1. Chapter 26
Maximum Flow
How do we transport the maximum amount data from source to sink?
Some of these slides are adapted from Lecture Notes of Kevin Wayne.

2. Contents Contents. Maximum flow problem. Minimum cut problem.
Max-flow min-cut theorem.
Augmenting path algorithm.
Capacity-scaling.
Shortest augmenting path.

3. Networks communication Network
telephone exchanges, computers, satellites
Nodes
Arcs
cables, fiber optics, microwave relays
Flow
voice, video, packets
circuits
gates, registers, processors
wires
current
mechanical
joints
rods, beams, springs
heat, energy
hydraulic
reservoirs, pumping stations, lakes
pipelines
fluid, oil
financial
stocks, currency
transactions
money
transportation
airports, rail yards, street intersections
highways, railbeds, airway routes
freight, vehicles, passengers
chemical
sites
bonds
energy
biological
genes
pathways
interactions

4. Maximum Flow & Minimum Cut
Max flow and min cut.
Two very rich algorithmic problems.
Cornerstone problem in combinatorial optimization.
Beautiful mathematical duality.
Nontrivial applications / reductions.
Network connectivity.
Bipartite matching.
Data mining.
Image processing.
Airline scheduling.
Project selection.
Network reliability.
Security of statistical data.
Distributed computing.
Egalitarian stable matching.
Computational biology.
Many many more . . .

5. Flow Network Flow network: G = (V, E, s, t, c).
(V, E) = weighted directed graph.
Two distinguished vertices (nodes):
s=source, t=sink.
c(u,v) = capacity of edge (arc) e=(u,v).
Capacity s 2 3 4 5 6 7 t 15 30 10 8 9

6. Positive Flows
A positive s-t flow is a func. p: V ´ V ® R that satisfies:
For all u, v Î V: 0  p(u,v)  c(u,v) (capacity constrain)
For all u  V–{s, t}: (flow conservation)
(flows out) (flows in) s 2 3 4 5 6 7 t 15 30 10 8 9
Capacity
Flow
* flow: entity generated at source, transmitted across edges, absorbed at sink
* flow conservation is analogous to Kirchoff's law

7. Positive Flows
A positive s-t flow is a func. p: V ´ V ® R that satisfies:
For all u, v Î V: 0  p(u,v)  c(u,v) (capacity constrain)
For all u  V–{s, t}: (flow conservation)
(flows out) (flows in)
MAX FLOW: positive s-t flow that maximizes net flow out of the source.
Value = 4
Capacity s 2 3 4 5 6 7 t 15 30 10 8 9
Flow
* flow conservation is analogous to Kirchoff's law

8. Positive Flows
A positive s-t flow is a func. p: V ´ V ® R that satisfies:
For all u, v Î V: 0  p(u,v)  c(u,v) (capacity constrain)
For all u  V–{s, t}: (flow conservation)
(flows out) (flows in)
MAX FLOW: find positive s-t flow that maximizes net flow out of the source. s 2 3 4 5 6 7 t 15 30 10 8 9 11 1
Value = 24
Capacity
Flow

9. Positive Flows
A positive s-t flow is a func. p: V ´ V ® R that satisfies:
For all u, v Î V: 0  p(u,v)  c(u,v) (capacity constrain)
For all u  V–{s, t}: (flow conservation)
(flows out) (flows in)
MAX FLOW: positive s-t flow that maximizes net flow out of the source. s 2 3 4 5 6 7 t 15 30 10 8 9 14 1
Value = 28
Capacity
Flow
* flow conservation is analogous to Kirchoff's law

10. Maximum flow problem Maximize the flow over a flow network. Capacitys 2 3 4 5 6 7 t 15 30 10 8 9 14 1
Value = 28
Capacity
Flow

11. Problem formulation

12. Net flow Net flow Graph G=(V,E) – a flow network
Directed, each edge has capacity c(u,v) ³ 0
Two special vertices: source s, and sink t
For any other vertex v, there is a path s®…®v®…®t
A (net) s-t flow is a func. f : V ´ V ® R that satisfies
Capacity constraint: For all u, v Î V: f(u,v) £ c(u,v)
Skew symmetry: For all u, v Î V: f(u,v) = – f(v,u)
Flow conservation: For all u Î V – {s, t}:

13. Cancellation of flows
Do we want to have flows going in both directions between two vertices?
No! such flows cancel (maybe partially) each other
Skew symmetry – notational convenience 13 11 5 4 15 10 14 19 3 s t 9 a b c d 6 13 11 5 4 15 10 14 19 3 s t 9 a b c d 2 6

