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10/4/10A. Smith; based on slides by E. Demaine, C. Leiserson, S. Raskhodnikova, K. Wayne
Adam Smith
Algorithm Design and Analysis
LECTURES 18-19Network Flows• Flows, cuts• Ford-Fulkerson• Min-cut/max-flow theorem
Detecting Negative Cycles
Theorem. Can find negative cost cycle in O(mn) time. Add new node t and connect all nodes to t with 0-cost edge. Check if OPT(n, v) = OPT(n-1, v) for all nodes v.
– if yes, then no negative cycles– if no, then extract cycle from shortest path from v to t
Last lecture we proved: if OPT(n, v) < OPT(n-1, v) then we can find a neg.-cost cycle on a path from v to t
Questions to think about: 1. Prove that if there is a negative-cost cycle,
then OPT(n, v) < OPT(n-1, v) for some v (Hint: what would happen ifwe kept running Bellman-
Ford for many iterations?)2. Why do we bother with the extra node t?
v
18 2
5 -23
-15
-11
6
t
0
0
0 0 0
A. Smith; based on slides by S. Raskhodnikova and K. Wayne
Network Flow and Linear Programming
10/15/2008
Soviet Rail Network, 1955
Reference: On the history of the transportation and maximum flow problems.Alexander Schrijver in Math Programming, 91: 3, 2002.
Maximum Flow and Minimum Cut
Max flow and min cut. Two very rich algorithmic problems. Cornerstone problems in combinatorial optimization. Beautiful mathematical duality.
Nontrivial applications / reductions. Data mining. Open-pit mining. Project selection. Airline scheduling. Bipartite matching. Image segmentation. Clustering Network connectivity.
Network reliability. Distributed computing. Egalitarian stable matching. Network intrusion detection. Multi-camera scene reconstruction. Data privacy. Many many more …
Flow network. Abstraction for material flowing through the edges. G = (V, E) = directed graph, no parallel edges. Two distinguished nodes: s = source, t = sink. c(e) = capacity of edge e.
Minimum Cut Problem
s
2
3
4
5
6
7
t
15
5
30
15
10
8
15
9
6 10
10
10 15 4
4
capacity
source sink
Def. An s-t cut is a partition (A, B) of V with s A and t B.
Def. The capacity of a cut (A, B) is:
Cuts
s
2
3
4
5
6
7
t
15
5
30
15
10
8
15
9
6 10
10
10 15 4
4
Capacity = 10 + 5 + 15 = 30
A
cap( A, B) c(e)e out of A
we don’t count
edges into A
s
2
3
4
5
6
7
t
15
5
30
15
10
8
15
9
6 10
10
10 15 4
4 A
Cuts
Def. An s-t cut is a partition (A, B) of V with s A and t B.
Def. The capacity of a cut (A, B) is:
cap( A, B) c(e)e out of A
Capacity = 9 + 15 + 8 + 30 = 62
Min s-t cut problem. Find an s-t cut of minimum capacity.
Minimum Cut Problem
s
2
3
4
5
6
7
t
15
5
30
15
10
8
15
9
6 10
10
10 15 4
4 A
Capacity = 10 + 8 + 10 = 28
Def. An s-t flow is a function f from E to real numbers that satisfies:
For each e E: [capacity] For each v V – {s, t}: [conservation]
Flows
4
0
0
0
0 0
0 4 4
0
0
0
0
0 f (e) c(e)
capacity
flow
s
2
3
4
5
6
7
t
15
5
30
15
10
8
15
9
6 10
10
10 15 4
4 0
4
veve
efef ofout into
)()(water flowing
from source to sink
Def. An s-t flow is a function f from E to real numbers that satisfies:
For each e E: [capacity] For each v V – {s, t}: [conservation]
Def. The value of a flow f is:
Flows
4
0
0
0
0 0
0 4 4
0
0
0
Value = 40
0 f (e) c(e)
capacity
flow
s
2
3
4
5
6
7
t
15
5
30
15
10
8
15
9
6 10
10
10 15 4
4 0
4
veve
efef ofout into
)()(
v( f ) f (e) e out of s
.water flowing
from source to sink
Def. An s-t flow is a function f from E to real numbers that satisfies:
For each e E: [capacity] For each v V – {s, t}: [conservation]
Def. The value of a flow f is:
Flows
10
6
6
11
1 10
3 8 8
0
0
0
11
capacity
flow
s
2
3
4
5
6
7
t
15
5
30
15
10
8
15
9
6 10
10
10 15 4
4 0
Value = 24
veve
efef ofout into
)()(
0 f (e) c(e)
v( f ) f (e) e out of s
.
4
water flowing from source to
sink
Max flow problem. Find s-t flow of maximum value.
Maximum Flow Problem
10
9
9
14
4 10
4 8 9
1
0 0
0
14
capacity
flow
s
2
3
4
5
6
7
t
15
5
30
15
10
8
15
9
6 10
10
10 15 4
4 0
Value = 28
Flow value lemma. Let f be any flow, and let (A, B) be any s-t cut. Then, the net flow across the cut is equal to the amount leaving s.
