dinic’ max flow algorithm - sjtudominik/teaching/material/dinic-slides.pdf · 4 2 2 1 1 1 3 3 3 3...

Dinic’ Max Flow AlgorithmSlides by Dominik Scheder

Part IDinic’ Algorithm in General Flow Networks

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s

t2 3

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Find out which edges lie on a shortest s-t-path,

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and forget the rest for the time being.

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and forget the rest for the time being.

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Now lets greedily route as much flow as possiblein this network!

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Find an s-t-path with depth-first search.

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Route as much flow through it as possible: 1 unit.

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(1)

(1)(1)

(1)

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(1)

(1)(1)

(1)

Update capacities.

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2 3 3

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Update capacities.

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Update capacities. Repeat!

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1

1 3

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2 3 3

s

t


3 1

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1 3

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2 3 3

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Find an s-t-path with depth-first search.Delete dead ends.

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2

1

1 3

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2 3 3

s

t


3

2

1

1 3

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3

23

2 3 3

s

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Find an s-t-path with depth-first search.Delete dead ends.Route flow.

3

2

1

1 3

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3

23

2 3 3

s

t


(1)

(1)

(1)

(1)

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1

1 3

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2 3 3

s

t


(1)

(1)

(1)

(1)

Update capacities.

2

1

1 3

33

3

13

2 3 3

s

t

Find an s-t-path with depth-first search.Delete dead ends.Route flow.Update capacities.

2

1

1 3

33

3

13

2 3 3

s

t

Find an s-t-path with depth-first search.Delete dead ends.Route flow.Update capacities. Repeat!

2

1

1 3

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3

13

2 3 3

s

t


1 3

33

3

13

2 3

s

t


1 3

33

3

13

2 3

s

t

Find an s-t-path with depth-first search.Delete dead ends.Route as much flow as possible.

1 3

33

3

13

2 3

s

t


(1)

(1)

(1)

(1)

1 3

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3

13

2 3

s

t


(1)

(1)

(1)

(1)

Update capacities.

2

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2

2 3

s

t

Find an s-t-path with depth-first search.Delete dead ends.Route as much flow as possible.Update capacities.

2

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3

2

2 3

s

t

Find an s-t-path with depth-first search.Delete dead ends.Route as much flow as possible.Update capacities. Repeat!

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(2)

(2) (2)

(2)

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(2)

(2) (2)

(2)

Update capacities.

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1

s

t

Find an s-t-path with depth-first search.Delete dead ends.Route as much flow as possible.Update capacities.

11

1

s

t

Find an s-t-path with depth-first search.Delete dead ends.Route as much flow as possible.Update capacities. Repeat.

11

1

s

t


t

Find an s-t-path with depth-first search.Delete dead ends.The source s has been deleted. The phase ends.

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Collect the flow of this phase.

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(1)

(1)(1)

(1)

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(1)

(1)(2)

(2)

(1)

(1)

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(1)

(1)(2)

(3)

(1)

(1)

(1)

(1) (1)

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(1)

(1)(2)

(3)

(1)

(1)

(1)

(1) (2)

(1)(1)

(1)

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(1)

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(1)

(1) (2)

(1)(1)

(1)

Observation: Only edges on a shortest s-t-path carry flow.

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(1)

(1)(2)

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(1)

(1) (2)

(1)(1)

(1)


Build residual network (and introduce back edges)!

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Build residual network (and introduce back edges)!

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dist(s, t) = 4 in the original network.

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Back edges go backwards. Not part of any length-4-path.

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Back edges go backwards. Not part of any length-4-path.

dist(s, t) > 4 in the residual network.

Runtime Analysis

dist(s, t) increases in every phase.


At most n− 2 phases.



Every phase performs a couple of depth-first searches.




How many?




How many?

How much time does each take?

Claim. Each phase performs at most m depth-first searches.


Proof.


Proof. There are at most m edges in the network.


Proof. There are at most m edges in the network.

After every depth-first search we delete at leastone edge.

Observation. The ith depth-first search takes time O(n+ di),where di is the number of edges deleted as dead ends.


