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

Definitions G=(V,E) is a directed graph capacity u(v,w) for every v,w V: If (v,w)  E then u(v,w) = 0 Two distinguished vertices s and t. s 3 4 1 a b 3 3 2 c d 4 3 t

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)  u(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

Flows and s-t cuts Let (X,X’) be a cut such that s X, t X’. Flow is the same across any cut: f(X,X’) = f(v,w) = f(v,w) - f(v,w) = |f| - 0 = |f| v X, w  X’ v X, w  V v X, w  X so |f|  cap(X,X’) = u(v,w) The value of the maximum flow is smaller than the minimum capacity of a cut.

More definitions The residual capacity of a flow is a function r on the edges such that r(v,w) = u(v,w) - f(v,w) a 2, 1 d Interpretation: We can push r(v,w) more flow from v to w by increasing f(v,w) and decreasing f(w,v)

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)

Example 1 3 3 4 1 3 3 1 1 3 1 3 1 2 2 1 1 2 2 2 1 3 4 1 3 3 3 The residual network A flow

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) = u(v,w) = cap(X,X’)

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

But we’ll go for the preflow/push method

Distance labels Defined with respect to residual capacities d(t) = 0, d(s) = n d(v) ≤ d(w) + 1 if r(v,w) > 0

Example (distance labels) 1 3 3 4 1 3 3 1 1 3 1 3 1 2 2 1 1 2 2 2 1 3 4 1 3 3 3 The residual network A flow

Example (distance labels) 1 6 3 3 4 1 3 3 1 1 2 2 3 1 3 1 2 2 1 1 2 2 2 1 1 1 3 4 1 3 3 3 The residual network A flow Is this a valid distance labeling ?

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

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)  u(v,w) For every w, 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)

Example (preflow) 2 2 2 3 1 3 2 s t 3 2 1 Nodes with positive excess are called active. The preflow push algorithm will try to push flow from active nodes towards the sink, relying on d( ).

Initialization (preflow) 3 4 2 1 3 3 4 4 3 4 1 3 3 2 4 3

Initialization (distance labels) 6 3 3 4 4 3 4 3 4 1 1 3 3 3 3 2 2 4 3 4 3 Recall: s must be disconnected from t when d(s) = n, and the labeling is valid…

Admissible arc in the residual graph v w d(v) = d(w) + 1

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) }

Correctness Lemma 1: The source is reachable from every active vertex in the residual network Proof: Assume that’s not the case: v s S Which means that no flow enters S -- A contradiction

Correctness (cont) Corollary: There is an outgoing residual arc incident with every active vertex Corollary: So we can push-relabel as long as there is an active vertex

Correctness (cont) Lemma 2: Distance labels only increase and remain valid at all times Proof: By induction on the number of push and relabel operations. For relabel this is clear by the definition of relabel For push: w v d(v) = d(w) + 1 so even if we add (w,v) to the residual network then it is still a valid labeling

Correctness (cont) Lemma 3: 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)

Complexity analysis

Another example

Complexity analysis Observation: d(v) increases when we relabel v ! Lemma: d(v) ≤ 2n-1 Proof: v v1 v2 s d(v) ≤ d(v1) + 1 ≤ d(v2) + 2 ..... ≤ d(s) + (n-1) = 2n-1

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

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

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)

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 All edges, not only those in R current edge

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.

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) became residual since it was scanned then when that happened d(w) = d(v) + 1  d(v) ≤ d(w) and this must be the case now

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)

Reducing the # of nonsaturating pushes 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) }

Example (FIFO order) 2 4 2 2 2 1 4

4 2 4 2 2 6 2 2 1 4

relabel 4 4 2 4 z 2 2 2 6 2 2 1 y w x 4 u v Q: z y 2 4 2 2 6 2 1 4

4 4 2 1 4 z 2 2 2 6 2 2 1 y w x 4 relabel u v Q: y z 2 1 4 2 2 6 2 1 4

4 push 4 2 1 4 z 2 2 2 2 6 push 2 2 1 y w x 4 u v Q: z y 2 1 4 2 2 2 6 2 1 4

4 2 2 1 4 z 2 2 2 2 2 6 2 2 push 1 y w x 4 2 u v Q: y u 2 1 4 2 2 2 6 2 1 4

2 4 2 2 1 4 z 2 2 2 2 2 6 2 2 1 y w x 4 2 u v Q: u z relabel 2 1 4 2 2 2 6 2 1 4

relabel 2 4 2 2 1 4 z 2 2 2 2 2 6 2 2 1 y w x 4 2 1 u v Q: z u 2 1 4 2 2 2 6 2 1 4 1

