CS 170 Lecture 20 Handout

CS 170:Algorithms

Prof Elchanan Mossel.Based on: Algorithms by Dasgupta Papadimitriou and Vazirani.Slides edited from a version created by Prof. Satish Rao.For UC-Berkeley CS170 Spring 2014 students use only.Do not re-post or distribute

Elchanan Mossel (UC Berkeley) CS 170:Spring 2014 April 3, 2014 1 / 16

CS 170: Algorithms

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Linear Programming today!!!!

No laptops please.

Thank you !!!Elchanan Mossel (UC Berkeley) CS 170:Spring 2014 April 3, 2014 2 / 16

Outline

Maximum Flow.

Problem.

Linear Program

Upper bound: Cut!

Residual Network

Algorithm

Minimum Cut.


Maximum flowFlow network G = (V ,E), source s, sink t V , capacities ce > 0.

a

c

d

e

s t

4

4 3

1 4

2

111

31

2

2

2

1

Find Flow: fe

1 0 fe ce. Capacity constraints. 3 = fs,c cs,c = 4.2 If u is not s or t(w ,u)E fwu = (u,w)E fuw . 3 = fs,c = fc,d + fc,e = 2+1.

Maximize: size(f ) = (s,u)E fsu. fsa+ fsc = 1+3 = 4Optimal? cad +ccd +cet = 1+1+2 = 4.The capacity of any s t cut gives an upper bound.


S-T cut.

An s t cut is a partition of V into S and T where s S and t T . Itscapacity is the total capacity of edges from S to T .


Do you know the definition?

Find Flow: fe

1 0 fe ce. Capacity constraints.2 If u is not s or t(w ,u)E fwu = (u,w)E fuw .

Valid or Invalid?

s

b

c

d

t

3211

2211

2323

s

b

c

d

t

3211

2211

22

2+1 6= 2


Algorithms.FindFlow: fe

1 0 fe ce. Capacity constraints.2 (w ,u)E fwu = (u,w)E fuw .

3 maximize su fsu.

Linear program!

Variables fe, linear constraints, linear optimization function.

Cool!

Note...

Integer? (Given integer capacities.)

Yes. There is an integer vertex solution!

Constraint matrix has every subdeterminant being 1, 0, 1.Vertex solution to linear program must be integral!


Ford-Fulkerson.

Simplex method.

Find s to t path with remaining capacity.

Add to flow variables along path.

Update remaining capacity.

Repeat.

a

c

d

e

s t

4

4 3

1 4

2

11

44X 31

11X 01

11X 0 1

22X 11

Uh oh...


Residual Capacity.Find s to t path with remaining capacity.Add to flow along path. Or reduce flow on reverse edge.Update remaining capacity.

Reduce re = ce feand add reverse ruv = fvu

Repeat.

s

a

b

t

1

1

1

1

1

11X 01

11X 01

11X 0 1

1

11

1

1

1X 0

1X 0

1

1

No remaining path. Uh oh! Optimal is 2! (At most 2 due to cut.)Add reverse arcs to indicate reverse capacity.


Bigger Example.

Find s to t path with remaining capacity.

Add to flow along path.

Update residual capacities: : re = ce fe; ruv = fvu.Repeat.

a

c

d

e

s t

4

4 3

1 4

2

11

44X 311

11X 011

11X 0 11

22X 111

44X 311

33X 211

2X 11X 0122

44X 311

1X 00X 1 1

4X 33X 212

3X 22X 112

11X 011

3X 22X 1233

4X 33X 2122


Check Result...

a

c

d

e

s t

4

4 3

1 4

2

111

31

2

2

2

1

Find Flow: fe

1 0 fe ce. Capacity constraints. 3 = fs,c cs,c = 4.2 If u is not s or t(w ,u)E fwu = (u,w)E fuw . 3 = fs,c = fc,d + fc,e = 2+1.

Maximize: size(f ) = (s,u)E fsu. fsa+ fsc = 1+3 = 4Optimal? cad +ccd +cet = 1+1+2 = 4.Any s t cut gives an upper bound.


Correctness.

1. Capacity Constraints: 0 fe ce.Only increase flow to ce.Or use reverse arcs decrease to 0.Flow values to be between 0 and ce.

2. Conservation Constraints:flow into v = flow out of v (if not s or t .)Algorithm adds flow, say f , to path from s to t .Each internal node has f in, and f out.


Optimality: upper bound.s-t Cut: V = ST and s S and t T .

s t

Lemma: Capacity of any s t cut is an upper bound on the flow.C(S,T ) - sum of capacities of all arcs from S to TC(S,T ) = e=(u,v):uS,VT ceFor valid flow:Flow out of (S) = Flow out of s.Flow into (T ) = Flow into t .

For any valid flow, f : E Z+, the flow out of S (into T )eST feeTS fe eST ceeTS 0 = C(S,T ). The value of any valid flow is at most C(S,T )!


Optimality: max flow = min cut.At termination of augmenting path algorithm.

No path with residual capacity!

Depth first search only starting at s does not reach t .

s tfe = ce

fe = 0

S be reachable nodes.

No arc with positive residual capacityleaving S

= All arcs leaving S are full.= No arcs into S have flow.Total flow leaving S is C(S,T ).

Valid flow = all that flow from source.Value of flow equals value of C(S,T ). and Optimal is C(S,T ). Flow is maximum!!Cut is minimum s t cut too!any flow any cut and this flow = this cut. Maximum flow and minimum s t cut!


Celebrated max flow -minimum cut theorem.

Theorem: In any flow network, the maximum s-t flow is equal to theminimum cut.

Celebrate!


Back to business: Algorithm Terminates?

It will!!

Flow keeps increasing.

How long?

One more unit every step!

O(mF ) time where F is size of flow.


Efficiency.

s

a

b

t

100000

1

100000100000

100000999991

999991

1

999991

999991

999991

999991

1


Edmonds-Karp

Augment along shortest path.

Breadth first search!

O(|V ||E |) augmentations.Analysis idea.

d(v) is distance to sink.

Only route flow on (u,v) if d(u) d(v).Only reverse flow on (u,v) if d(v) d(u).Maximum d(v) is |V |.Distances only go up. (To prove!)

Every augment removes edge at a distance.

O(|V ||E |) removals.O(|V ||E |2) time.


Bipartite Matching

Given a bipartite graph: B = (L,R,E) where E LR.

Cindy

Jin

Shelby

Katya

Chen

Ollie

Ivan

Jacque

Find largest subset of edges (matches) which are one to one.


Bipartite MatchingAlgorithm by Reduction.:

From matching problem produce flow problem.From flow solution produce matching solution.

C

J

S

K

C

O

I

J

s t

Max flow = Max Matching Size.Flow is not integer necessarily....

Augmenting path algorithm gives integer flow.


Review.

Maximum flow.

Greedy augment path...

Except reverse old decisions ..Reverse residual capacities.

Optimality?

Upper bound on optimal: s t cut!Find flow and s t cut with equal value!Optimal plus proof of optimality.

Application to bi-partite matchings.


Documents

CS 170 Lecture 20 Handout