Thus, the total running time of all (at most m) depth-firstsearches is:



O (∑m

i=1(n+ di))



O (∑m

i=1(n+ di)) = O (nm+∑m

i=1 di)



O (∑m

i=1(n+ di)) = O (nm+∑m

i=1 di)

= O (nm+m)



O (∑m

i=1(n+ di)) = O (nm+∑m

i=1 di)

= O (nm+m)

= O (nm)



O (∑m

i=1(n+ di)) = O (nm+∑m

i=1 di)

= O (nm+m)

= O (nm)

Thus, each phase takes O(nm) steps.



O (∑m

i=1(n+ di)) = O (nm+∑m

i=1 di)

= O (nm+m)

= O (nm)


Dinic’ algorithm performs at most n− 2 phases.



O (∑m

i=1(n+ di)) = O (nm+∑m

i=1 di)

= O (nm+m)

= O (nm)


Dinic’ algorithm performs at most n− 2 phases.

Dinic’ algorithm performs O(n2m) steps.

Theorem. Dinic’ algorithm finds a maximum flow inO(n2m) steps.

Part IIDinic’ Algorithm in Unit Capacity Networks

s

t

All capacities are 1

s

t


k := dist(s, t). Here, k = 4.

s

t


Ignore edges not on a shortest s-t-path.

k := dist(s, t). Here, k = 4.

s

t

Find s-t-path by depth-first search.

s

t


Route 1 unit of flow.

s

t



Update capacities.

s

t



Update capacities. This deletes k edges!

Observation. Let k = dist(s, t). Every capacity updateremoves k edges.


Thus, each phase performs at most m/k depth firstsearches.



Every depth-first search takes time O(k+ di), where di isthe number of edges deleted as dead ends.




Number of steps performed in phase stage is:

O(∑m/k

i=1 (k + di))





O(∑m/k

i=1 (k + di))= O

(m+

∑m/ki=1 di

)





O(∑m/k

i=1 (k + di))= O

(m+

∑m/ki=1 di

)= O(m).





O(∑m/k

i=1 (k + di))= O

(m+

∑m/ki=1 di

)= O(m).

There are at most n− 2 phases.

Theorem. On a network in which all capacities are 1,Dinic’ algorithm finds a maximum flow in O(nm) steps.


even better:

Lemma 1. On a network with unit capacities, Dinic’algorithm performs at most

√m phases, and thus

O(m1.5) steps.


even better:

Lemma 1. On a network with unit capacities, Dinic’algorithm performs at most


O(m1.5) steps.

Lemma 2. On a network with unit capacities, Dinic’algorithm performs at most O

(n2/3

)phases, and thus

O(n2/3m) steps.

Part IIINumber of Phases in Unit Capacity Networks

Lemma 1. On a network with unit capacities, Dinic’algorithm performs at most 2


O(m1.5) steps.



O(m1.5) steps.

Proof. Let t ∈ N, to be determined later.



O(m1.5) steps.


Consider Gt, the residual network after t phases.



O(m1.5) steps.



k := dist(s, t) ≥ t in Gt.



O(m1.5) steps.




Vi := {v | dist(s, v) = i in Gt}.



O(m1.5) steps.





s t...

......

...V0

V1 Vi Vi+1 Vk+1Vk


s t...

......

...V0

V1 Vi Vi+1 Vk+1Vk


s t...

......

...V0

V1 Vi Vi+1 Vk+1Vk

Ei := {(u, v) ∈ E | u ∈ Vi, v ∈ Vi+1}


s t...

......

...V0

V1 Vi Vi+1 Vk+1Vk

Ei := {(u, v) ∈ E | u ∈ Vi, v ∈ Vi+1}

|E0|+ |E1|+ . . .+ |Ek−1| ≤ m.


s t...

......

...V0

V1 Vi Vi+1 Vk+1Vk

Ei := {(u, v) ∈ E | u ∈ Vi, v ∈ Vi+1}

|E0|+ |E1|+ . . .+ |Ek−1| ≤ m.

|Ei| ≤ mk ≤mt for some i.


s t...

......

...V0

V1 Vi Vi+1 Vk+1Vk

Ei := {(u, v) ∈ E | u ∈ Vi, v ∈ Vi+1}

|E0|+ |E1|+ . . .+ |Ek−1| ≤ m.


S := V0 ∪ V1 ∪ . . . ∪ Vi.

is an s-t-cut!


s t...

......

