Combinatorial Optimization and Graph Theory ORCO Execution of the Push-Relabel Algorithm

Zolt´an Szigeti

1

4 t

s

2 4 3

4 b

Problem

Find in the network(D,g)ag-feasible (s,t)-flow ofmaximumvalue and an (s,t)-cut ofminimumcapacity!

(0,1)

(0,4) t

s

(0,2) (0,4) (3,3)

(4,4) b Initialization

Preflowx

(0,1)

(0,4) t

s

(0,2) (0,4) (3,3)

(4,4) b

s t

b Construction

Auxiliary graphD_x

(0,1)

(0,4) t

s

(0,2) (0,4) (3,3)

(4,4) b

s t

b

0 0

4

Initialization

Labellingℓ

(0,1)

(0,4) t

s

(0,2) (0,4) (3,3)

(4,4) b

s t

b

0 0

4

Looking for active vertex

f_x(a) =d_x⁻(a)−d_x⁺(a)

= (3 + 0)−(0 + 0) = 3>0

=⇒ais x-active

(0,1)

(0,4) t

s

(0,2) (0,4) (3,3)

(4,4) b

s t

b

0 0

4

Looking for tight arc leaving a

ℓ(a) = 06= 1 =ℓ(b) + 1 =ℓ(t) + 1 6= 5 =ℓ(s) + 1

=⇒noℓ-tight arc leavingaexists

(0,1)

(0,4) t

s

(0,2) (0,4) (3,3)

(4,4) b

s t

b

0 0

4

Relabel at a

ℓ^′(a) = min{ℓ(v) + 1 :av ∈A_x}

= min{0 + 1,0 + 1,4 + 1}= 1

(0,1)

(0,4) t

s

(0,2) (0,4) (3,3)

(4,4) b

s t

b

0 0

4

Relabel at a

ℓ^′(a) = min{ℓ(v) + 1 :av ∈A_x}

= min{0 + 1,0 + 1,4 + 1}= 1

(0,1)

(0,4) t

s

(0,2) (0,4) (3,3)

(4,4) b

s t

b

0 0

4

Relabel at a

ℓ^′(a) = min{ℓ(v) + 1 :av ∈A_x}

= min{0 + 1,0 + 1,4 + 1}= 1 arcatbecomes ℓ^′-tight

(0,1)

(0,4) t

s

(0,2) (0,4) (3,3)

(4,4) b

s t

b

0 0

4

Pushon at

ε =min{g(at),f_x(a)}

= min{4−0 + 0,3}= 3 x^′(at)=x(at) +ε= 0 + 3 = 3 x^′(e) =x(e)∀e∈A\ {at}

(0,1)

(0,4) t

s

(0,2) (3,4) (3,3)

(4,4) b

Pushon at

ε =min{g(at),f_x(a)}

= min{4−0 + 0,3}= 3 x^′(at)=x(at) +ε= 0 + 3 = 3 x^′(e) =x(e)∀e∈A\ {at}

ais notx^′-active anymore

(0,1)

(0,4) t

s

(0,2) (3,4) (3,3)

(4,4) b

Looking for active vertex

f_x(b) =d_x⁻(b)−d_x⁺(b)

= (4 + 0)−(0 + 0) = 4>0

=⇒bisx-active

(0,1)

(0,4) t

s

(0,2) (3,4) (3,3)

(4,4) b

s t

b Construction

Auxiliary graphD_x

(0,1)

(0,4) t

s

(0,2) (3,4) (3,3)

(4,4) b

s t

b

0 0

4

Previous labelling

(0,1)

(0,4) t

s

(0,2) (3,4) (3,3)

(4,4)

b 0

s t

b

0 4

Looking for tight arc leaving b

ℓ(b) = 06= 2 =ℓ(a) + 1 6= 1 =ℓ(t) + 1 6= 5 =ℓ(s) + 1

=⇒noℓ-tight arc leavingbexists

(0,1)

(0,4) t

s

(0,2) (3,4) (3,3)

(4,4)

b 0

s t

b

0 4

Relabel at b

ℓ^′(b) = min{ℓ(v) + 1 :bv ∈A_x}

= min{1 + 1,0 + 1,4 + 1}= 1

(0,1)

(0,4) t

s

(0,2) (3,4) (3,3)

(4,4)

b 1

s t

b

0 4

Relabel at b

ℓ^′(b) = min{ℓ(v) + 1 :bv ∈A_x}

= min{1 + 1,0 + 1,4 + 1}= 1

(0,1)

