CSE%20202%20-%20Algorithms

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CSE 202 - Algorithms

• Max Flow Problems

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Max Flow problem

• Input a directed graph with a non-negative
  weight (capacity) on each edge, and a designated
  source node s and sink node t.
• A flow assigns to each edge a number (from 0 to
  its capacity) such that for each node except s
  and t, the sum of the flow into the node equals
  sum of the flow out. The fine print this is
  slightly different from the books definition.
• The value of a flow is the sum of the flow out of
  the source (which equals total flow into the
  sink).
• Maximum-Flow problem find a flow with the
  maximum possible value.
• Note A max flow can assign 0 to each edge going
  into the source, and to each edge going out of
  the sink.

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Example
12
a
c
20
16
t
9
4
s
10
7
sink
4
d
13
source
b
14
12
a
c
14
10
0
t
4
s
2
2
4
d
8
This is wasteful! The texts formulation doesnt
even let it happen.
b
6
A flow of value 18
4
Leftover capacity
12-12
a
c
20-14
16-10
9-0
t
s
4-4
10-2
7-2
sink
4-4
d
13-8
source
b
14-6
0
New backedge showing we can reduce forward flow
by 10
a
c
6
6
12
14
10
9
t
2
s
8
5
2
4
8
6
0
d
5
b
8
Residual network
5
Augmenting Path
0
We can push 5 more through network on indicated
path (New flow has value 23.)
a
c
6
6
12
14
10
9
t
2
s
8
5
2
4
8
6
0
d
5
b
8
0
a
c
6
1
12
19
10
9
t
2
s
8
0
7
4
13
11
0
d
0
Now were stuck! We cant get any more across cut.
b
3
New residual network
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Duality Max Flow Min Cut

• Original problem had a cut of value 23. We cant
  possibly get more from source to sink.
• Theorem Max flow min cut
• Proof Obviously, any flow ? min cut.
• But if max flow lt min cut, there would be an
  augmenting path from source to sink, leading to a
  higher-valued flow.
• This is a contradiction. Thus, max flow min
  cut. QED.

12
a
c
20
16
t
9
4
s
10
7
4
d
13
b
14
7
Ford-Fulkerson Methods

• initialize flow to 0
• while (theres an augmenting path)
• update flow
• compute new residual graph
• If there are several augmenting paths, does it
  matter which we pick??

a
100
100
t
1
s
1
100
100
b
8
Edmonds-Karp Algorithm

• Always choose an augmenting path with as few
  edges as possible (say it has L edges).
• This might create a new backedge, e.g (v,u).
• This new edge cant be in a new path of L edges..
• (Handle multiple new backedges by induction
  proof.)
• Meanwhile, at least one original edge has been
  eliminated.
• Thus, there are at most E iterations using L
  long paths.

k long path
v
m long
v
t
t
s
s
n long
u
u
j long path
m ? j1 and n ? k1 so mn gt jk1 L
j1k is a minimal length path (L)
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Edmonds-Karp Algorithm

• Always choose an augmenting path with as few
  edges as possible.
• At most E iterations use L-edge augmenting path
• Longest augmenting path has V-1 edges.
• Thus, there are at most O(VE) augmentations.
• How long does it take to find and process a
  shortest augmenting path??

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Max Flow Algorithms

• Edmonds-Karp is O(VE2)
• Some faster algorithms based on Push-Relabel.
• E.g. Goldbergs algorithm is O(V2E)
• Start source at height V, all others at 0
• Add in flow from higher nodes to lower ones.
• If a non-sink node cant push all its incoming
  flow out, increase its height.
• When done, return excess back to source.
• Carefully choosing order gives O(V3) algorithm.
• Fastest known algorithm is
• O( min(V2/3, E1/2) E lg(1V2/E) lg(max capacity)
  ).

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An Application

• Maximum matching problem
• Given an (undirected) graph (V,E) find a
  maximum-sized set of disjoint edges.
• Not the same as a maximal set of disjoint edges.
• Amazingly, not NP-complete (but its not easy)
• Bipartite graph Graph such that you can
  partition the nodes V V1 U V2, and every edge
  goes between a node of V1 and one of V2.
• Maximum matching for a bipartite graph can be
  reduced to a max flow problem.

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Glossary (in case symbols are weird)

• ???????????????? ? ? ? ? ? ? ? ? ?
• ? subset ? element of ? infinity ? empty
  set
• ? for all ? there exists ? intersection ?
  union
• ? big theta ? big omega ? summation
• ? gt ? lt ? about equal
  
• ? not equal ? natural numbers(N)
• ? reals(R) ? rationals(Q) ? integers(Z)