14. Equivalence of Definitions
Define net flow in terms of positive flows
f(u, v) = p(u, v) - p(v, u)
How do we know f defined this way is a net flow?
Capacity constrain
Flow conservation
Skew symmetry

15. Equivalence of Definitions
Define positive flow in terms of net flow
p(u, v) = f(u, v) if f(v, u) >0
= 0 o.w.
How do we know p defined this way is a positive flow?
Capacity constrain
Flow conservation
In literature, 2 views of flow are used. The current version (3rd edition) of our book uses positive flow.

16. Flow conservation Define positive flow in terms of net flow
p(u, v) = f(u, v) if f(v, u) >0
= 0 o.w.

17. Implicit Summation Notation
Function over sets  summation:
Flow conservation for positive flow:
Flow conservation for net flow:

18. Flow Conservation Theorem
proof: ? s 2 3 4 5 6 7 t 15 30 10 8 9 14 1
Value = 28
Capacity
Flow 28 28

19. Lemma G = (V,E) is a flow network. f is a net flow in G.
f (X,X) = 0,  X, Y  V
f (X,Y) = − f (Y,X),  X, Y  V
f (XY, Z) = f (X,Z)+ f (Y,Z),  X,Y,ZV & XY=
f (Z, XY) = f (Z,X)+ f (Z,Y),  X,Y,ZV & XY=
Proof: …

20. Flow Conservation Theorem
proof:
Flow conservation

21. Cuts An s-t cut is a node partition (S, T) s.t. s  S, t  T.
The capacity of an s-t cut (S, T) is:
Min s-t cut: find an s-t cut of minimum capacity. s 2 3 4 5 6 7 t 15 30 10 8 9
Capacity = 30

22. Cuts An s-t cut is a node partition (S, T) s.t. s  S, t  T.
The capacity of an s-t cut (S, T) is:
Min s-t cut: find an s-t cut of minimum capacity. s 2 3 4 5 6 7 t 15 30 10 8 9
Capacity = 62

23. Cuts An s-t cut is a node partition (S, T) s.t. s  S, t  T.
The capacity of an s-t cut (S, T) is:
Min s-t cut: find an s-t cut of minimum capacity. s 2 3 4 5 6 7 t 15 30 10 8 9
Capacity = 28

24. Recall the example Capacity Value = 28 Flow 1 14 s 2 3 4 5 6 7 t 9 1015 30 10 8 9 14 1
Value = 28
Capacity
Flow

25. Flows and Cuts
Lemma 1. Let f be a net flow, and let (S, T) be a cut. Then, the net flow sent across the cut is equal to the amount reaching t. s 2 3 4 5 6 7 t 15 30 10 8 9
total amount of flow that leaves S minus amount flow that "swirls" back
Value = 24

26. Flows and Cuts
Lemma 1. Let f be a flow, and let (S, T) be a cut. Then, the net flow sent across the cut is equal to the amount reaching t. 10 6 4 8 s 2 3 5 7 t 15 30 9
Value = 24

27. Flows and Cuts
Lemma 1. Let f be a flow, and let (S, T) be a cut. Then, the net flow sent across the cut is equal to the amount reaching t. 10 6 4 8 s 2 3 5 7 t 15 30 9
Value = 24

28. Proof of the Lemma 1
For any flow f , the flow across any cut (S, T) is the flow value | f |, i.e.
f (S, T) = | f | = f (s, V).
Proof:
f (S, T) = f (S, V−S) = f (S, V) – f (S, S)
= f (s, V) + f (S-s, V)
= | f |.
(lemma 1)
(flow conservation)

29. Proof by Math Induction
For any flow f , the flow across any cut (S, T) is the flow value | f |, i.e. f (S, T) = | f | = f (s,V).
Proof: by induction on |S|.
Base case: S = { s } .
Inductive hypothesis: assume true for |S| < k.
consider cut (S, T) with |S| = k
S = S'  { v } for some v  s, t, |S' | = k-1  f(S', T') = | f |.
adding v to S' , change to the |f| ? s v t
After S'
Before S
skip

30. Flows and Cuts
Lemma 2. f is a flow & (S, T) is a cut. Then |f| c(S, T).
Proof.
Corollary. Let f be a flow, and let (S, T) be a cut. If |f| = c(S, T), then f is a max flow and (S, T) is a min cut. s t S T 7 6 8 4

