Applications of Network Flow
•Data mining
•Open-pit mining
•Bipartite matching
•Network reliability
•Baseball elimination
•Image segmentation
•Network connectivity
•Distributed computing
•Security of statistical data
•Egalitarian stable matching
•Multi-camera scene reconstruction
•Sensor placement for homeland security
•Many, many, more
Edge-Disjoint Paths
•A set of paths is edge-disjoint if their edge setsare disjoint, i.e., no two paths share an edge
– Multiple paths may go through some of the nodes
•The Edge-Disjoint Paths Problem is to find themaximum number of edge-disjoint s-t paths in G
•The maximum number of edge-disjoint s-t pathsis equal to the minimum number of edgeswhose removal separates s from t
Edge-Disjoint Paths
•The Ford-Fulkerson algorithm can be used to find amaximum set of edge-disjoint s-t paths in runtimeproportional to VE
•Given a graph G = (V,E) with source s and sink t,create a flow network with a capacity of 1 on eachedge
Maximum number of edge-disjoint paths = 3
s
t
Maximum Bipartite Matching
•The assignment/matching/marriage problem
•Goal: Find a maximum bipartite match
–Example: match L machines with R tasks to beperformed
–Example: match L applicants to R job offers
•Edge (x,y) represents “applicant x wants job y” and“company y wants applicant x”
•The goal is to make the maximum number ofmatches of student to jobs
–Multiple acceptance is forbidden!
Maximum Bipartite Matching
XY
XY
A bipartite graph. On the left is a match with cardinality 2,and on the right is a match with cardinality 3. The match onthe right is optimal because 3 > 2.
Maximum Bipartite Matching
•To find a maximum bipartite matching:construct a flow network, where a flowcorrespond to a match
1.Construct a flow network G’
2.Set all weights to 1 (initial capacities)
3.Run Ford-Fulkerson
4.Maximum bipartite matching = maximum flow
•Runtime is proportional to VE
Maximum Bipartite Matching
s
t
XYXY
The corresponding flow network G with maximum flow shown.Each edge in G has capacity of 1. Shaded edges have a flow of 1,and all other edges carry no flow.The shaded edges from X to Ycorrespond to those in a maximum matching of the bipartite graph.
G
G
Baseball Elimination
Four baseball teams are trying to finish in first place. Currently,each team has the following number of wins:
New York: 92Baltimore: 91Toronto: 91Boston: 90
There are five games left in the season; these consist of all possiblepairings of the four teams, except for New York and Boston.
Can Boston finish in first place, or at least tie for first?
Baseball Elimination
New York: 92 Baltimore: 91 Toronto: 91 Boston: 90
Boston can finish with at most 92 wins. Cumulatively, the otherthree teams have 274 wins currently, and their three gamesagainst each other will produce exactly three more wins, for afinal total of 277. But 277 wins divided by three teams meansthat one of them must end up with more than 92 wins.
Therefore, Boston cannot end up in first place, or even tie forfirst.
Baseball Elimination
Suppose each team has the following number of wins:
New York: 90 Baltimore: 88 Toronto: 87 Boston: 79
The remaining games are as follows: Boston still has four gamesagainst each of the other three teams. Baltimore has one moregame against each of New York and Toronto. New York and Torontostill have six games left to play against each other.
Is Boston eliminated?
Baseball Elimination
New York: 90 Baltimore: 88 Toronto: 87 Boston: 79
Boston can end with at most 91 wins. Together, New York andToronto already have 177 wins; their six remaining games willresult in a total of 183, and since 183/2 > 91, one of them mustend up with more than 91 wins. Therefore, Boston is eliminated.
Interestingly, in this instance, we cannot prove that Boston iseliminated by averaging all three teams. The three teams ahead ofBoston together have a total of 265 wins with 8 games left amongthem. This is a total of 273, and 273/3 = 91. Averaging over allthree teams does not prove that Boston could not tie for first. Youmust be careful to choose the correct set!
Baseball Elimination
In general, suppose we have a set S of teams, and for each x S,its current number of wins is wx. Also, for two teams x, y S, theystill have to play gxy games against each other. We also have aspecific team z, for which we want to know the best outcome.Let’s suppose that z requires m wins total to end up in first place.
Let S = S – {z}, and let g* = the total number of games leftbetween all pairs of teams in S. Now construct a flow network Gto determine whether z has been eliminated: include nodes s andt, a vertex uxy for each pair of teams x, y S with a non-zeronumber of games left to play against each other, and a node vx foreach team x S.
Baseball Elimination
Include the following edges:
•(s, uxy) with capacity gxy
•(uxy, vx) and (uxy, vy) with capacities gxy
•(vx, t) with capacity m - wx
If there is a flow of value at least g*, then it is possible for theoutcomes of all remaining games to yield a situation where noteam has more than m wins. Hence, if team z wins all itsremaining games, it can still achieve at least a tie for first place.
Team z has been eliminated if and only if the maximum flow in
G has value strictly less than g*.
Baseball Elimination
The maximum flow has value = 7, whereas g* = 6 + 1 + 1 = 8(g* is the total number of games left between teams in S).
Since max flow < g*, Boston cannot at least tie for first place.
wNewYork
= 90
wBaltimore = 88
wTorontowBoston
= 87
= 79
gNY-Tor = 6gNY-Bal = 1gBal-Tor = 1
Boston has 12 gamesleft, so m = 91
s
NY-Balt
Balt-Tor
NY
Balt
NY-Tor
1
1
6
91 - 88 = 3
Tor
91 - 87 = 4 t
91 - 90 = 1
6
6
1
1
1
1
Baseball Elimination
QUESTION: How many vertices and edges, respectively,are there in a flow network that is constructed todetermine whether one team is mathematicallyeliminated from a baseball league containing N teams?Assume the worst case (when there is a game remainingbetween every pair of teams in the league).
a)N and N2
b)N2 and N2
c)N2 and N3
d)N2 and N4
Network Flow Holy Grail
•Worst-case analysis is generally not useful forpredicting or comparing maxflow algorithmperformance in practice
•Current best in practice: Push-relabel methodwith gap relabeling runs in time proportional to
E3/2
Network Flow Holy Grail
Year
Method
Worst Case
Discovered By
1951
Simplex
E3 C
Dantzig
1955
Augmenting path
E C*
Ford-Fulkerson
1970
Shortest augmenting path
E3
Dinitz andEdmonds-Karp
1970
Fattest augmenting path
E2 log E log(E C)
Dinitz andEdmonds-Karp
1977
Blocking flow
E5/2
Cherkasky
1978
Blocking flow
E7/3
Galil
1983
Dynamic trees
E2 log E
Sleator and Tarjan
1985
Capacity scaling
E2 log C
Gabow
1997
Length function
E1/2 log E log C
Goldberg and Rao
2012
Compact network
E2 / log E
Orlin
?
?
E
?