...V0

V1 Vi Vi+1 Vk+1Vk

Ei := {(u, v) ∈ E | u ∈ Vi, v ∈ Vi+1}

|E0|+ |E1|+ . . .+ |Ek−1| ≤ m.


S := V0 ∪ V1 ∪ . . . ∪ Vi.

cap(S, V \ S) = |Ei| ≤ mt .

is an s-t-cut!


s t...

......

...V0

V1 Vi Vi+1 Vk+1Vk

Ei := {(u, v) ∈ E | u ∈ Vi, v ∈ Vi+1}

|E0|+ |E1|+ . . .+ |Ek−1| ≤ m.


S := V0 ∪ V1 ∪ . . . ∪ Vi.

cap(S, V \ S) = |Ei| ≤ mt .

maxflow(Gt) ≤ mt .

is an s-t-cut!

Gt := the residual network after t phases.



Every phase increases the flow by at least 1.




At most t+ mt phases.





At most 2√m phases (setting t =

√m).





At most 2√m phases (setting t =

√m).



O(m1.5) steps.


(n2/3

)phases, and thus

O(n2/3m) steps.


(n2/3

)phases, and thus

O(n2/3m) steps.

Proof.


(n2/3

)phases, and thus

O(n2/3m) steps.

Proof. Vi := {v | dist(s, v) = i in Gt}.

s t...

......

...V0

V1 Vi Vi+1 Vk+1Vk


(n2/3

)phases, and thus

O(n2/3m) steps.


s t...

......

...V0

V1 Vi Vi+1 Vk+1Vk

Vi is big if |Vi| ≥ 2n/k.


(n2/3

)phases, and thus

O(n2/3m) steps.


s t...

......

...V0

V1 Vi Vi+1 Vk+1Vk


At most n2n/k =k2 sets Vi can be big.

s t...

......

...V0

V1 Vi Vi+1 Vk+1Vk



s t...

......

...V0

V1 Vi Vi+1 Vk+1Vk



Thus, V0, . . . , Vk contains two consecutive “small” sets Vi, Vi+1.

......

Vi Vi+1

s t...

......

...V0

V1 Vi Vi+1 Vk+1Vk




|E(Vi, Vi+1)| ≤ |Vi| · |Vi+1| ≤ 4n2

k2 .

......

Vi Vi+1

s t...

......

...V0

V1 Vi Vi+1 Vk+1Vk




|E(Vi, Vi+1)| ≤ |Vi| · |Vi+1| ≤ 4n2

k2 .

......

Vi Vi+1

S := V0 ∪ V1 ∪ . . . Vi is an s-t-cut.

s t...

......

...V0

V1 Vi Vi+1 Vk+1Vk




|E(Vi, Vi+1)| ≤ |Vi| · |Vi+1| ≤ 4n2

k2 .

......

Vi Vi+1

S := V0 ∪ V1 ∪ . . . Vi is an s-t-cut.cap(S, V \ S) ≤ 4n

2

k2 ≤4n2

t2 .

s t...

......

...V0

V1 Vi Vi+1 Vk+1Vk




|E(Vi, Vi+1)| ≤ |Vi| · |Vi+1| ≤ 4n2

k2 .

......

Vi Vi+1

S := V0 ∪ V1 ∪ . . . Vi is an s-t-cut.cap(S, V \ S) ≤ 4n

2

k2 ≤4n2

t2 .maxflow(Gt) ≤


maxflow(Gt) ≤ 4n2

t2 .


maxflow(Gt) ≤ 4n2

t2 .


At most t+ 4n2

t2 phases.


maxflow(Gt) ≤ 4n2

t2 .


At most t+ 4n2

t2 phases.

At most 5n2/3 phases (setting t = n2/3).


maxflow(Gt) ≤ 4n2

t2 .


At most t+ 4n2

t2 phases.

At most 5n2/3 phases (setting t = n2/3).


(n2/3

)phases, and thus

O(n2/3m) steps.

Theorem. Dinic’ algorithm finds a maximum flow inO(n2m) steps.

Theorem. On a network with unit capacities, Dinic’algorithm finds a maximum flow in O(m1.5) steps.

Theorem. On a network with unit capacities, Dinic’algorithm finds a maximum flow in O(n2/3m) steps.

dinic’ max flow algorithm - sjtudominik/teaching/material/dinic-slides.pdf · 4 2 2 1 1 1 3 3 3 3...

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