31. Max Flow and Min Cut
Corollary. Let f be a flow, and let (S, T) be a cut. If |f| = c(S, T), then f is a max flow and (S, T) is a min cut. s 2 3 4 5 6 7 t 15 30 10 8 9 14 1
Cut capacity = 62
Flow value = 28

32. Max Flow and Min Cut
Corollary. Let f be a flow, and let (S, T) be a cut. If |f| = c(S, T), then f is a max flow and (S, T) is a min cut. s 2 3 4 5 6 7 t 15 30 10 8 9 14 1
Cut capacity = 28
Flow value = 28

33. Summary
Lemma 1. Let f be a flow, and let (S, T) be a cut. Then, the net flow sent across the cut is equal to the amount reaching t.
Lemma 2. f is a flow & (S, T) is a cut. Then |f| c(S, T).
Corollary. Let f be a flow, and let (S, T) be a cut. If |f| = c(S, T), then f is a max flow and (S, T) is a min cut.

34. Contents Maximum flow problem. Minimum cut problem.
Max-flow min-cut theorem.
Augmenting path algorithm.
Capacity-scaling.
Shortest augmenting path.

35. Max-Flow Min-Cut Theorem
MAX-FLOW MIN-CUT THEOREM (Ford-Fulkerson, 1956): In any network, the value of the max flow is equal to the value of the min cut.
Proof: ?
Flow value = 28 s 2 3 4 5 6 7 t 15 30 10 8 9 14 1
Cut capacity = 28
Later after the algorithm

36. Towards an Algorithm Find an s-t path where each edge e=(u,v) has
c(e) > f(e) and "augment" flow along the path.
Flow value = 0 s 4 2 5 3 t 10 13

37. Towards an Algorithm The design of this algorithm? A Greedy Algorithm
Find an s-t path where each arc e=(u,v) has
c(e) > f(e) and "augment" flow along the path.
Repeat until you get stuck. 4 5 4 4 4 4 10 s 10 2 13 3 10 t
The design of this algorithm?
A Greedy Algorithm
Flow value = 10

38. Towards an Algorithm
Find an s-t path where each arc has u(e) > f(e) and "augment" flow along the path.
Repeat until you get stuck.
The straight forward greedy algorithm fails. s 4 2 5 3 t 10 13
Flow value = 10
Flow value = 14 s 4 2 5 3 t 10 13 6

39. Residual Network
Let f be a flow on G = (V, E). The residual network Gf = (V, Ef ) is the graph with strictly positive residual capacities:
cf (u,v) = c(u,v)−f (u,v) > 0, (where f (u,v) >0)
Ef = {(u,v)VV| cf (u,v)>0}.
Gf contains edges of G that can admit more net flow.
(Gf GGT, where GT has G’s edges in reverse.) s 2 3 5 6 t 13 16 14 9 12 4 20 7 11 15 8 1 10
flow network and flow f
residual network s 2 3 5 6 t 11 4 15 12 8 1 7 9

40. Residual Arcs Original graph G = (V, E).
Flow f(u,v).
Arc e = (u, v)  E.
Residual graph: Gf = (V, Ef ).
Residual arcs e = (u, v) and eR = (v, u).
"Undo" flow sent. u v 17 6
Capacity
Flow u v 11
Residual capacity 6

41. Residual Graph & Augmenting Paths
Residual graph: Gf = (V, Ef ). s 4 2 5 3 t 10 13 G 10 s 4 2 5 3 t 10 Gf

42. Augmenting Path Augmenting path = path in residual graph. G Gf s 4 2 53 t 10 13 G Gf

43. Relation between flows on G & Gf
Lemma: Let f be a flow on G=(V,E), f ’ be a flow on the residual network Gf of G. Then the flow sum (f +f ’)(u,v)
 f (u,v)+f ’(u,v) is a flow on G and |f +f ’|=|f |+|f ’|.
Proof: f +f ’ is a flow on G
(flow conservation, capacity constrain, skew symmetry)
|f +f ’| = |f |+|f ’|

44. Augmenting Path Augmenting path = path in residual graph.
Max flow  no augmenting paths ??? s 4 2 5 3 t 10 13 G Gf
Flow value = 14

45. Max-Flow Min-Cut Theorem
Augmenting path theorem (Ford-Fulkerson, 1956): A flow f is a max flow if and only if there are no augmenting paths.
MAX-FLOW MIN-CUT THEOREM (Ford-Fulkerson, 1956): the value of the max flow is equal to the value of the min cut.
We prove both theorems simultaneously by showing the TFAE:
(i) f is a max flow.
(ii) There is no augmenting path relative to f.
(iii) There exists a cut (S, T) such that |f| = c(S, T).

