CSE 202: Design and Analysis of...

CSE 202: Design and Analysis of Algorithms

Lecture 8

Instructor: Kamalika Chaudhuri

Last Class: Max Flow Problem

An s-t flow is a function f: E R such that:- 0

Last Class: Flows and Cuts

s

v

t

u2/2

1/1

1/3

2/5

1/2

The Max Flow Problem: Given directed graph G=(V,E), source s, sink t, edge capacities c(e), find an s-t flow of maximum size

An s-t Cut partitions nodes into groups = (L, R)s.t. s in L, t in RCapacity of a cut (L, R) =

Property: For any flow f, any s-t cut (L, R), size(f)

Gf = (V, Ef) where Ef E U ER

For any (u,v) in E or ER,

cf(u,v) = c(u,v) – f(u,v) + f(v,u)[ignore edges with zero cf: don’t put them in Ef]

⊆

Last Class: Ford-Fulkerson Algorithm

1: Construct a residual graph Gf (“what’s left to take?”)

s

a

b

t

1

1

1

1

1

s

a

b

t

1

1

1

Example

G:

f:

Gf :

s

a

b

t

1

1

1

1

12: Find a path from s to t in Gf3: Increase flow along this path, as much as possible

FF Algorithm: Start with zero flow

Repeat:

Find a path from s to t along which flow

can be increased

Increase the flow along that path

In any iteration, we have some flow f and we are trying to improve it. How to do this?

FF algorithm gives us a valid flow. But is it the maximum possible flow?

Consider final residual graph Gf = (V, Ef)Let L = nodes reachable from s in Gf and let R = rest of nodes = V – LSo s L and t R

�

(u,v)∈E,u∈L,v∈R

c(u, v)

Analysis: Correctness

s

Lt

R

Edges from L to R must be at full capacity Edges from R to L must be emptyTherefore, flow across cut (L,R) is

Thus, size(f) = capacity(L,R)

Recall: for any flow and any cut,size(flow)

An Observation: Integrality

Integral Flows: A flow f is integral if f(e) is an integer for all e

Example:

s

a

b

v

1/1

0/1

1/1

0/1

t1/1

An Integral Flow

s

a

b

v

0.5/1

0.5/1

t1/1

0.5/1

0.5/1

A Fractional Flow

Property: If all edge capacities are integers, then, there is a max flow f which is integral.



Example:

s

a

b

v

1/1

0/1

1/1

0/1

t1/1

An Integral Flow

s

a

b

v

0.5/1

0.5/1

t1/1

0.5/1

0.5/1

A Fractional Flow




Example:

s

a

b

v

1/1

0/1

1/1

0/1

t1/1

An Integral Flow

s

a

b

v

0.5/1

0.5/1

t1/1

0.5/1

0.5/1

A Fractional Flow

Proof: If the edge capacities are integers, then, the FF algorithm always finds an integral flowThe FF algorithm also always finds a max flow.




Example:

s

a

b

v

1/1

0/1

1/1

0/1

t1/1

An Integral Flow

s

a

b

v

0.5/1

0.5/1

t1/1

0.5/1

0.5/1

A Fractional Flow

Note: All max flows are not necessarily integral flows!

Proof: If the edge capacities are integers, then, the FF algorithm always finds an integral flowThe FF algorithm also always finds a max flow.

Analysis: efficiency

A hillclimbing procedure

Flow size:

0

max flow

How many iterations are needed to reach the maximum flow?

Example:

Each iteration is fast (O(|E|) time).

s

a

b

t

106

106

106

106

1


Repeat:

Find a path from s to t along which flow can be increased


s

a

b

t

1

1

1

s

a

b

t

1

1

1

#iterations can be Max Capacity

Analysis: efficiency

A hillclimbing procedure

Flow size:

0

max flow

How many iterations are needed to reach the maximum flow?

Example:

Each iteration is fast (O(|E|) time).

s

a

b

t

106

106

106

106

1


Repeat:



#iterations can be Max Capacity (with integer capacities)

Example:

How to improve the efficiency?

