CSCI 4260 MATH 4150

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CSCI 4260 MATH 4150

• Network Flows

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Last lecture

• We covered maximum matchings in bipartite graphs.
• Today
• Network Flows
• But first some variations of the matching problem

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Recap maximum bipartite matching

• Given a bipartite graph G (X,Y,E)
• Find a set M of edges
• No two edges share an endpoint
• With maximum cardinality
• We presented an algorithm that searches for
  alternating paths from a given matching M
• If augmenting path found ? we find a bigger
  matching
• If not ? We showed that M is maximum by
  presenting a vertex cover of size M as a
  witness.

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Weighted bipartite matching

• Unweighted case maximize the number of edges in
  matching
• Weighted case maximize the sum of the weights in
  the matching
• Solvable in polynomial time using similar
  concepts.

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Stable matchings

• Instead of weights we are given preference
  rankings
• In other words, each node ranks available nodes
• Suppose we have matched pairs (x, b) and (y, a)
  but
• x prefers a to b and
• a prefers x to y
• We say the unmatched pair (x,a) is unstable.
• A stable matching is a matching with no unstable
  unmatched pair.
• Solvable in polynomial time.

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Matchings in general graphs

• Both weighted and unweighted cases can be solved
  in polynomial time.
• Initial algorithms by Edmonds.
• The running time is roughly cubic in n.

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Network Flow Problems

• A network is a directed graph with
• A capacity c(e) for each edge e
• A source vertex s
• And a terminal (sink) vertex t

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Flow

• A flow f assigns a value f(e) to each edge e
• f(v) total flow leaving v
• f-(v) total flow into v
• A flow is feasible if it satisfies
• Capacity constraints 0 ?f(e)?c(e) for each e
• Conservation constraints f(v) f-(v) for each
  node other than s and t

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Maximum Flow Problem

• The value of flow
• val(f) f-(t) - f(t)the amount of net flow
  into the sink
• The maximum flow problem is to find a feasible
  flow of maximum value

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Example
(1) 1
(1) 2
(0) 2
(1) 1
s
t
(0) 2
(1) 2
(1) 1
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Augmenting Paths

• Given a feasible flow f, for a network N, a path
  P is f-augmenting if for each edge e ? P
• If P follows e in the forward direction, then
  f(e) lt c(e).
• If P follows e in the backward direction, then
  f(e) gt 0

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Augmenting Paths cont

• Given a feasible flow f, for a network N, a path
  P is f-augmenting if for each edge e ? P
• If P follows e in the forward direction, then
  f(e) lt c(e). In this case, let ??(e) c(e)
  f(e)
• If P follows e in the backward direction, then
  f(e) gt 0. In this case, let ??(e) f(e).
• The tolerance of P min ??(e) e ?P

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Augmenting Path Example
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Lemma

• If P is an f-augmenting path with tolerance z,
  then changing flow by z on edges followed
  forward by P and by z on edges followed backward
  by P produces a feasible flow f with
  valueval(f) val(f) z

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Proof

• The definition of tolerance ensures that the
  capacity is never violated
• For conservation, there are four possibilities
• the netflow into the sink increases by z

P

-
-

-
-
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Maximum flow algorithm

• Similar to the maximum matching algorithm
• Search for an augmenting flow.
• If found, we create a bigger flow and repeat.
• If not we need a witness to show that our flow
  is maximum.

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Cuts

• In a network, a source/sink cut S,T consists of
  edges from a source set S to sink set T where
• S and T is a partitioning of the nodes
• s ? S
• t ? T
• The capacity of the cut cap(S,T) is the total
  capacities on the edges of S, T

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Intuitively

• Given an S, T cut, every s-t path must use at
  least one edge of the cut
• So the capacity of any flow is bounded by the
  capacity of the cut.
• Perhaps we can use cuts as witnesses
• Lets make this precise

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Net flow

• Let U be a set of vertices. Let
• f(U) total flow on edges leaving U
• f-(U) total flow on edges entering U
• Define net flow out of U as
• f(U)- f-(U)

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Lemma

• The net-flow out of U

Proof for each arc (x,y), lets look at the
contribution of (x,y) to lhs and rhs
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Moreover

• If f is a feasible flow and S, T is a
  source/sink cut, then the net flow out of S and
  the net flow into T are both equal to val(f).
• Net flow from S equals f(s)- f-(s)

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Corollary

• If f is a feasible flow and S, T is a
  source/sink cut, then val(f) ? cap(S,T)
• val(f) f(S)- f-(S) ? f(S)
• The capacity constraints require
• f(S) ? cap(S,T)

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Duality

• Therefore, for a given flow f, if we can find an
  S,T cut where the net flow out of S is equal to
  cap(f)
• f must be a max-flow

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Max Flow Algorithm

• Input a feasible flow f
• R s / reached set /
• S ? / searched set /

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Max Flow Algorithm Main Loop

• Pick v ? R S
• For each exiting edge (v,w) with f((v,w)) lt
  c((v,w)) and w?R,
• Add w to R
• For each entering edge (u,v) with f((u,v)) gt 0
  and u?R,
• add u to R
• After exploring all edges at v,
• Add v to S
• If the sink becomes in R
• Augmenting path found
• Else If R S
• Report S, V-S as a min-cut
• Otherwise, repeat

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Augmenting Path Example
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Min-cut/Max-flow Theorem

• In every network, the maximum value of a feasible
  flow equals the minimum capacity of a source-sink
  cut

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Proof

• If we reach the sink
• We find an augmenting flow, increase the value
• Eventually will reach the capacity of some cut
• The labeling algorithm terminates with S R
• Claim S,T is a source-sink cut with capacity
  val(f) where T V-S
• It is a cut because s?S and t ? S
• Note that when we apply the max-flow algorithm,
  no node in T is added to R
• This means that no edge from S to T has excess
  capacity andno edge from T to S has non-zero
  flow

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Recall our lemma

• val(f) f(S)- f-(S) ? f(S)
• Remember
• In other words f-(S) 0
• f(S) cap(S,T)

no edge from S to T has excess capacity andno
edge from T to S has non-zero flow
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A few remarks

• If the capacities are integers, the flow will
  also be integral
• We can compute maximum matchings in bipartite
  graphs using max-flows.