(0,4) t

s

(0,2) (3,4) (3,3)

(4,4)

b 1

s t

b

0 4

Relabel at b

ℓ^′(b) = min{ℓ(v) + 1 :bv ∈A_x}

= min{1 + 1,0 + 1,4 + 1}= 1 arcbt becomesℓ^′-tight

(0,1)

(0,4) t

s

(0,2) (3,4) (3,3)

(4,4)

b 1

s t

b

0 4

Pushon bt

ε =min{g(bt),f_x(b)}

= min{2−0 + 0,4}= 2 x^′(bt)=x(bt) +ε= 0 + 2 = 2

(0,1)

(0,4) t

s

(2,2) (3,4) (3,3)

(4,4) b

Pushon bt

ε =min{g(bt),f_x(b)}

= min{2−0 + 0,4}= 2 x^′(bt)=x(bt) +ε= 0 + 2 = 2

(0,1)

(0,4) t

s

(2,2) (3,4) (3,3)

(4,4) b

Pushon bt

ε =min{g(bt),f_x(b)}

= min{2−0 + 0,4}= 2 x^′(bt)=x(bt) +ε= 0 + 2 = 2

bis stillx^′-active

(0,1)

(0,4) t

s

(2,2) (3,4) (3,3)

(4,4) b

s t

b Construction

Auxiliary graphD_x

(0,1)

(0,4) t

s

(2,2) (3,4) (3,3)

(4,4) b

s t

b

0 1

4

Previous labelling

(0,1)

(0,4) t

s

(2,2) (3,4) (3,3)

(4,4)

b 1

s t

b

0 4

Looking for tight arc leaving b

ℓ(b) = 16= 2 =ℓ(a) + 1 6= 5 =ℓ(s) + 1

=⇒noℓ-tight arc leavingbexists

(0,1)

(0,4) t

s

(2,2) (3,4) (3,3)

(4,4)

b 1

s t

b

0 4

Relabel at b

ℓ^′(b) = min{ℓ(v) + 1 :bv ∈A_x}

= min{1 + 1,4 + 1}= 2

(0,1)

(0,4) t

s

(2,2) (3,4) (3,3)

(4,4)

b 2

s t

b

0 4

Relabel at b

ℓ^′(b) = min{ℓ(v) + 1 :bv ∈A_x}

= min{1 + 1,4 + 1}= 2

(0,1)

(0,4) t

s

(2,2) (3,4) (3,3)

(4,4)

b 2

s t

b

0 4

Relabel at b

ℓ^′(b) = min{ℓ(v) + 1 :bv ∈A_x}

= min{1 + 1,4 + 1}= 2 arcbabecomesℓ^′-tight

(0,1)

(0,4) t

s

(2,2) (3,4) (3,3)

(4,4)

b 2

s t

b

0 4

Pushon ba

ε =min{g(ba),f_x(b)}

= min{4−0 + 0,2}= 2 ε^′ =min{x(ab), ε}= min{0,2}= 0 x^′(ba)=x(ba) +ε−ε^′ = 0+2−0 = 2 x^′(ab)=x(ab)−ε^′= 0−0 = 0

(0,1)

(2,4) t

s

(2,2) (3,4) (3,3)

(4,4) b

Pushon ba

ε =min{g(ba),f_x(b)}

= min{4−0 + 0,2}= 2 ε^′ =min{x(ab), ε}= min{0,2}= 0 x^′(ba)=x(ba) +ε−ε^′ = 0+2−0 = 2 x^′(ab)=x(ab)−ε^′= 0−0 = 0

(0,1)

(2,4) t

s

(2,2) (3,4) (3,3)

(4,4) b

Pushon ba

ε =min{g(ba),f_x(b)}

= min{4−0 + 0,2}= 2 ε^′ =min{x(ab), ε}= min{0,2}= 0 x^′(ba)=x(ba) +ε−ε^′ = 0+2−0 = 2 x^′(ab)=x(ab)−ε^′= 0−0 = 0

(0,1)

(2,4) t

s

(2,2) (3,4) (3,3)

(4,4) b

Looking for active vertex

f_x′(a) =d⁻

x^′(a)−d⁺

x^′(a)

= (3 + 2)−(3 + 0) = 2>0

=⇒ais x^′-active

(0,1)

(2,4) t

s

(2,2) (3,4) (3,3)

(4,4) b

s t

b Construction

Auxiliary graphD_x

(0,1)

(2,4) t

s

(2,2) (3,4) (3,3)

(4,4) b

s t

b

0 2

4

Previous labelling

(0,1)