• Ford-Fulkerson Style Algorithms:• Edmonds Karp• Capacity Scaling

• Preflow-Push

Edmonds Karp

Bad Example:


Repeat:



Bad Path Sequence: (s, a, b, t), (s, b, a, t), (s, a, b, t),...

s

a

b

t

106

106

106

106

1

Edmonds Karp

EK Path Selection: Find the shortest path along which flow can be increased

(shortest path = shortest in terms of #edges)

Bad Example:


Repeat:




s

a

b

t

106

106

106

106

1

Edmonds Karp

EK Algorithm: Start with zero flow

Repeat:

Find the shortest path from s to t

along which flow can be increased


s

a

b

t

106 106

Iteration 1

f

Bad Example for FF:

s

a

b

t

106

106

106

106

1

Edmonds Karp


Repeat:




s

a

b

t

106 106

s

a

b

t

106

106

106

106

1Iteration 1

f Gf

Bad Example for FF:

s

a

b

t

106

106

106

106

1

Edmonds Karp


Repeat:




s

a

b

t

106 106

s

a

b

t

106106

s

a

b

t

106

106

106

106

1Iteration 1

f

Iteration 2

Gf

f

Bad Example for FF:

s

a

b

t

106

106

106

106

1

Edmonds Karp


Repeat:




s

a

b

t

106 106

s

a

b

t

106106

s

a

b

t

106

106

106

106

1

s

a

b

t

106

106

106

106

1

Iteration 1

f

Iteration 2

Gf

f Gf

Bad Example for FF:

s

a

b

t

106

106

106

106

1

Edmonds Karp



It can be shown that this requires only O(|V||E|) iterations (Proof not in this class)


Repeat:




Bad Example for FF:

s

a

b

t

106

106

106

106

1

Edmonds Karp



It can be shown that this requires only O(|V||E|) iterations (Proof not in this class)

Running Time: O(|V| |E|2)


Repeat:




Bad Example for FF:

s

a

b

t

106

106

106

106

1



• Preflow-Push

Capacity Scaling

Capacity Scaling: Find paths of high capacity first between s and t

Bad Example:


Repeat:




s

a

b

t

106

106

106

106

1

Capacity Scaling

Cmax = max capacity edge. Start with D = CmaxStart with zero flowWhile D >= 1, repeat:

Gf(D) = D-residual graphWhile there is a path from s to t in Gf(D) along which flow can be increased

Increase flow along that pathUpdate Gf(D)

D = D/2

D-Residual Graph: Subgraph of residual graph with only edges with capacity >= D

Example:

s

a

b

t

106

106

106

106

1

G

s

a

b

t

106

106

106

106

Gf(10)

For f = 0

Capacity Scaling: Correctness




D = D/2

D-Residual Graph: Subgraph of residual graph with only edges with capacity >= D

Property: If all edge capacities are integers, algorithm outputs a max flow

Proof: At D=1, Gf(D) = Gf. So on termination, Gf(D) has no more paths from s to t

Capacity Scaling: Running Time




D = D/2

D-Residual Graph: Subgraph of residual graph with only edges with capacity >= DProperty 1: While loop 1 is executed 1 + log2 Cmax times

1

2D scaling phase





D = D/2


1

2

Property 2: At the end of a D-scaling phase, size(max flow)





D = D/2


1

2D scaling phase

Proof: Let L = nodes reachable from s in Gf(D) and let R = rest of nodes = V – L

L

s

R

t






D = D/2


1

2D scaling phase


L

s

R

t#edges in Gf(D) in the (L, R) cut = 0






D = D/2


1

2D scaling phase


L

s

R

t#edges in Gf(D) in the (L, R) cut = 0#edges in Gf in the (L,R) cut





D = D/2


D scaling phase

Property 3: For any D, #iterations of loop 2 in the D-scaling phase





D = D/2

D scaling phase

Proof: After previous (2D-scaling) phase, size(max flow) = DProperty 1: While loop 1 is executed 1 + log2 Cmax times






D = D/2

D scaling phase

1

2





• Preflow-Push

Preflow-Push

Main Idea: - Each node has a label, which is a potential- Route flow from high to low potential

v

wLabels

Idea: Route flow along blue edges

Preflows

�

e into v

f(e)−�

e out of v

f(e) ≥ 0

Preflow: A function f: E R is a preflow if:1. Capacity Constraints: 0

Preflow-Push: Two Operations

v

wl

�

e into v

f(e)−�

e out of v

f(e) ≥ 0

Preflow: A function f: E R is a preflow if:1. Capacity Constraints: 0 0, for all w s.t (v, w) in Ef, h(w) >= h(v)Increase l(v) by 1

Pre-Flow Push: The Algorithm

Start with labeling: h(s) = n, h(t) = 0, h(v) = 0, for all other vStart with preflow f: f(e) = c(e) for e = (s, v), f(e) = 0, for all other edges e

While there is a node (other than t) with positive excessPick a node v with excess(v) > 0If there is an edge (v, w) in Ef such that push(v, w) can be applied

Push(v, w)Else

Relabel(v)

Push(v, w): Applies if excess(v) > 0, h(w) < h(v), (v, w) in Efq = min(excess(v), cf(v, w))Add q to f(v, w)

Relabel(v): Applies if excess(v) > 0, for all w s.t (v, w) in Ef, h(w) >= h(v)Increase h(v) by 1

Date post:	31-Jan-2021
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