Flows and Cuts
10
6
6
11
1 10
3 8 8
0
0
0
11
s
2
3
4
5
6
7
t
15
5
30
15
10
8
15
9
6 10
10
10 15 4
4 0
Value = 24
f (e)e out of A
f (e)e in to A
v( f )
4
A
Flow value lemma. Let f be any flow, and let (A, B) be any s-t cut. Then, the net flow across the cut is equal to the amount leaving s.
Flows and Cuts
10
6
6
1 10
3 8 8
0
0
0
11
s
2
3
4
5
6
7
t
15
5
30
15
10
8
15
9
6 10
10
10 15 4
4 0
f (e)e out of A
f (e)e in to A
v( f )
Value = 6 + 0 + 8 - 1 + 11 = 24
4
11
A
Flow value lemma. Let f be any flow, and let (A, B) be any s-t cut. Then, the net flow across the cut is equal to the amount leaving s.
Flows and Cuts
10
6
6
11
1 10
3 8 8
0
0
0
11
s
2
3
4
5
6
7
t
15
5
30
15
10
8
15
9
6 10
10
10 15 4
4 0
f (e)e out of A
f (e)e in to A
v( f )
Value = 10 - 4 + 8 - 0 + 10 = 24
4
A
Flows and Cuts
Flow value lemma. Let f be any flow, and let (A, B) be any s-t cut. Then
Proof.
f (e)e out of A
f (e) v( f )e in to A
.
v( f ) f (e)e out of s
v A f (e)
e out of v f (e)
e in to v
f (e)e out of A
f (e).e in to A
by flow conservation, all termsexcept v = s are 0
Question
Two problems• min cut• max flow
• How do they relate?
Flows and Cuts
Weak duality. Let f be any flow, and let (A, B) be any s-t cut. Then the value of the flow is at most the capacity of the cut.
Cut capacity = 30 Flow value 30
s
2
3
4
5
6
7
t
15
5
30
15
10
8
15
9
6 10
10
10 15 4
4
Capacity = 30
A
Big Optimization Idea #1:
Look for structural constraints, e.g.max flow min-cut
Weak duality. Let f be any flow. Then v(f) cap(A, B), for any s-t cut (A, B).
Pf.
▪
Flows and Cuts
v( f ) f (e)e out of A
f (e)e in to A
f (e)e out of A
c(e)e out of A
cap(A, B)
s
t
A B
7
6
8
4
Certificate of Optimality
Corollary. Let f be any flow, and let (A, B) be any s-t cut.If v(f) = cap(A, B), then f is a max flow and (A, B) is a min s-t cut.
Value of flow = 28Cut capacity = 28 Flow value 28
10
9
9
14
4 10
4 8 9
1
0 0
0
14
s
2
3
4
5
6
7
t
15
5
30
15
10
8
15
9
6 10
10
10 15 4
4 0A
Towards a Max Flow Algorithm
Greedy algorithm. Start with f(e) = 0 for all edge e E. Find an s-t path P where each edge has f(e) < c(e). Augment flow along path P. Repeat until you get stuck.
s
1
2
t
10
10
0 0
0 0
0
20
20
30
Flow value = 0
Towards a Max Flow Algorithm
Greedy algorithm. Start with f(e) = 0 for all edge e E. Find an s-t path P where each edge has f(e) < c(e). Augment flow along path P. Repeat until you get stuck.
s
1
2
t
20
Flow value = 20
10
10 20
30
0 0
0 0
0
X
X
X
20
20
20
Towards a Max Flow Algorithm
Greedy algorithm. Start with f(e) = 0 for all edge e E. Find an s-t path P where each edge has f(e) < c(e). Augment flow along path P. Repeat until you get stuck.
greedy = 20
s
1
2
t
20 10
10 20
30
20 0
0
20
20
opt = 30
s
1
2
t
20 10
10 20
30
20 10
10
10
20
locally optimality global optimality
How can we get from greedy to opt here? What if we
push water back across middle
edge?
Residual Graph
Original edge: e = (u, v) E. Flow f(e), capacity c(e).
Residual edge. e = (u, v) and eR = (v, u). Residual capacity:
If e in E:– unused capacity
If eR in E:– ability of sent less water, or “undo“ flow
Residual graph: Gf = (V, Ef ). Residual edges with positive residual capacity. Ef = {e : f(e) < c(e)} {eR : c(e) > 0}.
