linear programming – max flow – min cut orgad keller

Linear Programming –

Max Flow – Min Cut

Orgad Keller

Orgad Keller - Algorithms 2 2

We have an objective function:

Subject to constraints:

Linear Programming

1 1max ... n nc x c x

1,1 1 1, 1

,1 1 ,

...

...

, 0

n n

m m n n m

i

a x a x b

a x a x b

i x

Orgad Keller - Algorithms 2 - 3

Given all parameters

we want to find the optimal

Linear Programming

1,..., nx x, , , 1 ,1j i i ja c b i n j m


It is easier to present the problem with a matrix and vectors:

Linear Programming

max

. .

, 0i

c x

s t

Ax b

i x

1 1 1

1,1 1,

,1 ,

: ( ,..., ), ( ,..., ), ( ,..., )n n m

n

m m n

where x x x c c c b b b

a a

A

a a


Given the Primal Problem:

It’s Dual Problem is defined as:

The Dual Problem

max

. .

, 0i

c x

s t

Ax b

i x

1

min

. .: ( ,..., )

, 0

mT

i

b y

s twhere y y y

A y c

i y

1 1 1

1,1 1,

,1 ,

: ( ,..., ), ( ,..., ), ( ,..., )n n m

n

m m n

where x x x c c c b b b

a a

A

a a


Given a problem and it’s dual problem, then:

In other words, the optimal objective function’s value for the primal problem, is equal to the optimal objective function’s value for the dual problem.

Strong Duality Theorem

1 1

max minn m

i i j ji j

c x b y


As you remember from Algorithms I Given a directed graph , two

vertices and a capacity for each edge

We want to find a flow function

so that the flow value is maximal

Maximum Flow

:c E N

( , )G V E,s t V

( , )

v V s

f s v


But, we are subject to some rules: What goes in must come out:

Capacity restrictions:

Maximum Flow

( , ) , ( , ) ( , )u v E f u v f v u

, , ( , ) ( , )u V u V

v V s t f u v f v u

( , ) , ( , ) ( , )u v E f u v c u v


We’ll show by a simple example:

Max Flow with Linear Programming

s t

a

b32

13

1


We want to present the example in the form:

Subject to:


max c f

A f b


Formally:


saf

s t

a

b 32

13

1

sbf

abf

atf

btf

atfabfsaf

abfsbf

3211300

But we don’tpermit

equalities

saf sbfmax

btf


So we’ll add another edge, and change the problem’s representation a little.


s t

a

b 32

13

1

∞

saf

sbf

abf

atf

btf

atfabfsaf

abfsbf

3211300

saftsf 0

0sbf

tsfatf btf

btf

tsfmax


max

max 0 0 0 0 0 1

ts

sa

sb

ab

at

bt

ts

f

ff

fc f

f

f

f


1 0 0 0 0 0 3

0 1 0 0 0 0 2

0 0 1 0 0 0 1

0 0 0 1 0 0 1

0 0 0 0 1 0 3

1 0 1 1 0 0 0

0 1 1 0 1 0 0

1 1 0 0 0 1 0

0 0 0 1 1 1 0

sa

sb

ab

at

bt

ts

A b

ff

f

f

f

f

f


As you remember from Algorithms I Given a directed graph , two

vertices and a weight for each edge

We want to find a minimal-weight subset of edges such that if we’ll remove them, we won’t be able to travel from to .

Minimum Cut

:c E N

( , )G V E,s t V

s t


In other words:We’ll choose ,

where: , ,

such that the cut value,

is minimal.

Minimum Cut

,s tV V V

ss V tt V s tV V V

,

( , )s

t

u Vv V

c u v


Going back to the example:

Min Cut with Linear Programming

s t

a

b32

13

1



s t

a

b 32

13

1 What about:

Is that enough?We haven’t ensured paths from to are cut.

( , )

min ( , )

. .

( , ) , 0,1

uvu v E

uv

c u v y

s t

u v E y

This look like

And we know thatIP

IP NPhard

s t



s t

a

b 32

13

1 Beside a variable for every edge , we’ll want a variable for every vertex , such that:

uvy( , )u v

uz u V

1

0u s

u t

z u V

z u V



s t

a

b 32

13

1 Let’s take for instance: If is in the cut , that means

that If is not in the cut , that means

that either or or

So it is the same to constrain:

( , )a b( , )a b ( 1)aby

,s ta V b V ( 1, 0)a bz z ( , )a b ( 0)aby

1, 1a bz z

0ab a by z z

0, 0a bz z 0, 1a bz z


Formally:


0sa s ay z z

s t

a

b 32

13

1min3 2 3sa sb ab at bty y y y y

0sb s by z z

0ab a by z z

0at a ty z z

0bt b ty z z

1s tz z In order to ensure

ss V tt V


min3 2 3

min 3 2 1 1 3 0 0 0 0

sa sb ab at bt

sa

sb

ab

at

bt

a

b

s

t

y y y y y

y

y

y

yby

z

z

z

z


1 0 0 0 0 1 0 1 0 0

0 1 0 0 0 0 1 1 0 0

0 0 1 0 0 1 1 0 0 0

0 0 0 1 0 1 0 0 1 0

0 0 0 0 1 0 1 0 1 0

0 0 0 0 0 0 0 1 1 1

saT

sb

ab

at

bt

a

b

s

t

yA cy

y

y

y

z

z

z

z


We now see that the problems are dual So also according to the Strong Duality

Theorem, Max Flow = Min Cut

Max Flow – Min Cut Theorem


We want integer values for the variables, but this is LP, not IP!

So how do we know that the solution will yield integer values for the variables?

Theroem: If the constraints matrix is totally unimodular and the right hand side is comprised of integers, then it’s easy to find an integer solution.

Integer Values


Definition: A Unimodular matrix is a square matrix whose determinant is 0, 1 or -1.

Definition: A Totally Unimodular matrix is a matrix whose every non-singular square submatrix is unimodular.

Unimodular / Totally Unimodular

linear programming – max flow – min cut orgad keller

Documents

orgad keller algorithms

cut orgad keller slide

dual problem slide

minimum cut slide

linear programming max

maximal maximum flow

strong duality theorem

cut value