Fall’19 CSCE 629

Analysis of Algorithms

W, 10/23/19

Lecture 21

• Ford-Fulkerson algorithm

Credit: based on slides by A. Smith & K. Wayne

Guest lecture by Prof. Klappenecker

Recap: residue graph

§Original edge: ! = #, % ∈ '• Flow ((!), capacity +(!)

1

# %17/6Capacity Flow

§ Residual edge: “Undo” flow sent• ! = (#, %) and !/ = %, #• Residual capacity

# %

Residual Capacity

11

6

Residual capacity

+0 ! = 1+ ! − ( ! 3( ! ∈ '( ! 3( !/ ∈ '

§ Residual graph 40,5 = (6, '0,5)• Residual edges with positive residual capacity• '0,5 = !: ( ! < + ! ∪ {!/: ( ! > 0}

§ Augmenting path. Simple path > ↝ @ in residual graph 40. • Bottleneck capacity of an augmenting path A: minimum residual capacity of any edge in A.

Recap: Augmenting path theorem

2

Theorem. ! is a max flow iff. no augmenting paths (" ↝ $) in %&.

Proof. We show that the following are equivalent (' ⇒ ) ⇒ * ⇒ ')a. ! is a max flowb. There is no augmenting path (with respect to !)c. There exists a cut (,, .) such that *'0 ,, . = 2(!)

Max-flow min-cut theoremValue of max flow = capacity of min cut

a ⇔ c

Recap: proof of augmenting path theorem

3

§ a ⇒ b. We show contrapositive ¬b ⇒ ¬&

' ← bottleneck capacity of augmenting path P

For each ) ∈ +, ,′ ) = /,()) + ', ) ∈ 4,()) − ', )6 ∈ 4Pf.

• Exercise. Verify ,′ is a feasible flow (i.e., capacity and conservation hold).• 7 ,′ = 7 , + ' > 7(,) because only first edge in + leaves 9.

Lemma 1 (Augmented flow). Let + be an augmenting path with respect to ,. Then ,′ below is a feasible flow with 7 ,: > 7(,).

a. , is a max flowb. There is no augmenting path (with respect to ,)c. There exists a cut (;, <) such that =&> ;, < = 7(,)

Ford-Fulkerson augmenting-path algorithm

4

!"#$%&'(), +, ,). ← bottleneck capacity of PFor each 0 ∈ ,

If 0 ∈ 2 )3 0 = ) 0 + .

Else )3 06 = ) 06 − .return )3

89:;−8"<=%:>9&(?, @, A, +)

For each 0 ∈ 2 ) 0 ← 0, ?C ← D0@EFGHI JDHKℎ

While there is an augmenting path , in ?C) ← MGJN0OA(), +, ,)

Update ?Creturn )

Ford-Fulkerson Algorithm demo0

5

s 3

2

t

4

10 2 84

10106

59 10

(Capacity

Ford-Fulkerson algorithm demo1

6

s 3

2

t

4

10/0 2/0 8/04/0

10/0

10/06/059/0 10/0

)Flow

Capacity* + = 0

s 3

2

t

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10 2 84

10

10659 10

Residual graph )- Residual Capacity

An augmenting path: . − 2 − 5 − 1, 2 = 8

Ford-Fulkerson algorithm demo2

7

s 3

2

t

4

10/$ 2/0 8/$4/0

10/0

10/06/059/0 10/$

*+ , = 8

s 3

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t

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. 2 $4

10

10659/0 .

*/$

$

An augmenting path: 0 − 2 − 3 − 5 − 4, 5 = 2

Ford-Fulkerson algorithm demo3

8

s 3

2

t

4

10/$% 2/' 8/)4/0

10/0

10/06/059/' 10/$%

-. / = 10

s 3

2

t

4

' )4

10

10651

-2$%

$%'

An augmenting path: 3 − 3 − 5 − 4 − 7, 8 = 6

Ford-Fulkerson algorithm demo4

9

s 3

2

t

4

10/$% 2/' 8/)4/0

10/+

10/+6/+59/) 10/$%

./ 0 = 16

s 3

2

t

4

' )4

22+

51

.3$%

$%)+

+

An augmenting path: 4 − 3 − 2 − 4 − 7, 8 = 2

An augmenting path: ! − 3 − 5 − 2 − 4 − ', ( = 1

Ford-Fulkerson algorithm demo5

10

s 3

2

t

4

10/-. 8/04/1

10/0

10/06/359/0 10/-.

56 7 = 18

s 3

2

t

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1 01

113

51

58-.

-.00

01

2/.

Any more augmenting paths?

Ford-Fulkerson algorithm demo6

11

s 3

2

t

4

10/$% 2/% 8/(4/*

10/+

10/+6/-59/+ 10/$%

/0 1 = 19

s 3

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t

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3 $$

$$-

5

/4$%

$%++

+*

(

Cut (" = $, 3 , ' = (\A), ,-. ", ' = 19

Ford-Fulkerson algorithm demo7

12

s 3

2

t

4

10/34 2/4 8/74/9

10/:

10/:6/<59/: 10/34

=

> ? = 19 Is this a max flow?

Ford-Fulkerson algorithm: summary so far

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!"#$−!&'()#*"+While you can • Greedily push flow• Update residual graph

§Correctness. Augmenting path theorem.

§ Running time. Does it terminate at all?

Ford-Fulkerson algorithm: analysis

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§Assumption. All capacities are integers between 1 and ". § Invariant. Every flow value #(%) and every residual capacity '((%)

remains an integer throughout the algorithm.

§ Theorem. Ford-Fulkerson terminates in at most )" iterations. Pf. • Each augmentation increases flow value by at least 1.

• There are at most )" units of capacity leaving source *.

§ Integrality theorem. If all capacities are integers, then there exists a max flow # for which every flow value #(%) is an integer.

Running time + ,)" . Space +(, + )) Find an augmenting path in +(,) time (by BFS/DFS).

More to come on further concerns/improvements …