1 Maximum flow: The preflow/push method of Goldberg and Tarjan (87)

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Maximum flow: The preflow/push method of Goldberg and Tarjan

(87)

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Definitions

• G=(V,E) is a directed graph

• capacity c(v,w) for every v,w V: If (v,w) E then c(v,w) = 0

• Two distinguished vertices s and t.

s

t

a b

c d

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Definitions (cont)

A flow is a function on the edges which satisfies the following requirements

• f(v,w) = -f(w,v) skew symmetry

• f(v,w) c(v,w)

• For every v except s and t wf(v,w) = 0

The value of the flow |f| = wf(s,w)

The maxflow problem is to find f with maximum value

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Flows and s-t cutsLet (X,X’) be a cut such that s X, t X’.

st

f(X,X’) = f(v,w) = f(v,w) - f(v,w) = |f| - 0 = |f|

|f| cap(X,X’) = c(v,w)

Flow is the same across any cut:

so

The value of the maximum flow is smaller than the minimum capacity of a cut.

v X, w X’

v X, w V

v X, w X

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More definitionsThe residual capacity of a flow is a function r on the edges such that

r(v,w) = c(v,w) - f(v,w)

a

d2, 1

Interpretation: We can push r(v,w) more flow from v to w by increasing f(v,w) and decreasing f(w,v)

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More definitions (cont)We define the residual graph R on V such that there is an arc from v to w with capacity r(v,w) for every v and w such that r(v,w) > 0

An augmenting path p R is a path from s to t in R

r(p) = min r(v,w)(v,w) p

We can increase the flow by r(p)

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Example

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A flow The residual network

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Basic theorem

(1) f is max flow <==>

(2) There is no augmenting path in R <==>

(3) |f| = cap(X,X’) for some X

Proof. (3) ==> (1), (1) ==> (2) obvious

To prove (2) ==>(3) let X be all vertices reachable from s in R. By assumption t X. So (X,X’) is an s-t cut. Since there is no edge from X to X’ in R |f| = f(X,X’) = f(v,w) = c(v,w) = cap(X,X’)

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Augmenting path methods

Repeat the following step:

Find an augmenting path in R, increase the flow, update R

Stop when s and t are disconnected in R.

Need to be careful about how you choose those augmenting paths !

The best algorithm in this family is Dinic’s algorithm, that can be implemented in O(nmlog(n)) time

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But we’ll go for the preflow/push method

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Distance labels

• Defined with respect to residual capacities

• d(t) = 0• d(v) ≤ d(w) + 1 if r(v,w) > 0

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Example (distance labels)

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Distance labels – basic lemma

Lemma: d(v) is a lower bound on the length of the shortest path from v to the sink

Proof: Let the s.p. to the sink be:

v v1 v2 t

d(v) ≤ d(v1) + 1 ≤ d(v2) + 2 ..... ≤ d(t) + k = k

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Preflow (definition)

A preflow is a function on the edges which satisfies the following requirements

• f(v,w) = -f(w,v) skew symmetry

• f(v,w) c(v,w)

• For every v, except s and t, vf(v,w) ≥ 0

Let e(w) = vf(v,w) be the excess at the node v

(we’ll also have e(t) ≥ 0, and e(s) ≤ 0)

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Example (preflow)

Nodes with positive excess are called active.

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The preflow push algorithm will try to push flow from active nodes towards the sink, relying on d( ).

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Initialization (preflow)

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Initialization (distance labels)

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Note: s must be disconnected from t when d(s) = n, and the labeling is valid…

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Admissible arc in the residual graph

w

v

d(v) = d(w) + 1

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The preflow push algorithm

While there is an active node {

pick an active node v and push/relabel(v)

}Push/relabel(v) {

If there is an admissible arc (v,w) then {

push = min {e(v) , r(v,w)} flow from v to w

} else {

d(v) := min{d(w) + 1 | r(v,w) > 0} (relabel)

}

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Correctness

Lemma 1: The source is reachable from every active vertex in the residual network

Proof:

Which means that no flow enters S..

vs

S

Assume that’s not the case:

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Correctness (cont)

Corollary: There is an outgoing arc incident with every active vertex

so assuming distance labels are valid, we can always either push or relabel an active node.

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Correctness (cont)

Lemma 1: Distance labels remain valid at all times

Proof:

By induction on the number of pushing and relabeling operations.

For relabel this is clear by the definition of relabel

For push: v w

d(v) = d(w) + 1 so even if we add (w,v) to the residual network then it is still a valid labeling

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Correctness (cont)

Corollary: So we can push-relabel as long as there is an active vertex

Lemma: When (and if) the algorithm stops the preflow is a maximum flow

Proof:

It is a flow since there is no active vertex.

It is maximum since the sink is not reachable from the source in the residual network. (d(s) = n, and the labeling is valid)

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Complexity analysis

Another example

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Complexity analysis

Observation: d(v) increases when we relabel v !

Lemma: d(v) ≤ 2n-1

Proof:

vv1v2s

d(v) ≤ d(v1) + 1 ≤ d(v2) + 2 ..... ≤ d(s) + (n-1) = 2n-1

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Complexity analysis (cont)

Lemma: The # of relabelings is (2n-1)(n-2) < 2n2

Proof:

At most 2n-1 per each node other than s and t

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Complexity analysis (cont)

Def: Call a push saturating if min{e(v), r(v,w)} = r(v,w)

Lemma: The # of saturating pushes is at most 2nm

Proof: Before another saturating push on (v,w), we must push from w to v.

d(w) must increase by at least 2

Since d(w) ≤ 2n-1, this can happen at most n times

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Nonsaturating pushes

Lemma: The # of nonsaturating pushes is at most 4n2m

Proof:

Let Φ = Σv active d(v)

• Decreases (by at least one) by every nonsaturating push

• Increases by at most 2n-1 by a saturating push : total increase (2n-1)2nm

• Increases by each relabeling: total increase < (2n-1)(n-2)

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Implementation

Maintain a list of active nodes, so finding an active node is easy

Given an active node v, we need to decide if there is an admissible arc (v,w) to push on ?

v

current edge

All edges, not only those in R

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Current edge

v

current edge

If the current edge (v,w) is admissible, push on it (updating the list of active vertices)

Otherwise, advance the current edge pointer

if you are on the last edge, relabel v and set the current edge to be the first one.

