max flow min cut in higher dimensions · max ow min cut reformulation max ow 2 2 2 3 3 t 3 3 1 2 1...

GraphsAlgebra

Higher dimensions

Max flow min cut in higher dimensions

Art Duval1 Caroline Klivans2 Jeremy Martin3

1University of Texas at El Paso

2Brown University

3University of Kansas

12th Joint UTEP/NMSU Workshop on Mathematics,Computer Science, and Computational Sciences

University of Texas at El PasoOctober 27, 2012

Duval, Klivans, Martin Max flow min cut in higher dimensions

GraphsAlgebra

Higher dimensions

Max flow min cutReformulation

Max flow 2

2 2

3

3

T

3

3

1 2 1 2 1S

2

Given a graph G with source S , sink T , and edge capacities κ.

DefinitionFlow on G is an assignment of flow xe (non-negative number, anddirection) to each edge such that:

I net flow at each vertex, except S and T , is zero; and

I |xe | ≤ κe .

Value of flow is outflow(S) = inflow(T ).

What is maximum value of flow on G ? How can we be sure?


GraphsAlgebra

Higher dimensions


Max flow 2/2

2/2 2/2

2/3

2/3

T

2/3

2/3

0/1 0/2 0/1 0/2 0/1S

2/2




I |xe | ≤ κe .




GraphsAlgebra

Higher dimensions


Max flow

2/2 2/2

3/3

2/3

T

3/3

2/3

0/1 1/2 1/1 1/2 0/1S

2/2 2/2




I |xe | ≤ κe .




GraphsAlgebra

Higher dimensions


Min cut 2 2

2 2

3

3

T

3

3

1 2 1 2 1S

DefinitionCut is minimal set of edges whose removal disconnects S from T .Value of cut is

∑e∈cut κe .

Clearly, value(flow) ≤ value(cut), so max flow ≤ min cut.

Theorem (Classic max flow min cut)

Max flow = min cut.


GraphsAlgebra

Higher dimensions


Min cut 2/2

2/2 2/2

3/3

2/3

T

3/3

2/3

0/1 1/2 1/1 1/2 0/1S

2/2

DefinitionCut is minimal set of edges whose removal disconnects S from T .Value of cut is

∑e∈cut κe .

Clearly, value(flow) ≤ value(cut), so max flow ≤ min cut.

Theorem (Classic max flow min cut)

Max flow = min cut.


GraphsAlgebra

Higher dimensions


Add an extra edge 2 2

2 2

3

3

T

3

3

1 2 1 2 1S


I net flow at each vertex is zero; and

I |xe | ≤ κe .

Value of flow is x0.

DefinitionCut is minimal set of edges, including e0, whose removaldisconnects G . Value of cut is

∑e∈cut\e0

κe .


GraphsAlgebra

Higher dimensions


Add an extra edge

5

T

2/3

0/1 1/2 1/1 1/2 0/1S

2/2 2/2

2/2 2/2

3/3

2/33/3



I |xe | ≤ κe .



∑e∈cut\e0

κe .


GraphsAlgebra

Higher dimensions


Add an extra edge

5

T

3/3

2/3

0/1 1/2 1/1 1/2 0/1S

2/2 2/2

2/2 2/2

3/3

2/3



I |xe | ≤ κe .



∑e∈cut\e0

κe .


GraphsAlgebra

Higher dimensions

Boundary matrixLinear programming

Flows and boundary

2

4

3 (1 1 −1

)342

= 5.

Assign orientation to each edge (flow going “backwards” getsnegative value)

netflow(v) =∑v=e+

xe −∑

v=e−

xe =∑v∈e

(−1)ε(e,v)xe = (∂x)v

So net flow condition is ∂x = 0.


GraphsAlgebra

Higher dimensions


Cuts and coboundary

Assign 1 to every vertex in connected component with T , 0 toothers. Let yv be value at v . Edges in cut are those that haveboth 0 and 1 endpoints.

00

0

1 1

1

1

0

S T

Coboundary will do this: ∂T y (linear combination of rows of ∂)gives characteristic vector of cut.


GraphsAlgebra

Higher dimensions


Linear programming

Flow is now a linear program

I Find vector x (in edge space)

I ∂x = 0 (x is in flow space)

I −κe ≤ xe ≤ κe (can omit e0)

I max x0

The dual program is (can easily be reworked to say):

I Find vector y (in vertex space)

I Let u = ∂T y (in cut space)

I u0 = 1

I min∑

e κe |ue |Linear programming says the solutions are equal; with some effortwe can show the solution to the dual LP is the min cut problem.


GraphsAlgebra

Higher dimensions


Linear programming





I max x0




I u0 = 1

I min∑

e κe |ue |

Linear programming says the solutions are equal; with some effortwe can show the solution to the dual LP is the min cut problem.


GraphsAlgebra

Higher dimensions


Linear programming





I max x0




I u0 = 1

I min∑

e κe |ue |Linear programming says the solutions are equal; with some effortwe can show the solution to the dual LP is the min cut problem.


GraphsAlgebra

Higher dimensions

Translating from linear programmingSummary

Max flow in higher dimensionsExample: 2-dimensional complex; ∂ maps 2-dimensional cells(polygons) to edges.

I Find vector x (in polygon space)


I −κp ≤ xp ≤ κp (can omit p0)

I identify designated polygon p0; max x0

Find a 1-dimensional cycle on the complex, and attach a polygonface filling that cycle. We are trying to maximize circulation onthat designated polygon (around that cycle), while making allcirculation balance on each edge.


GraphsAlgebra

Higher dimensions


Min cut in higher dimensions


I Find vector y (in edge space)


I u0 = 1

I min∑

p κp|up|Linear programming says the solutions are equal; with some effortwe can show the solution to the dual LP is the min cut problem.

In particular, the support on this cut is a minimal set of polygonswhose removal introduces codimension-1 homology, e.g.,1-dimensional “circular” hole in 2-dimensional complex


GraphsAlgebra

Higher dimensions


Min cut in higher dimensions


I Find vector y (in edge space)


I u0 = 1

I min∑

p κp|up|Linear programming says the solutions are equal; with some effortwe can show the solution to the dual LP is the min cut problem.In particular, the support on this cut is a minimal set of polygonswhose removal introduces codimension-1 homology, e.g.,1-dimensional “circular” hole in 2-dimensional complex


GraphsAlgebra

Higher dimensions


Summary

Theorem (DKM)

The max circulation around a codimension-1 cycle (e.g.,1-dimensional cycle in 2-dimensional complex) equals the value ofa minimum cut containing the added face that fills in the cycle(e.g., polygon filling in 1-dimensional cycle).

Fine print:

I cut is minimal set of faces whose removal increasescodimension-1 homology

I cut vector is in span of row space of boundary matrix

I normalize cut vector by specifying its value is 1 on p0, theadded filling-in face

I cut vector might not be all 1’s and 0’s

I value of cut is inner product of capacities with cut vector


max flow min cut in higher dimensions · max ow min cut reformulation max ow 2 2 2 3 3 t 3 3 1 2 1...

Documents