Transcript Powerpoint slides

COMP 482: Design and
Analysis of Algorithms
Spring 2012
Lecture 19
Prof. Swarat Chaudhuri
Minimum Cut Problem
Min s-t cut problem. Find an s-t cut of minimum capacity.
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Capacity = 10 + 8 + 10
= 28
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Flows
Def. An s-t flow is a function that satisfies:
For each e  E:
0 £ f (e) £ c(e)
For each v  V – {s, t}: å f (e) = å f (e)
(capacity)
(conservation)
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e in to v
e out of v
Def. The value of a flow f is: v( f ) =
å f (e) .
e out of s
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Value = 24
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Maximum Flow Problem
Max flow problem. Find s-t flow of maximum value.
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Value = 28
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Augmenting Path Algorithm
Augment(f, c, P) {
b  bottleneck(P)
foreach e  P {
if (e  E) f(e)  f(e) + b
else
f(eR)  f(e) - b
}
return f
}
forward edge
reverse edge
Ford-Fulkerson(G, s, t, c) {
foreach e  E f(e)  0
Gf  residual graph
while (there exists augmenting path P) {
f  Augment(f, c, P)
update Gf
}
return f
}
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Running Time
Assumption. All capacities are integers between 1 and C.
Invariant. Every flow value f(e) and every residual capacities cf (e)
remains an integer throughout the algorithm.
Theorem. The algorithm terminates in at most v(f*)  nC iterations.
Pf. Each augmentation increase value by at least 1. ▪
Corollary. If C = 1, Ford-Fulkerson runs in O(mn) time.
Integrality theorem. If all capacities are integers, then there exists a
max flow f for which every flow value f(e) is an integer.
Pf. Since algorithm terminates, theorem follows from invariant. ▪
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Recap: Mobile computing
Consider a set of mobile computing clients in a certain town who each
needed to be connected to one of several base stations. We’ll
suppose there are n clients, with the position of each client
specified by its (x,y) coordinates. There are also k base stations,
each of them specified by its (x,y) coordinates as well.
For each client, we want to connect it to exactly one of the base
stations. We have a range parameter r—a client can only be
connected to a base station that is within distance r. There is also a
load parameter L—no more than L clients can be connected to a
single base station.
Your goal is to design a polynomial-time algorithm for the following
problem. Given the positions of the clients and base stations as well
as the range and load parameters, find the maximum number of
clients that can be connected simultaneously to a base station.
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7.3 Choosing Good Augmenting Paths
Ford-Fulkerson: Exponential Number of Augmentations
Q. Is generic Ford-Fulkerson algorithm polynomial in input size?
m, n, and log C
A. No. If max capacity is C, then algorithm can take C iterations.
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Choosing Good Augmenting Paths
Use care when selecting augmenting paths.
Some choices lead to exponential algorithms.
Clever choices lead to polynomial algorithms.
If capacities are irrational, algorithm not guaranteed to terminate!
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Goal: choose augmenting paths so that:
Can find augmenting paths efficiently.
Few iterations.
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Choose augmenting paths with: [Edmonds-Karp 1972, Dinitz 1970]
Max bottleneck capacity.
Sufficiently large bottleneck capacity.
Fewest number of edges.
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Capacity Scaling
Intuition. Choosing path with highest bottleneck capacity increases
flow by max possible amount.
Don't worry about finding exact highest bottleneck path.
Maintain scaling parameter .
Let Gf () be the subgraph of the residual graph consisting of only
arcs with capacity at least .
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Capacity Scaling
Scaling-Max-Flow(G, s, t, c) {
foreach e  E f(e)  0
  smallest power of 2 greater than or equal to C
Gf  residual graph
while (  1) {
Gf()  -residual graph
while (there exists augmenting path P in Gf()) {
f  augment(f, c, P)
update Gf()
}
   / 2
}
return f
}
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Capacity Scaling: Correctness
Assumption. All edge capacities are integers between 1 and C.
Integrality invariant. All flow and residual capacity values are integral.
Correctness. If the algorithm terminates, then f is a max flow.
Pf.
By integrality invariant, when  = 1  Gf() = Gf.
Upon termination of  = 1 phase, there are no augmenting paths. ▪
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Capacity Scaling: Running Time
Lemma 1. The outer while loop repeats 1 + log2 C times.
Pf. Initially C   < 2C.  decreases by a factor of 2 each iteration. ▪
Lemma 2. Let f be the flow at the end of a -scaling phase. Then the
proof on next slide
value of the maximum flow is at most v(f) + m .
Lemma 3. There are at most 2m augmentations per scaling phase.
Let f be the flow at the end of the previous scaling phase.
L2  v(f*)  v(f) + m (2).
Each augmentation in a -phase increases v(f) by at least . ▪
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Theorem. The scaling max-flow algorithm finds a max flow in O(m log C)
augmentations. It can be implemented to run in O(m2 log C) time. ▪
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Capacity Scaling: Running Time
Lemma 2. Let f be the flow at the end of a -scaling phase. Then value
of the maximum flow is at most v(f) + m .
Pf. (almost identical to proof of max-flow min-cut theorem)
We show that at the end of a -phase, there exists a cut (A, B)
such that cap(A, B)  v(f) + m .