u v 17
6
capacity
u v 11
residual capacity
6
residual capacity
flow
c f (e) c(e) f (e) if e E
f (e) if eR E
Ford-Fulkerson Algorithm
s
2
3
4
5 t 10
10
9
8
4
10
10 6 2
G:capacity
Ford-Fulkerson Algorithm
s
2
3
4
5 t 10
10
9
8
4
10
10 6 2
0
0
0
0 0 0
0
0
G:
Flow value = 0
0
flowcapacity
Ford-Fulkerson Algorithm
s
2
3
4
5 t 10
10
9
8
4
10
10 6 2
0
0
0
0 0 0
0
0
G:
s
2
3
4
5 t 10 9
4
10 6 2
Gf:10 8
10
8 8
8
X X
X
0
Flow value = 0
capacity
residual capacity
flow
Ford-Fulkerson Algorithm
s
2
3
4
5 t 10
10
9
8
4
10
10 6 2
8
0
0
0 0 8
8
0 0
G:
s
2
3
4
5 t 10
4
10 6 Gf:
8
8
8
9
22
2
10
210
X
X
X2X
Flow value = 8
0
Ford-Fulkerson Algorithm
s
2
3
4
5 t 10
10
9
8
4
10
10 6 2
10
0
0
0 2 10
8
2
G:
s
2
3
4
5 t
4
2
Gf:
10
810
2
10 7
10 6
X
66
6
X
X
8X
Flow value = 10
Ford-Fulkerson Algorithm
s
2
3
4
5 t 10
10
9
8
4
10
10 6 2
10
0
6
6 8 10
8
2
G:
s
2
3
4
5 t1
6 Gf:
10
8 10
8
6
6
6
4
4
4
2
X
8
2
8
X
X
0X
Flow value = 16
Ford-Fulkerson Algorithm
s
2
3
4
5 t 10
10
9
8
4
10
10 6 2
10
2
8
8 8 10
8
0
G:
s
2
3
4
5 t
6 2
Gf:
10
10
8
6
8
8
2
2 1
2
8 2
X
9
7 9
X
X
9X
X 3
Flow value = 18
Ford-Fulkerson Algorithm
s
2
3
4
5 t 10
10
9
8
4
10
10 6 2
10
3
9
9 9 10
7
0
G:
s
2
3
4
5 t 1 9
1
1 6 2
Gf:
10
7 10
6
9
9
3
1
Flow value = 19
Ford-Fulkerson Algorithm
s
2
3
4
5 t 10
10
9
8
4
10
10 6 2
10
3
9
9 9 10
7
0
G:
s
2
3
4
5 t 1 9
1
1 6 2
Gf:
10
7 10
6
9
9
3
1
Flow value = 19Cut capacity =
19
Augmenting Path Algorithm
Augment(f, c, P) { b bottleneck(P) foreach e P { if (e E) f(e) f(e) + b else f(e) f(e) - b } return f}
Ford-Fulkerson(G, s, t, c) { foreach e E f(e) 0 Gf residual graph
while (there is an s-t path P in Gf) { f Augment(f, c, P) update Gf
} return f}
forward edge
reverse edge
Min residual capacity of an edge in P
Ford-Fulkerson: Summary so far
Ford-Fulkerson: • While you can,
• Greedily push flow• Update residual graph
Lemma 1: This outputs a valid flow.
Proof: (Check conservation conditions… see book.)
Still to do:• Running time (in particular, termination!)• How good a flow?
Running Time
Assumption. All capacities are integers between 1 and C.
Invariant. Every flow value f(e) and every residual capacity cf (e)
remains an integer throughout the algorithm.
Proof: exercise.
Theorem. The algorithm terminates in at most v(f*) nC iterations.Pf. Each augmentation increase value by at least 1. ▪
Running time of Ford-Fulkerson: O(mnC). Space: O(m+n).Corollary. If C = 1, Ford-Fulkerson runs in O(mn) time.
Max-Flow Min-Cut Theorem
Augmenting path theorem. Flow f is a max flow iff there are no augmenting paths.
Max-flow min-cut theorem. [Elias-Feinstein-Shannon 1956, Ford-
Fulkerson 1956] The value of the max flow is equal to the value of the min s-t cut.
Pf. We prove both simultaneously by showing (i) -- (iii) are equivalent. (i) There exists an s-t cut (A, B) such that v(f) = cap(A, B). (ii) Flow f is a max flow, and (A,B) is min s-t cut. (iii) There is no augmenting path relative to f.
(i) (ii) This was the corollary to weak duality lemma.
(ii) (iii) We show contrapositive. Let f be a flow. If there exists an augmenting path, then we can
improve f by sending flow along path.
Proof of Max-Flow Min-Cut Theorem
(iii) (i) Let f be a flow with no augmenting paths. Let A be set of vertices reachable from s in residual graph. By definition of A, s A. By definition of f, t A. Observation: No edges of the residual graph go from A to B.
Claim 1: If e goes from A to B, then f(e) =c(e).Proof: Otherwise there would be residual capacity, and the residual graph would have an edge A to B.
Claim 2: If e goes from B to A, then f(e)=0.Proof: Otherwise residual edge would go from A to B.
v( f ) f (e)e out of A
f (e)e in to A
c(e)e out of A
cap(A, B)
original network
s
t
A B
Summary
Assumption. All capacities are integers between 1 and C.
Running time: The FF algorithm terminates in at most v(f*) nC iterations. Running time = O(mnC). Space: O(m+n).
Important special case: if C = 1, Ford-Fulkerson runs in O(mn) time.
Integrality theorem. If all capacities are integers, then there exists a max flow f for which every flow value f(e) is an integer.