(2,4) t

s

(2,2) (3,4) (3,3)

(4,4) b

s t

b

0 2

4

Looking for tight arc leaving a

ℓ(a) = 1 =ℓ(t) + 1

(0,1)

(2,4) t

s

(2,2) (3,4) (3,3)

(4,4) b

s t

b

0 2

4

Looking for tight arc leaving a

ℓ(a) = 1 =ℓ(t) + 1

=⇒arcatis ℓ-tight

(0,1)

(2,4) t

s

(2,2) (3,4) (3,3)

(4,4) b

s t

b

0 2

4

Pushon at

ε =min{g(at),f_x(a)}

= min{4−3 + 0,2}= 1 x^′(at)=x(at) +ε= 3 + 1 = 4

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4) b

Pushon at

ε =min{g(at),f_x(a)}

= min{4−3 + 0,2}= 1 x^′(at)=x(at) +ε= 3 + 1 = 4

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4) b

Pushon at

ε =min{g(at),f_x(a)}

= min{4−3 + 0,2}= 1 x^′(at)=x(at) +ε= 3 + 1 = 4 ais stillx^′-active

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4) b

s t

b Construction

Auxiliary graphD_x

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4) b

s t

b

0 2

4

Previous labelling

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4) b

s t

b

0 2

4

Looking for tight arc leaving a

ℓ(a) = 16= 3 =ℓ(b) + 1 6= 5 =ℓ(s) + 1

=⇒noℓ-tight arc leavingaexists

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4) b

s t

b

0 2

4

Relabel at a

ℓ^′(a) = min{ℓ(v) + 1 :av ∈A_x}

= min{2 + 1,4 + 1}= 3

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4) b

s t

b

0 2

4

Relabel at a

ℓ^′(a) = min{ℓ(v) + 1 :av ∈A_x}

= min{2 + 1,4 + 1}= 3

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4) b

s t

b

0 2

4

Relabel at a

ℓ^′(a) = min{ℓ(v) + 1 :av ∈A_x}

= min{2 + 1,4 + 1}= 3 arcabbecomesℓ^′-tight

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4) b

s t

b

0 2

4

Pushon ab

ε =min{g(ab),f_x(a)}

= min{1−0 + 2,1}= 1 ε^′ =min{x(ba), ε}= min{2,1}= 1 x^′(ab)=x(ab) +ε−ε^′= 0 + 1-1 = 0 x^′(ba)=x(ba)−ε^′= 2−1 = 1

(0,1)

(1,4) t

s

(2,2) (4,4) (3,3)

(4,4) b

Pushon ab

ε =min{g(ab),f_x(a)}

= min{1−0 + 2,1}= 1 ε^′ =min{x(ba), ε}= min{2,1}= 1 x^′(ab)=x(ab) +ε−ε^′= 0 + 1-1 = 0 x^′(ba)=x(ba)−ε^′= 2−1 = 1

(0,1)

(1,4) t

s

(2,2) (4,4) (3,3)

(4,4) b

Pushon ab

ε =min{g(ab),f_x(a)}

= min{1−0 + 2,1}= 1 ε^′ =min{x(ba), ε}= min{2,1}= 1 x^′(ab)=x(ab) +ε−ε^′= 0 + 1-1 = 0 x^′(ba)=x(ba)−ε^′= 2−1 = 1

ais notx^′-active anymore

(0,1)

(1,4) t

s

(2,2) (4,4) (3,3)

(4,4) b

Looking for active vertex

f_x′(b) =d⁻

x^′(b)−d⁺

x^′(b)

= (4 + 0)−(1 + 2) = 1>0

=⇒bisx^′-active

(0,1)

(1,4) t

s

(2,2) (4,4) (3,3)

(4,4) b

s t

b Construction

Auxiliary graphD_x

(0,1)

(1,4) t

s

(2,2) (4,4) (3,3)

(4,4) b

s t

b

0 2

4

Previous labelling

(0,1)

(1,4) t

s

(2,2) (4,4) (3,3)

(4,4)

b 2

s t

b

0 4

Looking for tight arc leaving b

ℓ(b) = 26= 4 =ℓ(a) + 1 6= 5 =ℓ(s) + 1

=⇒noℓ-tight arc leavingbexists

(0,1)

(1,4) t

s

(2,2) (4,4) (3,3)

(4,4)

b 2

s t

b

0 4

Relabel at b

ℓ^′(b) = min{ℓ(v) + 1 :bv ∈A_x}

= min{3 + 1,4 + 1}= 4

(0,1)

