1. Piyush Kumar (Lecture 3: Preflow Push)
Advanced Algorithms
Piyush Kumar
(Lecture 3: Preflow Push)
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3. Today
Preflow Push
Bipartite matching/Hall’s theorem
If time permits

4. Review: Augmenting Path
Maintains a flow in each iteration
Finds a path in the residual graph
Saturates the path to make progress.

5. Flows Def. An s-t flow is a function that satisfies:
For each e  E: (capacity)
For each v  V – {s, t}: (conservation)
Def. The value of a flow f is: s 2 3 4 5 6 7 t 15 30 10 8 9 4 4
source = where material originates, sink = where material goes
flow conservation = otherwise warehouse overfills or oil pipe bursts
flow conservation is analogous to Kirchoff's law
* flow: abstract entity generated at source, transmitted across edges, absorbed at sink
* assume no arcs enter s or leave t (makes a little cleaner, no loss of generality) 4 4
capacity
flow
Value = 4

6. A Preflow Def. An preflow is a function that satisfies:
For each e  E: (capacity)
For each v  V – {s, t}: (weak conservation)
Def. The excess of a preflow f at node v :
Required to be non-negative except s,t
source = where material originates, sink = where material goes
flow conservation = otherwise warehouse overfills or oil pipe bursts
flow conservation is analogous to Kirchoff's law
* flow: abstract entity generated at source, transmitted across edges, absorbed at sink
* assume no arcs enter s or leave t (makes a little cleaner, no loss of generality)
Algorithm maintains preflow (not flow)
Each vertex is associated with a height
Flow is “downhill”
Vertices with excess are sometimes “lifted/relabeled”.

7. A Preflow Total excess = flow out of s - flow into t Nodes with +ve1
7/10
Nodes with +ve
Excess are called
Active s
5/5
10/10 G:
2/2
3/3 2 t
e=9

8. Labelling/Height function
H: V -> N
Source and Sink condition
h(s) = n
h(t) = 0
Steepness condition
For all (v,w) in Ef , h(v) <= h(w)+1
Here Ef is the residual graph.
Invariant
Compatibility of preflows and labellings

9. Lemma
If s-t preflow f is compatible with the labelling h, then there is no s-t path in the residual graph Gf.
Cor: If s-t flow f is compatible with a labelling h, then f is a flow of maximum value.

10. Initial Preflow f h(v) = 0 for all v <> s h(s) = n
f(e) = ce for all e=(s,v)
f(e) = 0 for all other edges.
The algorithm gradually transforms a preflow to a flow. Why?

11. Lemma
An s-t preflow f is compatible with a labelling h, then there is no s-t path in the residual graph Gf.
Cor: An s-t flow f is maximum when its compatible with labelling h.

12. Preflow-Push Algorithm Initialization example
10/10
10/10 2 3 1 2 t s
e=10
e=10
Gf: 5 10 10 2 3 1 2 t 5

13. Basic Operation 1 Push forward edge
Suppose e(u)>0, cf(u,v)>0,
(u,v) in Ef
(u,v) is a forward edge h[u]= h[v]+1
Push as much flow across (u,v) as possible
r = min (e[u], c(u,v) – f(e))
Increase f(e) by r.

14. Basic Operation 2 Push backward edge
Suppose e(u)>0, cf(u,v)>0,
(u,v) in Ef
(u,v) is a backward edge h[u]= h[v]+1
Push as much flow across (u,v) as possible
r = min (e[u], f(e))
Decrease f(e) by r.
The preflow push algorithm will try to push from
Active nodes towards the sink
(maintaining compatibility with the labelling/heights of the nodes)

15. Basic Operation 3 Lift/Relabel
When no push can be done from overflowing vertex (except s,t)
Suppose e[u]>0, and for all (u,v) in Ef : h[u] £ h[v], u ¹ s, u ¹ t
Set h[u] = h[u] + 1

16. Preflow-Push Initialize
While there is a node v <> t with ef(v) > 0
If w such that push (f,h,v,w) is possible
Push(f,h,v,w)
Else
Relabel(f,h,v) E

17. Preflow push demo

18. Invariants The labels are >= 0 and integers
f is a preflow, if the capacities are integral, the preflows are integral.
The preflow f and labelling h are compatible.
The height of a node never decreases.

19. Lemma
Lef f be a preflow. If excessf(v) > 0 then there is a path in Gf from v to source s.
Pf: Let A denote all the nodes that have a path to s in Gf. B = V \ A.
What happens to edges inside B?
If e has only its head in B?
If e has only its tail in B? –fout(B)

20. Lemmas The height of a node is upper bounded by 2n-1.
Cor: Each node is relabeled at most 2n-1 times, hence total relabels is upper bounded by 2n2

21. Lemma: Saturating Pushes
Throughout the algorithm, the number of saturating push operations is at most 2mn.
Pf: Look at a saturating push along an edge. When can another saturating push happen thru the same edge?
(If it needs lifting, how many lifts can one do on such a node in total?)

22. Potential function arguments
Consider a joint bank account held by
A, B, and C
The account starts with an initial deposit of at most n2.
A only makes withdrawals, each for 1 or more.
B makes fewer than n2 deposits. Each deposit is for at most 1.
C makes fewer than nm withdrawals and deposits. Each deposit is for at most n.
The account never goes below 0.
What is the maximum number of withdrawals for A?

23. Lemma: Non-Saturating Pushes Potential Functions Argument
Lemma: The number of non-saturating push operations is bounded by 2mn2
Pf: Define:
A non saturating push decreases it by at least 1.
A saturating push increases it by at most 2n-1
A relabel increases it by exactly 1.
If we pick the active node at maximum height
For pushing, we can get O(n^3) non-sat. pushes.

24. Modifications A simple modification gives O(n3).
Theorem: The total running time of the algorithm is upper bounded by O(mn2)

25. References
R.K. Ahuja, T.L. Magnanti, and J.B. Orlin. Network Flows. Prentice Hall, (Reserved in Dirac)
K. Mehlhorn and S. Naeher. The LEDA Platform for Combinatorial and Geometric Computing. Cambridge University Press, pages