Advanced Algorithms
Piyush Kumar
(Lecture 3: Preflow Push)
Welcome to COT5405
Announcements
• Homework 1 is due this week.
• Scribing is worth 5% extra credit.
(Any scribers for today?)
Today
• Preflow Push
• Bipartite matching/Hall’s theorem
– If time permits
Review: Augmenting Path
• Maintains a flow in each iteration
• Finds a path in the residual graph
• Saturates the path to make progress.
Flows
• Def. An s-t flow is a function that satisfies:
0  f (e)  c(e)
– For each e  E:
(capacity)
– For each v  V – {s, t}:  f (e)   f (e) (conservation)
e in to v
e out of v
• Def. The value of aflow f is: v( f ) 

10
9

4 4
0
5
s
e out of s
0
2
4
 f (e) .
0
15
5
0
15
0
4
4
3
8
6
0
capacity
15
flow
0
4 0
6
15 0
0
4
30
10
7
10
t
0
10
Value = 4
A Preflow
• Def. An preflow is a function that satisfies:
0  f (e)  c(e)
– For each e  E:
(capacity)
– For each v  V – {s, t}:  f (e)   f (e) (weak conservation)
e in to v
e out of v
• Def. The excess ofa preflow f at node v :
– Required to be non-negative except s,t
e f (v ) 

e in s
1.
2.
3.
4.
f (e) 

f ( e) .
e out of s
Algorithm maintains preflow (not flow)
Each vertex is associated with a height
Flow is “downhill”
Vertices with excess are sometimes “lifted/relabeled”.
A Preflow
Total excess
= flow out of s
- flow into t
1
7/10
Nodes with +ve
Excess are called
Active
s
5/5
G:
10/10
2/2
2
e=9
3/3
t
Labelling/Height function
• H: V -> N
– Source and Sink condition
• h(s) = n
• h(t) = 0
Invariant
– Steepness condition
• For all (v,w) in Ef , h(v) <= h(w)+1
• Here Ef is the residual graph.
Compatibility of preflows and labellings
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.
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?
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.
s
G:
e=10
10/10
1
Preflow-Push Algorithm
Initialization example
10/10
2
2
e=10
5
3
t
s
Gf :
10
10
1
2
2
5
3
t
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.
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)
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  s, u  t
– Set h[u] = h[u] + 1
Preflow-Push
• Initialize
• While there is a node v <> t with ef(v) > 0
E
– If
w such that push (f,h,v,w) is possible
• Push(f,h,v,w)
– Else
• Relabel(f,h,v)
Preflow push demo
Invariants
1. The labels are >= 0 and integers
2. f is a preflow, if the capacities are
integral, the preflows are integral.
3. The preflow f and labelling h are
compatible.
4. The height of a node never
decreases.
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.
0   e f (v )    f in (v )  f out (v )
vB
.
vB
What happens to edges inside B?
If e has only its head in B?
If e has only its tail in B? –fout(B)
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
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?)
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?
Lemma: Non-Saturating Pushes
Potential Functions Argument
• Lemma: The number of non-saturating
push operations is bounded by 2mn2
• Pf: Define: 0  ( f , h)   h(v)
v:e f ( v )0
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.
Modifications
• A simple modification gives O(n3).
• Theorem: The total running time of
the algorithm is upper bounded by
O(mn2)
References
• R.K. Ahuja, T.L. Magnanti, and J.B.
Orlin. Network Flows. Prentice Hall,
1993. (Reserved in Dirac)
• K. Mehlhorn and S. Naeher. The
LEDA Platform for Combinatorial and
Geometric Computing. Cambridge
University Press, 1999. 1018 pages