4 2 2 3 4 z 2 2 2 2 2 6 2 2 1 y w x 4 1 u v Q: u z 2 3 4 2 2 2 6 2 1 4

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)

relabel 4 4 2 4 z 2 2 2 6 2 2 1 y w x 4 u v Q: z y 2 4 2 2 6 2 1 4

4 4 2 1 4 z 2 2 2 6 2 2 1 y w x 4 relabel u v Q: y z 2 1 4 2 2 6 2 1 4

End of pass 1 4 4 2 1 4 z 2 2 2 2 6 2 2 1 y w x 4 u v Q: z y 2 1 4 2 2 2 6 2 1 4

4 push 4 2 1 4 z 2 2 2 2 6 push 2 2 1 y w x 4 u v Q: z y 2 1 4 2 2 2 6 2 1 4

4 2 2 1 4 z 2 2 2 2 2 6 2 2 push 1 y w x 4 2 u v Q: y u 2 1 4 2 2 2 6 2 1 4

End of pass 2 2 4 2 2 1 4 z 2 2 2 2 2 6 2 2 1 y w x 4 2 u v Q: u z 2 1 4 2 2 2 6 2 1 4

2 4 2 2 1 4 z 2 2 2 2 2 6 2 2 1 y w x 4 2 u v Q: u z relabel 2 1 4 2 2 2 6 2 1 4

relabel 2 4 2 2 1 4 z 2 2 2 2 2 6 2 2 1 y w x 4 2 1 u v Q: z u 2 1 4 2 2 2 6 2 1 4 1

End of pass 3 4 2 2 3 4 z 2 2 2 2 2 6 2 2 1 y w x 4 1 u v Q: u z 2 3 4 2 2 2 6 2 1 4

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

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

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

Reminder: Admissible arc in the residual graph v w d(v) = d(w) + 1

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

A faster implementation Maintain a (dynamic) forest of some of the admissible current edges Active guys are among the roots

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) }

A faster implementation Q: v…..  discharge(v)  Treepush/relabel(v) v w

Case 1: (v,w) is admissible link(v,w,rf(v,w)), (v,c) = findmin(v), c = min(c,e(v)), addcost(v,-c) Let (u,c) = findmin(v) If c=0 cut(u) and repeat If e(v) > 0 and v is not a root then repeat

Case 2: (v,w) is not admissible If (v,w) is not the last edge then advance the current edge If (v,w) is the last edge we relabel v and perform cut(u) for every child u of v

Analysis O(1) work + O(1) tree operations in Treepush/relabel + O(1) work + O(1) tree operations per cut How many cuts do we do ? O(mn) (each charged to a saturating push or a relabel) How many times do we call Treepush/relabel ? O(mn), in each we either advance the current edge or do a link (there are O(mn) links since there are O(mn) cuts)

Analysis (Cont) 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)

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

Idea: Don’t let the trees to grow too large Case 1: (v,w) is admissible v We won’t do the link if we are about to create a too large tree (say larger than k) w link(v,w,rf(v,w)), (v,c) = findmin(v), c = min(c,e(v)), addcost(v,-c) Let (u,c) = findmin(v) If c=0 cut(u) and repeat If e(v) > 0 and v is not a root then repeat

If we are about to create a tree with at least k vertices Push from v to w min{e(v),rf(v,w)} flow (w,c) = findmin(w), c = min(c,e(w)), addcost(w,-c) Let (u,c) = findmin(w) If c=0 cut(u) and repeat If e(w) > 0 and w is not a root then repeat

What collapses in our analysis ? There are calls to Treepush/relabel that we cannot charge to links (or cuts) v w r So we cannot say that the # of Treepush/relabels is O(mn)

How do we recover ? v w r May assume that the push from v is not saturating..(there are only O(nm) saturating ones)  v is not active after such Treepush/relabel  We are going to bound the # of such Treepush/relabels by the # of times a node may become active

Activating nodes When does a node become active ? r In Treepush/relabel. We know that there are O(mn) of them except for nonsaturating ones, that do not link/cut.  So we can further focus on those problematic Treepush/relabels that make r active

Concluding Either Tv or Tr is large: ≥ k/2 We charge the large tree.  May assume that v becomes inactive and r becomes active and the push is not saturating and does not link so We charge the large tree.

Tv Tr v w r Since the root of the tree that we charge either becomes active or inactive each tree is charged at most twice in a phase If the tree did not exist at the beginning of the phase then further deliver the charge it to the link or cut that created it

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 that nodes get activated at most O(mn + n3/k) times  This also bounds the # of Treepush/relabels and the # of dynamic tree operations For k=n2/m we get the bound of O(mnlog(n2/m))