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Is this implementation correct?

Lemma: When we relabel v there is no admissible arc (v,w)

Proof: After we scanned (v,w) either (v,w) dropped off the residual network or d(v) ≤ d(w)

If d(v) ≤ d(w) then this must be the case now since v has not been relabeled.

If (v,w) got back on the list of v since it was scanned then when that happened d(w) = d(v) + 1 d(v) ≤ d(w) and this must be the case now

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Analysis

Lemma: The total time spent at v between two relabelings of v is Δv plus O(1) per push out of v

Summary: Since we relabel v at most (2n-1) times we get that the total work at v is O(nΔv) + O(1) per push out of v.

Summing over all vertices we get that the total time is O(nm) + #of pushes

O(n2m)

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Maintain the list of active vertices as a FIFO queue (Q)

Discharge the first vertex of the queue:

Discharge(v) {

While v is active and hasn’t been relabeled then push/relabel(v).

(If the loop stops because v is relabeled then add v to the end of Q)

}

Reducing the # of nonsaturating pushes

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Passes

Pass 1: Until you finish discharging all vertices initially in Q

Pass i: Until you finish discharging all vertices added to Q in pass (i-1)

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Analysis

Note that we still have the O(n2m) bound

How many passes are there ?

Let Φ = maxactive vd(v)

1) If the algorithm does not relabel during a pass then Φ decreases by at least 1 (each active node at the beginning of a pass moved its excess to a vertex with lower label)

2) If we relabel then Φ may increase by at most the maximum increase of a distance label

There are at most O(n2) passes of the second kind. These passes increase Φ by at most O(n2) There are at most O(n2) passes of the first kind

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Analysis (Cont)

So we have O(n2) passes

In each pass we have at most one nonsaturating push per vertex

O(n3) nonsaturating pushes

O(n3) total running time

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A faster implementation

Maintain a (dynamic) forest of some of the admissible current edges

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Reminder: Admissible arc in the residual graph

w

v

d(v) = d(w) + 1

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Active guys are among the roots

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At a high level the algorithm is almost the same

While there is an active node in Q {

Let v be the first in Q

discharge(v)

}

discharge(v) {

While v is active and hasn’t been relabeled then Treepush/relabel(v).

(If the loop stops because v is relabeled then add v to the end of Q)

}

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Q: v….. discharge(v) Treepush/relabel(v)

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w

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Case 1: (v,w) is admissible

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w

link(v,w,rf(v,w)),

(v,c) = findmin(v), addcost(v,-c)

Let (u,c) = findmin(v) If c=0 cut(u) and repeatIf e(v) > 0 and v is not a root then repeat

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Case 2: (v,w) is not admissible

v

w

a) If (v,w) is not the last edge then advance the current edge

b) If (v,w) is the last edge we relabel v and perform cut(u) for every child u of v

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Analysis

O(1) work + tree operations in Treepush/relabel +

O(1) work + tree operations per cut

How many cuts do we do ?

O(mn)

How many times do we call Treepush/relabel ?

O(mn) + # of times we add a vertex to Q

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Analysis (Cont)

How many times a vertex can become active ?

We can charge each new active vertex to a link or a cut

A vertex becomes active O(mn) times

Summary: we do O(mn) dynamic tree operations

We’ll see how to do those in O(log n) each so we get running time of O(mnlog n)

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Can we improve on that ?

Notice that we have not really used the fact that Q is a queue, any list would do !

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Idea: Don’t get the trees to grow too large

Case 1: (v,w) is admissible

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w

link(v,w,rf(v,w)),

(v,c) = findmin(v), addcost(v,-c)

Let (u,c) = findmin(v) If c=0 cut(u) and repeatIf e(v) > 0 and v is not a root then repeat

We won’t do the link if a too large tree is created (say larger than k)

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If we are about to create a tree with at least k vertices

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w

Push from v to w min{e(v),rf(v,w)} flow

(w,c) = findmin(w), addcost(w,-c)

Let (u,c) = findmin(w) If c=0 cut(u) and repeatIf e(w) > 0 and w is not a root then repeat

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What collapses in our analysis ?

Note r may become active and we cannot charge it to a link or a cut !

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w

So we cannot say that the # of insertions into Q is O(mn)

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How do we recover ?

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r

w

May assume that the push from v is not saturating..(there are only O(nm) saturating ones)

There cannot be another such push involving v as the pushing root and r as the “becoming active” node in a phase

Tv

Tr

Either Tv or Tr is large ≥ k/2

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v

r

w

Charge it to the large tree if it existed at the beginning of the phase

Tv

Tr

Charge it to the link or cut creating it if it did not exist at the beginning of the phase.

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Each link is charged once, a cut is charged twice O(mn) such charges over all phases.

At the beginning of a phase we have O(n/k) large trees, each charged once O(n3/k)

So we get O(mn + n3/k) dynamic tree operations each costs O(log k)

For k=n2/m we get O(mnlog(n2/m))

Documents

1 Maximum flow: The preflow/push method of Goldberg and Tarjan (87)