46. Proof We prove both simultaneously by showing the TFAE:
(i) f is a max flow.
(ii) There is no augmenting path relative to f.
(iii) There exists a cut (S, T) such that |f| = c(S, T).
(i)  (ii)
Let f be a flow. If there exists an augmenting path, then we can improve f by sending flow along path. (proof by contradiction)
(iii)  (i)
This was the Corollary to Lemma 2.

47. Proof (ii)  (iii) Let f be a flow with no augmenting paths.
Let S be set of vertices reachable from s in residual graph.
Let T=V−S. Is t  T. ?
(clearly s  S, and t  S by definition of f , o.w.  augmenting path)
For each u  S, v  T, f (u,v) = c(u,v). ?
(o.w. cf (u,v)=c(u,v)−f (u,v)>0, then v  S).
Therefore: ?
|f | = f (S,T) = c(S,T).
Residual Network s t S T

48. Ford-Fulkerson algorithm
Let G=(V, E) be a flow network.
Ford-Fulkerson(G,s,t,c)
01 for each edge (u,v) in E do
02 f(u,v) ¬ f(v,u) ¬ 0
03 while there exists a path p from s to t in residual
network Gf do
04 cf = min{cf(u,v): (u,v) is in p}
05 for each edge (u,v) in p do
f(u,v) ¬ f(u,v) + cf
f(v,u) ¬ -f(u,v)
08 return f
Cost?
(E)
(V+E)
depth first search or
breadth first search
Find path --- depth first search/breath first search (V+E)
(E)
( ? )

49. Performance
If capacities are integers, O(|E|f *), f * is max flow found by the algorithm (not a polynomial algorithm).
Example: 4
100
100 s 1 t
100
100 2

50. Performance Original network and flow Residual network 1 100 100 1 14 1
100
100 1 s 1 t 1
100
100 2 1 99 s 4 2 t
100
Original network and flow
Residual network

51. Performance Maximum flow: 200 Net flow: 2. 200 iterations possible. 14 1
Net flow: 2.
100
100 s 1 t
100
100 2
200 iterations possible.
Maximum flow: 200

52. Performance For noninteger capacities, might never terminate.
Need more efficient algorithm (Edmonds-Karp, 1972):
Augment along breadth-first augmenting path in Gf (shortest path)
Augment along max-capacity augmenting path in Gf (capacity-scaling)

53. BFS for Shortest Paths Data structure for BFS? S S S 2 1 3 Discovered
Finished
Discovered
Undiscovered
Data structure for BFS?

54. Edmonds-Karp Algorithm
100 s 4 2 t 1
100
100
Takes only two iterations.

55. Shortest Augmenting Path
Intuition: choosing path via breadth first search.
Easy to implement.
may implement by coincidence!
Finds augmenting path with fewest number of arcs.
FOREACH e  E
f(e)  0
Gf  residual graph
WHILE (there exists augmenting path)
//find such a path P by BFS
f  augment(f, P)
update Gf
RETURN f
ShortestAugmentingPath(V, E, s, t)
Cost?

56. Shortest Augmenting Path: Overview of Analysis
Lemma1. Throughout the algorithm, the length of the shortest path increases.
Proof ahead.
Lemma2. After at most |E| shortest path augmentations, the length of the shortest augmenting path increases at least by one.
Theorem. The shortest augmenting path algorithm runs in O(|V||E|2) time.
O(|V|+|E|) time to find a shortest augmenting path via BFS.
O(|E| + |E||V|) augmenting paths. (lemma 2)
(If there is an augmenting path, it must be a simple one.)
O(|V||E|) augmentations.
O((|V|+|E|)*(|V||E|)) = O(|V||E|2) running time.