(1,4) t

s

(2,2) (4,4) (3,3)

(4,4)

b 4

s t

b

0 4

Relabel at b

ℓ^′(b) = min{ℓ(v) + 1 :bv ∈A_x}

= min{3 + 1,4 + 1}= 4

(0,1)

(1,4) t

s

(2,2) (4,4) (3,3)

(4,4)

b 4

s t

b

0 4

Relabel at b

ℓ^′(b) = min{ℓ(v) + 1 :bv ∈A_x}

= min{3 + 1,4 + 1}= 4 arcbabecomesℓ^′-tight

(0,1)

(1,4) t

s

(2,2) (4,4) (3,3)

(4,4)

b 4

s t

b

0 4

Pushon ba

ε =min{g(ba),f_x(b)}

= min{4−1 + 0,1}= 1 ε^′ =min{x(ab), ε}= min{0,1}= 0 x^′(ba)=x(ba) +ε−ε^′ = 1 + 1-0 = 2 x^′(ab)=x(ab)−ε^′= 0−0 = 0

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4) b

Pushon ba

ε =min{g(ba),f_x(b)}

= min{4−1 + 0,1}= 1 ε^′ =min{x(ab), ε}= min{0,1}= 0 x^′(ba)=x(ba) +ε−ε^′ = 1 + 1-0 = 2 x^′(ab)=x(ab)−ε^′= 0−0 = 0

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4) b

Pushon ba

ε =min{g(ba),f_x(b)}

= min{4−1 + 0,1}= 1 ε^′ =min{x(ab), ε}= min{0,1}= 0 x^′(ba)=x(ba) +ε−ε^′ = 1 + 1-0 = 2 x^′(ab)=x(ab)−ε^′= 0−0 = 0

bis notx^′-active anymore

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4) b

Looking for active vertex

f_x′(a) =d⁻

x^′(a)−d⁺

x^′(a)

= (3 + 2)−(4 + 0) = 1>0

=⇒ais x^′-active

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4) b

s t

b Construction

Auxiliary graphD_x

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4)

b 4

s t

b

0 4

Previous labelling

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4)

b 4

s t

b

0 4

Looking for tight arc leaving a

ℓ(a) = 36= 5 =ℓ(b) + 1 6= 5 =ℓ(s) + 1

=⇒noℓ-tight arc leavingaexists

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4)

b 4

s t

b

0 4

Relabel at a

ℓ^′(a) = min{ℓ(v) + 1 :av ∈A_x}

= min{4 + 1,4 + 1}= 5

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4)

b 4

s t

b

0 4

Relabel at a

ℓ^′(a) = min{ℓ(v) + 1 :av ∈A_x}

= min{4 + 1,4 + 1}= 5

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4)

b 4

s t

b

0 4

Relabel at a

ℓ^′(a) = min{ℓ(v) + 1 :av ∈A_x}

= min{4 + 1,4 + 1}= 5 arcas becomesℓ^′-tight

(0,1)

(2,4) t

s

(2,2) (4,4) (3,3)

(4,4)

b 4

s t

b

0 4

Pushon as

ε =min{g(as),f_x(a)}

= min{0−0 + 3,1}= 1 ε^′ =min{x(sa), ε}= min{3,1}= 1 x^′(sa)=x(sa)−ε^′= 3−1 = 2

(0,1)

(2,4) t

s

(2,2) (4,4) (2,3)

(4,4) b

Pushon as

ε =min{g(as),f_x(a)}

= min{0−0 + 3,1}= 1 ε^′ =min{x(sa), ε}= min{3,1}= 1 x^′(sa)=x(sa)−ε^′= 3−1 = 2

(0,1)

(2,4) t

s

(2,2) (4,4) (2,3)

(4,4) b

Pushon as

ε =min{g(as),f_x(a)}

= min{0−0 + 3,1}= 1 ε^′ =min{x(sa), ε}= min{3,1}= 1 x^′(sa)=x(sa)−ε^′= 3−1 = 2

Looking for active vertex

(0,1)

(2,4) t

s

(2,2) (4,4) (2,3)

(4,4) b

Optimal flow

=⇒x is ag-feasible(s,t)-flow of maximum value

s (2,4) (0,1) t (2,2) (4,4) (2,3)

(4,4) b

s

4

t

b

0 4

Optimal flow

=⇒x is ag-feasible(s,t)-flow of maximum value

Optimal cut

ℓdoes not take1

Z ={v :ℓ(v)>1} is an(s,t)-cut of minimum capacity