Choose A to be the set of nodes reachable from s in Gf().
By definition of A, s  A.
By definition of f, t  A.
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original network
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7.5 Bipartite Matching
Matching
Matching.
Input: undirected graph G = (V, E).
M  E is a matching if each node appears in at most edge in M.
Max matching: find a max cardinality matching.
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Bipartite Matching
Bipartite matching.
Input: undirected, bipartite graph G = (L  R, E).
M  E is a matching if each node appears in at most one edge in M.
Max matching: find a max cardinality matching.
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Bipartite Matching
Bipartite matching.
Input: undirected, bipartite graph G = (L  R, E).
M  E is a matching if each node appears in at most edge in M.
Max matching: find a max cardinality matching.
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Bipartite Matching
Max flow formulation.
Create digraph G' = (L  R  {s, t}, E' ).
Direct all edges from L to R, and assign unit capacity.
Add source s, and unit capacity edges from s to each node in L.
Add sink t, and unit capacity edges from each node in R to t.
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Bipartite Matching: Proof of Correctness
Theorem. Max cardinality matching in G = value of max flow in G'.
Pf. 
Given max matching M of cardinality k.
Consider flow f that sends 1 unit along each of k paths.
f is a flow, and has cardinality k. ▪
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Bipartite Matching: Proof of Correctness
Theorem. Max cardinality matching in G = value of max flow in G'.
Pf. 
Let f be a max flow in G' of value k.
Integrality theorem  k is integral and can assume f is 0-1.
Consider M = set of edges from L to R with f(e) = 1.
– each node in L and R participates in at most one edge in M
– |M| = k: consider cut (L  s, R  t) ▪
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Bipartite Matching: Running Time
Which max flow algorithm to use for bipartite matching?
Generic augmenting path: O(m val(f*) ) = O(mn).
Capacity scaling: O(m2 log C ) = O(m2).
Shortest augmenting path: O(m n1/2).
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Non-bipartite matching.
Structure of non-bipartite graphs is more complicated, but
well-understood. [Tutte-Berge, Edmonds-Galai]
Blossom algorithm: O(n4). [Edmonds 1965]
Best known: O(m n1/2).
[Micali-Vazirani 1980]
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Q1: Edge Disjoint Paths
Disjoint path problem. Given a digraph G = (V, E) and two nodes s and t,
find the max number of edge-disjoint s-t paths.
Def. Two paths are edge-disjoint if they have no edge in common.
Ex: communication networks.
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Edge Disjoint Paths
Disjoint path problem. Given a digraph G = (V, E) and two nodes s and t,
find the max number of edge-disjoint s-t paths.
Def. Two paths are edge-disjoint if they have no edge in common.
Ex: communication networks.
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Answer: Edge Disjoint Paths
Max flow formulation: assign unit capacity to every edge.
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Theorem. Max number edge-disjoint s-t paths equals max flow value.
Pf. 
Suppose there are k edge-disjoint paths P1, . . . , Pk.
Set f(e) = 1 if e participates in some path Pi ; else set f(e) = 0.
Since paths are edge-disjoint, f is a flow of value k. ▪
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Edge Disjoint Paths
Max flow formulation: assign unit capacity to every edge.
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Theorem. Max number edge-disjoint s-t paths equals max flow value.
Pf. 
Suppose max flow value is k.
Integrality theorem  there exists 0-1 flow f of value k.
Consider edge (s, u) with f(s, u) = 1.
– by conservation, there exists an edge (u, v) with f(u, v) = 1
– continue until reach t, always choosing a new edge
Produces k (not necessarily simple) edge-disjoint paths. ▪
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can eliminate cycles to get simple paths if desired
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Q2: Project Selection
can be positive or negative
Projects with prerequisites.
Set P of possible projects. Project v has associated revenue pv.
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some projects generate money: create interactive e-commerce interface,
redesign web page
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others cost money: upgrade computers, get site license
Set of prerequisites E. If (v, w)  E, can't do project v and unless
also do project w.
A subset of projects A  P is feasible if the prerequisite of every
project in A also belongs to A.
Project selection. Choose a feasible subset of projects to maximize
revenue.
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Answer: Project Selection: Prerequisite Graph
Prerequisite graph.
Include an edge from v to w if can't do v without also doing w.
{v, w, x} is feasible subset of projects.
{v, x} is infeasible subset of projects.
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Project Selection: Min Cut Formulation
Min cut formulation.
Assign capacity  to all prerequisite edge.
Add edge (s, v) with capacity -pv if pv > 0.
Add edge (v, t) with capacity -pv if pv < 0.
For notational convenience, define ps = pt = 0.
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Project Selection: Min Cut Formulation
Claim. (A, B) is min cut iff A  { s } is optimal set of projects.
Infinite capacity edges ensure A  { s } is feasible.
Max revenue because:
cap(A, B) =
å p v + å (- p v )
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vÎ B: pv > 0
=
vÎ A: pv < 0
åpv - åpv
v: pv > 0
vÎ A
constant
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Open Pit Mining
Open-pit mining. (studied since early 1960s)
Blocks of earth are extracted from surface to retrieve ore.
Each block v has net value pv = value of ore - processing cost.
Can't remove block v before w or x.
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