57. Shortest Augmenting Path: Analysis
Level graph of (V, E, s).
For each vertex v, define (v) to be the length (number of edges) of shortest path from s to v.
LG = (V, EG) is subgraph of G that contains only those edges (v, w)  E
with (w) = (v) + 1. s 2 3 5 6 t G: s 2 3 5 6 t
 = 0
 = 1
 = 2
 = 3
LG:

58. Shortest Augmenting Path: Analysis
Level graph of (V, E, s).
For each vertex v, define (v) to be the length of shortest path from s to v.
L = LG = (V, F) is subgraph of G that contains only those edges (v,w)  E
with (w) = (v) + 1.
Compute in O(|V|+|E|) time using BFS, deleting back and side edges.
P is a shortest s-v path in G if and only if it is an s-v path in L. 2 5 L: s 3 6 t
 = 0
 = 1
 = 2
 = 3

59. Shortest Augmenting Path: Analysis
Lemma1. the length of the shortest path increases.
Let f and f' be flow before and after a shortest path augmentation.
Let L and L' be level graphs of residual network Gf and Gf '
Only back edges added to Gf '
path with back edge has length greater than previous length 2 5 L s 3 6 t
 = 0
 = 1
 = 2
 = 3 s 2 3 5 6 t L'

60. Shortest Augmenting Path: Analysis
Lemma2. After at most |E| shortest path augmentations, the length of the shortest augmenting path strictly increases.
At least one edge (the bottleneck edge) is deleted from L after each augmentation.
No new edges added to L until length of shortest path to t strictly increases. 2 5 L
Length=3 s 3 6 t
Shortest path (length the same). Each one path must have at least one different edge  at most |E| shortest path
 = 0
 = 1
 = 2
 = 3 2 5 L'
Length=3 s 3 6 t

61. Shortest Augmenting Path: Review of Analysis
Theorem. The shortest augmenting path algorithm runs in O(|V||E|2) time.
O(|V|+|E|) time to find a shortest augmenting path via BFS.
O(|V||E|) augmentations.
O((|V|+|E|)*(|V||E|)) = O(|V||E|2) running time.
More rigorous proof is given in our textbook.
Other approaches:
Augment along max-capacity augmenting path in Gf (capacity-scaling)

62. Capacity Scaling Intuition: choosing path with highest bottleneck
capacity increases flow by max possible amount.
In practice, don't worry about finding exact highest bottleneck path.
Maintain scaling parameter .
Let Gf () be the subgraph of the residual graph consisting of only arcs with capacity at least .
110 s 4 2 t 1
170
102
122 Gf
Gf (100)

63. ScalingMaxFlow(V, E, s, t)
Capacity Scaling
Assumption: all capacities are integers between 0 & U.
FOREACH e  E, f(e)  0
(smallest power of 2 greater than or equal to U)/2
WHILE (  1)
Gf()  -residual graph
WHILE (there exists augmenting path P in Gf())
f  augment(f, P)
update Gf()
   / 2
RETURN f
ScalingMaxFlow(V, E, s, t)
Theorem. The algorithm runs in O(|E|2 log (2U) ) time.

64. Capacity Scaling: Analysis
L1. If all edges capacities are integers, then throughout the algorithm, all flow and residual capacity values remain integers.
Thus,  = 1  Gf() = Gf , so upon termination f is a max flow.
L2. The outer while loop repeats 1 + log2 U times.
Initially U/2   < U, and  decreases by a factor of 2 each iteration.
L3. Let f be the flow at the end of a -scaling phase. Then value of the maximum flow is at most | f |+| E |. (proof ahead)
L4. There are at most 2|E| augmentations per scaling phase.
Let f be the flow at the end of the previous scaling phase.
L3  |f*|  | f | + |E| (2).
Each augmentation in a -phase increases | f | by at least .
Theorem. The algorithm runs in O(|E|2 log (2U) ) time.

65. Capacity Scaling: Analysis
L3. Let f be the flow at the end of a -scaling phase. Then value of the maximum flow is at most |f| + |E| .
We show that at the end of a -phase, there exists a cut (S, T) such that c(S, T)  | f | + |E| .
Choose S to be the set of nodes reachable from s in Gf().
clearly s  S, and t  S by definition of S S T t
intuition: f is max flow in Gf(). Adding back arc with capacity less than  can only increase capacity of cut by at most m. s
Original Network

66. Choosing Good Augmenting Paths: Summary
Method
Augmentations
Augmenting path
|V|U
Max capacity
|E| log U
Capacity scaling
Shortest path
|E||V|
Running time
|E||V|U
|E| log U (|E| + |V| log |V|)
|E|2 log U
|E|2|V|
Improved shortest path
|E||V|2
Improved capacity scaling
|E||V| log U
First 4 rules assume edge capacities are between 0 and U.

67. |E|3/2 log (|V|2 / |E|) log U |E||V|2/3 log (|V|2 / |E|) log U
History
Year
Discoverer
Method
Big-Oh
1951
Dantzig
Simplex
|E||V|2U
1955
Ford, Fulkerson
Augmenting path
|E||V|U
1970
Edmonds-Karp
Shortest path
|E|2|V|
1970
Dinitz
Shortest path
|E||V|2
1972
Edmonds-Karp, Dinitz
Capacity scaling
|E|2 log U
1973
Dinitz-Gabow
Capacity scaling
|E||V| log U
1974
Karzanov
Preflow-push
|V|3
1983
Sleator-Tarjan
Dynamic trees
|E||V| log |V|
M=|E|, n=|V|
1986
Goldberg-Tarjan
FIFO preflow-push
|E||V| log (|V|2 / |E|)
. . .
. . .
. . .
. . .
1997
Goldberg-Rao
Length function
|E|3/2 log (|V|2 / |E|) log U |E||V|2/3 log (|V|2 / |E|) log U

69. Network with multiple sources and sinks15 14 19 30 s1 t1 a b c d 11 1 9 13 s2 t2 e f 5 4 12 23 s3 t3 g h k t s  
Can be reduced to a single source & single sink max problem

70. Maximum bipartite matching
Bipartite graph G=(V,E):
Undirected
V = V1V2, V1V2=.
e=(u, v)E, uV1 and vV2
Complete bipartite: u V1,
u is joined with every v V2
Matching: set of edges with no vertex included twice
Maximum matching: a matching of maximum cardinality
i.e. no matching contains more edges

71. Applications Dating assignment Job placement L – women, R – men.
An edge between vertices: they have a chance to be “compatible” (can be matched)
Job placement
Companies make job offers
Students have job choices
An edge between vertices: the student can fit the job

72. Example
Problem: Given N men and M women, find a matching between men and women.
Participants list members of opposite sex they like to date.
Zeus
Amy
Diane
Yancey
Clare
Xavier
Erika
Wyatt
Bertha
Victor
Diana
Man
1st
2nd
3rd
Men’s Preference List
Can we find a perfect match? What is the max match?

73. Students Preference List
Example
Problem: Job placement
Eliza
IBM
Google
Sun
Dave
Apple
Dell
Carol HP
Bob
Alice
Student
1st
2nd
3rd
Students Preference List
Frank
Can we find a perfect match? What is the max match?

74. Maximum bipartite matching
How can we reformulate this problem to become a max-flow problem?

75. Maximum bipartite matching
How can we reformulate this problem to become a max-flow problem?
Keep all edges & vertices
Add source connected to all vertices in one set
Add sink connected to all vertices in another set
Set all capacities to 1
Theorem: The set of edges in the maxflow solution gives the solution to the maximum matching problem.
Proof homework

76. Matching 1 Eliza Sun Dave Apple Carol IBM Bob Dell Alice HP StudentF A I D H S G t s
Eliza
Sun
Dave
Apple
Carol
IBM
Bob
Dell
Alice HP
Student
1st
2nd
3rd
Solution
Frank
Google

77. Lemma
Lemma 1:
Let G=(V,E), V = V1V2, V1V2=.
Let G’=(V’,E’), V’ = V{s,t}, E’=…. flow network
Then:
If M is a matching, then there is an integer valued flow f on G’ with |f|=|M|.
If f is an integer valued flow on G’, then there is a matching M on G with cardinality |M|=|f|
Proof:
M →f, s.t.|f|=|M|
If (u,v)M, define: f(s,u)=f(u,v)=f(v,t)=1; f(u,s)=f(v,u)=f(t,v)=-1; (u,v)M, f(u,v)=0,
then f is a flow and |f|=|M| ?
f →M, s.t.|M|=|f|
since f is an integer valued flow, f(u,v)=1 or 0.
|M|=f(V1’, V2’) = |f|
(skew symmetry, capacity constraints, flow conservation)

78. Integrality Theorem
Integrality Theorem: If the capacity of a flow network is an integer function, maximum flow f produced by Ford-Fulkerson method is an integer-valued flow.
Proof: homework
Corollary: If |M| is the cardinality of a maximum matching M, then the value of a maximum flow f = |M|.
Proof: ?
Show contrapositive.
suppose M is a max matching and f is not a max flow,
i.e. f ’,s.t.|f ’|>|f|
Then M’, s.t.|M’|=|f ’|>|f|=|M|
Contradicting our assumption that M is a maximum matching.